Chapter 12: Aldehydes, Ketones and Carboxylic Acids

Before the middle of the nineteenth century, making a good mirror was dangerous work. Glass was backed with a shining film of tin–mercury amalgam, and the workers who spread it slowly poisoned themselves breathing mercury vapour.

Around the 1830s the German chemist Justus von Liebig found a completely different route. He took a solution of a silver salt, made it alkaline, added ammonia, and then poured in a sugar solution. Within minutes the inside of the glass vessel had turned into a smooth, brilliant mirror of metallic silver.

Nothing had been painted on. The silver had been pulled out of the solution by the sugar itself. Glucose carries a –CHO group, and that group gives up electrons so readily that it hands them to silver ions, reducing them to a film of silver metal, while the –CHO group is itself oxidized. Replace that –CHO with the C=O of a ketone, tucked between two carbon chains, and nothing happens at all — no mirror, no silver.

That single structural difference, a hydrogen atom sitting on the carbon of the C=O group, is the thread running through this chapter. Aldehydes, ketones and carboxylic acids are all built around the same carbon–oxygen double bond, as shown in Fig. 12.0, yet they smell different, boil at different temperatures, and behave differently with the same reagent.

Diagram

Fig. 12.0 – A round-bottomed flask with a mirror-bright silvered inner surface, and beside it the three general structures R–CHO, R–CO–R′ and R–COOH labelled formyl group, ketonic carbonyl group and carboxyl group

NoteCuriosity Corner

Q1. An aldehyde and a ketone both contain the same C=O group. Why does only the aldehyde reduce silver ions to a silver mirror? Q2. Acetic acid and propanal have almost the same molecular mass, yet acetic acid boils about 70 K higher. What could a single small molecule be doing to hold on to its neighbours so tightly? Q3. In chloroacetic acid the chlorine atom is not even attached to the acidic hydrogen, yet the compound is a much stronger acid than acetic acid. How can a distant atom change how easily a proton leaves? Q4. When benzaldehyde is heated with concentrated alkali, one half of it becomes an alcohol and the other half becomes the salt of an acid. Why should identical molecules take two opposite paths in the same flask?

The chapter ahead will give you everything you need to reason out each of these questions for yourself.

12.1 Introduction

Draw the structures of the following compounds and classify them on the basis of the C–O single bond and the C=O double bond present in them: ethyl alcohol, acetaldehyde, o-nitrophenol, diethyl ether, isopropyl alcohol, acetone. Also answer: what are carbonyl compounds?

Moving from a Single Bond to a Double Bond

In the previous chapter you studied organic compounds that contain a carbon–oxygen single bond. In this chapter we study organic compounds that contain a carbon–oxygen double bond, \(>\mathrm{C=O}\), which is called the carbonyl group. It is one of the most important functional groups in organic chemistry.

A functional group is the small cluster of atoms that decides how a molecule behaves chemically. This is why the carbonyl group repays close study: once you understand how one \(>\mathrm{C=O}\) reacts, you can predict the behaviour of a very large number of compounds built around it.

The two atoms of this group have names of their own. The carbon of the double bond is called the carbonyl carbon and the oxygen is called the carbonyl oxygen, as marked in Fig. 12.1a.

Diagram

Fig. 12.1a – General carbonyl compound R–C(=O)–R′ with an arrow to the oxygen labelled “carbonyl oxygen” and an arrow to the central carbon labelled “carbonyl carbon”

Three Families Built on One Group

Both aldehydes and ketones contain a carbon–oxygen double bond as their functional group. For this reason they are together called carbonyl compounds.

NoteDefinition

Carbonyl compounds are organic compounds whose functional group is the carbon–oxygen double bond, \(>\mathrm{C=O}\); aldehydes and ketones belong to this class.

In an aldehyde, the carbonyl carbon is bonded to at least one hydrogen atom apart from an alkyl or aryl group. The functional group of aldehydes is therefore \(\mathrm{-CHO}\), which is called the formyl group or the aldehydic carbonyl group. Here an alkyl group means a group derived from an alkane, such as \(\mathrm{CH_3-}\) or \(\mathrm{CH_3CH_2-}\), while an aryl group means a group derived from an aromatic ring, such as \(\mathrm{C_6H_5-}\).

Q. Why is the \(\mathrm{-CHO}\) group always found at the end of a carbon chain and never in the middle of it?

A. The carbonyl carbon of an aldehyde must carry a hydrogen atom. That leaves only one of its remaining bonds free for a carbon chain, so an aldehyde carbon can be joined to at most one other carbon. A carbon in the middle of a chain is joined to two carbons, so it can never be an aldehyde carbon.

In a ketone, on the other hand, the carbonyl carbon is bonded to two alkyl or aryl groups. These two groups may be identical (\(\mathrm{R = R}\)) or different (\(\mathrm{R \neq R'}\)). This group is called the ketonic carbonyl group. Notice the single structural difference from an aldehyde: the hydrogen on the carbonyl carbon has been replaced by a second carbon-containing group.

The functional group of carboxylic acids is \(\mathrm{-COOH}\), called the carboxyl group. Because of the \(\mathrm{-OH}\) group bonded to the \(>\mathrm{C=O}\) group, carboxylic acids are distinct from aldehydes and ketones. In other words, a carboxyl group is a carbonyl group and a hydroxyl group sitting on the same carbon atom, and this combination behaves so differently from either group alone that carboxylic acids are treated as a separate family.

aldehydes vs ketones vs carboxylic acids

Feature Aldehyde Ketone Carboxylic acid
Functional group \(\mathrm{-CHO}\) \(>\mathrm{C=O}\) \(\mathrm{-COOH}\)
Name of the group Formyl or aldehydic carbonyl group Ketonic carbonyl group Carboxyl group
Carbonyl carbon is bonded to At least one H atom, plus an alkyl or aryl group Two alkyl or aryl groups One \(\mathrm{-OH}\) group, plus H or an alkyl/aryl group
General formula \(\mathrm{R\!-\!CHO}\) \(\mathrm{R\!-\!CO\!-\!R'}\) \(\mathrm{R\!-\!COOH}\)
Position in the chain Always at a chain end Always within the chain Always at a chain end

The word aldehyde is a contraction of the Latin phrase alcohol dehydrogenatus, meaning “alcohol deprived of hydrogen” — a reminder that these compounds were first obtained by removing hydrogen from alcohols.

Acetone, the simplest ketone, is one of the most widely used solvents in industry and in the laboratory. It dissolves greases, resins and many organic compounds, and it is the active ingredient of common nail-polish remover. It is also extremely flammable and its vapour is heavier than air, so it must always be kept away from open flames and hot plates, and used in a well-ventilated space.

Formaldehyde, \(\mathrm{H\!-\!CHO}\), has two hydrogen atoms attached to its carbonyl carbon. Explain, using the definitions in this section, why it is classified as an aldehyde and not as a ketone.

12.2 Classification of Aldehydes, Ketones and Carboxylic Acids

Aldehydes, ketones and carboxylic acids are classified according to the nature of the carbon skeleton bonded to the \(>\mathrm{C=O}\) group. In other words, we do not look at the carbonyl group itself — that is the same in every case — but at what kind of carbon framework it is attached to. If that framework is an open chain of carbon atoms, the compound is called aliphatic; if it is a benzene ring or another aromatic ring, the compound is called aromatic. The complete classification scheme is summarised in FC 12.2a.

Flowchart

FC 12.2a – Classification of carbonyl compounds: Carbonyl-containing compounds → (branch 1) Aldehydes, R–CHO → aliphatic (acetaldehyde) / aromatic (benzaldehyde); (branch 2) Ketones, R–CO–R′ → aliphatic → simple (acetone) / mixed (ethyl methyl ketone), and aromatic (acetophenone, benzophenone); (branch 3) Carboxylic acids, R–COOH → aliphatic → mono (acetic acid) / di (oxalic acid) / tri (citric acid), and aromatic (benzoic acid)

12.2.1 Classification of Aldehydes

Aldehydes are classified as aliphatic and aromatic aldehydes.

Aliphatic Aldehydes

The compounds in which the \(\mathrm{-CHO}\) group (formyl group) is attached directly to an \(sp^3\) hybridized carbon atom, that is, to a saturated carbon atom, are called aliphatic aldehydes. An \(sp^3\) hybridized carbon is one that forms four single bonds and no multiple bonds, so this condition simply says that the carbon holding the \(\mathrm{-CHO}\) group is a plain, saturated chain carbon.

There is one exception. Formaldehyde, \(\mathrm{H\!-\!CHO}\), is also classified as an aliphatic aldehyde even though its \(\mathrm{-CHO}\) group is not attached to any carbon at all. For example:

\[\mathrm{CH_3\!-\!CHO} \qquad \mathrm{CH_3\!-\!CH_2\!-\!CHO}\]

\[\text{(Acetaldehyde)} \qquad \text{(Propionaldehyde)}\]

The general formula of an aliphatic aldehyde is drawn in Fig. 12.2a, where R may be a hydrogen atom or an alkyl group.

Diagram

Fig. 12.2a – General formula of an aliphatic aldehyde: R–C(=O)–H with the –CHO portion shaded and labelled “general formula (R = H or alkyl group)”

Aromatic Aldehydes

The compounds in which the \(\mathrm{-CHO}\) group is attached directly to an aromatic ring are called aromatic aldehydes. The word “directly” is important here: the carbon of the \(\mathrm{-CHO}\) group must be bonded to a ring carbon itself, with no chain carbon in between. Three examples are shown in Fig. 12.2b.

Diagram

Fig. 12.2b – Three aromatic aldehydes: benzaldehyde (C₆H₅–CHO); salicylaldehyde (benzene ring with –CHO and an –OH on the adjacent carbon); p-nitrobenzaldehyde (benzene ring with –CHO and –NO₂ at the para position); each labelled with its name

Classify the following as aliphatic and aromatic aldehydes: phenylacetaldehyde (\(\mathrm{C_6H_5\!-\!CH_2\!-\!CHO}\)), cinnamaldehyde (\(\mathrm{C_6H_5\!-\!CH{=}CH\!-\!CHO}\)), cyclohexanecarbaldehyde, and 2-methylbenzaldehyde.

12.2.2 Classification of Ketones

Ketones are classified as aliphatic and aromatic ketones.

Aliphatic Ketones

The compounds in which the \(>\mathrm{C=O}\) group is attached to two alkyl groups are called aliphatic ketones. In a ketone, the carbonyl group is also called the oxo group or the ketonic carbonyl group; its general formula is shown in Fig. 12.2c, where R and \(\mathrm{R'}\) are alkyl groups that may be identical or different.

Diagram

Fig. 12.2c – General formula of an aliphatic ketone: R–C(=O)–R′ with an arrow to the C=O labelled “oxo or ketonic carbonyl group” and the caption “R, R′ = alkyl group, identical or different”

Simple and Mixed Ketones

On the basis of the types of alkyl groups bonded to the carbonyl carbon, aliphatic ketones are further classified as simple and mixed ketones.

Simple or symmetrical ketones are the ketones in which both the alkyl groups bonded to the carbonyl carbon are identical. Because the two halves of such a molecule are the same, the molecule is symmetrical about the carbonyl carbon. For example:

\[\mathrm{H_3C\!-\!CO\!-\!CH_3} \qquad \mathrm{H_5C_2\!-\!CO\!-\!C_2H_5}\]

\[\text{(Dimethyl ketone) (Acetone)} \qquad \text{(Diethyl ketone)}\]

Mixed or unsymmetrical ketones are the ketones in which the two alkyl groups bonded to the carbonyl carbon are different. For example:

\[\mathrm{H_5C_2\!-\!CO\!-\!CH_3} \qquad \mathrm{H_5C_2\!-\!CO\!-\!CH_2\!-\!CH_2\!-\!CH_3}\]

\[\text{(Ethyl methyl ketone)} \qquad \text{(Ethyl n-propyl ketone)}\]

Aromatic Ketones

The compounds in which a \(>\mathrm{C=O}\) group is attached to either two aryl groups, or to one aryl and one alkyl group, are called aromatic ketones. Two common examples are drawn in Fig. 12.2d.

Diagram

Fig. 12.2d – Two aromatic ketones: benzophenone (two benzene rings joined through a C=O) and acetophenone (a benzene ring joined through a C=O to a CH₃ group); label their trivial names diphenyl ketone and methyl phenyl ketone

Q. Acetophenone has an ordinary methyl group on one side of its carbonyl carbon. Why is it still called an aromatic ketone?

A. The definition allows an aromatic ketone to carry either two aryl groups or one aryl and one alkyl group. It is enough that at least one aromatic ring is bonded directly to the carbonyl carbon, which is true in acetophenone.

Classify the following as simple and mixed ketones: benzophenone, acetone, butanone, acetophenone.

Aldehydes and ketones are responsible for many flavours and odours you will readily recognise: benzaldehyde gives the bitter almond flavour, vanillin the vanilla flavour, cinnamaldehyde the cinnamon flavour, and camphor its characteristic fragrance. The structures of many important biological compounds also contain a carbonyl moiety — progesterone and testosterone, the female and male sex hormones respectively, are two examples.

Butyraldehyde is used in margarine and other foods for its buttery odour. Acetophenone has the smell of pistachio and is used in ice-cream, while muscone has a musky aroma and is used in perfumes. The butter flavour of popcorn comes from butane-2,3-dione.

12.2.3 Classification of Carboxylic Acids

Carboxylic acids are classified as aliphatic and aromatic carboxylic acids.

Aliphatic Carboxylic Acids

The organic compounds in which the carboxyl (\(\mathrm{-COOH}\)) group is bonded to an alkyl group are called aliphatic carboxylic acids, or fatty acids. The second name is a reminder of where they were first obtained: the higher members of this series were isolated from natural fats and oils.

As with aldehydes, there is one exception to the definition. Formic acid, \(\mathrm{H\!-\!COOH}\), is also classified as an aliphatic carboxylic acid, even though its \(\mathrm{-COOH}\) group is not attached to any carbon. For example:

\[\mathrm{H_3C\!-\!COOH} \qquad \mathrm{H_3C\!-\!CH_2\!-\!COOH}\]

\[\text{(Acetic acid)} \qquad \text{(Propionic acid)}\]

The general formula of a carboxylic acid, with the carboxyl functional group marked, is given in Fig. 12.2e, where R is a hydrogen atom or an alkyl group.

Diagram

Fig. 12.2e – General formula of a carboxylic acid: R–C(=O)–OH with the –COOH portion boxed and an arrow labelled “carboxyl functional group”; caption “R = H or alkyl group”

Classification by Number of Carboxyl Groups

A molecule need not be limited to one carboxyl group. Depending on the number of \(\mathrm{-COOH}\) groups present, carboxylic acids are classified as mono, di, tri carboxylic acids and so on. For example, propionic acid, \(\mathrm{H_3C\!-\!CH_2\!-\!COOH}\), is a monocarboxylic acid; oxalic acid, \(\mathrm{HOOC\!-\!COOH}\), is a dicarboxylic acid; and citric acid, whose structure is drawn in Fig. 12.2f, is a tricarboxylic acid.

Diagram

Fig. 12.2f – Structure of citric acid: a central carbon bearing –OH and –COOH, with a –CH₂–COOH group above and a –CH₂–COOH group below; label “citric acid (a tricarboxylic acid)”

mono- vs di- vs tri-carboxylic acids

Feature Monocarboxylic acid Dicarboxylic acid Tricarboxylic acid
Number of \(\mathrm{-COOH}\) groups One Two Three
Example Propionic acid Oxalic acid Citric acid
Structure \(\mathrm{H_3C\!-\!CH_2\!-\!COOH}\) \(\mathrm{HOOC\!-\!COOH}\) see Fig. 12.2f
Aromatic Carboxylic Acids

These are the compounds in which one or more carboxyl groups (\(\mathrm{-COOH}\)) are attached directly to the aromatic ring. Three examples are shown in Fig. 12.2g.

Diagram

Fig. 12.2g – Three aromatic carboxylic acids: benzoic acid (ring + COOH); phthalic acid (ring with two COOH groups on adjacent carbons); acetyl salicylic acid or aspirin (ring with COOH and an OCOCH₃ group on the adjacent carbon); label each with its name

Once again, the word “directly” decides the classification, so a compound whose ring and carboxyl group are separated by a chain carbon needs a name of its own.

NoteDefinition

The aromatic compounds in which the \(\mathrm{-COOH}\) group is not attached directly to the ring are called side-chain aromatic acids; for example, phenyl acetic acid, \(\mathrm{C_6H_5\!-\!CH_2\!-\!COOH}\).

Salicylaldehyde and salicylic acid take their names from the Latin salix, meaning willow, because they were obtained from willow bark — a traditional remedy for pain and fever. Aspirin, shown in Fig. 12.2g, is acetyl salicylic acid, a modified form of that same acid.

Carboxylic Acids in Nature

Carboxylic acids are widely distributed in nature; they are found in both plants and animals. L-lactic acid is present in curd and citric acid is found in citrus fruits such as lemons. Acetic acid is the key ingredient of vinegar.

Classify each of the following compounds, giving a reason for your answer.

(i) \(\mathrm{C_6H_5\!-\!CO\!-\!C_2H_5}\) (ii) \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CHO}\) (iii) \(\mathrm{C_6H_5\!-\!CH_2\!-\!CH_2\!-\!COOH}\)

  1. The carbonyl carbon carries no hydrogen atom, so the compound is a ketone. One of the two groups attached to the carbonyl carbon is a phenyl (aryl) group and the other is an ethyl (alkyl) group. A \(>\mathrm{C=O}\) group carrying one aryl and one alkyl group defines an aromatic ketone.

  2. The carbonyl carbon carries a hydrogen atom, so the compound is an aldehyde. The \(\mathrm{-CHO}\) group is attached to a saturated, \(sp^3\) hybridized chain carbon and there is no ring, so it is an aliphatic aldehyde.

  3. The functional group is \(\mathrm{-COOH}\), so the compound is a carboxylic acid. The molecule does contain a benzene ring, but the \(\mathrm{-COOH}\) group is separated from the ring by two \(\mathrm{-CH_2-}\) carbons and is therefore not attached directly to the ring. It is not an aromatic carboxylic acid in the strict sense; it is a side-chain aromatic acid.

Draw the structure of a mixed aliphatic ketone containing five carbon atoms in an unbranched chain, and name the two alkyl groups attached to its carbonyl carbon.

12.3 Nomenclature of Aldehydes, Ketones and Carboxylic Acids

12.3.1 Nomenclature of Aldehydes and Carboxylic Acid

The names of aldehydes and carboxylic acids are related to each other. There are two systems of naming aldehydes and carboxylic acids: trivial and IUPAC.

Trivial Names of Aldehydes and Carboxylic Acids

Trivial names of aliphatic aldehydes are derived from the corresponding trivial names of carboxylic acids. Here the ending ‘ic acid’ of the carboxylic acid is replaced by the ending ‘aldehyde’. Acetic acid therefore gives acetaldehyde, and propionic acid gives propionaldehyde.

In the case of substituted aliphatic carboxylic acids and aldehydes, the position of the substituent is indicated by labelling the carbon serially as \(\alpha\), \(\beta\), \(\gamma\) and so on. The carbon atom adjacent to the carbonyl carbon is labelled as \(\alpha\), the next one as \(\beta\), and so on, as shown in Fig. 12.3a. Note that the carbonyl carbon itself gets no Greek letter — the counting starts from its neighbour. Some trivial names formed in this way are listed in Table 12.1.

Diagram

Fig. 12.3a – A four-carbon chain ending in –COOH, with the carbon next to the –COOH carbon labelled α, the next labelled β and the last labelled γ; alongside, the same labelling on a chain ending in –CHO

Which suffix appears in the IUPAC names of aldehydes and carboxylic acids?

A series of straight-chain dicarboxylic acids is commercially known by these common names: \(\mathrm{HOOC\!-\!COOH}\) (oxalic acid), \(\mathrm{HOOC\!-\!CH_2\!-\!COOH}\) (malonic acid), \(\mathrm{HOOC\!-\!(CH_2)_2\!-\!COOH}\) (succinic acid), \(\mathrm{HOOC\!-\!(CH_2)_3\!-\!COOH}\) (glutaric acid) and \(\mathrm{HOOC\!-\!(CH_2)_4\!-\!COOH}\) (adipic acid). A series of lower fatty acids is likewise known commercially as \(\mathrm{CH_3COOH}\) (acetic acid), \(\mathrm{CH_3CH_2COOH}\) (n-propionic acid), \(\mathrm{CH_3(CH_2)_2COOH}\) (n-butyric acid), \(\mathrm{CH_3(CH_2)_3COOH}\) (n-valeric acid) and \(\mathrm{CH_3(CH_2)_4COOH}\) (n-caproic acid).

The trivial names of carboxylic acids are often derived from the Latin names of their original natural source. Formic acid was obtained from red ants (formica means ant), acetic acid from acetum (vinegar), propionic acid from basic fat (propion means first fat), and butyric acid from butter (butyrum means butter).

IUPAC Names of Aldehydes and Carboxylic Acids

According to the IUPAC system, the name of an aliphatic aldehyde is derived from the name of the corresponding alkane by replacing the ending ‘e’ of the alkane with ‘al’. An aldehyde is therefore named as an alkanal, as shown in Table 12.1. The IUPAC name of an aliphatic carboxylic acid is derived from the name of the corresponding alkane by replacing the ending ‘e’ of the alkane with ‘oic acid’.

\[\text{Alkane} \longrightarrow \text{Alkanal}\]

\[\text{Alkane} \longrightarrow \text{Alkanoic acid}\]

The longest chain including the \(\mathrm{-CHO}\) or \(\mathrm{-COOH}\) group is identified as the parent chain. Numbering of the chain is done by giving number 1 to the \(\mathrm{-CHO}\) or \(\mathrm{-COOH}\) carbon. The name of the substituent is included along with its locant, that is, the number of the carbon on which it sits.

\[^4\mathrm{CH_3}\!-\!^3\mathrm{CH_2}\!-\!^2\mathrm{CH_2}\!-\!^1\mathrm{CHO} \ ;\qquad ^4\mathrm{CH_3}\!-\!^3\mathrm{CH_2}\!-\!^2\mathrm{CH_2}\!-\!^1\mathrm{COOH}\]

The aldehyde (\(\mathrm{-CHO}\)) group and the carboxyl (\(\mathrm{-COOH}\)) group are always present at the end of the parent straight chain. The complete naming procedure is set out in FC 12.3a.

