Chapter 11: Alcohols, Phenols and Ethers
Before the middle of the nineteenth century, surgery was something a patient simply had to endure while fully awake. Operations were carried out at desperate speed, because there was no reliable way to stop the pain. And even when the surgery itself succeeded, the wound very often turned septic afterwards and the patient died anyway. Within about twenty years, two ordinary-looking liquids changed both of these problems — and both liquids belong to this chapter.
The first was diethyl ether. In 1846, in a hospital in Boston, William Morton demonstrated to a room full of sceptical surgeons that a patient breathing ether vapour could be operated upon without feeling the knife. News of the demonstration travelled across the world within months, and ether became the first widely used general anaesthetic.
The second was phenol. In the 1860s, the Scottish surgeon Joseph Lister argued that deaths after surgery were caused by something living entering the open wound. He began soaking dressings and instruments in carbolic acid — an older name for phenol — and his patients started to survive. Phenol became the ancestor of modern antiseptic practice.
Both molecules look almost too simple for such an effect. An ether is two carbon groups joined through an oxygen atom; phenol is an –OH group sitting directly on a benzene ring. Add the alcohols, where that same –OH sits on a saturated carbon, and you have the three families of this chapter — three ways of placing a single oxygen atom, and three strikingly different sets of chemical behaviour.
Diagram
Fig. 11.0 – Three side-by-side panels showing the general structures R–OH (alcohol, with –OH on a saturated sp³ carbon), Ar–OH (phenol, with –OH drawn directly on a benzene ring) and R–O–R′ (ether, oxygen bridging two carbon groups); label each panel with its family name and highlight the oxygen atom in each.
Keep these five questions in mind as you read — the chapter gives you everything you need to reason out each one.
11.1 Introduction
Every compound studied in this chapter is built in the same way: a hydrocarbon skeleton with a single oxygen atom worked into it. What decides the whole personality of the molecule is where that oxygen sits and what it is bonded to. Place it between a carbon and a hydrogen and you get the hydroxyl group; place it between two carbons and you get an ether linkage.
Alcohols
An alcohol is a compound in which a hydroxyl group, –OH, is joined to a saturated carbon atom. A saturated carbon is one that is linked to its neighbours only by single bonds — it is \(sp^3\) hybridised and carries no double or triple bond. The general skeleton is shown in Fig. 11.1.
Diagram
Fig. 11.1 – A carbon atom drawn with three plain bonds and a fourth bond to –OH; shade or box the –OH group and label it “hydroxyl group”; label the carbon “saturated (sp³) carbon”.
Hydroxy compounds of aromatic rings
The hydroxyl group can also be carried by an aromatic compound, and here it can occupy two quite different positions. Because of this, aromatic hydroxy compounds are of two types — phenols and aromatic alcohols.
In a phenol, the –OH group is fixed directly on a carbon atom of the benzene ring itself. Phenol, \(\mathrm{C_6H_5OH}\), and naphthol are examples. In an aromatic alcohol, the ring is present but the –OH group sits on a carbon of the side chain attached to the ring, not on the ring carbon. Phenylmethanol (benzyl alcohol), \(\mathrm{C_6H_5CH_2OH}\), is the simplest example. Compare the two carefully in Fig. 11.2.
Diagram
Fig. 11.2 – Left panel labelled “Phenols”: benzene ring with –OH on a ring carbon, and naphthalene with –OH on a ring carbon. Right panel labelled “Aromatic alcohol”: benzene ring bearing a –CH₂–OH side chain; draw an arrow to the ring carbon in the left panel and to the side-chain carbon in the right panel.
The quickest way to place any oxygen-containing compound of this chapter is to ask two questions in order, as set out in FC 11.1.
Flowchart
FC 11.1 – Classifying a monofunctional oxygen compound: start → is the oxygen bonded to two carbon atoms? → if yes → ether; if no (it carries an O–H bond) → is that O–H attached directly to an aromatic ring carbon? → if yes → phenol; if no → is a ring present in a side chain? → if yes → aromatic alcohol; if no → alcohol.
Ethers
Ethers are compounds in which an oxygen atom is bonded to two alkyl groups, or to two aryl groups, or to one alkyl and one aryl group. There is no O–H bond at all — the oxygen is fully occupied by two carbon atoms. Their general formulae are written as
\[\mathrm{R\!-\!O\!-\!R'},\qquad \mathrm{R\!-\!O\!-\!Ar},\qquad \mathrm{Ar\!-\!O\!-\!Ar'}\]
where R and R′ stand for alkyl groups and Ar and Ar′ for aryl (aromatic) groups.
Ethers are described as organic oxides, and also as the anhydrides of alcohols. The word anhydride means “without water”: if two molecules of an alcohol are imagined to lose one molecule of water between them, what is left is an ether linkage. This is only a way of seeing the relationship on paper, but it is a useful one, and Section 11.5 will show that ethers really can be made from alcohols by removing water.
Epoxides
An ether oxygen need not lie in an open chain. In an epoxide, the ethereal oxygen is part of a three-membered ring, so the ring contains two carbon atoms and one oxygen atom. The simplest member is ethylene oxide, named 1,2-epoxyethane, drawn in Fig. 11.3.
Diagram
Fig. 11.3 – General three-membered epoxide ring (–C–C– bridged by O) beside the specific structure of ethylene oxide (H₂C–CH₂ bridged by O); label them “epoxide (cyclic ether)” and “ethylene oxide / 1,2-epoxyethane”.
11.2 Classification
A single family name such as “alcohol” covers thousands of compounds, and they do not all behave alike. Chemists therefore sort them into smaller groups, using features that actually control the chemistry — how many hydroxyl groups a molecule carries, what kind of carbon the hydroxyl group sits on, and, for ethers, whether the two groups on either side of the oxygen are the same. We shall take alcohols and phenols first, and then ethers.
11.2.1 Mono, di, tri and polyhydric compounds
Alcohols and phenols are classified as monohydric, dihydric, trihydric or polyhydric compounds according to whether their molecules contain one, two, three or more hydroxyl groups. Representative structures of each type are drawn in Fig. 11.4.
Diagram
Fig. 11.4 – Two rows of examples. Top row (alcohols): CH₃–CH₂–OH labelled monohydric; HOCH₂–CH₂OH labelled dihydric; HOCH₂–CH(OH)–CH₂OH labelled trihydric. Bottom row (phenols): benzene ring with one OH; benzene ring with OH groups on adjacent carbons; benzene ring with three OH groups. Label the three columns “Monohydric alcohols/phenols”, “Dihydric alcohols/phenols”, “Trihydric alcohols/phenols”.
Ethanol, \(\mathrm{CH_3CH_2OH}\), and phenol, \(\mathrm{C_6H_5OH}\), are monohydric. Ethane-1,2-diol (ethylene glycol), \(\mathrm{HOCH_2\!-\!CH_2OH}\), is dihydric, and propane-1,2,3-triol (glycerol), \(\mathrm{HOCH_2\!-\!CH(OH)\!-\!CH_2OH}\), is trihydric. The same counting applies to phenols: a benzene ring carrying two –OH groups is a dihydric phenol, one carrying three is a trihydric phenol.
Classifying monohydric alcohols by hybridisation
Monohydric alcohols are sorted further on the basis of the hybridisation state of the carbon atom to which the hydroxyl group is attached. Recall that a carbon atom joined to its neighbours by four single bonds is \(sp^3\) hybridised, while a carbon atom taking part in a carbon–carbon double bond is \(sp^2\) hybridised. This one difference changes the whole environment of the –OH group.
a. Alcohols containing the \(sp^3\mathrm{C\!-\!OH}\) bond. In these alcohols the –OH group is attached to an \(sp^3\) hybridised carbon atom of an alkyl group. They are classified further as primary (\(1^0\)), secondary (\(2^0\)) and tertiary (\(3^0\)) alcohols, according as the –OH group is attached to a primary, a secondary or a tertiary carbon atom respectively, as shown in Fig. 11.5.
A carbon atom is called primary if it is bonded to only one other carbon atom, secondary if it is bonded to two other carbon atoms, and tertiary if it is bonded to three. So the label of the alcohol is simply the label of the carbon that carries the –OH group.
Diagram
Fig. 11.5 – Three structures side by side: R–CH₂–OH with the two H atoms shown and the carbon marked “primary carbon”, labelled 1⁰ alcohol; R–CH(R)–OH marked “secondary carbon”, labelled 2⁰ alcohol; R–C(R)(R)–OH marked “tertiary carbon”, labelled 3⁰ alcohol.
Each of these three types can, in addition, be allylic or benzylic, if the \(sp^3\) carbon carrying the –OH group is further bonded to an \(sp^2\) carbon. Both types are drawn in Fig. 11.6.
Diagram
Fig. 11.6 – Left: H₂C=CH–CH₂–OH with an arrow to the sp³ carbon bearing OH and a note “bonded to a C=C carbon”, labelled “allylic alcohol”. Right: benzene ring–CH₂–OH with an arrow to the sp³ carbon bearing OH and a note “bonded to an aromatic ring carbon”, labelled “benzylic alcohol”.
- Allylic alcohols: here the –OH group is attached to an \(sp^3\) hybridised carbon atom which is further bonded to a carbon–carbon double bond. An allylic alcohol may itself be primary, secondary or tertiary.
- Benzylic alcohols: here the –OH group is attached to an \(sp^3\) hybridised carbon atom which is further bonded to an aromatic ring. A benzylic alcohol too may be primary, secondary or tertiary.
\(\mathrm{H_2C{=}CH\!-\!CH_2\!-\!OH}\), \(\quad \mathrm{H_2C{=}CH\!-\!CH(CH_3)\!-\!OH}\), \(\quad \mathrm{H_2C{=}CH\!-\!C(CH_3)_2\!-\!OH}\), \(\quad \mathrm{C_6H_5\!-\!CH_2\!-\!OH}\), \(\quad \mathrm{C_6H_5\!-\!CH(CH_3)\!-\!OH}\)
b. Alcohols containing the \(sp^2\mathrm{C\!-\!OH}\) bond. In these alcohols the –OH group is attached to an \(sp^2\) hybridised carbon atom which is itself part of a carbon–carbon double bond. Such alcohols are known as vinylic alcohols. For example,
\[\mathrm{CH_2{=}CH\!-\!OH} \qquad \text{(vinyl alcohol)}\]
The complete scheme for alcohols, from the first division down to the last, is collected in FC 11.2.
Flowchart
FC 11.2 – Classification of alcohols and phenols: compound → count the –OH groups → monohydric / dihydric / trihydric / polyhydric; branch from monohydric alcohol → hybridisation of the C bearing –OH → sp³C–OH → primary (1⁰) / secondary (2⁰) / tertiary (3⁰) → each may in addition be allylic (next to C=C) or benzylic (next to aromatic ring); other branch → sp²C–OH → vinylic alcohol.
11.2.2 Classification of Ethers
Ethers are classified according to whether the two alkyl or aryl groups bonded to the oxygen atom are the same or different. When the two groups are identical the ether is symmetrical, also called a simple ether; when they differ it is unsymmetrical, also called a mixed ether. For example:
\[\mathrm{R\!-\!O\!-\!R} \ / \ \mathrm{Ar\!-\!O\!-\!Ar} \qquad \text{symmetrical ethers (simple ethers)}\]
\[\mathrm{CH_3\!-\!O\!-\!CH_3} \ / \ \mathrm{C_6H_5\!-\!O\!-\!C_6H_5}\]
\[\mathrm{R\!-\!O\!-\!R'} \ / \ \mathrm{Ar\!-\!O\!-\!Ar'} \qquad \text{unsymmetrical ethers (mixed ethers)}\]
\[\mathrm{CH_3\!-\!O\!-\!C_2H_5} \ / \ \mathrm{C_6H_5\!-\!O\!-\!CH_3}\]
This distinction is worth remembering, because Section 11.5 will show that the method used to prepare an ether in the laboratory depends on exactly this point — a symmetrical ether can be made from a single alcohol, while an unsymmetrical one needs two different starting materials.
11.3 Nomenclature
A compound may be classified correctly and still be useless in conversation until it has a name that everyone reads the same way. This section sets out how alcohols, phenols and ethers are named — first by the older common systems that survive in laboratories and catalogues, and then by the IUPAC system, which is the one an examiner expects.
11.3.1 Alcohols
There are three systems of nomenclature of monohydric alcohols.
a. Common/trivial names
The common or trivial name of an alcohol is obtained by writing the name of the alkyl group bonded to the –OH group, followed by the word alcohol. Thus \(\mathrm{CH_3\!-\!OH}\) is methyl alcohol and \(\mathrm{CH_3CH_2\!-\!OH}\) is ethyl alcohol.
For higher alkyl groups, more than one arrangement of the carbon skeleton is possible, so the names include prefixes such as normal (n-), iso, secondary (sec-) and tertiary (tert- or t-) to say which arrangement is meant. Several examples are collected in Table 11.1.
b. Carbinol system
In this system, alcohols are treated as derivatives of methyl alcohol, \(\mathrm{CH_3OH}\), which is given the special name carbinol. The carbon atom carrying the –OH group is called the carbinol carbon. The alkyl groups attached to this carbon are named in alphabetical order, and then the suffix carbinol is added. For example, the compound shown in Fig. 11.7 is named methyl carbinol.