Flowchart

FC 12.3a – Writing the IUPAC name of an aliphatic aldehyde or carboxylic acid: identify the longest carbon chain that includes the –CHO or –COOH carbon → number that chain giving locant 1 to the –CHO or –COOH carbon → name the parent alkane and replace the final ‘e’ with ‘al’ (aldehyde) or ‘oic acid’ (acid) → add each substituent as a prefix with its locant, in alphabetical order → assemble the full name

Q. Why does the carbon of the \(\mathrm{-CHO}\) or \(\mathrm{-COOH}\) group always receive the number 1, without our having to check the other end of the chain?

A. Both these groups can occur only at the end of a chain, because their carbon is already using its bonds on oxygen and hydrogen and has just one bond left for the chain. Being terminal, that carbon is either the first or the last atom, and IUPAC gives the principal group the lowest possible locant — so it is always numbered 1.

Chains Carrying Two Groups

When two \(\mathrm{-CHO}\) groups are present at the two ends of the chain, the ending ‘e’ of the alkane is retained and the suffix ‘dial’ is added to the name of the parent aldehyde. In the case of dicarboxylic acids, ‘dioic acid’ is added to the name of the parent alkane. Retaining the ‘e’ here keeps the name easy to pronounce and signals at once that two groups are present.

Naming Ring Compounds: Carbaldehyde and Carboxylic Acid

In IUPAC nomenclature, an alicyclic compound in which the \(\mathrm{-CHO}\) group is attached directly to the ring is named as a carbaldehyde. The suffix ‘carbaldehyde’ is added after the full name of the parent cycloalkane structure. Similarly, an alicyclic compound having a carboxyl group directly attached to the alicyclic ring is named as a cycloalkane carboxylic acid. An alicyclic compound is a ring compound that is not aromatic, such as cyclohexane.

Substituted Aromatic Aldehydes and Carboxylic Acids

When two or more different functional groups are attached to a ring, the higher priority group (Std. XI Chemistry Textbook, Chapter 14, sec. 14.4.7) is given the lower number. When the \(\mathrm{-CHO}\) group appears as a substituent, the prefix ‘formyl’ is used in the IUPAC name. Three worked examples of such names are drawn in Fig. 12.3b.

Diagram

Fig. 12.3b – Three substituted aromatic compounds with names: a benzene ring with CHO at C1, CH₃ at C3 and OH at C4, labelled 4-hydroxy-3-methylbenzaldehyde; a ring with COOH at C1, OH at C3 and CH₃ at C4, labelled 3-hydroxy-4-methylbenzoic acid; a ring with COOH at C1 and CHO at C2, labelled 2-formylbenzoic acid

Trivial and IUPAC names of some aldehydes and carboxylic acids are given in Table 12.1.

Table 12.1 Trivial and IUPAC names of carboxylic acids and aldehydes

Structure (acid) Trivial name IUPAC name Structure (aldehyde) Trivial name IUPAC name
\(\mathrm{H\!-\!COOH}\) Formic acid Methanoic acid \(\mathrm{H\!-\!CHO}\) Formaldehyde Methanal
\(\mathrm{CH_3\!-\!COOH}\) Acetic acid Ethanoic acid \(\mathrm{CH_3\!-\!CHO}\) Acetaldehyde Ethanal
\(\mathrm{CH_3\!-\!CH_2\!-\!COOH}\) Propionic acid Propanoic acid \(\mathrm{CH_3\!-\!CH_2\!-\!CHO}\) Propionaldehyde Propanal
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!COOH}\) n-Butyric acid Butanoic acid \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CHO}\) Butyraldehyde Butanal
\(\mathrm{(CH_3)_2CH\!-\!COOH}\) Isobutyric acid (\(\alpha\)-methylpropionic acid) 2-Methylpropanoic acid \(\mathrm{(CH_3)_2CH\!-\!CHO}\) Isobutyraldehyde (\(\alpha\)-methylpropionaldehyde) 2-Methylpropanal
\(\mathrm{CH_2{=}CH\!-\!COOH}\) Acrylic acid Prop-2-enoic acid \(\mathrm{CH_2{=}CH\!-\!CHO}\) Acrolein Prop-2-enal
\(\mathrm{HOOC\!-\!COOH}\) Oxalic acid Ethanedioic acid \(\mathrm{CHO\!-\!CHO}\) Oxaldehyde (Glyoxal) Ethanedial
\(\mathrm{C_6H_5\!-\!COOH}\) Benzoic acid (Benzenecarboxylic acid) Benzoic acid \(\mathrm{C_6H_5\!-\!CHO}\) Benzaldehyde Benzaldehyde (Benzenecarbaldehyde)
\(\mathrm{2\text{-}CH_3\!-\!C_6H_4\!-\!COOH}\) o-Toluic acid 2-Methylbenzoic acid \(\mathrm{2\text{-}CH_3\!-\!C_6H_4\!-\!CHO}\) o-Tolualdehyde 2-Methylbenzaldehyde
\(\mathrm{2\text{-}OH\!-\!C_6H_4\!-\!COOH}\) Salicylic acid 2-Hydroxybenzoic acid \(\mathrm{2\text{-}OH\!-\!C_6H_4\!-\!CHO}\) Salicylaldehyde 2-Hydroxybenzaldehyde
\(\mathrm{1,2\text{-}C_6H_4(COOH)_2}\) Phthalic acid Benzene-1,2-dicarboxylic acid \(\mathrm{1,2\text{-}C_6H_4(CHO)_2}\) Phthalaldehyde Benzene-1,2-dicarbaldehyde
\(\mathrm{C_6H_{11}\!-\!COOH}\) Cyclohexylcarboxylic acid Cyclohexanecarboxylic acid \(\mathrm{C_6H_{11}\!-\!CHO}\) Cyclohexane aldehyde Cyclohexanecarbaldehyde

Write the IUPAC name of \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CHO}\).

The longest chain that includes the \(\mathrm{-CHO}\) carbon contains four carbon atoms, so the parent alkane is butane and the parent name is butanal. Numbering begins at the \(\mathrm{-CHO}\) carbon, which becomes C-1. Counting along, C-2 is the \(\mathrm{-CH_2-}\) group, C-3 carries a methyl branch, and C-4 is the terminal \(\mathrm{-CH_3}\). The substituent is a methyl group at C-3.

IUPAC name: 3-methylbutanal.

Write the IUPAC name of \(\mathrm{CH_3\!-\!CH_2\!-\!CH(CH_3)\!-\!COOH}\).

Adipic acid, the six-carbon dicarboxylic acid listed above, is manufactured on an enormous industrial scale because it is one of the two building blocks of nylon-6,6. Its two \(\mathrm{-COOH}\) groups, one at each end of the chain, let it link repeatedly with a diamine to build the long polymer chain that is drawn into fibre.

12.3.2 Trivial and IUPAC Names of Ketones

Trivial Names of Ketones

The trivial names of aliphatic ketones are based on the names of the alkyl groups or aryl groups attached to the carbonyl carbon. The names of the alkyl or aryl groups are written in alphabetical order, followed by the word ketone. Thus a carbonyl carbon carrying an ethyl group and a methyl group gives ethyl methyl ketone.

In the case of substituted aliphatic ketones, the position of the substituent is indicated by labelling the carbon serially as \(\alpha\), \(\beta\), \(\gamma\) and so on. The carbon atom adjacent to the carbonyl carbon is labelled as \(\alpha\) and the next one is \(\beta\), and so on — the same convention you met for aldehydes and acids. Names of aromatic ketones are based on a phenone, as seen in Table 12.2.

IUPAC Names of Ketones

The IUPAC names of aliphatic ketones are derived from the names of the corresponding alkanes by replacing the ending ‘e’ of the alkane with ‘one’. They are named as alkanone. The longest chain of carbon atoms containing the ketonic carbonyl group is numbered from the end closer to the carbonyl carbon.

\[\text{Alkane} \longrightarrow \text{Alkanone}\]

Q. Why must a ketone’s name carry a locant, such as pentan-2-one, when an aldehyde’s name does not?

A. A ketonic carbonyl group can sit at any inner position of the chain, so the position has to be stated. An aldehyde group can only be terminal, so its position is already fixed at carbon 1 and needs no number.

When two \(>\mathrm{C=O}\) groups are present, the ending ‘e’ of the alkane is retained and the suffix ‘dione’ is added to the name of the parent ketone, indicating the locants of the ketonic carbonyl groups. In the case of polyfunctional ketones, the higher priority group is given the lower number. When the ketonic carbonyl is a lower priority group, it is named as ‘oxo’, preceded by the locant. In alicyclic ketones, the carbonyl carbon is numbered as 1. These conventions can all be traced in Table 12.2.

how each family is signalled in an IUPAC name

Family Suffix for one group Suffix for two groups Suffix on an alicyclic ring Prefix when it is the lower priority group
Aldehyde -al -dial -carbaldehyde formyl
Ketone -one -dione -one (carbonyl carbon numbered 1) oxo
Carboxylic acid -oic acid -dioic acid -carboxylic acid —

Table 12.2 Trivial and IUPAC names of some ketones

Sr. No. Compound Trivial name IUPAC name
1 \(\mathrm{CH_3\!-\!CO\!-\!CH_3}\) Dimethyl ketone (Acetone) Propanone
2 \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH_3}\) Ethyl methyl ketone Butanone
3 \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) Methyl n-propyl ketone Pentan-2-one
4 \(\mathrm{CH_3\!-\!CH_2\!-\!CO\!-\!CH_2\!-\!CH_3}\) Diethyl ketone Pentan-3-one
5 \(\mathrm{CH_3\!-\!CHBr\!-\!CO\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) \(\alpha\)-Bromoethyl n-propyl ketone 2-Bromohexan-3-one
6 \(\mathrm{(CH_3)_2C{=}CH\!-\!CO\!-\!CH_3}\) Mesityl oxide 4-Methylpent-3-en-2-one
7 \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CO\!-\!CH_2\!-\!CH_3}\) — Hexane-2,4-dione
8 \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CHO}\) — 3-Oxobutanal
9 \(\mathrm{C_6H_5\!-\!CO\!-\!C_6H_5}\) Diphenyl ketone (Benzophenone) Benzophenone
10 \(\mathrm{C_6H_5\!-\!CO\!-\!CH_3}\) Methyl phenyl ketone (Acetophenone) Acetophenone
11 Cyclohexanone ring bearing \(\mathrm{-Cl}\) at C2 and \(\mathrm{-CH_3}\) at C4 — 2-Chloro-4-methylcyclohexanone

Entry 8 in Table 12.2 is worth a second look. The molecule contains both a \(>\mathrm{C=O}\) and a \(\mathrm{-CHO}\) group; the aldehyde group has the higher priority, so it takes the suffix and locant 1, and the ketonic carbonyl is demoted to the prefix ‘oxo’ at C-3, giving 3-oxobutanal.

Write the IUPAC name of \(\mathrm{CH_3\!-\!CH_2\!-\!CO\!-\!CH_2\!-\!CH(CH_3)\!-\!CH_3}\).

The longest chain containing the ketonic carbonyl group has six carbon atoms, so the parent name is hexanone. The chain must be numbered from the end closer to the carbonyl carbon. Counting from the left-hand \(\mathrm{CH_3}\) places the carbonyl carbon at position 3; counting from the right would place it at position 4. Numbering from the left is therefore correct. The methyl branch then falls on C-5.

IUPAC name: 5-methylhexan-3-one.

Write the IUPAC name of \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CO\!-\!CH_3}\).

Write IUPAC names for the compounds (i) to (v) drawn in Fig. 12.3c.

Diagram

Fig. 12.3c – Five structures to be named: (i) a branched chain bearing two Br atoms and a terminal –CHO; (ii) a branched chain containing two ketonic C=O groups; (iii) an unbranched chain with a terminal –CHO; (iv) a benzene ring carrying –CHO and a –CH₃ substituent; (v) a cyclohexanone ring carrying a –CH₃ substituent

Draw structures for the following: (a) 2-methylpentanal, (b) hexan-2-one.

12.4 Preparation of Aldehydes and Ketones

12.4.1 General Methods of Preparation of Aldehydes and Ketones

By Oxidation of Alcohols

Aldehydes and ketones are prepared by the oxidation of primary and secondary alcohols respectively (see Chapter 11). The pairing is worth fixing in your mind: a primary alcohol has two hydrogen atoms on its carbon-bearing \(\mathrm{-OH}\) group, and removing two of the hydrogens from that carbon and its \(\mathrm{-OH}\) leaves a \(\mathrm{-CHO}\) group. A secondary alcohol has only one such hydrogen, so oxidation can go no further than a \(>\mathrm{C=O}\) flanked by two carbon groups, which is a ketone.

What is the reagent which oxidizes primary alcohols to only aldehydes and does not oxidize aldehydes further into carboxylic acid?

By Dehydrogenation of Alcohols

This method has industrial application. Aldehydes and ketones are prepared by passing the vapours of primary and secondary alcohols respectively over hot copper powder (see Chapter 11). Here no oxidising agent is consumed at all; hydrogen is simply pulled off the alcohol at the hot metal surface, which is why the method suits large-scale continuous production.

From Hydrocarbons

What is ozonolysis? What is the role of zinc dust in the ozonolysis process?

By ozonolysis: An alkene reacts with ozone to give an ozonide, which on decomposition with zinc dust and water gives aldehydes and/or ketones (see Std. XI Chemistry Textbook, Chapter 15). Whether the product is an aldehyde or a ketone depends on what the doubly bonded carbons were carrying: a carbon holding a hydrogen becomes a \(\mathrm{-CHO}\) group, while a carbon holding two alkyl groups becomes a ketonic carbonyl.

By hydration of alkynes: Alkynes react with water in the presence of 40% sulfuric acid and 1% mercuric sulfate to give aldehydes or ketones (see Std. XI Chemistry Textbook, Chapter 15).

12.4.2 Other Methods of Preparation of Aldehydes and Ketones

Some methods of preparation of aldehydes and ketones involve common starting functional groups but different types. In other words, the same starting compound can be steered towards an aldehyde or towards a ketone, and what decides the outcome is the reagent chosen. The routes covered in this subsection are collected in FC 12.4a.

Flowchart

FC 12.4a – Routes to aldehydes and ketones from a common starting material: acyl chloride R–COCl → (H₂, Pd–BaSO₄) → aldehyde R–CHO, or → (R₂Cd) → ketone R′–CO–R; nitrile R–C≡N → (SnCl₂/HCl, then H₃O⁺) or (DIBAl-H, then H₃O⁺) → aldehyde R–CHO, or → (R′MgX, dry ether, then H₃O⁺) → ketone R–CO–R′; arene → (Etard / CrO₃ / side-chain chlorination / Gatterman–Koch) → aromatic aldehyde, or → (Friedel–Crafts acylation) → aromatic ketone; ester R–CO–O–R′ → (DIBAl-H at 195 K) → aldehyde

From Acyl Chlorides: Preparation of Aldehydes

Aldehydes and ketones both can be obtained from acyl chloride, but the reactions involved are different.

An acyl chloride is reduced to the corresponding aldehyde by hydrogen using a palladium catalyst poisoned with barium sulfate. This reaction is known as Rosenmund reduction.

\[\mathrm{R\!-\!COCl} \xrightarrow{\ \mathrm{H_2},\ \mathrm{Pd\text{-}BaSO_4}\ } \mathrm{R\!-\!CHO} + \mathrm{HCl}\]

\[\text{(Acyl chloride)} \qquad\qquad \text{(Aldehyde)}\]

Pd–BaSO₄ (poisoned palladium) (A palladium catalyst deliberately deactivated by barium sulfate → used with hydrogen gas on an acyl chloride → its lowered activity lets the reduction stop cleanly at the aldehyde instead of running on to the alcohol. With unpoisoned palladium the same reduction does not stop at the aldehyde stage.)

The reaction is named after the German chemist Karl Wilhelm Rosenmund, who developed it in the early twentieth century as a controlled way of stopping a reduction halfway.

Write the structure of the product formed on Rosenmund reduction of ethanoyl chloride and benzoyl chloride.

What is the alcohol formed when benzoyl chloride is reduced with pure palladium as the catalyst?

From Acyl Chlorides: Preparation of Ketones

Preparation of aliphatic ketones from acyl chloride. Ketones are obtained from acyl chloride by reaction with dialkyl cadmium, which is prepared by the treatment of cadmium chloride with a Grignard reagent.

\[2\,\mathrm{R\!-\!MgX} + \mathrm{CdCl_2} \longrightarrow \mathrm{R_2Cd} + 2\,\mathrm{Mg(X)Cl}\]

\[2\,\mathrm{R'\!-\!COCl} + \mathrm{R_2Cd} \longrightarrow 2\,\mathrm{R'\!-\!CO\!-\!R} + \mathrm{CdCl_2}\]

\[\text{(Acyl chloride)} \qquad\qquad \text{(Ketone)}\]

For example,

\[2\,\mathrm{CH_3\!-\!COCl} + \mathrm{(CH_3)_2Cd} \longrightarrow 2\,\mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{CdCl_2}\]

\[\text{(Ethanoyl chloride)} \quad \text{(Dimethyl cadmium)} \quad \text{(Propanone, acetone)}\]

\[2\,\mathrm{C_6H_5\!-\!COCl} + \mathrm{(CH_3)_2Cd} \longrightarrow \mathrm{C_6H_5\!-\!CO\!-\!CH_3} + \mathrm{CdCl_2}\]

\[\text{(Benzoyl chloride)} \quad \text{(Dimethyl cadmium)} \quad \text{(Acetophenone)}\]

Preparation of aromatic ketones from acyl chloride. Aromatic ketones are prepared by Friedel–Crafts acylation reaction (Std. XI Chemistry Textbook, Chapter 15, sec. 15.4.6).

getting an aldehyde or a ketone from the same starting material

Starting material Reagent that gives an aldehyde Reagent that gives a ketone
Acyl chloride, \(\mathrm{R\!-\!COCl}\) \(\mathrm{H_2}\) with \(\mathrm{Pd}\)–\(\mathrm{BaSO_4}\) (Rosenmund reduction) Dialkyl cadmium, \(\mathrm{R_2Cd}\)
Nitrile, \(\mathrm{R\!-\!C{\equiv}N}\) \(\mathrm{SnCl_2}\)/\(\mathrm{HCl}\), then \(\mathrm{H_3O^{+}}\) (Stephen reaction); or DIBAl-H, then \(\mathrm{H_3O^{+}}\) Grignard reagent in dry ether, then \(\mathrm{H_3O^{+}}\)
Arene Etard reaction, \(\mathrm{CrO_3}\) route, side-chain chlorination, Gatterman–Koch Friedel–Crafts acylation
From Nitriles: Preparation of Aldehydes

Aldehydes and ketones both can be obtained from nitriles, but by different reactions.

Nitriles are reduced to imine hydrochloride by stannous chloride in the presence of hydrochloric acid, which on acid hydrolysis gives the corresponding aldehyde. This reaction is called the Stephen reaction.

\[\mathrm{R\!-\!C{\equiv}N} + 2[\mathrm{H}] \xrightarrow{\ \mathrm{SnCl_2},\ \mathrm{HCl}\ } \mathrm{R\!-\!HC{=}NH\cdot HCl}\]

\[\text{(Alkane nitrile)} \qquad\qquad \text{(imine hydrochloride)}\]

\[\mathrm{R\!-\!HC{=}NH\cdot HCl} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{R\!-\!CHO} + \mathrm{NH_4Cl}\]

\[\text{(Aldehyde)}\]

Here \(2[\mathrm{H}]\) stands for the two hydrogen atoms supplied by the reducing system; the triple bond is reduced only as far as a \(\mathrm{C{=}N}\) double bond, and water then replaces the nitrogen in the hydrolysis step. For example,

\[\mathrm{H_3C\!-\!C{\equiv}N} + 2[\mathrm{H}] \xrightarrow[\text{(reduction)}]{\ \mathrm{SnCl_2},\ \mathrm{HCl}\ } \mathrm{CH_3\!-\!HC{=}NH\cdot HCl}\]

\[\text{(Ethanenitrile)} \qquad\qquad \text{(ethanimine hydrochloride)}\]

\[\mathrm{CH_3\!-\!HC{=}NH\cdot HCl} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{CH_3\!-\!CHO} + \mathrm{NH_4Cl}\]

\[\text{(Ethanal)}\]

\[\mathrm{C_6H_5\!-\!C{\equiv}N} + 2[\mathrm{H}] \xrightarrow{\ \mathrm{SnCl_2},\ \mathrm{HCl}\ } \mathrm{C_6H_5\!-\!HC{=}NH\cdot HCl}\]

\[\text{(Benzonitrile)} \qquad\qquad \text{(Benzanimine hydrochloride)}\]

\[\mathrm{C_6H_5\!-\!HC{=}NH\cdot HCl} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!CHO} + \mathrm{NH_4Cl}\]

\[\text{(Benzaldehyde)}\]

Alternatively, nitriles are also reduced by diisobutylaluminium hydride (DIBAl-H), or \(\mathrm{AlH(i\text{-}Bu)_2}\), to imines, followed by acid hydrolysis to aldehydes. An advantage of this method is that a double or triple bond present in the same molecule is not reduced.

\[\mathrm{H_3C\!-\!CH{=}CH\!-\!CH_2\!-\!C{\equiv}N} \xrightarrow[\ \mathrm{H_3O^{+}}\ ]{\ \mathrm{AlH(i\text{-}Bu)_2}\ } \mathrm{H_3C\!-\!CH{=}CH\!-\!CH_2\!-\!CHO}\]

\[\text{(Pent-3-enenitrile)} \qquad\qquad \text{(Pent-3-enal)}\]

Notice that the \(\mathrm{C{=}C}\) double bond at position 3 survives untouched — that is exactly the selectivity being claimed.