Diagram
Fig. 11.7 – Structure H₃C–CH(H)–OH drawn with all bonds shown; arrow pointing to the central carbon labelled “carbinol carbon”; caption “Methyl carbinol”.
c. IUPAC system
According to the IUPAC system, alcohols are named as alkanols. The ending ‘e’ in the name of the parent alkane, alkene or alkyne is replaced by the suffix ‘ol’. So ethane becomes ethanol, and propane becomes propanol.
For naming a polyhydric alcohol, the ‘e’ at the end of the alkane name is retained, the ending ‘ol’ is added, and the number of –OH groups is indicated by the prefix di, tri, etc., placed before ‘ol’. The positions of the –OH groups are shown by appropriate locants — the numbers that tell you which carbon atom of the chain carries each group. For example, \(\mathrm{HOCH_2\!-\!CH_2OH}\) is ethane-1,2-diol (see Table 11.1).
Cyclic alcohols are named by using the prefix cyclo with the parent alkane, and the carbon carrying the –OH group is taken as C-1. Two examples are drawn in Fig. 11.8.
Diagram
Fig. 11.8 – Left: cyclobutane ring with –OH on one carbon, labelled “Cyclobutanol”. Right: cyclopentane ring with –OH on C-1 and a –C₂H₅ group on C-2, labelled “2-Ethylcyclopentanol”.
The full naming procedure is set out step by step in FC 11.3.
Flowchart
FC 11.3 – IUPAC naming of an alcohol: select the longest carbon chain that contains the carbon bearing –OH → number the chain from the end that gives the –OH carbon the lower locant → name the parent alkane and replace the final ‘e’ by ‘ol’ → insert the locant of the –OH group before ‘ol’ → if more than one –OH is present, retain the ‘e’ and use di/tri before ‘ol’ with all locants → add the names and locants of any other substituents as prefixes in alphabetical order; branch: if the chain is a ring → add the prefix cyclo and number the ring from the carbon carrying –OH.
Table 11.1 Common/Trivial and IUPAC Names of some alcohols
| Structural formula | Common/Trivial Name | IUPAC Name |
|---|---|---|
| \(\mathrm{H_3C\!-\!OH}\) | Methyl alcohol | Methanol |
| \(\mathrm{H_3C\!-\!CH_2\!-\!OH}\) | Ethyl alcohol | Ethanol |
| \(\mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!OH}\) | n-Propyl alcohol | Propan-1-ol |
| \(\mathrm{H_3C\!-\!CH_2\!-\!CH(OH)\!-\!CH_3}\) | sec-Butyl alcohol | Butan-2-ol |
| \(\mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_2\!-\!OH}\) | Isobutyl alcohol | 2-Methylpropan-1-ol |
| \(\mathrm{(H_3C)_3C\!-\!OH}\) | tert-Butyl alcohol | 2-Methylpropan-2-ol |
| \(\mathrm{HO\!-\!H_2C\!-\!CH_2\!-\!OH}\) | Ethylene glycol | Ethane-1,2-diol |
| \(\mathrm{HO\!-\!H_2C\!-\!CH(OH)\!-\!CH_2\!-\!OH}\) | Propylene glycerol | Propane-1,2,3-triol |
| \(\mathrm{H_3C\!-\!CH{=}CH\!-\!CH_2OH}\) | Crotonyl alcohol | But-2-en-1-ol |
Diagram
Fig. 11.9 – Four labelled structures drawn as the answers: (i) benzene ring numbered 1–6 with OH at C-1 and C₂H₅ at C-2 and C-5; (ii) H₂C=CH–CH₂–OH with carbons numbered 3, 2, 1; (iii) CH₃–CH(O–CH₃)–CH₃; (iv) benzene ring bearing –CH₂–OH.
11.3.2 Nomenclature of phenols
The hydroxyl derivative of benzene is called phenol. Its IUPAC name is benzenol, but the common name phenol is also accepted by IUPAC, and it is the name almost always used.
When the ring carries other substituents as well, the common names use the prefixes ortho, meta and para to describe where the second group sits relative to the –OH group. The IUPAC system does the same job with the locants 2-, 3-, 4- and so on, counting from the carbon that carries the –OH group. Ortho corresponds to position 2, meta to position 3 and para to position 4. Several examples are given in Table 11.2.
Diagram
Fig. 11.10 – Benzene ring with –OH at C-1, ring carbons numbered 1 to 6, and the positions 2, 3 and 4 additionally labelled ortho, meta and para respectively.
Table 11.2 Common and IUPAC names of some phenols
| Structural formula | Common name | IUPAC Name |
|---|---|---|
| \(\mathrm{C_6H_5\!-\!OH}\) | Phenol | Benzenol/Phenol |
| \(\mathrm{2\text{-}CH_3\!-\!C_6H_4\!-\!OH}\) | o-Cresol | 2-Methylphenol |
| \(\mathrm{4\text{-}NO_2\!-\!C_6H_4\!-\!OH}\) | p-Nitrophenol | 4-Nitrophenol |
| \(\mathrm{1,2\text{-}C_6H_4(OH)_2}\) | Catechol | Benzene-1,2-diol |
| \(\mathrm{1,3\text{-}C_6H_4(OH)_2}\) | Resorcinol | Benzene-1,3-diol |
| \(\mathrm{1,4\text{-}C_6H_4(OH)_2}\) | Hydroquinone/quinol | Benzene-1,4-diol |
| \(\mathrm{1,3,5\text{-}C_6H_3(OH)_3}\) | Phloroglucinol | Benzene-1,3,5-triol |
| \(\mathrm{1,2,3\text{-}C_6H_3(OH)_3}\) | Pyrogallol | Benzene-1,2,3-triol |
11.3.3 Nomenclature of Ethers
In the common system, an ether is named by writing the names of the two alkyl groups attached to the oxygen atom in alphabetical order, followed by the word ether. If both alkyl groups are the same, the prefix di- is used. Thus \(\mathrm{CH_3\!-\!O\!-\!CH_2CH_3}\) is ethyl methyl ether, and \(\mathrm{CH_3\!-\!O\!-\!CH_3}\) is dimethyl ether.
According to the IUPAC system, ethers are named as alkoxyalkanes (see Table 11.3). The larger alkyl group is considered to be the parent alkane. The name of the smaller alkane, converted into an alkoxy group, is prefixed to it along with its locant. For example, the compound drawn in Fig. 11.11 is named 2-methoxypropane: the three-carbon propane chain is the parent, and the \(\mathrm{-OCH_3}\) group sits on C-2.