DIBAl-H, \(\mathrm{AlH(i\text{-}Bu)_2}\) (A bulky, mild aluminium hydride → used in a dry solvent, and for esters at a low temperature of 195 K → converts a nitrile to an imine, and an ester to an aldehyde, while leaving carbon–carbon double and triple bonds in the same molecule untouched. Reach for it when the target is an aldehyde and the molecule contains other reducible groups you want to keep.)

Identify A and B in the following sequence and name the reaction.

\[\mathrm{CH_3\!-\!CH_2\!-\!C{\equiv}N} \xrightarrow{\ \mathrm{SnCl_2},\ \mathrm{HCl}\ } \mathbf{A} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathbf{B}\]

The reagent pair \(\mathrm{SnCl_2}\)/\(\mathrm{HCl}\) acting on a nitrile is the signature of the Stephen reaction. In the first step the \(\mathrm{C{\equiv}N}\) triple bond is reduced only as far as \(\mathrm{C{=}N}\), and the product is isolated as its hydrochloride.

\[\mathbf{A} = \mathrm{CH_3\!-\!CH_2\!-\!HC{=}NH\cdot HCl} \quad \text{(propanimine hydrochloride)}\]

Acid hydrolysis then replaces the nitrogen by oxygen, releasing ammonium chloride.

\[\mathbf{B} = \mathrm{CH_3\!-\!CH_2\!-\!CHO} \quad \text{(propanal)}\]

The overall change converts propanenitrile into propanal; the carbon skeleton is unchanged, since the nitrile carbon becomes the aldehyde carbon.

From Nitriles: Preparation of Ketones

Ketones are prepared by reacting nitriles with a Grignard reagent in dry ether as solvent, followed by acid hydrolysis.

\[\mathrm{H_3C\!-\!C{\equiv}N} + \mathrm{H_3CMgCl} \xrightarrow{\ \text{dry ether}\ } \mathrm{CH_3\!-\!C(CH_3){=}NMgCl}\]

\[\text{(Ethanenitrile)}\]

\[\mathrm{CH_3\!-\!C(CH_3){=}NMgCl} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{NH_3} + \mathrm{Mg(Cl)OH}\]

\[\text{(Acetone)}\]

\[\mathrm{C_6H_5\!-\!C{\equiv}N} + \mathrm{C_6H_5\!-\!MgBr} \xrightarrow{\ \text{dry ether}\ } \mathrm{C_6H_5\!-\!C(C_6H_5){=}NMgBr}\]

\[\text{(Benzonitrile)}\]

\[\mathrm{C_6H_5\!-\!C(C_6H_5){=}NMgBr} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!CO\!-\!C_6H_5} + \mathrm{NH_3} + \mathrm{Mg(Br)OH}\]

\[\text{(Benzophenone)}\]

The alkyl or aryl group of the Grignard reagent becomes the second group on the carbonyl carbon, so the ketone you obtain can be designed by choosing the Grignard reagent.

Q. Why must the solvent for a Grignard reaction be dry ether rather than, say, ethanol or moist ether?

A. The carbon atom bonded to magnesium in a Grignard reagent is strongly electron rich and reacts instantly with any acidic hydrogen. Water or an alcohol would simply protonate it and destroy the reagent before it could attack the nitrile, so the ether must be scrupulously dry.

What is the product formed when benzonitrile is treated with methylmagnesium bromide in dry ether, followed by acid hydrolysis? Name it.

The Grignard reagent adds its methyl group to the nitrile carbon, giving the magnesium salt of an imine.

\[\mathrm{C_6H_5\!-\!C{\equiv}N} + \mathrm{CH_3MgBr} \xrightarrow{\ \text{dry ether}\ } \mathrm{C_6H_5\!-\!C(CH_3){=}NMgBr}\]

Acid hydrolysis of this salt replaces the nitrogen by oxygen:

\[\mathrm{C_6H_5\!-\!C(CH_3){=}NMgBr} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!CO\!-\!CH_3} + \mathrm{NH_3} + \mathrm{Mg(Br)OH}\]

The product is \(\mathrm{C_6H_5\!-\!CO\!-\!CH_3}\), an aromatic ketone named acetophenone (methyl phenyl ketone).

From Aromatic Hydrocarbons

Aromatic aldehydes and ketones are both prepared from aromatic hydrocarbons, but by different methods.

For the aldehydes there is a difficulty to overcome. Strong oxidizing agents transform a \(\mathrm{-CH_3}\) group bonded to an aromatic ring straight into a carboxyl group (\(\mathrm{-COOH}\)), overshooting the aldehyde stage. For obtaining an aromatic aldehyde from a methyl arene, the following special methods are therefore used.

Etard Reaction

The methyl group in methyl benzene (or a methyl arene) is oxidized by the oxidizing agent chromyl chloride in carbon disulfide as solvent, to form a chromium complex, from which the corresponding benzaldehyde is obtained on acid hydrolysis. This reaction is known as the Etard reaction. The two stages are drawn in Fig. 12.4a.

\[\mathrm{C_6H_5\!-\!CH_3} + \mathrm{CrO_2Cl_2} \xrightarrow{\ \mathrm{CS_2}\ } \text{chromium complex} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!CHO}\]

Diagram

Fig. 12.4a – Etard reaction: toluene ring with a CH₃ group plus CrO₂Cl₂ in CS₂ giving the chromium complex, a benzene ring bearing a CH with two OCr(OH)Cl₂ groups, which on H₃O⁺ gives benzaldehyde; label toluene, chromyl chloride, chromium complex, benzaldehyde

The key idea is that the chromium is held back at an intermediate stage as a stable complex; only when that complex is hydrolysed is the carbon released, and it is released as \(\mathrm{-CHO}\), not \(\mathrm{-COOH}\).

Oxidation of Methyl Arene Using CrO₃

Methylarene is converted into a benzylidene diacetate on treatment with chromium oxide in acetic anhydride at 273–278 K. The diacetate derivative on acid hydrolysis gives the corresponding aldehyde. The sequence is shown in Fig. 12.4b.

\[\mathrm{C_6H_5\!-\!CH_3} + \mathrm{CrO_3} + \mathrm{(CH_3CO)_2O} \xrightarrow{\ 273\text{–}278\ \text{K}\ } \mathrm{C_6H_5\!-\!CH(OCOCH_3)_2}\]

\[\text{(Methylbenzene)} \qquad \text{(Acetic anhydride)} \qquad \text{(benzylidene diacetate)}\]

\[\mathrm{C_6H_5\!-\!CH(OCOCH_3)_2} \xrightarrow[\ \mathrm{H_3O^{+}}\ ]{\ \Delta\ } \mathrm{C_6H_5\!-\!CHO}\]

\[\text{(Benzaldehyde)}\]

Diagram

Fig. 12.4b – Oxidation of methylbenzene with CrO₃ in acetic anhydride at 273–278 K to give benzylidene diacetate (ring with CH carrying two OCOCH₃ groups), which on Δ/H₃O⁺ gives benzaldehyde

Here the low temperature and the acetic anhydride together do the protecting: as soon as the carbon is oxidized it is capped as a diacetate, so it cannot be attacked further.

Side Chain Chlorination of Toluene

Side chain chlorination of toluene gives benzal chloride, which on acid hydrolysis at 373 K gives benzaldehyde. Benzaldehyde is manufactured commercially by this method.

\[\mathrm{C_6H_5\!-\!CH_3} \xrightarrow{\ \mathrm{Cl\!-\!Cl},\ h\nu\ } \mathrm{C_6H_5\!-\!CHCl_2} \xrightarrow[\ \Delta\ 373\ \text{K}\ ]{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!CHO}\]

\[\text{(Toluene)} \qquad \text{(Benzal chloride)} \qquad \text{(Benzaldehyde)}\]

The light, written \(h\nu\) over the first arrow, is what sends the chlorination to the side chain rather than to the ring. Two chlorine atoms on the same carbon are unstable towards water, so hydrolysis converts that carbon directly into \(\mathrm{-CHO}\).

Gatterman–Koch Formylation of Arene

Benzene or substituted benzene is treated under high pressure with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride to give benzaldehyde or substituted benzaldehyde.

\[\mathrm{C_6H_6} \xrightarrow[\ \text{high pressure}\ ]{\ \mathrm{CO},\ \mathrm{HCl},\ \text{Anhyd.}\ \mathrm{AlCl_3}\ } \mathrm{C_6H_5\!-\!CHO}\]

\[\text{(Benzene)} \qquad\qquad \text{(Benzaldehyde)}\]

This is the one method in the group that does not need a methyl group to start with: the \(\mathrm{-CHO}\) group is installed directly on the ring.

Draw the structure of the product formed by the combination of carbon monoxide and HCl.

Preparation of Aromatic Ketones from Hydrocarbons

Aromatic ketones are prepared by Friedel–Crafts acylation of arene (refer to sec. 12.4.2 (a) ii and Std. XI Chemistry Textbook, Chapter 15, sec. 15.4.6).

Name the compounds which are used for the preparation of benzophenone by the Friedel–Crafts acylation reaction. Draw their structures.

12.4.3 Preparation of Aldehydes Only from Esters

Aliphatic or aromatic esters are reduced to aldehydes by using diisobutylaluminium hydride, DIBAl-H, or \(\mathrm{AlH(i\text{-}Bu)_2}\). The reaction is usually carried out at 195 K to prevent further reduction of the aldehyde produced.

\[\mathrm{R\!-\!CO\!-\!O\!-\!R'} \xrightarrow[\ \mathrm{H_3O^{+}}\ ]{\ \mathrm{AlH(i\text{-}Bu)_2}\ } \mathrm{R\!-\!CO\!-\!H}\]

\[\text{(Ester)} \qquad\qquad \text{(Aldehyde)}\]

The low temperature is not a detail to be skimmed over. An aldehyde is itself easily reduced, so if the mixture is allowed to warm, the reaction continues past the aldehyde to the alcohol. Cooling to 195 K slows that second step enough for the aldehyde to be isolated.

The ester \(\mathrm{CH_3\!-\!CH_2\!-\!CO\!-\!O\!-\!CH_3}\) is treated with \(\mathrm{AlH(i\text{-}Bu)_2}\) at 195 K, followed by acid hydrolysis. Write the structure and IUPAC name of the organic product.

Two moles of propanoyl chloride are treated with dimethyl cadmium. Write the balanced equation for the reaction and give the IUPAC name of the ketone formed.

Several of the reagents in this section are chosen for the job they do rather than for their safety. Mercuric sulfate in alkyne hydration, cadmium chloride and dialkyl cadmium in the ketone synthesis, chromyl chloride and chromium oxide in the arene oxidations, and carbon disulfide as solvent are all either toxic, highly flammable, or both. Modern process chemistry works hard to replace such reagents — mercury and cadmium compounds especially, because they are cumulative poisons and persist in the environment — with catalytic routes that use safer metals and generate less waste. Whenever any of these is used in a teaching laboratory, it is handled in a fume hood, with gloves, and with a plan for collecting the waste rather than pouring it away.

Identify the reagents necessary to achieve each of the following transformations: (i) benzoyl chloride to benzaldehyde; (ii) phenylacetylene to acetophenone; (iii) methyl 4-nitrobenzoate to 4-nitrobenzaldehyde.

12.5 Preparation of Carboxylic Acids

A carboxylic acid can be reached from several different starting materials. Broadly, either a group that already contains a carbon–oxygen or carbon–nitrogen multiple bond is hydrolysed, or a carbon skeleton is oxidized until the carbon in question carries the \(\mathrm{-COOH}\) group. The six routes of this section are gathered in FC 12.5a.

Flowchart

FC 12.5a – Routes to carboxylic acids: nitrile R–C≡N → (dil. acid) → amide R–CONH₂ → (dil. acid) → R–COOH; acyl chloride R–COCl → (H₂O) → R–COOH; anhydride → (H₂O) → 2 R–COOH; ester → (dil. H₂SO₄) → R–COOH, or → (dil. NaOH, then conc. HCl) → R–COOH; alkyl benzene → (alk./acidic KMnO₄ or dil. HNO₃ or chromic acid) → aromatic acid; alkene → (KMnO₄ in dil. H₂SO₄) → R–COOH; Grignard reagent R–MgX → (dry ice in dry ether) → R–COOMgX → (dil. HCl) → R–COOH

12.5.1 From Nitriles and Amides

Alkyl nitriles or aryl nitriles on acid hydrolysis give amides. Amides on further acid hydrolysis give the corresponding carboxylic acids. Hydrolysis is carried out by using dilute mineral acids like dilute sulfuric acid or dilute hydrochloric acid.

The first molecule of water adds across the \(\mathrm{C{\equiv}N}\) bond to give an unstable intermediate, which rearranges to the amide:

\[\mathrm{R\!-\!C{\equiv}N} + \mathrm{H_2O} \longrightarrow [\,\mathrm{R\!-\!C(OH){=}NH}\,] \longrightarrow \mathrm{R\!-\!CO\!-\!NH_2}\]

\[\text{(Nitrile)} \qquad\qquad \text{(Amide)}\]

\[\mathrm{R\!-\!CO\!-\!NH_2} \xrightarrow[\ \text{dil. HCl}\ ]{\ \mathrm{H_2O}\ } \mathrm{R\!-\!COOH} + \mathrm{NH_4Cl}\]

\[\text{(carboxylic acid)}\]

Written as a single overall change, the two stages become

\[\mathrm{R\!-\!C{\equiv}N} + 2\,\mathrm{H_2O} + \text{dil. HCl} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!COOH} + \mathrm{NH_4Cl}\]

Two molecules of water are needed because two separate additions take place — one to reach the amide and one to replace the nitrogen. The nitrogen leaves as ammonium chloride because the medium is acidic.

12.5.2 From Acyl Chloride and Anhydrides

Acyl chlorides on hydrolysis with water give carboxylic acids. This method is useful for the preparation of aliphatic as well as aromatic acid.

\[\mathrm{R\!-\!COCl} + \mathrm{H_2O} \longrightarrow \mathrm{R\!-\!COOH} + \mathrm{H\!-\!Cl}\]

\[\text{(Acyl chloride)} \qquad \text{(Carboxylic acid)}\]

Anhydrides on hydrolysis with water give carboxylic acids. Since an anhydride is, in effect, two acid molecules joined with the loss of one water molecule, putting that water back releases both acid molecules.

\[\mathrm{R\!-\!CO\!-\!O\!-\!CO\!-\!R} + \mathrm{H_2O} \longrightarrow 2\,\mathrm{R\!-\!CO\!-\!O\!-\!H}\]

\[\text{(Anhydride)} \qquad\qquad \text{(Carboxylic acid)}\]

12.5.3 From Esters

Carboxylic acids can be obtained from esters either by acid hydrolysis or by alkaline hydrolysis.

Acid Hydrolysis of Ester

Esters on hydrolysis with a dilute mineral acid like dilute HCl or dilute \(\mathrm{H_2SO_4}\) give the corresponding carboxylic acid.

\[\mathrm{R\!-\!CO\!-\!O\!-\!R'} + \mathrm{H_2O} \xrightarrow[\ \Delta\ ]{\ \text{dil.}\ \mathrm{H_2SO_4}\ } \mathrm{R\!-\!CO\!-\!O\!-\!H} + \mathrm{R'\!-\!OH}\]

\[\text{(Ester)} \qquad\qquad \text{(Carboxylic acid)} \qquad \text{(Alcohol)}\]

Note that the acid here is a catalyst, not a reactant: it is not consumed, and the reaction is reversible, which is why an excess of water is used to drive it forward.

Alkaline Hydrolysis of Ester

Alkaline hydrolysis of ester using dilute alkali like dilute NaOH or dilute KOH forms a solution of the water soluble sodium or potassium salt of the acid (carboxylate). On acidification with concentrated HCl, the free acid is formed.

\[\mathrm{H_5C_2\!-\!CO\!-\!O\!-\!CH_3} + \text{dil. NaOH} \xrightarrow{\ \Delta\ } \mathrm{H_5C_2\!-\!CO\!-\!O^{-}Na^{+}} + \mathrm{CH_3\!-\!OH}\]

\[\text{(Methyl propanoate)} \qquad\qquad \text{(Sodium propanoate)}\]

\[\mathrm{H_5C_2\!-\!CO\!-\!O^{-}Na^{+}} + \mathrm{H_2O} \xrightarrow[\ \text{Conc. HCl}\ ]{\ \mathrm{H^{+}}\ } \mathrm{H_5C_2\!-\!CO\!-\!O\!-\!H} + \mathrm{NaOH}\]

\[\text{(Sodium propanoate)} \qquad\qquad \text{(Propanoic acid)}\]

Q. Why does alkaline hydrolysis need a second step with concentrated HCl, while acid hydrolysis gives the acid straight away?

A. In alkali the acid cannot survive as an acid — as soon as it forms, the alkali neutralises it to the carboxylate salt. Adding a strong acid afterwards puts the proton back on the carboxylate and releases the free carboxylic acid.

acid hydrolysis vs alkaline hydrolysis of an ester

Feature Acid hydrolysis Alkaline hydrolysis
Reagent Dilute HCl or dilute \(\mathrm{H_2SO_4}\) Dilute NaOH or dilute KOH
Immediate organic products Carboxylic acid + alcohol Carboxylate salt + alcohol
Reversible? Yes No — the salt cannot re-esterify
Extra step needed None Acidification with concentrated HCl to free the acid
Special name — Saponification

The sodium or potassium salts of higher fatty acids are known as soaps. Hence alkaline hydrolysis of esters is called saponification (Std. XI Chemistry Textbook, Chapter 16).

NoteDefinition

Saponification is the alkaline hydrolysis of an ester, in which the ester is heated with a dilute alkali to give the sodium or potassium salt of the acid (the carboxylate) together with the alcohol.

Soap is made by exactly this reaction on a very large scale. Natural fats and oils are esters of long-chain fatty acids with glycerol; boiling them with sodium hydroxide splits every ester linkage and leaves the sodium salts of those fatty acids — soap — along with glycerol, which is recovered and sold separately. Potassium hydroxide gives softer, more soluble soaps, which is why it is preferred for liquid and shaving soaps.

12.5.4 From Alkyl Benzene

Aromatic carboxylic acids can be prepared by oxidation of alkyl benzene with dilute \(\mathrm{HNO_3}\) or alkaline/acidic \(\mathrm{KMnO_4}\) or chromic acid. The entire alkyl chain, regardless of its length, is oxidized to a carboxyl group. A tertiary alkyl substituent on benzene, however, is not oxidized.

For example, both toluene and cumene give the same product, benzoic acid, as shown in Fig. 12.5a.

\[\mathrm{C_6H_5\!-\!CH_3} \xrightarrow[\ \Delta\ ]{\ \mathrm{KMnO_4}\text{, }\mathrm{KOH}\ } \mathrm{C_6H_5\!-\!COO^{-}K^{+}} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!COOH}\]

\[\text{(Toluene)} \qquad \text{(Potassium benzoate)} \qquad \text{(Benzoic acid)}\]

\[\mathrm{C_6H_5\!-\!CH(CH_3)_2} \xrightarrow[\ \Delta\ ]{\ \mathrm{KMnO_4}\text{, }\mathrm{KOH}\ } \mathrm{C_6H_5\!-\!COO^{-}K^{+}} \xrightarrow{\ \mathrm{H_3O^{+}}\ } \mathrm{C_6H_5\!-\!COOH}\]

\[\text{(Cumene)} \qquad \text{(Potassium benzoate)} \qquad \text{(Benzoic acid)}\]

Diagram

Fig. 12.5a – Two parallel schemes: toluene (ring + CH₃) and cumene (ring + CH(CH₃)₂), each with KMnO₄/KOH and Δ giving potassium benzoate (ring + COO⁻K⁺), which with H₃O⁺ gives benzoic acid; label all four compounds

Q. Cumene has a three-carbon side chain, yet it gives benzoic acid with only one carbon outside the ring. Where do the other two carbons go?

A. The oxidation attacks the side chain at the carbon joined to the ring and burns away the rest of the chain as carbon dioxide and water. Only the carbon directly bonded to the ring survives, and it survives as \(\mathrm{-COOH}\) — which is why chain length makes no difference to the product.

alkaline \(\mathrm{KMnO_4}\) on an alkyl benzene (A powerful oxidising agent → used hot, in alkaline solution, followed by acidification → converts any alkyl side chain on a benzene ring, however long, into a single \(\mathrm{-COOH}\) group. It cannot oxidize a tertiary alkyl group such as the one in tert-butylbenzene, because that carbon carries no hydrogen atom for the oxidant to remove. Use it when you need the ring intact and the side chain cut back to an acid.)

12.5.5 From Alkenes

Carboxylic acids can also be prepared by the oxidation of alkenes by \(\mathrm{KMnO_4}\) in dilute \(\mathrm{H_2SO_4}\). The double bond is cleaved and each doubly bonded carbon that carries hydrogen ends up as a carboxyl carbon.

\[\mathrm{H_5C_6\!-\!CH{=}CH_2} \xrightarrow[\ \Delta\ ]{\ \mathrm{KMnO_4},\ \text{dil.}\ \mathrm{H_2SO_4}\ } \mathrm{H_5C_6\!-\!COOH}\]

\[\text{(Phenyl ethene)} \qquad\qquad \text{(Benzoic acid)}\]

When the double bond lies inside a ring, cleaving it does not give two separate molecules; the ring simply opens out into a chain carrying a \(\mathrm{-COOH}\) group at each end, as drawn in Fig. 12.5b.

\[\text{Cyclohexene} \xrightarrow[\ \Delta\ ]{\ \mathrm{KMnO_4},\ \text{dil.}\ \mathrm{H_2SO_4}\ } \mathrm{HOOC\!-\!(CH_2)_4\!-\!COOH}\]

\[\text{(Adipic acid)}\]

Diagram

Fig. 12.5b – Cyclohexene ring with KMnO₄/dil. H₂SO₄ and Δ opening to an open chain bearing COOH at both ends, labelled adipic acid

12.5.6 From Grignard Reagent

A Grignard reagent in dry ether solvent is added to solid carbon dioxide (dry ice) to give a complex, which on acid hydrolysis gives the corresponding carboxylic acid.

\[\mathrm{R\!-\!Mg\!-\!X} + \mathrm{O{=}C{=}O} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!COOMgX}\]

\[\text{(Alkyl magnesium halide)} \quad \text{(dry ice)} \quad \text{(complex)}\]

\[\mathrm{R\!-\!COOMgX} \xrightarrow[\ \text{dil. HCl}\ ]{\ \mathrm{HOH}\ } \mathrm{R\!-\!COOH} + \mathrm{Mg(X)OH}\]

\[\text{(carboxylic acid)}\]

This route is valuable because it adds a carbon atom: the carbon of the carbon dioxide becomes the carboxyl carbon, so the acid formed always has one carbon more than the alkyl halide you started from.

Organomagnesium halides of the type \(\mathrm{R\!-\!Mg\!-\!X}\) are named after the French chemist Victor Grignard, who developed them at the very beginning of the twentieth century and was awarded the Nobel Prize in Chemistry for the work.

Collect information about dry ice and ice from the internet. Draw the chemical structures of dry ice and regular ice. Prepare a chart of the uses of dry ice.

How would you prepare 2-methylpropanoic acid, \(\mathrm{(CH_3)_2CH\!-\!COOH}\), starting from 2-bromopropane?

The target acid has four carbon atoms, while 2-bromopropane has three. One carbon must therefore be added, and the Grignard–dry ice route does exactly that.

First the halide is converted into its Grignard reagent in dry ether:

\[\mathrm{(CH_3)_2CH\!-\!Br} + \mathrm{Mg} \xrightarrow{\ \text{dry ether}\ } \mathrm{(CH_3)_2CH\!-\!MgBr}\]

The Grignard reagent is then added to solid carbon dioxide:

\[\mathrm{(CH_3)_2CH\!-\!MgBr} + \mathrm{O{=}C{=}O} \xrightarrow{\ \text{dry ether}\ } \mathrm{(CH_3)_2CH\!-\!COOMgBr}\]

Acid hydrolysis of the complex liberates the acid:

\[\mathrm{(CH_3)_2CH\!-\!COOMgBr} \xrightarrow[\ \text{dil. HCl}\ ]{\ \mathrm{HOH}\ } \mathrm{(CH_3)_2CH\!-\!COOH} + \mathrm{Mg(Br)OH}\]

The product is 2-methylpropanoic acid, with the new carboxyl carbon supplied by the carbon dioxide.

Write the two steps by which methyl butanoate is converted into butanoic acid using dilute NaOH.

Step 1 — alkaline hydrolysis gives the sodium salt of the acid together with the alcohol:

\[\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CO\!-\!O\!-\!CH_3} + \text{dil. NaOH} \xrightarrow{\ \Delta\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CO\!-\!O^{-}Na^{+}} + \mathrm{CH_3\!-\!OH}\]

Step 2 — acidification of the sodium salt with concentrated HCl sets the free acid free:

\[\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CO\!-\!O^{-}Na^{+}} \xrightarrow{\ \text{Conc. HCl}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!COOH} + \mathrm{NaCl}\]

The organic product is butanoic acid. The alcohol released in step 1 is methanol, since the ester was the methyl ester.

tert-Butylbenzene, \(\mathrm{C_6H_5\!-\!C(CH_3)_3}\), is heated with alkaline \(\mathrm{KMnO_4}\) and the mixture is then acidified. State whether benzoic acid is obtained, and justify your answer in terms of the structure of the side chain.

Predict the products (name and structure) in the following reactions: \[\mathrm{CH_3CH_2CN} \xrightarrow[\ \Delta\ ]{\ \text{dil. HCl}\ } ? \qquad \mathrm{CH_3\!-\!CONH_2} \xrightarrow[\ \Delta\ ]{\ \text{dil. HCl}\ } ?\] \[\mathrm{C_6H_5\!-\!CH_2\!-\!CH_3} \xrightarrow[\ \Delta\ ]{\ \text{alk. }\mathrm{KMnO_4}\ } ? \qquad \mathrm{C_6H_5\!-\!COO\!-\!C_2H_5} \xrightarrow[\ \Delta\ ]{\ \text{dil. }\mathrm{H_2SO_4}\ } ?\] \[\mathrm{CH_3MgBr} \xrightarrow[\ \text{(ii) dil. HCl}\ ]{\ \text{(i) dry ice/dry ether}\ } ?\]

12.6 Physical Properties

12.6.1 Nature of Intermolecular Forces

The carbonyl bond (\(>\mathrm{C=O}\)) in aldehydes and ketones is a polar covalent bond. Oxygen is much more electronegative than carbon, so the shared electrons of the double bond are pulled towards oxygen, leaving the oxygen slightly negative (\(\delta-\)) and the carbon slightly positive (\(\delta+\)). Every molecule therefore behaves as a small permanent dipole.

As a result, these compounds contain dipole–dipole forces of attraction, illustrated in Fig. 12.1. The molecules orient in such a way as to have oppositely polarized atoms facing each other, so that the positive end of one molecule sits next to the negative end of the next.

Diagram

Fig. 12.1 – Dipole–dipole attraction in carbonyl compounds: two >C=O groups drawn head to tail, with δ⁻ marked on each oxygen and δ⁺ on each carbon, and dotted lines showing the attraction between the δ⁻ oxygen of one molecule and the δ⁺ carbon of the other

The carboxyl group of a carboxylic acid contains an O–H bond which is responsible for the formation of hydrogen bonding. Thus, carboxylic acids have the strongest intermolecular forces of attraction (Fig. 12.3 in section 12.6.4).

Q. Aldehydes and ketones have an oxygen atom carrying lone pairs. Why can they not form hydrogen bonds with each other, the way alcohols and acids do?

A. A hydrogen bond needs both a donor and an acceptor. The carbonyl oxygen is a good acceptor, but an aldehyde or ketone has no hydrogen attached to oxygen, so it has no donor of its own. Between two carbonyl molecules there is nothing to donate, and the strongest force available is the dipole–dipole attraction. This is exactly why they can still hydrogen bond with water, which supplies the donor hydrogen.

12.6.2 Physical State and Boiling Points of Aldehydes and Ketones

Formaldehyde is a gas at room temperature and has an irritating odour. Acetaldehyde is an extremely volatile, colourless liquid. Higher aldehydes have a pleasant odour. Acetone is a liquid at room temperature and has a pleasant odour, but most of the higher ketones have bland odours.

Increasing boiling points in the homologous series of aldehydes and ketones are listed in Table 12.4. A homologous series is a family of compounds differing by one \(\mathrm{-CH_2-}\) unit at a time, so the table lets us see the effect of chain length alone.

Table 12.4 Boiling points of aldehydes and ketones

Number of carbon atoms Aldehyde Boiling point Ketone Boiling point
1 Methanal 252 K —– ——–
2 Ethanal 294 K ——- ———
3 Propanal 319 K Propanone 329 K
4 Butanal 348 K Butan-2-one 353 K
5 Pentanal 376 K Pentan-2-one 375 K
6 Hexanal 392 K Hexan-2-one 400 K

As the chain grows, the boiling point rises steadily. The dipole–dipole force does not change much from one member to the next, but the growing hydrocarbon chain increases the surface over which weak van der Waals forces can act, and more energy is needed to pull the molecules apart. Notice also that the first two ketone entries are blank: a ketone needs a carbonyl carbon flanked by two carbon groups, so the smallest possible ketone has three carbon atoms.

boiling points of the aldehyde and ketone homologous series, plotted from Table 12.4 in Fig. 12.6a (Read Fig. 12.6a and answer: (i) Does the boiling point rise by roughly equal steps for each extra carbon atom, or by increasing steps? (ii) For which carbon number do the aldehyde and the ketone curves come closest together or cross? (iii) State one reason why adding a \(\mathrm{-CH_2-}\) unit to the chain should raise the boiling point at all, given that the carbonyl group is unchanged.)

Diagram

Fig. 12.6a – Line graph of boiling point in K (y-axis, 240 to 410) against number of carbon atoms (x-axis, 1 to 6), with one line for the aldehydes methanal to hexanal and a second line for the ketones propanone to hexan-2-one, each point labelled with the compound name

12.6.3 Solubility of Aldehydes and Ketones

The oxygen atom of the \(>\mathrm{C=O}\) group can involve itself in hydrogen bonding with a water molecule, as shown in Fig. 12.2. Here water supplies the hydrogen and the carbonyl oxygen supplies the lone pair, so the hydrogen bond that was impossible between two carbonyl molecules becomes possible between a carbonyl molecule and water.

Diagram

Fig. 12.2 – Hydrogen bonding in carbonyl compound and water: a >C=O group with δ⁻ on the oxygen and δ⁺ on the carbon, and a dotted hydrogen bond from that oxygen to the δ⁺ hydrogen of an H–O–H molecule

As a result of this, the lower aldehydes and ketones are water soluble; acetaldehyde and acetone are common examples. As the molecular mass increases, the proportion of the hydrocarbon part of the molecule increases, which cannot form hydrogen bonds, and the water solubility decreases.

Acetone mixes with water in all proportions, but hexan-2-one is only slightly soluble in water. Explain this difference in terms of the two parts of each molecule.

12.6.4 Physical State, Boiling Points and Solubilities of Carboxylic Acids

Lower aliphatic carboxylic acids up to nine carbon atoms are colourless liquids with irritating odours. The higher homologues are colourless, odourless, wax-like solids and have low volatility. Boiling points of the lower carboxylic acids are listed in Table 12.5.

Table 12.5 Increasing boiling points of carboxylic acids

Name Formula Boiling point in K
Formic acid \(\mathrm{HCOOH}\) 373 K
Acetic acid \(\mathrm{CH_3COOH}\) 391 K
Propionic acid \(\mathrm{CH_3CH_2COOH}\) 414 K
Butyric acid \(\mathrm{CH_3CH_2CH_2COOH}\) 437 K
Valeric acid \(\mathrm{CH_3CH_2CH_2CH_2COOH}\) 460 K
Why Carboxylic Acids Boil So High

Carboxylic acids have higher boiling points than those of alkanes, ethers, alcohols, aldehydes and ketones of comparable mass, as Table 12.6 shows. The reason is that, in the liquid phase, carboxylic acids form a dimer in which two molecules are held together by two hydrogen bonds. The acidic hydrogen of one molecule forms a hydrogen bond with the carbonyl oxygen of the other molecule, as drawn in Fig. 12.3.

Diagram

Fig. 12.3 – Dimer of acetic acid: two acetic acid molecules facing each other in a ring, the O–H hydrogen of each molecule dotted across to the carbonyl oxygen of the other, with δ⁺ and δ⁻ marked; caption “two molecules held by two hydrogen bonds”

This doubles the size of the molecule, resulting in an increase in the intermolecular van der Waals forces, which in turn results in a high boiling point. In the case of acetic acid, dimers exist even in the gas phase (Fig. 12.3).

Table 12.6 Variation of boiling point with functional group

Compound Family Molecular mass Boiling point Strength of intermolecular forces
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) Alkane 58 272 K increases down the column
\(\mathrm{CH_3\!-\!O\!-\!CH_2\!-\!CH_3}\) Ether 60 281 K
\(\mathrm{CH_3\!-\!CH_2\!-\!CHO}\) Aldehyde 58 322 K
\(\mathrm{CH_3\!-\!CO\!-\!CH_3}\) Ketone 58 329 K
\(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH}\) Alcohol 60 370 K
\(\mathrm{CH_3\!-\!COOH}\) Carboxylic acid 60 391 K

Every compound in Table 12.6 has a molecular mass of 58 or 60, so mass is effectively held constant and the whole 119 K spread of boiling points is the work of the functional group alone.

strength of intermolecular force and the resulting boiling point (Relative strength of intermolecular force: hydrogen bond > dipole–dipole attraction > van der Waals force. Hence boiling points, for compounds of comparable mass: carboxylic acids > alcohols > ketones > aldehydes > ethers > alkanes. Reason: acids form two hydrogen bonds per pair as a dimer; alcohols form single hydrogen bonds; ketones and aldehydes have only dipole–dipole attraction, slightly stronger in ketones because two alkyl groups make the carbonyl more polar; ethers and alkanes are left with van der Waals forces alone.)

Q2. Acetic acid and propanal have almost the same molecular mass, yet acetic acid boils about 70 K higher. What could a single small molecule be doing to hold on to its neighbours so tightly?

A. It is pairing up. Propanal has only dipole–dipole attraction between its molecules, because it has no O–H hydrogen to donate. Acetic acid does have that hydrogen, and it uses it to lock onto the carbonyl oxygen of a neighbouring acid molecule while that neighbour does exactly the same in return. The two molecules are therefore held by two hydrogen bonds at once, forming the dimer of Fig. 12.3. Boiling now means separating a particle of double the size held by two strong hydrogen bonds, not a single molecule held by weak dipole forces — hence the jump from 322 K for propanal to 391 K for acetic acid in Table 12.6.

The weak attractions between non-polar molecules are named after the Dutch physicist Johannes Diderik van der Waals, who first accounted for them in his work on the behaviour of real gases and was awarded the Nobel Prize in Physics for it.

Formaldehyde and acetaldehyde are inconvenient to store as such, so they are supplied in other forms. Formaldehyde is available commercially as a solid polymer called paraformaldehyde, \(\mathrm{HO\!-\![CH_2\!-\!O]_n\!-\!H}\), and as trioxane, \(\mathrm{(CH_2O)_3}\), which has a cyclic structure; both are convenient sources of formaldehyde for chemical reactions. The aqueous solution of formaldehyde gas is called formalin, which is used for the preservation of biological and anatomical specimens. When dry formaldehyde is required, it is obtained by heating paraformaldehyde or trioxane. Acetaldehyde is likewise conveniently used as its solid trimer, paraldehyde, and its tetramer, metaldehyde.

Solubility of Carboxylic Acids

Lower aliphatic carboxylic acids containing up to four carbons are miscible with water due to the formation of intermolecular hydrogen bonds between carboxylic acid molecules and solvent water molecules. The solubility of carboxylic acids in water decreases with increase in molecular mass.

Higher carboxylic acids are practically insoluble in water due to the increased hydrophobic (water hating) interaction of the hydrocarbon part with water. Aromatic acids like benzoic acid are also practically insoluble in water at room temperature. Water insoluble carboxylic acids are soluble in less polar organic solvents like ether, alcohol, benzene and so on.

Arrange the following four compounds, all of nearly equal molecular mass, in increasing order of boiling point and justify the order: butane, ethyl methyl ether, propanal, propan-1-ol.

The molecular masses are 58, 60, 58 and 60, so mass can be set aside and the order is decided entirely by intermolecular forces.

Butane is non-polar and has only van der Waals forces, so it is held together most weakly and boils lowest, at 272 K.

Ethyl methyl ether has a slightly polar C–O–C linkage but no hydrogen bonded to oxygen, so its forces remain essentially van der Waals with a weak dipole contribution; it boils at 281 K.

Propanal has a strongly polar \(>\mathrm{C=O}\) group and therefore genuine dipole–dipole attraction, which is stronger than the previous two; it boils at 322 K.

Propan-1-ol carries an O–H group, so its molecules hydrogen bond to one another. Hydrogen bonding is the strongest of the three forces here, and it boils highest, at 370 K.

Increasing order of boiling point: butane < ethyl methyl ether < propanal < propan-1-ol.

12.7 Polarity of Carbonyl Group

The polarity of a carbonyl group originates from the higher electronegativity of oxygen relative to carbon, as well as from resonance effects, as shown in Fig. 12.4.

Two separate influences are therefore at work on the same bond. The first is the inductive pull: because oxygen holds electrons more tightly than carbon does, the shared electrons of the \(\sigma\) bond are drawn towards oxygen. The second is resonance: the \(\pi\) electrons of the double bond can shift completely onto oxygen, giving a structure in which oxygen carries a full negative charge and carbon a full positive charge.

Diagram

Fig. 12.4 – Polarity of carbonyl group in four parts: (A) >C=O with an arrow on the bond showing electron shift to O, δ⁻ on O and δ⁺ on C; (B) the major resonance structure with the intact C=O and two lone pairs on O; (C) the minor resonance structure with a C–O single bond, O carrying a negative charge and three lone pairs, and C carrying a positive charge; (D) the resonance hybrid drawn with a dashed partial double bond, δ⁻ on O and δ⁺ on C

Structure (C) is described as the minor contributor because separating charge costs energy, while structure (B), with no separated charge, is the major contributor. The true molecule is neither one nor the other; it is the resonance hybrid (D), in which the oxygen carries a partial negative charge and the carbon a partial positive charge.

Q. If structure (C) is the minor contributor, why does it matter at all?

A. A minor contributor still shifts the average electron distribution of the hybrid. Even a small contribution from (C) leaves measurable positive charge on the carbon, and it is that partial positive charge, not any full charge, which decides how the molecule reacts.

The carbonyl carbon has positive polarity, as seen in structures (A) and (D). Therefore, it is electron deficient. As a result, this carbon atom is electrophilic (electron loving) and is susceptible to attack by a nucleophile (\(\mathrm{Nu\!:^{-}}\)).

NoteDefinition

An electrophile is a species that is short of electrons and therefore seeks a pair of electrons; the carbonyl carbon, bearing a partial positive charge, is an electrophilic centre.

NoteDefinition

A nucleophile is a species carrying a lone pair or a negative charge, which donates that electron pair to an electron-deficient centre; nucleophiles attack the carbonyl carbon, not the carbonyl oxygen.

The paired terms nucleophile and electrophile were introduced by the English chemist Christopher Ingold in the 1930s, as part of the effort to describe organic reactions in terms of where electrons come from and where they go.

Draw the structure of propanone and indicate its polarity.

12.7.1 Reactivity of Aldehydes and Ketones

The reactivity of aldehydes and ketones is due to the polarity of the carbonyl group, which results in the electrophilicity of carbon. In general, aldehydes are more reactive than ketones toward nucleophilic attack. This can be well explained in terms of both the electronic effects and the steric effect.

Influence of Electronic Effects

Alkyl groups have an electron donating inductive effect (\(+\mathrm{I}\)). A ketone has two electron donating alkyl groups bonded to the carbonyl carbon, which are responsible for decreasing its positive polarity and its electrophilicity. In contrast, aldehydes have only one electron donating group bonded to the carbonyl carbon. This makes aldehydes more electrophilic than ketones.

The logic is worth stating plainly: electron density pushed towards the carbonyl carbon partly cancels its \(\delta+\) charge. Two alkyl groups push twice as much as one, so the ketone’s carbonyl carbon is the less hungry of the two for a nucleophile’s electron pair.

Steric Effects

Two bulky alkyl groups in a ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.

On the other hand, a nucleophile can easily attack the carbonyl carbon in an aldehyde, because it has one alkyl group and is less crowded, or sterically less hindered. Hence aldehydes are more easily attacked by nucleophiles. The two situations are compared in Fig. 12.7a.

Diagram

Fig. 12.7a – Side by side: a ketone R–C(=O)–R with δ⁻ on O and δ⁺ on C, arrows from both R groups pushing electrons towards the carbon and the two bulky R groups shown blocking an approaching nucleophile; and an aldehyde R–C(=O)–H with only one such arrow and a clear path for the approaching nucleophile

So the two effects point the same way. Electronically the aldehyde carbon carries more positive charge, and geometrically it is easier to reach — and reactivity towards nucleophilic addition depends on both.

Aromatic aldehydes are less reactive than aliphatic aldehydes in nucleophilic addition reactions. This is due to the electron-donating resonance effect of the aromatic ring, which makes the carbonyl carbon less electrophilic. The ring is able to feed \(\pi\) electron density into the carbonyl group in a way that a simple alkyl group cannot.

aldehydes vs ketones towards nucleophilic addition

Feature Aldehyde Ketone
Groups on carbonyl carbon One alkyl or aryl group and one H Two alkyl or aryl groups
Electron donating (\(+\mathrm{I}\)) groups present One Two
Positive polarity of carbonyl carbon Higher Lower
Crowding around carbonyl carbon Less hindered More hindered
Reactivity towards nucleophiles More reactive Less reactive

reactivity of carbonyl compounds towards nucleophilic addition (As groups are added to the carbonyl carbon → reactivity falls: \(\mathrm{HCHO} > \mathrm{CH_3CHO} > \mathrm{CH_3COCH_3}\), because each alkyl group both donates electron density into the electron-deficient carbon and crowds the space around it. Replacing an alkyl group by an aryl group ↓ reactivity further within the same family — an aromatic aldehyde is less reactive than an aliphatic aldehyde, and an aromatic ketone less reactive than an aliphatic ketone — because the ring feeds \(\pi\) electron density into the carbonyl by resonance.)

Which of the two, acetone or acetophenone, is attacked more readily by a nucleophile? Give reasons.

Both compounds are ketones, so the comparison rests on what the two groups attached to the carbonyl carbon are doing.

In acetone, \(\mathrm{CH_3\!-\!CO\!-\!CH_3}\), the carbonyl carbon carries two methyl groups. Each donates electron density by the \(+\mathrm{I}\) effect, reducing the \(\delta+\) charge on that carbon.

In acetophenone, \(\mathrm{C_6H_5\!-\!CO\!-\!CH_3}\), one of those groups is replaced by a phenyl ring. Besides its own bulk, the ring donates \(\pi\) electron density to the carbonyl carbon by resonance, and this resonance donation lowers the positive charge on the carbon much more effectively than an inductive push does. The phenyl group is also larger than a methyl group, so it blocks the approach of the nucleophile more.

Both the electronic and the steric effect therefore make acetophenone the less electrophilic and more hindered of the two. Acetone is attacked more readily by a nucleophile.

Draw the two resonance structures of the carbonyl group of ethanal, and use them to explain why a nucleophile attacks the carbon atom of the group rather than the oxygen atom.

12.8 Chemical Properties of Aldehydes and Ketones

12.8.1 Laboratory Tests for Aldehydes and Ketones

Aldehydes are easily oxidized to carboxylic acids and therefore act as reducing agents toward mild oxidizing agents. Ketones do not have a hydrogen atom directly attached to the carbonyl carbon. Hence they are not oxidized by mild oxidizing agents.