Diagram
Fig. 11.11 – Structure CH₃–O–CH(CH₃)–CH₃ with the propane chain carbons numbered 1, 2, 3 and the –OCH₃ group marked at C-2; caption “2-Methoxypropane”.
Some alkoxy groups
| Group | Name |
|---|---|
| \(\mathrm{CH_3\!-\!CH_2O\!-}\) | Ethoxy |
| \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!O\!-}\) | n-Propoxy |
| \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!O\!-}\) | Isopropoxy |
Table 11.3 Common and IUPAC Names of some Ethers
| Structural formula | Common Name | IUPAC Name |
|---|---|---|
| \(\mathrm{H_3C\!-\!O\!-\!CH_3}\) | Dimethyl ether | Methoxymethane |
| \(\mathrm{H_3C\!-\!O\!-\!CH_2\!-\!CH_3}\) | Ethyl methyl ether | Methoxyethane |
| \(\mathrm{H_3C\!-\!O\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) | Methyl n-propyl ether | 1-Methoxypropane |
| \(\mathrm{C_6H_5\!-\!O\!-\!CH_3}\) | Methyl phenyl ether (Anisole) | Methoxybenzene |
| \(\mathrm{C_6H_5\!-\!O\!-\!CH_2\!-\!CH_2\!-\!CH_3}\) | Phenyl n-propyl ether | 1-Propoxybenzene |
| A cyclobutane ring carrying two \(\mathrm{CH_3}\) groups on one carbon and \(\mathrm{-OCH_3}\) on the next | – | 2-Methoxy-1,1-dimethylcyclobutane |
Diagram
Fig. 11.12 – Cyclobutane ring with two CH₃ groups on C-1 and an –OCH₃ group on C-2, with ring positions numbered; caption “2-Methoxy-1,1-dimethylcyclobutane”.
Diagram
Fig. 11.13 – Four structures for the student to name: (i) a cyclopentane ring carrying –OH groups on two adjacent carbons; (ii) H₃C–CH(OCH₂CH₃)–CH₂–CH₃; (iii) benzene ring bearing –CH₂–CH₂–OH; (iv) benzene ring with –OH on one carbon and –CH₃ on the carbon in the meta position.
11.4 Alcohols and Phenols
11.4.1 Preparation of alcohols
Alcohols can be built from several different starting materials — halides, alkenes, carbonyl compounds and organometallic reagents. Each route places the –OH group in a predictable position, so choosing the right method is really a matter of deciding where you want the oxygen to end up.
a. From alkyl halides
An alkyl halide is hydrolysed to an alcohol by warming it with aqueous alkali, or with moist silver oxide (refer to Section 10.6.2). The halide ion is displaced by the hydroxide ion, which is a stronger nucleophile — that is, a species that donates a lone pair of electrons to an electron-deficient carbon atom.
\[\mathrm{R\!-\!X} + \mathrm{KOH}\text{(aq)} \rightarrow \mathrm{R\!-\!OH} + \mathrm{KX}\]
\[2\,\mathrm{R\!-\!X} + \mathrm{Ag_2O} + \mathrm{H_2O} \rightarrow 2\,\mathrm{R\!-\!OH} + 2\,\mathrm{AgX}\]
b. By acid catalyzed hydration of alkenes
An alkene reacts with sulfuric acid to produce an alkyl hydrogen sulfate, which on hydrolysis gives the alcohol (refer to Std XI Chemistry Textbook, Section 15.2.4). Taking propene as the example:
\[\mathrm{CH_3\!-\!CH{=}CH_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{CH_3\!-\!CH(OSO_3H)\!-\!CH_3}\]
\[\mathrm{CH_3\!-\!CH(OSO_3H)\!-\!CH_3} + \mathrm{H_2O} \xrightarrow{\ \Delta\ } \mathrm{CH_3\!-\!CH(OH)\!-\!CH_3} + \mathrm{H_2SO_4}\]
This reaction follows Markownikoff’s rule, so the –OH group lands on the carbon that already carries fewer hydrogen atoms.
The mechanism of acid catalyzed hydration of an alkene involves three steps, drawn with curly arrows in Fig. 11.14. First, water is protonated by the acid to give the hydronium ion, \(\mathrm{H_3O^+}\), which supplies the proton to the alkene.
Step 1: Formation of a carbocation intermediate. The \(\pi\) electrons of the double bond attack the proton. One carbon gains the hydrogen, and the other is left with only six electrons and a positive charge — a carbocation.
Step 2: Nucleophilic attack of \(\mathrm{H_2O}\) on the carbocation. A lone pair on the oxygen of a water molecule attacks the positively charged carbon, giving a protonated alcohol.
Step 3: Deprotonation. A second water molecule removes the extra proton, releasing the neutral alcohol and regenerating \(\mathrm{H_3O^+}\).
Diagram
Fig. 11.14 – Three-step mechanism of acid-catalysed hydration with curly arrows: (Step 1) H–O–H + H⁺ → H₃O⁺, then π electrons of C=C attacking H of H₃O⁺ to give carbocation + H₂O; (Step 2) lone pair of H₂O attacking the C⁺ to give protonated alcohol; (Step 3) H₂O removing the proton to give the alcohol + H₃O⁺.
Flowchart
FC 11.4 – Acid-catalysed hydration of an alkene: alkene + H₃O⁺ → protonation of the double bond → carbocation intermediate (Markownikoff orientation: H adds to the carbon with more H atoms) → nucleophilic attack by H₂O → protonated alcohol → loss of H⁺ → alcohol; note that the catalyst H⁺ is regenerated at the end.
c. Hydroboration – Oxidation of alkenes
With diborane, \(\mathrm{B_2H_6}\), an alkene undergoes an addition reaction called hydroboration to give a trialkylborane, \(\mathrm{R_3B}\). This on oxidation with hydrogen peroxide in alkaline medium gives the alcohol (refer to Std XI Chemistry Textbook, Section 15.2.4).
\[6\,\mathrm{R\!-\!CH{=}CH_2} + \mathrm{B_2H_6} \xrightarrow{\ \mathrm{THF}\ } 2\,(\mathrm{R\!-\!CH_2\!-\!CH_2})_3\mathrm{B}\]
\[(\mathrm{R\!-\!CH_2\!-\!CH_2})_3\mathrm{B} + 3\,\mathrm{H_2O_2} \xrightarrow{\ \mathrm{OH^-}\ } 3\,\mathrm{R\!-\!CH_2\!-\!CH_2\!-\!OH} + \mathrm{H_3BO_3}\]
This is an antimarkownikoff hydration of the alkene: the –OH group ends up on the less substituted carbon, exactly opposite to the acid catalyzed route. The two methods therefore give two different alcohols from the same alkene, and Fig. 11.15 sets them side by side.