The structural reason is worth pausing on. Oxidising an aldehyde means converting \(\mathrm{-CHO}\) into \(\mathrm{-COOH}\), and the hydrogen sitting on the carbonyl carbon is what makes this possible without breaking any carbon–carbon bond. A ketone has no such hydrogen, so oxidising it would mean breaking a strong C–C bond, and a mild oxidant cannot do that.

On the basis of this difference in reactivity, aldehydes and ketones are distinguished by the following tests. The whole scheme is set out as a decision path in FC 12.8a.

Flowchart

FC 12.8a – Distinguishing an aldehyde from a ketone: unknown carbonyl compound → add Schiff’s reagent; if pink/red/magenta colour appears → aldehyde, if no colour → not an aldehyde → boil with Tollens’ reagent; if silver mirror forms → aldehyde, if no mirror → boil with Fehling solution; if red precipitate of Cu₂O forms → aldehyde (note: α-hydroxy ketone also positive), if no precipitate → add freshly prepared sodium nitroprusside and basify with NaOH; if red colour appears → ketone

Tests Given by Only Aldehydes

Schiff test. When an alcoholic solution of an aldehyde is treated with a few drops of Schiff’s reagent, a pink or red or magenta colour appears. This confirms the presence of the aldehydic (\(\mathrm{-CHO}\)) group.

Schiff’s reagent (A colourless solution obtained by passing sulfur dioxide gas, the oxidant, through the magenta coloured solution of p-rosaniline hydrochloride → used as a few drops added to an alcoholic solution of the unknown → the magenta colour returns if an aldehyde is present, which is why the test is read as a colour restoration.)

Tollens’ test or silver mirror test. When an aldehyde is boiled with Tollens’ reagent (ammoniacal silver nitrate), a silver mirror is formed. The aldehyde is oxidized to the carboxylate ion by Tollens’ reagent, and the \(\mathrm{Ag^{+}}\) ion is reduced to \(\mathrm{Ag}\).

\[\mathrm{R\!-\!CHO} + 2\,\mathrm{Ag(NH_3)_2^{+}} + 3\,\mathrm{OH^{-}} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!COO^{-}} + 2\,\mathrm{Ag}\downarrow + 4\,\mathrm{NH_3}\uparrow + 2\,\mathrm{H_2O}\]

\[\text{(aldehyde)} \qquad \text{(Tollens' reagent)} \qquad \text{(carboxylate)} \quad \text{(Silver mirror)}\]

Two changes happen in the same equation, and it is worth naming them. The carbon of \(\mathrm{-CHO}\) gains an oxygen, so it is oxidized; the silver in \(\mathrm{Ag(NH_3)_2^{+}}\) goes from the \(+1\) state to free metal, so it is reduced. A reaction in which one species is oxidized while another is simultaneously reduced is a redox reaction.

Tollens’ reagent (Ammoniacal silver nitrate → prepared by mixing a few drops of \(\mathrm{AgNO_3}\) solution with a few mL of dilute sodium hydroxide solution, when a brown precipitate forms, and then dissolving that precipitate by adding dilute ammonium hydroxide → used hot, to detect an aldehyde group by the deposition of a bright silver mirror on clean glass.)

Fehling test. When a mixture of an aldehyde and Fehling solution is boiled in hot water, a red precipitate of cuprous oxide is formed.

\[\mathrm{R\!-\!CHO} + 2\,\mathrm{Cu^{2+}} + 5\,\mathrm{OH^{-}} \xrightarrow{\ \text{boil}\ } \mathrm{R\!-\!COO^{-}} + \mathrm{Cu_2O}\downarrow + 3\,\mathrm{H_2O}\]

\[\text{(Aldehyde)} \quad \text{(Fehling solution)} \qquad \text{(carboxylate ion)} \quad \text{(red ppt)}\]

An aldehyde is oxidized to the carboxylate ion by Fehling solution, and the \(\mathrm{Cu^{2+}}\) ion is reduced to the \(\mathrm{Cu^{+}}\) ion. It may be noted that an \(\alpha\)-hydroxy ketone also gives this test positive, so a positive Fehling test alone does not prove that a compound is an aldehyde.

Fehling solution (A mixture of two separately stored solutions, Fehling A and Fehling B, combined just before use → Fehling A is prepared by dissolving crystals of copper sulfate in concentrated sulfuric acid, and Fehling B by dissolving sodium potassium tartarate in sodium hydroxide solution → used boiling, to detect an aldehyde by the red precipitate of cuprous oxide.)

the four laboratory tests

Test Reagent Positive observation Given by
Schiff test Schiff’s reagent Pink, red or magenta colour Aldehydes only
Tollens’ test Ammoniacal silver nitrate Silver mirror on the glass Aldehydes only
Fehling test Fehling A + Fehling B Red precipitate of \(\mathrm{Cu_2O}\) Aldehydes; also \(\alpha\)-hydroxy ketones
Sodium nitroprusside test Sodium nitroprusside, then NaOH Red colour in solution Ketones

Q1. An aldehyde and a ketone both contain the same C=O group. Why does only the aldehyde reduce silver ions to a silver mirror?

A. Because the test is not really about the C=O group at all — it is about the hydrogen sitting on the carbonyl carbon. Depositing silver requires the organic molecule to give up electrons, that is, to be oxidized. For an aldehyde this only means converting \(\mathrm{-CHO}\) into \(\mathrm{-COO^{-}}\), which the hydrogen on that carbon allows without disturbing the carbon skeleton. A ketone has two carbon groups and no such hydrogen, so being oxidized would mean tearing apart a strong C–C bond — far beyond what a mild reagent like \(\mathrm{Ag(NH_3)_2^{+}}\) can do. So the aldehyde hands over its electrons, the silver ions accept them and plate out as the mirror, and the ketone simply sits there unchanged.

Fehling’s test was for many years the standard clinical check for sugar in urine. Glucose carries an aldehyde group, so it reduces the blue \(\mathrm{Cu^{2+}}\) of Fehling solution to the red precipitate of cuprous oxide; sugars that behave this way are still called reducing sugars. Simple colour-change strips have replaced the boiling tube in hospitals, but the chemistry being read is the same oxidation.

Simple hydrocarbons, ethers, ketones and alcohols do not get oxidized by Tollens’ reagent. Explain why.

Why is benzaldehyde NOT oxidized by Fehling solution?

Laboratory Test for Ketonic Group: Sodium Nitroprusside Test

When a freshly prepared sodium nitroprusside solution is added to a ketone, the mixture is shaken well and basified by adding sodium hydroxide solution drop by drop, a red colour appears in the solution, which indicates the presence of the ketonic (\(>\mathrm{C=O}\)) group.

\[\mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{OH^{-}} \longrightarrow \mathrm{CH_3\!-\!CO\!-\!CH_2^{-}} + \mathrm{H_2O}\]

\[\text{(Acetone)}\]

\[[\mathrm{Fe(CN)_5NO}]^{2-} + \mathrm{CH_3\!-\!CO\!-\!CH_2^{-}} \longrightarrow [\mathrm{Fe(CN)_5NO(CH_3\!-\!CO\!-\!CH_2)}]^{3-}\]

\[\text{(Nitroprusside ion)} \qquad\qquad \text{(Red colouration)}\]

The anion of the ketone formed by the alkali reacts with the nitroprusside ion to form a red coloured complex, which indicates the presence of the ketonic group.

12.8.2 Chemical Reactions of Aldehydes and Ketones with Nucleophile

In all these reactions the nucleophilic reagent brings about the reaction by attacking the positively polarized electrophilic carbonyl carbon in aldehydes and ketones.

Addition of Hydrogen Cyanide (H–CN)

Hydrogen cyanide, a weak acid, adds across the carbon–oxygen double bond in aldehydes and ketones to produce compounds called cyanohydrins. The negative part of the reagent, \(\mathrm{{}^{-}CN}\), attacks the electrophilic carbon of the carbonyl group. The reaction requires either acid or base as a catalyst.

\[\mathrm{R(R')C{=}O} + \mathrm{HCN} \longrightarrow \mathrm{R(R')C(OH)\!-\!CN}\]

\[\text{(Aldehyde when R' = H; ketone when R' = alkyl/aryl)} \qquad \text{(cyanohydrin)}\]

For example,

\[\mathrm{CH_3\!-\!CHO} + \mathrm{H\!-\!CN} \longrightarrow \mathrm{CH_3\!-\!CH(OH)\!-\!CN}\]

\[\text{(Acetaldehyde)} \qquad \text{(acetaldehyde cyanohydrin)}\]

\[\mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{H\!-\!CN} \longrightarrow \mathrm{CH_3\!-\!C(CH_3)(OH)\!-\!CN}\]

\[\text{(Acetone)} \qquad \text{(Acetone cyanohydrin)}\]

Notice what has happened to the double bond: the \(\mathrm{{}^{-}CN}\) has joined the carbon and the \(\mathrm{H}\) has joined the oxygen, so the \(\pi\) bond is replaced by two new \(\sigma\) bonds. This is the general pattern of every nucleophilic addition in this section.

Cyanohydrin formation is a ‘step-up’ reaction, as a new carbon–carbon single bond is formed. The \(\mathrm{-C{\equiv}N}\) group can then be converted into \(\mathrm{-COOH}\), \(\mathrm{-CH_2\!-\!NH_2}\) and so on. Therefore, cyanohydrins are used as intermediates in step-up synthesis — that is, in building a molecule with one more carbon atom than the one you began with.

Addition of NaHSO₃ (Sodium Bisulphite)

Aldehydes and ketones react with a saturated aqueous solution of sodium bisulfite to give a crystalline precipitate of the sodium bisulfite adduct (addition compound). For example,

\[\mathrm{CH_3\!-\!CHO} + \mathrm{NaHSO_3} \longrightarrow \mathrm{CH_3\!-\!CH(OH)\!-\!SO_3Na}\]

\[\text{(Acetaldehyde)} \qquad \text{(Acetaldehyde sodium bisulfite adduct)}\]

\[\mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{NaHSO_3} \longrightarrow \mathrm{CH_3\!-\!C(CH_3)(OH)\!-\!SO_3Na}\]

\[\text{(Acetone)} \qquad \text{(Acetone sodium bisulfite adduct)}\]

The sodium bisulfite addition product can be split easily to regenerate aldehydes and ketones on treatment with dilute acid or base. This makes the reaction a practical purification tool: the carbonyl compound is converted into its crystalline adduct, which is filtered off free of impurities that do not react, and then decomposed to give back the pure aldehyde or ketone.

Sodium bisulfite is the sodium salt of sulfurous acid; write down its detailed bond structure.

Addition of Alcohols

An aldehyde reacts with one molecule of an anhydrous monohydric alcohol in the presence of dry hydrogen chloride to give an alkoxyalcohol known as a hemiacetal, which further reacts with one more molecule of anhydrous monohydric alcohol to give a geminal dialkoxy compound known as an acetal. “Geminal” means that both alkoxy groups sit on the same carbon atom.

Step 1:

\[\mathrm{R\!-\!CHO} + \mathrm{R'\!-\!OH} \ \underset{\text{dil. HCl}}{\overset{\text{dry HCl}}{\rightleftharpoons}}\ \mathrm{R\!-\!CH(OH)\!-\!OR'}\]

\[\text{(Aldehyde)} \qquad\qquad \text{(Hemiacetal) unstable}\]

Step 2:

\[\mathrm{R\!-\!CH(OH)\!-\!OR'} + \mathrm{R'\!-\!OH} \ \underset{\text{dil. HCl}}{\overset{\text{dry HCl}}{\rightleftharpoons}}\ \mathrm{R\!-\!CH(OR')_2} + \mathrm{H_2O}\]

\[\text{(Hemiacetal)} \qquad\qquad \text{(Acetal) stable}\]

For example, with ethanol:

Step 1:

\[\mathrm{CH_3\!-\!CHO} + \mathrm{H_5C_2\!-\!OH} \ \underset{\text{dil. HCl}}{\overset{\text{dry HCl}}{\rightleftharpoons}}\ \mathrm{H_3C\!-\!CH(OH)\!-\!OC_2H_5}\]

\[\text{(Acetaldehyde)} \qquad\qquad \text{(Hemiacetal) unstable}\]

Step 2:

\[\mathrm{H_3C\!-\!CH(OH)\!-\!OC_2H_5} + \mathrm{H_5C_2\!-\!OH} \ \underset{\text{dil. HCl}}{\overset{\text{dry HCl}}{\rightleftharpoons}}\ \mathrm{H_3C\!-\!CH(OC_2H_5)_2} + \mathrm{H_2O}\]

\[\text{(Hemiacetal, unstable)} \qquad \text{(1,1-Diethoxyethane) (stable)}\]

Similarly, ketones react with alcohol in the presence of an acid catalyst to form a hemiketal and a ketal.

Ketones react with 1,2- or 1,3-diols in the presence of dry hydrogen chloride to give five- or six-membered cyclic ketals, as shown in Fig. 12.8a. Because both \(\mathrm{-OH}\) groups belong to the same molecule, the second one is always close at hand once the first has reacted, which is why the ring closes so readily.

Diagram

Fig. 12.8a – A ketone R₂C=O plus ethane-1,2-diol (HO–CH₂–CH₂–OH) with dry HCl/dil. HCl giving a five-membered cyclic ketal: the former carbonyl carbon bearing both R groups and two ring oxygens joined through –CH₂–CH₂–; label ketone, ethane-1,2-diol, cyclic ketal

The reaction can be reversed by treating the cyclic ketal with aqueous HCl to regenerate the ketone.

Acetals and ketals are hydrolysed with aqueous mineral acids to give the corresponding aldehydes and ketones respectively.

An organic molecule containing both an alcohol and a carbonyl group can undergo an intramolecular reaction with dry HCl to form cyclic hemiacetals or hemiketals.

A cyclic ketal is used as a protecting group for a ketone in reactions of a multifunctional compound to which the ketone is sensitive. The strategy is simple and widely used: convert the ketone into its cyclic ketal, carry out the reaction on the other functional group, then hydrolyse the ketal with aqueous acid to bring the ketone back unchanged.

Predict the product of the following reaction: a p-bromo substituted aromatic ketone treated with \(\mathrm{HO\!-\!CH_2\!-\!CH_2\!-\!OH}\), followed by Mg in dry ether.

Addition of Grignard Reagent

Aldehydes and ketones on reaction with alkyl magnesium halide followed by acid hydrolysis give alcohols (refer to Chapter 11, sec. 11.4.1 d). The alkyl group of the Grignard reagent is the nucleophile here, and it attacks the carbonyl carbon exactly as \(\mathrm{{}^{-}CN}\) did.

Nucleophilic Addition–Elimination with Ammonia Derivatives

Aldehydes and ketones undergo addition–elimination with some ammonia derivatives (\(\mathrm{NH_2\!-\!Z}\)) to give a product containing \(\mathrm{C{=}N}\) bonds (imines). The reaction is reversible and takes place in a weakly acidic medium. The substituted imine is called a Schiff’s base.

\[>\mathrm{C{=}O} + \mathrm{NH_2\!-\!Z} \ \rightleftharpoons\ [\,>\mathrm{C(OH)\!-\!NH\!-\!Z}\,] \xrightarrow{\ -\mathrm{H_2O}\ } >\mathrm{C{=}N\!-\!Z} + \mathrm{H_2O}\]

\[\text{(Aldehyde or Ketone)} \qquad\qquad \text{(imine)}\]

The name of the reaction describes it exactly: the nitrogen first adds to the carbonyl carbon, and a molecule of water is then eliminated to create the new double bond.

NoteDefinition

A Schiff’s base is the substituted imine, containing a \(\mathrm{C{=}N}\) bond, formed when an aldehyde or ketone reacts with an ammonia derivative \(\mathrm{NH_2\!-\!Z}\) in a weakly acidic medium, with the loss of one molecule of water.

Where \(\mathrm{Z} = \mathrm{-R}\), \(\mathrm{-Ar}\), \(\mathrm{-NH_2}\), \(\mathrm{-NHC_6H_5}\), \(\mathrm{-NHCONH_2}\), \(\mathrm{-NHC_6H_3(NO_2)_2}\).

Some important reactions of this type are listed in Table 12.7.

Table 12.7 Nucleophilic addition–elimination reactions of aldehydes and ketones with ammonia derivatives

Sr. No. Aldehyde (R′ = H) / Ketone (R′ ≠ H) Ammonia derivative \(\mathrm{NH_2\!-\!Z}\) Product (\(-\mathrm{H_2O}\)): imine, a crystalline derivative
1 \(\mathrm{R(R')C{=}O}\) \(\mathrm{NH_2\!-\!OH}\) (Hydroxyl amine) \(\mathrm{R(R')C{=}N\!-\!OH}\) (oxime)
2 \(\mathrm{R(R')C{=}O}\) \(\mathrm{NH_2\!-\!NH_2}\) (Hydrazine) \(\mathrm{R(R')C{=}N\!-\!NH_2}\) (hydrazone)
3 \(\mathrm{R(R')C{=}O}\) \(\mathrm{NH_2\!-\!NH\!-\!C_6H_5}\) (Phenyl hydrazine) \(\mathrm{R(R')C{=}N\!-\!NH\!-\!C_6H_5}\) (phenylhydrazone)
4 \(\mathrm{R(R')C{=}O}\) \(\mathrm{NH_2\!-\!NH\!-\!CONH_2}\) (Semicarbazide) \(\mathrm{R(R')C{=}N\!-\!NH\!-\!CONH_2}\) (semicarbazone)
5 \(\mathrm{R(R')C{=}O}\) \(\mathrm{H_2N\!-\!NH\!-\!C_6H_3(NO_2)_2}\) (2,4-Dinitrophenyl hydrazine) \(\mathrm{R(R')C{=}N\!-\!NH\!-\!C_6H_3(NO_2)_2}\) (2,4-Dinitrophenylhydrazone)

For example,

\[\mathrm{CH_3\!-\!CHO} + \mathrm{NH_2\!-\!OH} \ \rightleftharpoons\ [\,\mathrm{H_3C\!-\!CH(OH)\!-\!NH\!-\!OH}\,] \xrightarrow{\ -\mathrm{H_2O}\ } \mathrm{H_3C\!-\!CH{=}N\!-\!OH}\]

\[\text{(Acetaldehyde)} \quad \text{(Hydroxyl amine)} \qquad\qquad \text{(Acetaldoxime)}\]

All aldehydes and ketones give similar reactions. The resulting products have high molecular mass and are crystalline solids. These reactions are, therefore, useful for characterization of the original aldehydes and ketones — a liquid that is hard to identify becomes a solid with a sharp melting point that can be matched against known values.

In strongly acidic medium, the nitrogen atom of the ammonia derivative \(\mathrm{H_2N\!-\!Z}\) is protonated to form the \(\mathrm{(H_3N^{+}\!-\!Z)}\) ion, which is no longer a nucleophile. This is why the medium must be only weakly acidic: a little acid helps the water leave, but too much destroys the attacking species itself.

Draw the structures of (i) the semicarbazone of cyclohexanone, (ii) the imine formed in the reaction between 2-methylhexanal and ethyl amine, (iii) the 2,4-dinitrophenylhydrazone of acetaldehyde.

Write the structure of the product formed when cyclohexanone reacts with hydroxylamine, and name the class of compound it belongs to.

Haloform Reaction

This reaction is given by acetaldehyde, all methyl ketones (\(\mathrm{CH_3\!-\!CO\!-\!R}\)) and all alcohols containing the \(\mathrm{CH_3\!-\!(CHOH)-}\) group. When an alcohol or methyl ketone is warmed with sodium hydroxide and iodine, a yellow precipitate is formed. Here the reagent sodium hypoiodite is produced in situ, that is, generated in the reaction mixture itself rather than added ready-made.

During the reaction, the sodium salt of a carboxylic acid is formed, which contains one carbon atom less than the substrate. The methyl group is converted into haloform. For example, acetone is oxidized by sodium hypoiodite to give the sodium salt of acetic acid and a yellow precipitate of iodoform.

\[\mathrm{H_3C\!-\!CO\!-\!CH_3} + 3\,\mathrm{NaOI} \xrightarrow[\ \Delta\ ]{\ \mathrm{NaOH},\ \mathrm{I_2}\ } \mathrm{CHI_3}\downarrow + \mathrm{H_3C\!-\!CO\!-\!ONa} + 2\,\mathrm{NaOH}\]

\[\text{(Acetone)} \quad \text{(Sodium hypoiodite)} \quad \text{(Iodoform)} \quad \text{(Sodium acetate)}\]

Q. Why does the acid formed in a haloform reaction always have one carbon less than the compound you started with?

A. The three hydrogens of the methyl group are replaced one by one by iodine, and the resulting \(\mathrm{-CI_3}\) group is then cleaved off the molecule as iodoform, \(\mathrm{CHI_3}\). That carbon leaves with the yellow precipitate, so the carboxylate that remains is one carbon shorter.

Three points are worth remembering about this test. First, if a \(\mathrm{C{=}C}\) bond is present in a given aldehyde, ketone or methyl ketone, it is not attacked by the hypohalite. Second, non-methyl ketones do not give a positive iodoform test. Third, secondary alcohols having the \(\mathrm{CH_3\!-\!CHOH-}\) group give a positive iodoform test, because the reagent first oxidizes them to a \(\mathrm{CH_3\!-\!CO-}\) group, which subsequently forms iodoform.

Which of the following gives a positive iodoform test — butan-2-one or pentan-3-one? Write the equation for the positive case.

The test requires a \(\mathrm{CH_3\!-\!CO-}\) group, that is, a methyl group attached directly to the carbonyl carbon.

Butan-2-one is \(\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH_3}\). Its carbonyl carbon does carry a methyl group, so it is a methyl ketone and gives a positive test:

\[\mathrm{CH_3\!-\!CO\!-\!CH_2\!-\!CH_3} + 3\,\mathrm{NaOI} \xrightarrow[\ \Delta\ ]{\ \mathrm{NaOH},\ \mathrm{I_2}\ } \mathrm{CHI_3}\downarrow + \mathrm{CH_3\!-\!CH_2\!-\!CO\!-\!ONa} + 2\,\mathrm{NaOH}\]

The yellow precipitate is iodoform and the salt formed is sodium propanoate, which has one carbon less than butan-2-one.

Pentan-3-one is \(\mathrm{CH_3\!-\!CH_2\!-\!CO\!-\!CH_2\!-\!CH_3}\). Both groups on the carbonyl carbon are ethyl groups, so there is no \(\mathrm{CH_3\!-\!CO-}\) unit and the test is negative.

Which of the following give a positive iodoform test: ethanol, propan-1-ol, propan-2-ol, pentan-3-one? Give a reason for each.

Write the chemical reactions taking place when propan-2-ol is treated with iodine and sodium hydroxide.

Aldol Condensation

When acetaldehyde is treated with dilute NaOH, the following reaction is observed: \[2\,\mathrm{CH_3\!-\!CHO} \xrightarrow{\ \text{dil. NaOH}\ } \mathrm{CH_3\!-\!CH(OH)\!-\!CH_2\!-\!CHO}\] (i) What are the functional groups in the product? (ii) Will there be another product formed during the same reaction? Deduce the answer by doing an atomic audit of reactant and product. (iii) Is this an addition reaction or a condensation reaction?

Aldehydes containing at least one \(\alpha\)-hydrogen atom undergo a reaction in the presence of dilute alkali (dilute NaOH, KOH or \(\mathrm{Na_2CO_3}\)) as catalyst to form \(\beta\)-hydroxy aldehydes (aldol). This reaction is known as the aldol reaction. Formation of the aldol is an addition reaction.

NoteDefinition

The aldol reaction is the reaction in which two molecules of an aldehyde (or ketone) carrying at least one \(\alpha\)-hydrogen combine in the presence of dilute alkali to give a \(\beta\)-hydroxy aldehyde (or \(\beta\)-hydroxy ketone), called an aldol.

Q. What exactly is an \(\alpha\)-hydrogen, and how do I count them?

A. The \(\alpha\)-carbon is the carbon immediately next to the carbonyl carbon, and any hydrogen atoms attached to that carbon are the \(\alpha\)-hydrogens. In ethanal, \(\mathrm{CH_3\!-\!CHO}\), the methyl carbon is \(\alpha\) and carries three \(\alpha\)-hydrogens. In benzaldehyde, \(\mathrm{C_6H_5\!-\!CHO}\), the carbon next to \(\mathrm{-CHO}\) belongs to the ring and carries no hydrogen at all, so benzaldehyde has none.

The aldol formed from an aldehyde having \(\alpha\)-hydrogens undergoes subsequent elimination of a water molecule on warming, giving rise to an \(\alpha,\beta\)-unsaturated aldehyde.

\[2\,\mathrm{R\!-\!CH_2\!-\!CHO} \xrightarrow{\ \text{aq. NaOH}\ } \mathrm{R\!-\!CH_2\!-\!CH(OH)\!-\!CH(R)\!-\!CHO}\]

\[\text{(aldehyde)} \qquad\qquad \text{(aldol)}\]

\[\mathrm{R\!-\!CH_2\!-\!CH(OH)\!-\!CH(R)\!-\!CHO} \xrightarrow[\ \text{warm}\ ]{\ -\mathrm{H_2O}\ } \mathrm{R\!-\!CH_2\!-\!CH{=}C(R)\!-\!CHO}\]

\[\text{(aldol)} \qquad\qquad (\alpha,\beta\text{ - unsaturated aldehyde})\]

For example:

\[2\,\mathrm{CH_3\!-\!CHO} \xrightarrow{\ \text{dil. NaOH}\ } \mathrm{CH_3\!-\!CH(OH)\!-\!CH_2\!-\!CHO} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \Delta\ } \mathrm{CH_3\!-\!CH{=}CH\!-\!CHO}\]

\[\text{(Ethanal)} \qquad \text{(3-Hydroxybutanal, an Aldol)} \qquad \text{(But-2-enal)}\]

The overall reaction is called aldol condensation. It is a nucleophilic addition–elimination reaction.

Ketones containing at least two \(\alpha\)-hydrogens also undergo the aldol condensation reaction and give an \(\alpha,\beta\)-unsaturated ketone. For example:

\[2\,\mathrm{H_3C\!-\!CO\!-\!CH_3} \xrightarrow{\ \mathrm{Ba(OH)_2}\ } \mathrm{H_3C\!-\!C(CH_3)(OH)\!-\!CH_2\!-\!CO\!-\!CH_3}\]

\[\text{(Propanone)} \qquad \text{(4-Hydroxy-4-methylpentan-2-one, a Ketol)}\]

\[\mathrm{H_3C\!-\!C(CH_3)(OH)\!-\!CH_2\!-\!CO\!-\!CH_3} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \Delta\ } \mathrm{H_3C\!-\!C(CH_3){=}CH\!-\!CO\!-\!CH_3}\]

\[\text{(4-Methylpent-3-en-2-one)}\]

Flowchart

FC 12.8b – Aldol condensation in two stages: carbonyl compound with at least one α-hydrogen → dilute alkali removes an α-hydrogen → carbanion → attacks the carbonyl carbon of a second molecule (addition) → β-hydroxy carbonyl compound, the aldol → warm, –H₂O (elimination) → α,β-unsaturated carbonyl compound; branch: no α-hydrogen → no aldol reaction possible

Cross aldol condensation refers to the aldol condensation that takes place between two different aldehydes or ketones. If both aldehydes or ketones contain two \(\alpha\)-hydrogen atoms each, then a mixture of four products is formed. This happens because each molecule can act either as the one that loses an \(\alpha\)-hydrogen or as the one that is attacked, giving two self-condensation products and two cross products.

For example, a mixture of ethanal and propanal on reaction with dilute alkali followed by heating gives a mixture of four products (Fig. 12.5).

Diagram

Fig. 12.5 – Cross aldol condensation: ethanal (CH₃CHO) plus propanal (CH₃CH₂CHO) with i. dil. NaOH, ii. heat, branching into “self condensation” giving crotonaldehyde (but-2-enal) and 2-methylpent-2-enal, and “cross condensation” giving 2-methylbut-2-enal and pent-2-enal; label all four products

Ketones can also be used as one of the components in cross aldol condensation.

Write the aldol condensation of propanal, naming the aldol and the final product.

Propanal is \(\mathrm{CH_3\!-\!CH_2\!-\!CHO}\). Its \(\alpha\)-carbon is the \(\mathrm{-CH_2-}\) group next to \(\mathrm{-CHO}\), and it carries two \(\alpha\)-hydrogens, so the reaction is possible.

In the addition step, the \(\alpha\)-carbon of one molecule bonds to the carbonyl carbon of the second:

\[2\,\mathrm{CH_3\!-\!CH_2\!-\!CHO} \xrightarrow{\ \text{dil. NaOH}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH(OH)\!-\!CH(CH_3)\!-\!CHO}\]

Numbering from the \(\mathrm{-CHO}\) carbon, this aldol is 3-hydroxy-2-methylpentanal.

On warming, water is eliminated between the \(\mathrm{-OH}\) carbon and the \(\alpha\)-carbon:

\[\mathrm{CH_3\!-\!CH_2\!-\!CH(OH)\!-\!CH(CH_3)\!-\!CHO} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \Delta\ } \mathrm{CH_3\!-\!CH_2\!-\!CH{=}C(CH_3)\!-\!CHO}\]

The final product is 2-methylpent-2-enal, an \(\alpha,\beta\)-unsaturated aldehyde.

Observe the following reaction: benzaldehyde and 1-phenylethanone (acetophenone), on treatment with dilute base and heat, give 1,3-diphenylprop-2-en-1-one (benzal acetophenone). Will this reaction give a mixture of products like a cross aldol reaction?

Cannizzaro Reaction

This reaction is given only by aldehydes having no \(\alpha\)-hydrogen atom. Such aldehydes undergo a self-oxidation and reduction reaction on heating with concentrated alkali. This is an example of a disproportionation reaction. In the Cannizzaro reaction, one molecule of an aldehyde is reduced to an alcohol and at the same time a second molecule is oxidized to a carboxylic acid salt.

NoteDefinition

A disproportionation reaction is one in which the same substance is simultaneously oxidized and reduced, so that one portion of it ends up in a higher oxidation state and another portion in a lower one.

For example, formaldehyde and benzaldehyde:

\[2\,\mathrm{H\!-\!CHO} + \mathrm{NaOH}\ (50\%) \xrightarrow{\ \Delta\ } \mathrm{H\!-\!CO\!-\!O^{-}Na^{+}} + \mathrm{CH_3\!-\!OH}\]

\[\text{(Formaldehyde)} \quad \text{(sodium hydroxide)} \quad \text{(Sodium formate)} \quad \text{(Methanol)}\]

\[2\,\mathrm{C_6H_5\!-\!CHO} \xrightarrow[\ \Delta\ ]{\ \mathrm{KOH}\ \text{in MeOH}\ } \mathrm{C_6H_5\!-\!CO\!-\!O^{-}K^{+}} + \mathrm{C_6H_5\!-\!CH_2\!-\!OH}\]

\[\text{(Benzaldehyde)} \qquad \text{(Potassium benzoate)} \quad \text{(Phenylmethanol / benzyl alcohol)}\]

Q4. When benzaldehyde is heated with concentrated alkali, one half of it becomes an alcohol and the other half becomes the salt of an acid. Why should identical molecules take two opposite paths in the same flask?

A. Because there is no third reagent to do the work, so the molecules must use each other. An aldehyde with an \(\alpha\)-hydrogen would react through that hydrogen and give an aldol; benzaldehyde has none, since the carbon next to its \(\mathrm{-CHO}\) group is a ring carbon carrying no hydrogen. What is left open to it is a hydride transfer: one molecule passes a hydrogen, with its pair of electrons, to the carbonyl carbon of a neighbour. The molecule that gives the hydrogen away is oxidized and ends up as benzoate; the molecule that receives it is reduced and ends up as benzyl alcohol. The two molecules are identical to begin with, but only one of them can be the donor in any given pair — and that is precisely what disproportionation means.

The reaction is named after the Italian chemist Stanislao Cannizzaro, who reported it in the middle of the nineteenth century. He is remembered just as much for a second contribution: at the Karlsruhe Congress of 1860 he set out a clear and consistent method of assigning atomic weights, which helped end years of confusion over molecular formulae.

Can isobutyraldehyde undergo the Cannizzaro reaction? Explain.

Cross Cannizzaro reaction. When a mixture of formaldehyde and a non-enolisable aldehyde (an aldehyde with no \(\alpha\)-hydrogen) is treated with a strong base, formaldehyde is oxidized to formic acid while the other non-enolisable aldehyde is reduced to an alcohol. Formic acid forms sodium formate with NaOH. On acidification, sodium formate is converted into formic acid. For example:

\[\mathrm{H\!-\!CHO} + \mathrm{C_6H_5\!-\!CHO} \xrightarrow[\ \text{ii. } \mathrm{H_3O^{+}}\ ]{\ \text{i. conc. NaOH}\ } \mathrm{C_6H_5\!-\!CH_2\!-\!OH} + \mathrm{H\!-\!COOH}\]

\[\text{(Formaldehyde)} \quad \text{(Benzaldehyde)} \qquad \text{(Phenylmethanol)} \quad \text{(formic acid)}\]

Formaldehyde takes the oxidation role because its carbonyl carbon, carrying only hydrogens, gives up a hydride most readily of all.

What happens when 2,2-dimethylpropanal, \(\mathrm{(CH_3)_3C\!-\!CHO}\), is heated with concentrated sodium hydroxide? Write the products.

First check for \(\alpha\)-hydrogens. The carbon next to the \(\mathrm{-CHO}\) group is bonded to three methyl groups and to the \(\mathrm{-CHO}\) carbon, so it carries no hydrogen at all. With no \(\alpha\)-hydrogen, aldol condensation is impossible and the Cannizzaro reaction takes place instead.

Two molecules disproportionate: one is reduced to the primary alcohol and the other is oxidized to the sodium salt of the acid.

\[2\,\mathrm{(CH_3)_3C\!-\!CHO} + \mathrm{NaOH} \xrightarrow{\ \Delta\ } \mathrm{(CH_3)_3C\!-\!CH_2\!-\!OH} + \mathrm{(CH_3)_3C\!-\!CO\!-\!O^{-}Na^{+}}\]

The products are 2,2-dimethylpropan-1-ol and sodium 2,2-dimethylpropanoate.

Predict, with a reason, whether pentan-3-one can undergo aldol condensation.

12.8.3 Oxidation and Reduction Reactions of Aldehydes and Ketones

Oxidation by Dilute HNO₃, KMnO₄ and K₂Cr₂O₇

Aldehydes are oxidized to the corresponding carboxylic acids by an oxidant such as dilute nitric acid, potassium permanganate, and sodium or potassium dichromate in acidic medium.

\[\mathrm{R\!-\!CHO} \xrightarrow[\ \text{dil. } \mathrm{H_2SO_4}\ ]{\ \mathrm{K_2Cr_2O_7}\ } \mathrm{R\!-\!COOH}\]

\[\text{(Aldehyde)} \qquad \text{(Carboxylic acid)}\]

Ketones resist oxidation due to the strong CO–C bond, but they are oxidized by strong oxidizing agents such as \(\mathrm{CrO_3}\), alkaline \(\mathrm{KMnO_4}\) or hot concentrated \(\mathrm{HNO_3}\) to a mixture of carboxylic acids having a smaller number of carbon atoms than the starting ketone. Thus, oxidation of ketones is accompanied by breaking of the C–C bond.

\[\mathrm{R\!-\!CO\!-\!R'} \xrightarrow{\ \mathrm{CrO_3}\ } \mathrm{R\!-\!COOH} + \mathrm{R'\!-\!COOH}\]

\[\text{(Ketone)} \qquad\qquad \text{(Carboxylic acids)}\]

For example,

\[\mathrm{H_3C\!-\!CO\!-\!CH_3} \xrightarrow{\ \mathrm{CrO_3}\ } \mathrm{H_3C\!-\!COOH}\]

\[\text{(Acetone)} \qquad \text{(Acetic Acid)}\]

\[\mathrm{H_3C\!-\!CO\!-\!C_2H_5} \xrightarrow{\ \mathrm{CrO_3}\ } \mathrm{H_3C\!-\!COOH} + \mathrm{H_5C_2\!-\!COOH}\]

\[\text{(Butan-2-one)} \qquad \text{(Ethanoic acid)} \quad \text{(Propanoic acid)}\]

Because the C–C bond can break on either side of the carbonyl group, an unsymmetrical ketone gives a mixture of acids rather than a single product — which is why this oxidation is of little use in synthesis and mainly of interest as a contrast to the easy oxidation of aldehydes.

Clemmensen and Wolf-Kishner Reduction

The carbonyl group of aldehydes and ketones is reduced to the methylene group (\(\mathrm{-CH_2-}\)) on treatment with zinc–amalgam and concentrated hydrochloric acid (Clemmensen reduction), or with hydrazine followed by heating with sodium or potassium hydroxide in a high boiling solvent like ethylene glycol (Wolf-Kishner reduction). In both the reactions, oxygen is replaced by two hydrogen atoms.

Clemmensen reduction:

\[>\mathrm{C{=}O} + 4[\mathrm{H}] \xrightarrow[\ \Delta\ ]{\ \mathrm{Zn\text{-}Hg},\ \text{conc. HCl}\ } \mathrm{{-}CH_2{-}} + \mathrm{H_2O}\]

\[\text{(Carbonyl group in aldehydes and ketones)} \qquad \text{(Methylene group)}\]

For example:

\[\mathrm{H_3C\!-\!CO\!-\!CH_3} + 4[\mathrm{H}] \xrightarrow[\ \Delta\ ]{\ \mathrm{Zn\text{-}Hg},\ \text{conc. HCl}\ } \mathrm{H_3C\!-\!CH_2\!-\!CH_3} + \mathrm{H_2O}\]

\[\text{(Acetone)} \qquad\qquad \text{(propane)}\]

\[\mathrm{CH_3\!-\!CH_2\!-\!CHO} + 4[\mathrm{H}] \xrightarrow[\ \Delta\ ]{\ \mathrm{Zn\text{-}Hg},\ \text{conc. HCl}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_3} + \mathrm{H_2O}\]

\[\text{(Propanal)} \qquad\qquad \text{(Propane)}\]

Wolf-Kishner reduction:

\[>\mathrm{C{=}O} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \mathrm{H_2N\!-\!NH_2}\ } >\mathrm{C{=}N\!-\!NH_2} \xrightarrow[\ \Delta\ ]{\ \mathrm{KOH},\ \mathrm{HO\text{-}CH_2\text{-}CH_2\text{-}OH}\ } \mathrm{{-}CH_2{-}} + \mathrm{N_2}\]

\[\text{(Carbonyl group)} \qquad \text{(Hydrazone)} \qquad\qquad \text{(Methylene group)}\]

The first stage is nothing new — it is the addition–elimination of Table 12.7, with hydrazine as the ammonia derivative. The second stage expels the nitrogen as \(\mathrm{N_2}\) gas, and it is the escape of this very stable gas that drives the reaction forward.

For example:

\[\mathrm{C_2H_5\!-\!CHO} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \mathrm{H_2N\!-\!NH_2}\ } \mathrm{C_2H_5\!-\!CH{=}N\!-\!NH_2} \xrightarrow[\ \Delta\ ]{\ \mathrm{KOH},\ \mathrm{HO\text{-}CH_2\text{-}CH_2\text{-}OH}\ } \mathrm{CH_3\!-\!CH_2\!-\!CH_3} + \mathrm{N_2}\]

\[\text{(Propanal)} \qquad \text{(Hydrazone)} \qquad\qquad \text{(Propane)}\]

\[\mathrm{C_6H_5\!-\!CO\!-\!C_2H_5} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \mathrm{H_2N\!-\!NH_2}\ } \mathrm{C_6H_5\!-\!C(C_2H_5){=}N\!-\!NH_2} \xrightarrow[\ \Delta\ ]{\ \mathrm{KOH},\ \mathrm{HO\text{-}CH_2\text{-}CH_2\text{-}OH}\ } \mathrm{C_6H_5\!-\!CH_2\!-\!C_2H_5} + \mathrm{N_2}\]

\[\text{(Ethyl phenyl ketone)} \qquad \text{(Hydrazone)} \qquad\qquad \text{(n-Propyl benzene)}\]

Wolf-Kishner reduction is used to synthesize straight chain alkyl substituted benzenes, which is not possible by the Friedel–Crafts alkylation reaction.

Clemmensen vs Wolf-Kishner reduction

Feature Clemmensen reduction Wolf-Kishner reduction
Reagents Zinc–amalgam and concentrated HCl Hydrazine, then KOH or NaOH in ethylene glycol
Medium Strongly acidic Strongly basic
Intermediate — Hydrazone, \(>\mathrm{C{=}N\!-\!NH_2}\)
Change brought about \(>\mathrm{C{=}O} \rightarrow \mathrm{-CH_2-}\) \(>\mathrm{C{=}O} \rightarrow \mathrm{-CH_2-}\)
Suited to Compounds stable in acid Compounds stable in base

12.8.4 Electrophilic Substitution Reactions

Aromatic aldehydes and ketones undergo electrophilic substitution reactions such as nitration, sulfonation and halogenation. The aldehydic (\(\mathrm{-CHO}\)) and ketonic (\(>\mathrm{C=O}\)) groups are electron-withdrawing by inductive as well as resonance effects. They deactivate the benzene ring at the ortho- and para- positions. This results in the formation of the meta-product.

Deactivation means the ring reacts more slowly than benzene itself, because electron density has been pulled out of it. That withdrawal is felt most strongly at the ortho and para positions, leaving the meta position comparatively the richest in electrons and therefore the site the incoming electrophile attacks.

For example,

\[\mathrm{C_6H_5\!-\!CHO} + \mathrm{HO\!-\!NO_2} \xrightarrow{\ \text{conc. } \mathrm{H_2SO_4}\ } \mathrm{3\text{-}NO_2\!-\!C_6H_4\!-\!CHO} + \mathrm{H_2O}\]

\[\text{(Benzaldehyde)} \quad \text{(conc. } \mathrm{HNO_3}\text{)} \qquad \text{(m-Nitrobenzaldehyde)}\]

Diagram

Fig. 12.8b – Nitration of benzaldehyde: a benzene ring bearing –CHO reacting with conc. HNO₃ and conc. H₂SO₄ to give the same ring with –NO₂ at the meta position, plus H₂O; mark the ortho, meta and para positions on the starting ring

12.9 Chemical Properties of Carboxylic Acids

12.9.1 Acidic Character of Carboxylic Acids

The carboxyl group (\(\mathrm{-COOH}\)) imparts acidic character to carboxylic acids. A carboxyl group is made of an \(\mathrm{-OH}\) group bonded to a carbonyl group. In aqueous solution the H atom in the OH of the carboxyl group dissociates as a proton, and the carboxylate ion is formed as the conjugate base.

\[\mathrm{R\!-\!COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{R\!-\!COO^{-}} + \mathrm{H_3O^{+}}\]

\[\text{(carboxylate ion)}\]

The conjugate base is simply what is left of the acid after the proton has gone. How strong an acid is depends on how comfortable that leftover species is, because a stable conjugate base means the equilibrium above lies further to the right.

The carboxylate ion is resonance stabilized by two equivalent resonance structures, as shown in Fig. 12.9a.

Diagram

Fig. 12.9a – Resonance in the carboxylate ion: structure (i) R–C(=O)–O⁻ with a curly arrow moving the π electrons and the lone pair, structure (ii) R–C(–O⁻)=O as the equivalent mirror image, and the resonance hybrid drawn with two identical dashed partial C–O bonds each carrying half a negative charge

The carboxylate ion has two resonance structures (i) and (ii), and both of them are equivalent to each other (refer to Std. XI Chemistry Textbook, Chapter 14). This gives good resonance stabilization to the carboxylate ion, which in turn gives acidic character to carboxylic acids. Because the two structures are identical in energy, the negative charge is shared equally between the two oxygen atoms instead of being trapped on one.