Diagram
Fig. 11.15 – Propene at the centre with two arrows: upward path labelled “(i) conc. H₂SO₄ (ii) H₂O” leading to propan-2-ol (Markownikoff); downward path labelled “(i) B₂H₆–THF (ii) H₂O₂, OH⁻” leading to propan-1-ol (anti-Markownikoff); mark the carbon that receives the –OH in each case.
- \(\mathrm{CH_3\!-\!CH{=}CH_2} \xrightarrow[\text{(ii) } \mathrm{H_2O_2},\ \mathrm{OH^-}]{\text{(i) } \mathrm{B_2H_6\!-\!THF}} \ ?\)
- 1-Methylcyclohexene \(\xrightarrow[\text{(ii) } \mathrm{H_2O}]{\text{(i) conc. } \mathrm{H_2SO_4}} \ ?\)
d. By reduction of carbonyl compounds
i. By reduction of aldehydes and ketones. Aldehydes on reduction by \(\mathrm{H_2/Ni}\) or \(\mathrm{LiAlH_4}\) give primary alcohols (\(1^0\)). Similarly, ketones on reduction with \(\mathrm{H_2/Ni}\) or \(\mathrm{LiAlH_4}\) give secondary alcohols (\(2^0\)). In each case two hydrogen atoms are added across the carbon–oxygen double bond.
\[\mathrm{R\!-\!CHO} \xrightarrow{\ \mathrm{H_2/Ni\ or\ Pd},\ \Delta\ } \mathrm{R\!-\!CH_2\!-\!OH} \qquad (1^0\ \text{alcohol})\]
\[\mathrm{R\!-\!CHO} \xrightarrow{\ \text{(i) } \mathrm{LiAlH_4};\ \text{(ii) } \mathrm{H_3O^+}\ } \mathrm{R\!-\!CH_2\!-\!OH}\]
\[\mathrm{R\!-\!CO\!-\!R} \xrightarrow{\ \mathrm{H_2/Ni\ or\ Pd},\ \Delta\ } \mathrm{R\!-\!CH(OH)\!-\!R} \qquad (2^0\ \text{alcohol})\]
\[\mathrm{R\!-\!CO\!-\!R} \xrightarrow{\ \text{(i) } \mathrm{LiAlH_4};\ \text{(ii) } \mathrm{H_3O^+}\ } \mathrm{R\!-\!CH(OH)\!-\!R}\]
e. By addition of Grignard reagent to aldehydes and ketones
A Grignard reagent, \(\mathrm{R\!-\!Mg\!-\!X}\), reacts with an aldehyde or a ketone to form an adduct, which on hydrolysis with dilute acid gives the corresponding alcohol.
The reason the reaction works lies in the polarity of the two bonds involved, shown in Fig. 11.16. In the carbonyl group, oxygen is far more electronegative than carbon, so the carbon carries a partial positive charge (\(\delta^+\)). In the Grignard reagent the magnesium is much less electronegative than carbon, so there the carbon is partially negative (\(\delta^-\)). The electron-rich carbon of the reagent therefore attacks the electron-poor carbon of the carbonyl group.
Diagram
Fig. 11.16 – Grignard addition: C=O with δ⁺ on C and δ⁻ on O, R–Mg–X with δ⁻ on R and δ⁺ on Mg, curly arrow from R to the carbonyl carbon and from the C=O π bond to O; product drawn as the adduct –C(OMgX)(R)–, then an arrow labelled H₃O⁺ giving –C(OH)(R)– + Mg(OH)X.
\[\mathrm{R'\!-\!CHO} + \mathrm{R\!-\!Mg\!-\!X} \xrightarrow{\ \text{dry ether}\ } \mathrm{R'\!-\!CH(OMgX)\!-\!R} \qquad \text{(adduct)}\]
\[\mathrm{R'\!-\!CH(OMgX)\!-\!R} \xrightarrow{\ \mathrm{H_3O^+}\ } \mathrm{R'\!-\!CH(OH)\!-\!R} + \mathrm{Mg(OH)X}\]
This reaction is useful in the synthesis of a wide variety of alcohols, because the class of alcohol obtained depends only on the carbonyl compound chosen (see Table 11.4).
Table 11.4 Preparation of alcohols by Grignard reagent
| Aldehyde/ketone | Grignard reagent | Final product | Type of alcohol |
|---|---|---|---|
| \(\mathrm{H\!-\!CHO}\) (formaldehyde) | \(\mathrm{R\!-\!MgX}\) | \(\mathrm{R\!-\!CH_2OH}\) | \(1^0\) |
| \(\mathrm{R'\!-\!CHO}\) (aldehyde) | \(\mathrm{R\!-\!MgX}\) | \(\mathrm{R\!-\!CH(OH)\!-\!R'}\) | \(2^0\) |
| \(\mathrm{R'\!-\!CO\!-\!R''}\) (ketone) | \(\mathrm{R\!-\!MgX}\) | \(\mathrm{R\!-\!C(OH)(R')\!-\!R''}\) | \(3^0\) |
An epoxide behaves in the same way, but because its strained three-membered ring opens on attack, the chain is lengthened by two carbon atoms and a primary alcohol results.
\[\text{ethylene oxide} + \mathrm{R\!-\!MgX} \xrightarrow{\ \text{dry ether}\ } [\mathrm{R\!-\!CH_2\!-\!CH_2\!-\!OMgX}] \xrightarrow{\ \mathrm{H_3O^+}\ } \mathrm{R\!-\!CH_2CH_2\!-\!OH} + \mathrm{Mg(OH)X}\]
11.4.2 Preparation of phenol
The –OH group cannot simply be pushed on to a benzene ring the way it can be attached to an alkyl chain, because an aromatic ring is electron-rich and resists attack by a nucleophile. Every method below therefore works indirectly, by first putting some other group on the ring and then replacing it.
a. From chlorobenzene (Dow Process)
Chlorobenzene is fused with NaOH at high temperature and pressure (623 K and 300 atm) followed by treatment with dilute HCl to obtain phenol, as shown in Fig. 11.17.
\[\mathrm{C_6H_5Cl} \xrightarrow{\ \mathrm{NaOH},\ 623\ \text{K}/300\ \text{atm}\ } \mathrm{C_6H_5O^-Na^+} \xrightarrow{\ \mathrm{H_3O^+}\ } \mathrm{C_6H_5OH}\]
Diagram
Fig. 11.17 – Dow process drawn with ring structures: chlorobenzene → (NaOH, 623 K/300 atm) → sodium phenoxide → (H₃O⁺) → phenol; label each structure.
The severe conditions are the point worth noticing: they tell you how unwilling an aryl halide is to undergo substitution compared with an alkyl halide, which is hydrolysed by warm aqueous alkali.
b. From Cumene
This is the commercial method of preparation of phenol. Cumene (isopropylbenzene) on air oxidation in presence of Co-naphthenate at 423 K gives cumene hydroperoxide, which on decomposition with dilute acid gives phenol, with acetone as a valuable by-product.
\[\mathrm{C_6H_5\!-\!CH(CH_3)_2} + \mathrm{O_2} \xrightarrow{\ \text{Co-naphthenate},\ 423\ \text{K}\ } \mathrm{C_6H_5\!-\!C(CH_3)_2\!-\!O\!-\!O\!-\!H}\]
\[\mathrm{C_6H_5\!-\!C(CH_3)_2\!-\!O\!-\!O\!-\!H} \xrightarrow{\ \text{dil. HCl},\ \Delta\ } \mathrm{C_6H_5OH} + \mathrm{(CH_3)_2C{=}O}\]
Diagram
Fig. 11.18 – Cumene process with ring structures: cumene + O₂ → (Co-naphthenate, 423 K) → cumene hydroperoxide → (dil. HCl, Δ) → phenol + acetone; label cumene, cumene hydroperoxide, phenol and acetone.
Flowchart
FC 11.5 – Cumene process for phenol: benzene + propene → cumene → air oxidation (Co-naphthenate, 423 K) → cumene hydroperoxide → acid decomposition (dil. HCl, Δ) → phenol + acetone (by-product recovered and sold).
c. From benzene sulfonic acid
Benzene sulfonic acid on neutralization by NaOH gives sodium benzene sulfonate, which on fusion with solid NaOH at 573 K gives sodium phenoxide; reaction with dilute acid then gives phenol.
\[\mathrm{C_6H_5SO_3H} \xrightarrow{\ \mathrm{NaOH}\ } \mathrm{C_6H_5SO_3Na} \xrightarrow{\ \mathrm{NaOH},\ 573\ \text{K}\ } \mathrm{C_6H_5O^-Na^+} \xrightarrow{\ \text{dil. HCl}\ } \mathrm{C_6H_5OH}\]
Diagram
Fig. 11.19 – Benzene sulfonic acid route with ring structures: benzene sulphonic acid → (NaOH) → sodium benzene sulphonate → (NaOH, 573 K) → sodium phenoxide → (dil. HCl) → phenol.
d. From aniline
Aniline is treated with nitrous acid (\(\mathrm{NaNO_2} + \mathrm{HCl}\)) at low temperature to obtain benzene diazonium chloride, which on hydrolysis gives phenol (also refer to Chapter 13 for this reaction).
\[\mathrm{C_6H_5NH_2} + \mathrm{HNO_2} \xrightarrow{\ \mathrm{NaNO_2/HCl},\ 273\ \text{K}\ } \mathrm{C_6H_5N_2^+Cl^-}\]
\[\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{\ \mathrm{H_2O},\ \Delta\ } \mathrm{C_6H_5OH} + \mathrm{N_2}\uparrow + \mathrm{HCl}\]
Diagram
Fig. 11.20 – Aniline route with ring structures: aniline + HNO₂ → (NaNO₂/HCl, 273 K) → benzene diazonium chloride → (H₂O, Δ) → phenol + N₂↑ + HCl.
The low temperature of 273 K matters: the diazonium salt is unstable and decomposes if the mixture is allowed to warm, which is exactly what is wanted in the second step but must be prevented in the first.
11.4.3 Physical properties of alcohols and phenols
a. Nature of intermolecular forces
Alcohols and phenols are very polar molecules due to the presence of –OH groups. Oxygen is strongly electronegative, so it pulls the shared pair of the O–H bond towards itself, leaving the hydrogen with a partial positive charge and the oxygen with a partial negative charge.
Because of this, the polar –OH groups are held together by strong intermolecular forces, namely hydrogen bonding. A hydrogen bond forms when the \(\delta^+\) hydrogen of one molecule is attracted to a lone pair on the \(\delta^-\) oxygen of a neighbouring molecule, as drawn in Fig. 11.21.
Diagram
Fig. 11.21 – Intermolecular hydrogen bonding: two R–O–H molecules with δ⁺ on H and δ⁻ on O joined by a dotted line labelled “intermolecular hydrogen bonding”, drawn once for alcohols (R–O–H) and once for phenols (Ar–O–H).
b. Physical State
Lower alcohols are colourless, toxic liquids having a characteristic alcoholic odour. Pure phenol is a colourless, toxic, low melting solid having a characteristic carbolic or phenolic odour.
c. Boiling Points
The boiling points of alcohols and phenols increase with increase in their molecular mass (Table 11.5). A larger molecule has a larger surface over which van der Waals forces can act, so more energy is needed to pull the molecules apart.
d. Solubility
Phenols and lower alcohols (having up to three carbons) show appreciable solubility in water due to their ability to form intermolecular hydrogen bonding with water molecules (see Table 11.5). The arrangement is shown in Fig. 11.22.
Diagram
Fig. 11.22 – Intermolecular hydrogen bonding of R–OH and Ar–OH with water: an alcohol molecule and a phenol molecule each shown hydrogen bonded to two water molecules, with δ⁺/δ⁻ marked and the hydrogen bonds drawn as dotted lines labelled “hydrogen bond”.
Table 11.5 M.P/B.P and solubilities of some alcohols and phenols
| Name | Formula | M.P. (\(^0\)C) | B.P. (\(^0\)C) | Solubility (g/100 g \(\mathrm{H_2O}\)) |
|---|---|---|---|---|
| Methyl alcohol | \(\mathrm{H_3C\!-\!OH}\) | −97 | 65 | 0.793 |
| Ethyl alcohol | \(\mathrm{H_3C\!-\!CH_2\!-\!OH}\) | −115 | 78 | 0.789 |
| n-Propyl alcohol | \(\mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!OH}\) | −126 | 97 | 0.804 |
| Isopropyl alcohol | \(\mathrm{H_3C\!-\!CH(CH_3)\!-\!OH}\) | −86 | 83 | 0.789 |
| n-Butyl alcohol | \(\mathrm{H_3C\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!OH}\) | −90 | 118 | 0.810 |
| Isobutyl alcohol | \(\mathrm{H_3C\!-\!CH(CH_3)\!-\!CH_2\!-\!OH}\) | −108 | 108 | 0.802 |
| sec-Butyl alcohol | \(\mathrm{H_3C\!-\!CH_2\!-\!CH(CH_3)\!-\!OH}\) | −114 | 99.5 | 0.806 |
| tert-Butyl alcohol | \(\mathrm{(H_3C)_3C\!-\!OH}\) | 25.5 | 83 | 0.789 |
| Phenol | \(\mathrm{C_6H_5\!-\!OH}\) | 41 | 182 | 9.3 |
| p-Cresol | \(\mathrm{4\text{-}CH_3\!-\!C_6H_4\!-\!OH}\) | 35 | 202 | 2.3 |
| o-Nitrophenol | \(\mathrm{2\text{-}NO_2\!-\!C_6H_4\!-\!OH}\) | 45 | 217 | 0.2 |
| p-Nitrophenol | \(\mathrm{4\text{-}NO_2\!-\!C_6H_4\!-\!OH}\) | 114 | – | 1.7 |
Diagram
Fig. 11.23 – Bar chart of the Table 11.5 data: boiling point (°C) plotted for methyl, ethyl, n-propyl, n-butyl, isobutyl, sec-butyl and tert-butyl alcohol, phenol, p-cresol and o-nitrophenol, with a second, differently shaded bar beside each giving its solubility in g/100 g H₂O; label both axes.