What is the numerical parameter used to express acid strength?

acid dissociation constant \(K_a\) and \(\mathrm{p}K_a\) (\(K_a\) is the equilibrium constant for the dissociation of the acid, with units of \(\text{mol dm}^{-3}\) for a monoprotic acid; \(\mathrm{p}K_a = -\log K_a\) and is a pure number with no unit. Reading them: a lower \(K_a\) value means a higher \(\mathrm{p}K_a\) and a weaker acid; a higher \(K_a\) value means a lower \(\mathrm{p}K_a\) and a stronger acid. Because of the logarithm, a difference of one unit in \(\mathrm{p}K_a\) means a tenfold difference in strength.)

Alcohols (R-OH), phenols (Ar-OH) and carboxylic acids (R-COOH) can all undergo ionization of the O-H bond to give away a proton \(\mathrm{H^{+}}\); yet they have different \(\mathrm{p}K_a\) values, which are 16, 10 and 4.5 respectively. Explain.

The \(\mathrm{p}K_a\) value is indicative of acid strength. The lower the \(\mathrm{p}K_a\) value, the stronger the acid. Alcohols, phenols and carboxylic acids all involve ionization of an O-H bond. But their different \(\mathrm{p}K_a\) values indicate that their acid strengths are different. This is because the resulting conjugate bases are stabilized to different extents.

Acid (HA) | Conjugate base (\(\mathrm{A^{-}}\)) | Electronic effect | Stabilization / destabilization |

|—|—|—|—|

\(\mathrm{R\!-\!O\!-\!H}\) | \(\mathrm{R \rightarrow O^{-}}\) | \(+\mathrm{I}\) effect of R group | destabilization of conjugate base |
\(\mathrm{Ar\!-\!O\!-\!H}\) | \(\mathrm{Ar\!-\!O^{-}}\) | \(-\mathrm{R}\) effect of Ar group | stabilization of conjugate base is moderate because all the resonance structures are not equivalent to each other |
\(\mathrm{R\!-\!CO\!-\!O\!-\!H}\) | \(\mathrm{R\!-\!CO\!-\!O^{-}}\) | \(-\mathrm{R}\) effect of C=O group | stabilization is good because all the resonance structures are equivalent to each other |

As the conjugate base of the carboxylic acid is best stabilized, among the three, carboxylic acids are the strongest and have the lowest \(\mathrm{p}K_a\) value. As the conjugate base of alcohols is destabilized, alcohols are the weakest acids and have the highest \(\mathrm{p}K_a\) value. As the conjugate base of phenols is moderately stabilized, phenols are moderately acidic and have an intermediate \(\mathrm{p}K_a\) value.

Influence of Electronic Effects on Acidity of Carboxylic Acids

All the carboxylic acids do not have the same \(\mathrm{p}K_a\) value. The structure of ‘R’ in R-COOH has an influence on the acid strength of carboxylic acids. Various haloacetic acids illustrate this point very well (Tables 12.8 and 12.9).

Table 12.8: \(\mathrm{p}K_a\) values of haloacetic acids

Acid \(\mathrm{p}K_a\) Acid strength
\(\mathrm{F\!-\!CH_2\!-\!COOH}\) 2.56 decreases down the column
\(\mathrm{Cl\!-\!CH_2\!-\!COOH}\) 2.86
\(\mathrm{Br\!-\!CH_2\!-\!COOH}\) 2.90
\(\mathrm{I\!-\!CH_2\!-\!COOH}\) 3.18
\(\mathrm{CH_3\!-\!COOH}\) 4.76

Halogens are electronegative atoms and exert an electron withdrawing inductive effect (\(-\mathrm{I}\) effect). The negatively charged carboxylate ion in the conjugate base of a haloacetic acid gets stabilized by the \(-\mathrm{I}\) effect of the halogen, which is responsible for diffusing the native charge, as drawn in Fig. 12.9b.

Diagram

Fig. 12.9b – Conjugate base of a haloacetic acid, X–CH₂–COO⁻, with arrowheads placed on the C–X and C–C bonds pointing towards the halogen to show the electron withdrawing –I effect pulling density away from the negatively charged carboxylate end

The higher the electronegativity of the halogen, the greater is the stabilization of the conjugate base, the stronger is the acid, and the smaller is the \(\mathrm{p}K_a\) value. This is exactly the order in Table 12.8: fluorine, the most electronegative, gives the lowest \(\mathrm{p}K_a\), and acetic acid, with no halogen at all, the highest.

Q. The halogen is nowhere near the O–H bond that breaks. How can it affect whether the proton leaves?

A. Acid strength is decided by the stability of the conjugate base, not by the O–H bond alone. Once the proton has left, a negative charge is sitting on the carboxylate group, and an electron withdrawing halogen a bond or two away pulls some of that charge towards itself and spreads it out. A spread-out charge is a more comfortable charge, so the anion is more stable and the equilibrium of dissociation shifts further to the right.

Q3. In chloroacetic acid the chlorine atom is not even attached to the acidic hydrogen, yet the compound is a much stronger acid than acetic acid. How can a distant atom change how easily a proton leaves?

A. It works through the conjugate base. When acetic acid loses its proton, the resulting \(\mathrm{CH_3\!-\!COO^{-}}\) has a methyl group that pushes electron density towards the already negative carboxylate — an unhelpful \(+\mathrm{I}\) effect that makes the anion less stable. Replace one hydrogen of that methyl group by chlorine and the direction of the pull reverses: chlorine is electronegative and withdraws electron density along the chain of \(\sigma\) bonds, so the negative charge of \(\mathrm{Cl\!-\!CH_2\!-\!COO^{-}}\) is spread over a larger region instead of being concentrated on two oxygens. The more stable anion means the acid gives up its proton more readily — and the numbers in Table 12.8 confirm it, with \(\mathrm{p}K_a\) falling from 4.76 for acetic acid to 2.86 for chloroacetic acid, nearly a hundredfold rise in strength.

Compare the two conjugate bases \(\mathrm{CH_3\!-\!COO^{-}}\) (a) and \(\mathrm{Cl\!-\!CH_2\!-\!COO^{-}}\) (b) and answer: (i) Indicate the inductive effects of the \(\mathrm{CH_3-}\) group in (a) and the \(\mathrm{Cl-}\) group in (b) by putting arrowheads in the middle of the appropriate covalent bonds. (ii) Which species is stabilized by the inductive effect, (a) or (b)? (iii) Which species is destabilized by the inductive effect, (a) or (b)?

  1. Compare the \(\mathrm{p}K_a\) values and arrange the following in increasing order of acid strength: \(\mathrm{Cl_3CCOOH}\), \(\mathrm{ClCH_2COOH}\), \(\mathrm{CH_3COOH}\), \(\mathrm{Cl_2CHCOOH}\). (ii) Draw the structures of the conjugate bases of monochloroacetic acid and dichloroacetic acid. Which one is more stabilized by the \(-\mathrm{I}\) effect?
Acidity of Aromatic Carboxylic Acids

Benzoic acid is the simplest aromatic acid. From the \(\mathrm{p}K_a\) value of benzoic acid (4.2) we understand that it is stronger than acetic acid (\(\mathrm{p}K_a\) 4.76). The \(sp^2\) hybrid carbon of the aromatic ring exerts an electron withdrawing inductive effect (\(-\mathrm{I}\) effect), which stabilizes the conjugate base and increases the acid strength of aromatic acids. An \(sp^2\) carbon holds its electrons more tightly than an \(sp^3\) carbon does, because it has a greater share of s character, and this is what makes the ring carbon a withdrawing group rather than a donating one.

Table 12.9 illustrates that the more the number of electron withdrawing substituents, the higher is the acid strength.

Table 12.9 \(\mathrm{p}K_a\) values of chloroacetic acids

Name Structure \(\mathrm{p}K_a\) Acid strength
Monochloroacetic acid \(\mathrm{Cl\!-\!CH_2\!-\!COOH}\) 2.86 increases down the column
Dichloroacetic acid \(\mathrm{Cl_2CH\!-\!COOH}\) 1.26
Trichloroacetic acid \(\mathrm{Cl_3C\!-\!COOH}\) 0.6

Arrange the following acids in order of their decreasing acidity: \(\mathrm{CH_3\!-\!CH(Cl)\!-\!CH_2\!-\!COOH}\), \(\mathrm{CCl_3\!-\!CH_2\!-\!COOH}\), \(\mathrm{CH_3COOH}\).

Electron-withdrawing groups like \(\mathrm{-Cl}\), \(\mathrm{-CN}\) and \(\mathrm{-NO_2}\) increase the acidity of substituted benzoic acids, while electron-donating groups like \(\mathrm{-CH_3}\), \(\mathrm{-OH}\), \(\mathrm{-OCH_3}\) and \(\mathrm{-NH_2}\) decrease the acidity of substituted benzoic acids. The three compounds in Fig. 12.9c make the comparison directly.

Diagram

Fig. 12.9c – Three benzene rings side by side: 4-nitrobenzoic acid (COOH at C1, NO₂ at C4, pKa = 3.41), benzoic acid (COOH only, pKa = 4.2), and 4-methylbenzoic acid (COOH at C1, CH₃ at C4, pKa = 4.4); label each with its name and pKa value

acid strength of carboxylic acids (As electron withdrawing power near the carboxyl group increases →: acid strength increases and \(\mathrm{p}K_a\) falls, because the conjugate base is better stabilized. As electron donating power increases →: acid strength decreases and \(\mathrm{p}K_a\) rises, because the anion is destabilized. Hence \(\mathrm{Cl_3CCOOH} > \mathrm{Cl_2CHCOOH} > \mathrm{ClCH_2COOH} > \mathrm{CH_3COOH}\) among the chloroacetic acids, and 4-nitrobenzoic acid > benzoic acid > 4-methylbenzoic acid among the substituted benzoic acids.)

Using the data in Fig. 12.9c, explain why 4-nitrobenzoic acid (\(\mathrm{p}K_a\) 3.41) is a stronger acid than benzoic acid (\(\mathrm{p}K_a\) 4.2), while 4-methylbenzoic acid (\(\mathrm{p}K_a\) 4.4) is weaker than benzoic acid.

All three acids ionize by the same route, giving a benzoate ion. The comparison therefore rests on what the substituent at position 4 does to that anion.

The \(\mathrm{-NO_2}\) group is strongly electron withdrawing. It pulls electron density away from the ring and, through it, away from the carboxylate group, spreading the negative charge of the conjugate base more widely. The more stable anion means a stronger acid, and the \(\mathrm{p}K_a\) falls from 4.2 to 3.41.

The \(\mathrm{-CH_3}\) group is electron donating through its \(+\mathrm{I}\) effect. It pushes electron density towards a group that is already negatively charged, which destabilizes the conjugate base. The weaker acid has the higher \(\mathrm{p}K_a\), 4.4.

Order of acid strength: 4-nitrobenzoic acid > benzoic acid > 4-methylbenzoic acid.

Arrange the following carboxylic acids in order of increasing acidity: m-nitrobenzoic acid, trichloroacetic acid, benzoic acid, \(\alpha\)-chlorobutyric acid.

12.9.2 Laboratory Tests for Carboxyl (–COOH) Group

The presence of the \(\mathrm{-COOH}\) group in carboxylic acids is identified by the following tests.

Litmus Test (valid for water soluble substances)

The aqueous solution of an organic compound containing the \(\mathrm{-COOH}\) group turns blue litmus red, which indicates the presence of an acidic functional group. It may be noted that aqueous solutions of water soluble phenols also turn blue litmus red, so this test alone cannot separate the two.

Sodium Bicarbonate Test

When sodium bicarbonate is added to an organic compound containing the \(\mathrm{-COOH}\) group, a brisk effervescence of carbon dioxide gas is evolved. A water insoluble acid goes into solution and gives a precipitate on acidification with concentrated HCl. This indicates the presence of the \(\mathrm{-COOH}\) group.

\[\mathrm{R\!-\!COOH} + \mathrm{NaHCO_3}\text{(aq)} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ -\mathrm{CO_2}\ } \mathrm{R\!-\!COONa}\text{(aq)}\]

\[\text{(water insoluble)}\]

\[\mathrm{R\!-\!COONa}\text{(aq)} \xrightarrow{\ \mathrm{HCl}\ } \mathrm{R\!-\!COOH}\downarrow + \mathrm{NaCl}\text{(aq)}\]

Phenol does not evolve \(\mathrm{CO_2}\) gas with sodium bicarbonate. Hence carboxylic acid and phenol are distinguished by this test.

telling a carboxylic acid from a phenol

Test Carboxylic acid Phenol
Blue litmus (aqueous solution) Turns red Turns red
Sodium bicarbonate Brisk effervescence of \(\mathrm{CO_2}\) No effervescence
Ester test with ethanol and conc. \(\mathrm{H_2SO_4}\) Fruity smell of ester No such smell
Ester Test

One drop of concentrated sulfuric acid is added to a mixture of the given organic compound containing the \(\mathrm{-COOH}\) group and one mL of ethanol; the reaction mixture is heated for 5 minutes in a hot water bath. After this, the hot solution is poured into a beaker containing water. A fruity smell of ester confirms the presence of the carboxylic acid.

\[\mathrm{R\!-\!COOH} + \mathrm{C_2H_5OH} \xrightarrow[\ \text{warm}\ ]{\ \mathrm{H^{+}}\ } \mathrm{R\!-\!COO\!-\!C_2H_5} + \mathrm{H_2O}\]

\[\text{(ester)}\]

12.9.3 Formation of Acyl Chloride

Reaction with \(\mathrm{PCl_3}\), \(\mathrm{PCl_5}\), \(\mathrm{SOCl_2}\). Carboxylic acids on heating with \(\mathrm{PCl_3}\), \(\mathrm{PCl_5}\) or \(\mathrm{SOCl_2}\) give the corresponding acyl chlorides. Thionyl chloride (\(\mathrm{SOCl_2}\)) is preferred because the byproducts formed are in the gaseous state, so they can easily escape from the reaction mixture. In this reaction the \(\mathrm{-OH}\) group of \(\mathrm{-COOH}\) is replaced by \(\mathrm{-Cl}\).

\[\mathrm{R\!-\!COOH} + \mathrm{SOCl_2} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!COCl} + \mathrm{SO_2}\uparrow + \mathrm{HCl}\uparrow\]

\[\text{(Carboxylic acid)} \qquad \text{(acyl chloride)}\]

\[3\,\mathrm{R\!-\!COOH} + \mathrm{PCl_3} \xrightarrow{\ \Delta\ } 3\,\mathrm{R\!-\!COCl} + \mathrm{P(OH)_3}\]

\[\mathrm{R\!-\!COOH} + \mathrm{PCl_5} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!COCl} + \mathrm{POCl_3} + \mathrm{HCl}\]

thionyl chloride, \(\mathrm{SOCl_2}\) (A chlorinating agent for the \(\mathrm{-OH}\) of a carboxyl group → used with heating, on a carboxylic acid → replaces \(\mathrm{-OH}\) by \(\mathrm{-Cl}\) to give the acyl chloride. Its advantage over \(\mathrm{PCl_3}\) and \(\mathrm{PCl_5}\) is purity of product: both byproducts, \(\mathrm{SO_2}\) and \(\mathrm{HCl}\), are gases and leave the flask by themselves, whereas the phosphorus reagents leave \(\mathrm{P(OH)_3}\) or \(\mathrm{POCl_3}\) behind to be separated.)

Fill in the blanks and rewrite the balanced equations. (i) \(\mathrm{CH_3COOH}\) + thionyl chloride \(\xrightarrow{\ \Delta\ }\) ….. + ….. + ….. (ii) \(\mathrm{CH_3\!-\!CH_2\!-\!COOH}\) + ….. \(\xrightarrow{\ \Delta\ }\) ….. + ….. + \(\mathrm{H_3PO_3}\) (iii) \(\mathrm{C_6H_5\!-\!COOH}\) + ….. \(\xrightarrow{\ \Delta\ }\) ….. + phosphorous oxychloride + \(\mathrm{HCl}\) (iv) \(\mathrm{CH_3\!-\!COOH}\) + phosphorous trichloride \(\xrightarrow{\ \Delta\ }\) ….. + ….. (v) \(\mathrm{CH_3\!-\!COOH} \xrightarrow{\ \mathrm{NH_3}\ }\) ….. \(\xrightarrow{\ \Delta\ }\) ….. (vi) ….. \(\xrightarrow{\ \mathrm{NH_3}\ }\) ….. \(\xrightarrow{\ \Delta\ } \mathrm{C_6H_5\!-\!CONH_2}\)

12.9.4 Reaction with Ammonia: Formation of Amide

Carboxylic acids react with ammonia to form the ammonium carboxylate salt, which on further strong heating at high temperature decomposes to give the acid amide.

\[\mathrm{R\!-\!COOH} + \mathrm{NH_3} \rightleftharpoons \mathrm{R\!-\!COO^{-}NH_4^{+}}\]

\[\text{(Carboxylic acid)} \qquad \text{(ammonium carboxylate)}\]

\[\mathrm{R\!-\!COO^{-}NH_4^{+}} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \Delta\ } \mathrm{R\!-\!CONH_2}\]

\[\text{(Acid amide)}\]

The first step is an ordinary acid–base reaction; the amide appears only in the second step, when heating drives a molecule of water out of the salt.

Acid amides can also be prepared by reacting acid chloride with ammonia.

\[\mathrm{R\!-\!COCl} + \mathrm{NH_3} \longrightarrow \mathrm{R\!-\!CONH_2} + \mathrm{HCl}\]

\[\text{(Acyl chloride)} \qquad \text{(Acid amide)}\]

Phthalic acid, the benzene-1,2-dicarboxylic acid you met earlier, takes this chemistry one step further. With two molecules of ammonia it forms ammonium phthalate; heating with loss of two molecules of water gives phthalamide, and further loss of ammonia closes a five-membered ring to give phthalimide. Phthalimide is not a curiosity: it is the starting material of the Gabriel synthesis, a standard laboratory route to pure primary amines.

12.9.5 Formation of Acid Anhydride

What is the term used for the elimination of a water molecule?

Mono carboxylic acids on heating with a strong dehydrating agent like \(\mathrm{P_2O_5}\) or concentrated \(\mathrm{H_2SO_4}\) give acid anhydrides. The reaction is reversible. Anhydrides are readily hydrolyzed back to acids on reaction with water.

\[2\,\mathrm{R\!-\!CO\!-\!O\!-\!H} \ \underset{+\mathrm{H_2O}}{\overset{\mathrm{P_2O_5},\ \Delta}{\rightleftharpoons}}\ \mathrm{R\!-\!CO\!-\!O\!-\!CO\!-\!R}\]

\[\text{(Carboxylic acid)} \qquad\qquad \text{(Acid anhydride)}\]

The name anhydride is built from Greek roots meaning “without water” — an exact description of a compound formed by removing a water molecule from two acid molecules, and one that also tells you how to undo it.

A better yield of acid anhydride is obtained by heating sodium carboxylate with acyl chloride.

\[\mathrm{R\!-\!COO^{-}Na^{+}} + \mathrm{R\!-\!COCl} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!CO\!-\!O\!-\!CO\!-\!R} + \mathrm{NaCl}\]

\[\text{(sodium carboxylate)} \quad \text{(acyl chloride)} \qquad \text{(acid anhydride)}\]

Acyl chloride and the sodium salt of the acid are prepared by reacting the carboxylic acid separately with thionyl chloride and sodium hydroxide respectively.

\[\mathrm{R\!-\!COOH} \xrightarrow{\ \mathrm{SOCl_2}\ } \mathrm{R\!-\!COCl}\]

\[\mathrm{R\!-\!COOH} \xrightarrow{\ \mathrm{NaOH}\ } \mathrm{R\!-\!COO^{-}Na^{+}}\]

\[\mathrm{R\!-\!COCl} + \mathrm{R\!-\!COO^{-}Na^{+}} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!CO\!-\!O\!-\!CO\!-\!R} + \mathrm{NaCl}\]

\[\text{(Anhydride)}\]

12.9.6 Decarboxylation of Carboxylic Acids

Which molecule is eliminated in a decarboxylation?

Sodium salts of carboxylic acids on heating with soda lime give hydrocarbons which contain one carbon atom less than the carboxylic acid. Soda lime is a mixture of sodium hydroxide and calcium oxide; the calcium oxide keeps the mixture dry and easy to handle at the high temperature required.

\[\mathrm{R\!-\!COO^{-}Na^{+}} + \mathrm{NaOH} \xrightarrow[\ \Delta\ ]{\ \mathrm{CaO}\ } \mathrm{R\!-\!H} + \mathrm{Na_2CO_3}\]

For example,

\[\mathrm{CH_3\!-\!COO^{-}Na^{+}} + \mathrm{NaOH} \xrightarrow[\ \Delta\ ]{\ \mathrm{CaO}\ } \mathrm{CH_4} + \mathrm{Na_2CO_3}\]

\[\text{(Sodium acetate)} \qquad\qquad \text{(Methane)}\]

The carbon that is lost is the carboxyl carbon itself, which departs in the carbonate, so a two-carbon acid salt gives a one-carbon hydrocarbon.

Sodium propanoate is heated with soda lime. Write the balanced equation for the reaction and name the hydrocarbon formed.

12.9.7 Reduction of Carboxylic Acids

Carboxylic acids are reduced to primary alcohols by a powerful reducing agent like lithium aluminium hydride. A carboxylic acid can also be reduced by diborane; diborane does not reduce \(\mathrm{-COOR}\), \(\mathrm{-NO_2}\) or \(\mathrm{-X}\). Note that sodium borohydride (\(\mathrm{NaBH_4}\)) does not reduce the \(\mathrm{-COOH}\) group.

\[\mathrm{R\!-\!COOH} + \mathrm{LiAlH_4} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!CH_2OH}\]

\(\mathrm{LiAlH_4}\) versus \(\mathrm{NaBH_4}\) (Lithium aluminium hydride is a powerful reducing agent → used in dry ether, since it reacts violently with water → it reduces even the stubborn \(\mathrm{-COOH}\) group all the way to a primary alcohol. Sodium borohydride is much milder and will not touch a \(\mathrm{-COOH}\) group at all. Diborane occupies a middle position: it reduces \(\mathrm{-COOH}\) but leaves \(\mathrm{-COOR}\), \(\mathrm{-NO_2}\) and \(\mathrm{-X}\) untouched, which makes it the reagent of choice when those groups must survive.)

Complete the following, giving the structure and IUPAC name of each organic product:

(i) \(\mathrm{CH_3\!-\!CH_2\!-\!COOH} + \mathrm{SOCl_2} \xrightarrow{\ \Delta\ }\) ?

(ii) \(\mathrm{CH_3\!-\!CH_2\!-\!COOH} + \mathrm{LiAlH_4} \xrightarrow{\ \text{dry ether}\ }\) ?

The seven reactions of this section are collected in FC 12.9a, which is worth memorising as a single picture of what the carboxyl group can be turned into.

Flowchart

FC 12.9a – Reactions of a carboxylic acid R–COOH: → (H₂O) → R–COO⁻ + H₃O⁺, acidic character; → (NaHCO₃) → R–COONa + CO₂↑, laboratory test; → (C₂H₅OH, conc. H₂SO₄) → ester; → (SOCl₂ or PCl₃ or PCl₅) → acyl chloride R–COCl; → (NH₃, then Δ) → amide R–CONH₂; → (P₂O₅, Δ) → anhydride R–CO–O–CO–R; → (as sodium salt, NaOH/CaO, Δ) → hydrocarbon R–H + Na₂CO₃; → (LiAlH₄, dry ether) → primary alcohol R–CH₂OH

12.10 Summary

  1. Aldehydes, ketones and carboxylic acids are all built around the carbon–oxygen double bond, \(>\mathrm{C=O}\), called the carbonyl group. In an aldehyde the carbonyl carbon carries at least one hydrogen, giving the formyl group \(\mathrm{-CHO}\); in a ketone it carries two alkyl or aryl groups; in a carboxylic acid it carries an \(\mathrm{-OH}\) group, giving the carboxyl group \(\mathrm{-COOH}\).

  2. Each family is classified by the carbon skeleton attached to the carbonyl group. Aldehydes and ketones are aliphatic or aromatic; aliphatic ketones are further simple (symmetrical) or mixed (unsymmetrical). Formaldehyde and formic acid are counted as aliphatic by exception. Carboxylic acids are aliphatic or aromatic, and are mono, di or tri carboxylic acids by the number of \(\mathrm{-COOH}\) groups. An aromatic compound whose \(\mathrm{-COOH}\) is not joined directly to the ring is a side-chain aromatic acid.

  3. In the IUPAC system an aldehyde is an alkanal, a ketone an alkanone, and an acid an alkanoic acid; two groups give -dial, -dione and -dioic acid. A group on an alicyclic ring gives carbaldehyde or cycloalkane carboxylic acid. As a lower-priority substituent, \(\mathrm{-CHO}\) becomes the prefix formyl and a ketonic \(>\mathrm{C=O}\) becomes oxo. Trivial names use \(\alpha\), \(\beta\), \(\gamma\) for positions, counting from the carbon next to the carbonyl carbon.

  4. Aldehydes and ketones are prepared by oxidation or dehydrogenation of primary and secondary alcohols respectively, by ozonolysis of alkenes, and by hydration of alkynes with 40% sulfuric acid and 1% mercuric sulfate.

  5. From an acyl chloride, hydrogen with a palladium catalyst poisoned by barium sulfate gives an aldehyde (Rosenmund reduction), while dialkyl cadmium gives a ketone. From a nitrile, \(\mathrm{SnCl_2}\)/HCl followed by hydrolysis gives an aldehyde (Stephen reaction), and a Grignard reagent followed by hydrolysis gives a ketone. DIBAl-H, \(\mathrm{AlH(i\text{-}Bu)_2}\), reduces a nitrile to an aldehyde without touching C=C or C≡C bonds, and reduces an ester to an aldehyde at 195 K.

  6. Aromatic aldehydes are made from methyl arenes by the Etard reaction with chromyl chloride in \(\mathrm{CS_2}\), by \(\mathrm{CrO_3}\) in acetic anhydride at 273–278 K, and by side chain chlorination of toluene followed by hydrolysis at 373 K, which is the commercial route to benzaldehyde. Gatterman–Koch formylation puts \(\mathrm{-CHO}\) directly on the ring using CO and HCl with anhydrous \(\mathrm{AlCl_3}\). Aromatic ketones come from Friedel–Crafts acylation.

  7. Carboxylic acids are obtained by acid hydrolysis of nitriles and amides, of acyl chlorides and anhydrides, and of esters. Alkaline hydrolysis of an ester gives the carboxylate salt, which is acidified to free the acid; this reaction is called saponification and it is how soaps are made.

  8. Oxidation of an alkyl benzene with dilute \(\mathrm{HNO_3}\), alkaline or acidic \(\mathrm{KMnO_4}\), or chromic acid converts the whole side chain, however long, into one \(\mathrm{-COOH}\) group; a tertiary alkyl substituent is not oxidized. Alkenes are oxidized by \(\mathrm{KMnO_4}\) in dilute \(\mathrm{H_2SO_4}\) to acids. A Grignard reagent added to dry ice and then hydrolysed gives an acid with one carbon more than the halide used.

  9. The carbonyl bond is polar, so aldehydes and ketones have dipole–dipole attraction and dissolve in water by hydrogen bonding through the carbonyl oxygen; solubility falls as the hydrocarbon part grows. Carboxylic acids form dimers held by two hydrogen bonds, which doubles the effective molecular size and gives them the highest boiling points. For comparable mass: carboxylic acids > alcohols > ketones > aldehydes > ethers > alkanes, following H-bond > dipole–dipole > van der Waals.

  10. The polarity of \(>\mathrm{C=O}\) arises from the electronegativity of oxygen and from resonance, leaving the carbon electron deficient and electrophilic, open to attack by nucleophiles. Aldehydes are more reactive than ketones because a ketone’s two alkyl groups both reduce the positive charge on the carbon (\(+\mathrm{I}\) effect) and crowd it sterically. Aromatic aldehydes are less reactive than aliphatic ones because the ring donates electron density by resonance.

  11. Aldehydes are detected by the Schiff test (magenta colour), Tollens’ test (silver mirror) and the Fehling test (red \(\mathrm{Cu_2O}\) precipitate, also given by \(\alpha\)-hydroxy ketones). Ketones are detected by the sodium nitroprusside test (red colour). In Tollens’ and Fehling’s tests the aldehyde is oxidized to the carboxylate ion while \(\mathrm{Ag^{+}}\) or \(\mathrm{Cu^{2+}}\) is reduced.

  12. With nucleophiles, aldehydes and ketones add HCN to give cyanohydrins (a step-up reaction that builds a new C–C bond), add \(\mathrm{NaHSO_3}\) to give crystalline adducts used for purification, add alcohols to give hemiacetals and acetals (hemiketals and ketals from ketones; cyclic ketals from 1,2- or 1,3-diols, useful as protecting groups), and add Grignard reagents to give alcohols.

  13. With ammonia derivatives \(\mathrm{NH_2\!-\!Z}\) in weakly acidic medium they undergo addition–elimination, losing water to give imines: oximes, hydrazones, phenylhydrazones, semicarbazones and 2,4-dinitrophenylhydrazones. These crystalline solids are used to characterize the parent compound. In strongly acidic medium the derivative is protonated and stops being a nucleophile.

  14. Acetaldehyde, all methyl ketones and all alcohols containing the \(\mathrm{CH_3\!-\!(CHOH)-}\) group give the haloform reaction with \(\mathrm{NaOH}\) and \(\mathrm{I_2}\), producing yellow iodoform and a carboxylate with one carbon less. Non-methyl ketones give a negative test, and a \(\mathrm{C{=}C}\) bond present in the molecule is not attacked.

  15. Carbonyl compounds with at least one \(\alpha\)-hydrogen (two for ketones) undergo the aldol reaction with dilute alkali to give a \(\beta\)-hydroxy aldehyde or ketone, which on warming loses water to give an \(\alpha,\beta\)-unsaturated product. The overall change is aldol condensation, a nucleophilic addition–elimination. Cross aldol condensation between two different partners, each with two \(\alpha\)-hydrogens, gives a mixture of four products.

  16. Aldehydes with no \(\alpha\)-hydrogen undergo the Cannizzaro reaction with concentrated alkali, a disproportionation in which one molecule is reduced to an alcohol and another oxidized to the carboxylate salt. In the cross Cannizzaro reaction with a non-enolisable aldehyde, formaldehyde is the one oxidized.

  17. Aldehydes are oxidized to carboxylic acids by dilute \(\mathrm{HNO_3}\), \(\mathrm{KMnO_4}\) or \(\mathrm{K_2Cr_2O_7}\). Ketones resist oxidation, and strong oxidants such as \(\mathrm{CrO_3}\) break a C–C bond to give a mixture of shorter acids. Both aldehydes and ketones are reduced to \(\mathrm{-CH_2-}\) by Zn–Hg with concentrated HCl (Clemmensen) or by hydrazine with KOH in ethylene glycol (Wolf-Kishner), the latter being the route to straight-chain alkyl benzenes that Friedel–Crafts alkylation cannot give.

  18. Aromatic aldehydes and ketones undergo electrophilic substitution such as nitration, sulfonation and halogenation. The \(\mathrm{-CHO}\) and \(>\mathrm{C=O}\) groups withdraw electrons inductively and by resonance, deactivating the ortho and para positions, so the meta product is formed.

  19. Carboxylic acids are acidic because the carboxylate ion left behind is stabilized by two equivalent resonance structures. Lower \(\mathrm{p}K_a\) means a stronger acid. Electron withdrawing groups (\(-\mathrm{I}\)) such as halogens, \(\mathrm{-NO_2}\) and \(\mathrm{-CN}\) stabilize the conjugate base and raise acid strength; electron donating groups such as \(\mathrm{-CH_3}\), \(\mathrm{-OH}\), \(\mathrm{-OCH_3}\) and \(\mathrm{-NH_2}\) lower it. Benzoic acid (\(\mathrm{p}K_a\) 4.2) is stronger than acetic acid (\(\mathrm{p}K_a\) 4.76) because of the \(-\mathrm{I}\) effect of the \(sp^2\) ring carbon.

  20. The \(\mathrm{-COOH}\) group is identified by the litmus test, by brisk effervescence of \(\mathrm{CO_2}\) with sodium bicarbonate — which phenols do not give — and by the fruity smell of an ester with ethanol and concentrated \(\mathrm{H_2SO_4}\).

  21. Carboxylic acids give acyl chlorides with \(\mathrm{SOCl_2}\), \(\mathrm{PCl_3}\) or \(\mathrm{PCl_5}\), thionyl chloride being preferred because its byproducts are gases; amides with ammonia followed by heating; anhydrides with \(\mathrm{P_2O_5}\) or by heating sodium carboxylate with an acyl chloride; hydrocarbons with one carbon less on decarboxylation of their sodium salts with soda lime; and primary alcohols on reduction with \(\mathrm{LiAlH_4}\) or diborane, though not with \(\mathrm{NaBH_4}\).

  22. The chapter as a whole moves through one connected cycle of interconversions, summarised in FC 12.S1.

Flowchart

FC 12.S1 – The chapter as one interconversion cycle: primary alcohol → (oxidation/dehydrogenation) → aldehyde → (oxidation) → carboxylic acid; secondary alcohol → (oxidation) → ketone; carboxylic acid → (SOCl₂) → acyl chloride → (H₂/Pd–BaSO₄) → aldehyde, or → (R₂Cd) → ketone; carboxylic acid → (LiAlH₄) → primary alcohol, closing the loop; side branches: aldehyde/ketone → (HCN) → cyanohydrin, → (NaHSO₃) → adduct, → (R′OH) → acetal/ketal, → (NH₂–Z) → imine, → (dil. alkali, α-H present) → aldol → α,β-unsaturated carbonyl compound, → (conc. alkali, no α-H) → Cannizzaro products, → (Zn–Hg/HCl or H₂N–NH₂/KOH) → –CH₂–; carboxylic acid → (NH₃, Δ) → amide, → (P₂O₅) → anhydride, → (soda lime, Δ) → hydrocarbon

12.11 Objective Questions

  1. In the resonating structures A and B shown in Fig. E12.1, the number of unshared electrons in the valence shell present on oxygen respectively are
    1. 2, 4
    2. 2, 6
    3. 4, 6
    4. 6, 4

Diagram

Fig. E12.1 – Two resonating structures: (A) a carbonyl group drawn as >C=O with its lone pairs on oxygen; (B) the charge-separated form with a C–O single bond, a positive charge on carbon and a negative charge on oxygen

  1. In the Wolf-Kishner reduction, alkyl aryl ketones are reduced to alkyl benzenes. During this change, ketones are first converted into
    1. acids
    2. alcohols
    3. hydrazones
    4. alkenes
  2. Aldol condensation is
    1. electrophilic substitution reaction
    2. nucleophilic substitution reaction
    3. elimination reaction
    4. addition–elimination reaction
  3. Which one of the compounds shown in Fig. E12.2 has the lowest acidity?

Diagram

Fig. E12.2 – Four benzene rings each bearing –COOH: (a) with –NO₂ at the meta position; (b) with –Cl at the ortho position; (c) unsubstituted benzoic acid; (d) with –OCH₃ at the para position

  1. Diborane reduces
    1. ester group
    2. nitro group
    3. halo group
    4. acid group
  2. Benzaldehyde does NOT show a positive test with
    1. Schiff reagent
    2. Tollens’ reagent
    3. Sodium bisulphite solution
    4. Fehling solution

12.12 Check Your Concepts

  1. Assertion: Formaldehyde does not undergo aldol condensation. Reason: Formaldehyde has no \(\alpha\)-hydrogen atom. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.

  2. A ketone contains the same carbonyl group as an aldehyde, yet it is not oxidized by Tollens’ reagent. Explain the structural reason for this difference.

  3. Explain why benzaldehyde is less reactive than ethanal towards nucleophilic addition reactions.

  4. Propanal is warmed with dilute NaOH and the product is then heated. Write both products and name each of them.

  5. Distinguish between the members of each of the following pairs using a single chemical test: (a) ethanal and propanone; (b) benzoic acid and phenol; (c) pentan-2-one and pentan-3-one.

  6. Complete the following reaction and name the product: \(\mathrm{CH_3\!-\!CO\!-\!CH_3} + \mathrm{NH_2\!-\!NH\!-\!CONH_2} \longrightarrow\) ?

  7. Arrange the following in decreasing order of boiling point and justify the order: ethanoic acid, propan-1-ol, propanone, butane.

  8. Why is the Rosenmund reduction carried out with palladium poisoned by barium sulfate rather than with pure palladium? What would be obtained if pure palladium were used?

  9. An organic compound A gives a silver mirror with Tollens’ reagent and a yellow precipitate with \(\mathrm{I_2}\) and NaOH. On oxidation, A gives an acid containing the same number of carbon atoms as A. Identify A and explain each observation.

  10. Why does a carboxylic acid have a higher boiling point than an alcohol of comparable molecular mass, even though both can form hydrogen bonds?

  11. Explain how the sodium bicarbonate test distinguishes a carboxylic acid from a phenol, and write the equation involved.

  12. Which of \(\mathrm{CH_3COOH}\), \(\mathrm{F\!-\!CH_2\!-\!COOH}\) and \(\mathrm{I\!-\!CH_2\!-\!COOH}\) would you expect to be the strongest acid? Justify your answer using the idea of conjugate base stability.

  13. Predict the major organic product formed when benzaldehyde is treated with concentrated \(\mathrm{HNO_3}\) in the presence of concentrated \(\mathrm{H_2SO_4}\), and explain why that particular position of the ring is attacked.

  14. What is the action of DIBAl-H on (a) pent-3-enenitrile and (b) an ester at 195 K? State the special advantage of this reagent in each case.

  15. Match each reagent in column I with the observation it gives in column II.

Column I Column II
a. Tollens’ reagent i. red colour in solution
b. Fehling solution ii. magenta colour
c. sodium nitroprusside with NaOH iii. silver mirror
d. Schiff’s reagent iv. red precipitate of cuprous oxide
  1. Why is cyanohydrin formation described as a ‘step-up’ reaction? Name two functional groups into which the \(\mathrm{-C{\equiv}N}\) group of a cyanohydrin can subsequently be converted.

  2. State what happens when acetic acid is treated with each of the following, writing the equation in each case: (a) thionyl chloride, (b) \(\mathrm{P_2O_5}\) on heating, (c) ammonia followed by strong heating.

  3. Explain why a cyclic ketal is used as a protecting group for a ketone, and state how the ketone is recovered at the end of the synthesis.

  4. Complete the sequence and identify A and B: \[\mathrm{C_6H_5\!-\!CH_3} \xrightarrow{\ \mathrm{Cl_2},\ h\nu\ } \mathbf{A} \xrightarrow[\ 373\ \text{K}\ ]{\ \mathrm{H_3O^{+}}\ } \mathbf{B}\] State also why this route is used commercially.

  5. Why does the oxidation of an unsymmetrical ketone such as butan-2-one with \(\mathrm{CrO_3}\) give a mixture of two acids rather than a single product?

  6. Assertion: Sodium borohydride cannot be used to convert a carboxylic acid into a primary alcohol. Reason: Sodium borohydride is a milder reducing agent than lithium aluminium hydride. State whether both statements are correct and whether the reason explains the assertion.

  7. n-Propylbenzene cannot be prepared satisfactorily by Friedel–Crafts alkylation. Name the reaction studied in this chapter that can be used instead, and write the sequence of steps starting from ethyl phenyl ketone.

  8. Arrange the following in increasing order of reactivity towards nucleophilic addition and give one reason for the position of each: propanone, benzaldehyde, ethanal, methanal.

  9. An aldehyde X has no \(\alpha\)-hydrogen. State which of the following reactions X can and cannot undergo, giving a reason in each case: (a) aldol condensation, (b) Cannizzaro reaction, (c) reaction with 2,4-dinitrophenylhydrazine, (d) Tollens’ test.

  10. Explain why the entire alkyl side chain of an alkyl benzene is oxidized to a single \(\mathrm{-COOH}\) group by alkaline \(\mathrm{KMnO_4}\), while a tertiary alkyl substituent is left untouched.

  11. What are aromatic ketones?

  12. Is phenyl acetic acid an aromatic carboxylic acid?

  13. Write the reaction showing the conversion of ethanenitrile into ethanol.

  14. Predict the product of the following reaction: \[\mathrm{CH_3\!-\!CH_2\!-\!COOCH_3} \xrightarrow[\ \text{ii. } \mathrm{H_3O^{+}}\ ]{\ \text{i. } \mathrm{AlH(i\text{-}Bu)_2}\ } ?\]

  15. Name the product obtained by reacting toluene with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride.

  16. Write the reaction showing the conversion of benzonitrile into benzoic acid.

  17. Name the product obtained by the oxidation of 1,2,3,4-tetrahydronaphthalene with acidified potassium permanganate.

  18. What is formalin?

  19. Arrange the following compounds in the increasing order of their boiling points: formaldehyde, ethane, methyl alcohol.

  20. Acetic acid is prepared from methyl magnesium bromide and dry ice in the presence of dry ether. Name the compound which serves not only as a reagent but also as a cooling agent in the reaction.

  21. Observe the following equation for the reaction of Tollens’ reagent with an aldehyde. How do we know that a redox reaction has taken place? Explain. \[\mathrm{R\!-\!CHO} + 2\,\mathrm{Ag(NH_3)_2^{+}} + \mathrm{OH^{-}} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!COO^{-}} + 2\,\mathrm{Ag}\downarrow + 4\,\mathrm{NH_3} + 2\,\mathrm{H_2O}\]

  22. Formic acid is stronger than acetic acid. Explain.

  23. What is the action of hydrazine on cyclopentanone in the presence of KOH in ethylene glycol?

  24. Write the reaction showing the conversion of acetaldehyde into acetaldehyde dimethyl acetal.

  25. Aldehydes are more reactive toward nucleophilic addition reactions than ketones. Explain.

  26. Write the reaction showing the action of the following reagents on propanenitrile:

  1. dilute NaOH
  2. dilute HCl.
  1. Arrange the carboxylic acids shown in Fig. E12.3 in increasing order of their acidic strength and justify your answer.

Diagram

Fig. E12.3 – Three acids: cyclohexanecarboxylic acid (a six-membered all-carbon ring bearing –COOH); an oxygen-containing six-membered ring with –COOH on the carbon adjacent to the ring oxygen; and an oxygen-containing six-membered ring with –COOH on a carbon one position further from the ring oxygen

  1. Write a note on:
  1. Cannizaro reaction
  2. Stephen reaction.
  1. What is the action of the following reagents on toluene?
  1. Alkaline \(\mathrm{KMnO_4}\), dilute HCl and heat
  2. \(\mathrm{CrO_2Cl_2}\) in \(\mathrm{CS_2}\)
  3. Acetyl chloride in the presence of anhydrous \(\mathrm{AlCl_3}\).
  1. Write the IUPAC names of the structures shown in Fig. E12.4.

Diagram

Fig. E12.4 – Two structures: (a) a cyclohexanone ring carrying a methyl substituent; (b) an open chain with a –COOH group at each end joined directly to one another

  1. Write the reaction showing the conversion of p-bromoisopropyl benzene into p-isopropyl benzoic acid (3 steps).

  2. Write the reaction showing the aldol condensation of cyclohexanone.

  3. Draw and complete the reaction scheme of Fig. E12.5, which starts with acetaldehyde. In each empty box, write the structural formula of the organic compound that would be formed.

Diagram

Fig. E12.5 – Reaction scheme with CH₃CHO at the centre: an arrow labelled HCN leading left to an empty box; an arrow labelled “reduction” leading right to an empty box; an arrow labelled “dilute H₂SO₄, heat” leading down from the HCN product to an empty box; an arrow labelled “Tollens’ reagents” leading down from CH₃CHO to an empty box; an arrow labelled “Conc. H₂SO₄, heat” leading down to the given compound CH₂=CHCO₂H; and an arrow labelled “Cold dilute KMnO₄/H⁺” leading right from it to a final empty box