Diagram
Fig. 11.24 – Left: o-nitrophenol with a dotted line from the phenolic H to an oxygen of the adjacent –NO₂ group, labelled “intramolecular hydrogen bonding in o-nitrophenol”. Right: p-nitrophenol with dotted lines from its –OH and –NO₂ groups to surrounding water molecules, labelled “intermolecular hydrogen bonding in p-nitrophenol and water”.
11.4.4 Chemical properties of Alcohols and Phenols
a. Laboratory tests of alcohols and phenols
i. Litmus test. Water soluble alcohols and phenols can be tested with litmus paper. An aqueous solution of an alcohol is neutral to litmus — neither blue nor red litmus changes colour. Aqueous solutions of phenols turn blue litmus red. Thus, phenols have acidic character.
ii. Reaction with bases. The acid strength of phenols being very low, phenols cannot react with \(\mathrm{NaHCO_3}\) but do react with NaOH.
\[\mathrm{Ar\!-\!OH} + \mathrm{NaHCO_3}\text{(aq)} \rightarrow \text{No reaction}\]
Phenols dissolve in aqueous NaOH by forming water soluble sodium phenoxide, and are reprecipitated (regenerated) on acidification with HCl.
\[\mathrm{Ar\!-\!OH} + \mathrm{NaOH}\text{(aq)} \rightarrow \mathrm{Ar\!-\!O^-Na^+}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
\[\mathrm{Ar\!-\!O^-Na^+}\text{(aq)} + \mathrm{HCl}\text{(aq)} \rightarrow \mathrm{Ar\!-\!OH}\downarrow + \mathrm{NaCl}\text{(aq)}\]
Sodium bicarbonate, sodium hydroxide and sodium metal are increasingly strong bases, and this graded strength is what makes the test informative. Weak and strong acids can be distinguished from each other qualitatively by testing their reactivity towards bases of different strengths. A weak acid does not react with a weak base; it requires a stronger base instead. Hence phenols react with NaOH but not with \(\mathrm{NaHCO_3}\). A strong acid shows high reactivity towards weak as well as strong bases. For example, HCl is a strong acid and reacts with both \(\mathrm{NaHCO_3}\) and NaOH:
\[\mathrm{HCl}\text{(aq)} + \mathrm{NaHCO_3}\text{(aq)} \rightarrow \mathrm{H_2O}\text{(l)} + \mathrm{NaCl}\text{(aq)} + \mathrm{CO_2}\uparrow\]
\[\mathrm{HCl}\text{(aq)} + \mathrm{NaOH}\text{(aq)} \rightarrow \mathrm{NaCl}\text{(aq)} + \mathrm{H_2O}\text{(l)}\]
Alcohols show no acidic character in aqueous solution; thus, alcohols do not react with either aqueous \(\mathrm{NaHCO_3}\) or aqueous NaOH. The very weak acidic character of an alcohol is revealed only in its reaction with an active metal. When alcohols are treated with a very strong base such as the alkali metal Na or K, they react to give sodium or potassium alkoxide with liberation of hydrogen gas.
\[2\,\mathrm{R\!-\!OH} + 2\,\mathrm{Na} \rightarrow 2\,\mathrm{R\!-\!O^-Na^+} + \mathrm{H_2}\uparrow\]
Liberation of \(\mathrm{H_2}\) gas is used to detect the presence of an alcoholic –OH group in a molecule.
iii. Characteristic test for phenols. Phenols react with neutral ferric chloride solution to give a deep (purple/violet/green) colouration of ferric phenoxide.
\[3\,\mathrm{Ar\!-\!OH} + \mathrm{FeCl_3} \rightarrow (\mathrm{Ar\!-\!O})_3\mathrm{Fe} + 3\,\mathrm{HCl}\]
The ferric chloride must be neutral, since an acidic or alkaline solution suppresses the colour.
iv. Distinguishing test for alcohols (Lucas test). Primary, secondary and tertiary alcohols can be distinguished from each other in the laboratory using Lucas reagent (conc. HCl and \(\mathrm{ZnCl_2}\)). The reaction involved is:
\[\mathrm{R\!-\!OH} \xrightarrow{\ \mathrm{HCl}/\mathrm{ZnCl_2}\ } \mathrm{R\!-\!Cl}\]
Alcohols are soluble in Lucas reagent but the product alkyl chloride is not. Hence the clear solution becomes turbid when the product starts forming. Tertiary alcohols react fast and the reagent turns turbid instantaneously. Secondary alcohols turn the reagent turbid slowly. Primary alcohols turn the reagent turbid only on heating.
Flowchart
FC 11.6 – Lucas test: alcohol + Lucas reagent (conc. HCl + ZnCl₂) at room temperature → does turbidity appear at once? → if yes → tertiary alcohol; if no → does turbidity appear within a few minutes? → if yes → secondary alcohol; if no → turbidity only on heating → primary alcohol.
b. Reactions due to breaking of the O–H bond
i. Acidic character of alcohols and phenols. From the laboratory tests it is understood that in aqueous medium phenols show weak acidic character while alcohols are neutral. It is clear, therefore, that the reactivity of alcohols and phenols towards ionization of the O–H bond in them is different. The reason behind this difference lies in the extent of stabilization of their respective conjugate bases by electronic effects.
Remember the general rule: an acid ionises readily only if the ion left behind — its conjugate base — is stable. Anything that spreads out the negative charge stabilises that ion; anything that concentrates the charge destabilises it.
Ionization of alcohols is represented by the following equilibrium:
\[\mathrm{R\!-\!OH} + \mathrm{H_2O} \rightleftharpoons \mathrm{R\!-\!O^-} + \mathrm{H_3O^+}\] \[\text{(alcohol)} \qquad\qquad \text{(alkoxide)}\]
The electron donating inductive effect (+I effect) of the alkyl group destabilizes the alkoxide ion, the conjugate base of the alcohol, because it pushes still more electron density on to an oxygen that is already negatively charged. As a result the alcohol does not ionize much in water, and behaves like a neutral compound in aqueous medium.
Ionization of phenol is represented by the equilibrium shown in Fig. 11.25.
\[\mathrm{C_6H_5OH} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_6H_5O^-} + \mathrm{H_3O^+}\]
Diagram
Fig. 11.25 – Ionization of phenol and resonance stabilization of the phenoxide ion: phenol + H₂O ⇌ phenoxide + H₃O⁺, followed by the five resonance structures I–V of the phenoxide ion with curly arrows showing delocalization of the negative charge on to the ortho and para ring carbons and back on to oxygen.
The phenoxide ion, the conjugate base of phenol, is resonance stabilized by delocalization of the negative charge. Instead of sitting entirely on the oxygen atom, the charge is shared with the ortho and para carbons of the ring. Therefore phenol ionizes in aqueous medium to a moderate extent, and thereby shows a weak acidic character.
Diagram
Fig. 11.26 – The six resonance structures I–VI of the p-nitrophenoxide ion, with curly arrows showing the negative charge moving from the phenoxide oxygen round the ring and finally on to the oxygen atoms of the –NO₂ group in structure VI.
ii. Esterification. Alcohols and phenols form esters by reaction with carboxylic acids, acid halides and acid anhydrides. The reaction between an alcohol or a phenol with a carboxylic acid to form an ester is called esterification.
Esterification of an alcohol or a phenol is carried out in the presence of concentrated sulphuric acid. The reaction is reversible and can be shifted in the forward direction by removing water as soon as it is formed — an application of Le Chatelier’s principle, since removing a product drives the equilibrium towards the products.
\[\mathrm{R\!-\!OH} + \mathrm{HO\!-\!CO\!-\!R'} \rightleftharpoons \mathrm{R\!-\!O\!-\!CO\!-\!R'} + \mathrm{H_2O} \qquad (\mathrm{H^+})\] \[\text{(alcohol)} \qquad \text{(acid)} \qquad\qquad \text{(ester)}\]
\[\mathrm{Ar\!-\!OH} + \mathrm{HO\!-\!CO\!-\!R'} \rightleftharpoons \mathrm{Ar\!-\!O\!-\!CO\!-\!R'} + \mathrm{H_2O} \qquad (\mathrm{H^+})\] \[\text{(phenol)} \qquad \text{(acid)} \qquad\qquad \text{(ester)}\]
Alcohols and phenols react with acid anhydrides in presence of an acid catalyst to form esters.
\[\mathrm{R\!-\!OH} + \mathrm{R'\!-\!CO\!-\!O\!-\!CO\!-\!R'} \rightleftharpoons \mathrm{R\!-\!O\!-\!CO\!-\!R'} + \mathrm{R'\!-\!COOH} \qquad (\mathrm{H^+})\]
\[\mathrm{Ar\!-\!OH} + \mathrm{R'\!-\!CO\!-\!O\!-\!CO\!-\!R'} \rightleftharpoons \mathrm{Ar\!-\!O\!-\!CO\!-\!R'} + \mathrm{R'\!-\!COOH} \qquad (\mathrm{H^+})\]
The reaction of alcohols and phenols with an acid chloride is carried out in the presence of pyridine, a base, which neutralizes the HCl produced. If the HCl were left in the mixture it would slow the reaction down and attack the product.
\[\mathrm{R\!-\!OH} + \mathrm{Cl\!-\!CO\!-\!R'} \xrightarrow{\ \text{pyridine}\ } \mathrm{R'\!-\!CO\!-\!OR} + \mathrm{HCl}\]
\[\mathrm{Ar\!-\!OH} + \mathrm{Cl\!-\!CO\!-\!R'} \xrightarrow{\ \text{pyridine}\ } \mathrm{R'\!-\!CO\!-\!O\!-\!Ar} + \mathrm{HCl}\]
Acetyl derivatives. The \(\mathrm{CH_3\!-\!CO\!-}\) group is called the acetyl group. The acetate esters of alcohols or phenols are also called ‘acetyl derivatives’ of alcohols or phenols respectively. The number of alcoholic or phenolic –OH groups in a given compound can be deduced from the number of acetyl groups introduced into it as a result of acetylation — each –OH group takes up exactly one acetyl group, so counting acetyl groups counts hydroxyl groups.
Aspirin, a well known generic medicine, is an acetyl derivative of salicylic acid formed by its acetylation using acetic anhydride, as shown in Fig. 11.27.
Diagram
Fig. 11.27 – Salicylic acid (benzene ring with –COOH and an adjacent –OH) + acetic anhydride (CH₃–CO–O–CO–CH₃) → (H⁺) → aspirin (ring with –COOH and an adjacent –O–CO–CH₃) + acetic acid; label the reactants and label the product “Aspirin / acetyl salicylic acid”.
c. Reaction due to breaking of the C–O bond in alcohols
i. Reaction with hydrogen halides. Alcohols react with hydrogen halides to form alkyl halides (refer to Chapter 10, Section 10.3.1).
\[\mathrm{R\!-\!OH} + \mathrm{HX} \rightarrow \mathrm{R\!-\!X} + \mathrm{H_2O}\]
In general, tertiary alcohols react rapidly with hydrogen halides; secondary alcohols react somewhat slower; and primary alcohols, even more slowly. The order of reactivity of hydrogen halides is
\[\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\]
HCl reacts only in the presence of anhydrous \(\mathrm{ZnCl_2}\). No catalyst is required in the case of HBr and HI.
ii. Reaction with phosphorous halide. Alcohols react with phosphorous pentachloride (\(\mathrm{PCl_5}\)) and phosphorous trihalides (\(\mathrm{PX_3}\)) to form alkyl halides (refer to Chapter 10, Section 10.3.1).
\[\mathrm{R\!-\!OH} + \mathrm{PCl_5} \rightarrow \mathrm{R\!-\!Cl} + \mathrm{POCl_3} + \mathrm{HCl}\]
\[3\,\mathrm{R\!-\!OH} + \mathrm{PX_3} \rightarrow 3\,\mathrm{R\!-\!X} + \mathrm{H_3PO_3}\]
iii. Dehydration of alcohols to alkenes. An alcohol, when treated with concentrated sulphuric acid or phosphoric acid or alumina, undergoes dehydration to form an alkene and water (refer to Std XI Chemistry Textbook, Section 15.2). The reaction gives the more substituted alkene as the major product, in accordance with Saytzeff rule.
According to the commonly accepted mechanism, dehydration involves the following three steps, drawn in Fig. 11.28.
- Formation of the protonated alcohol, \(\mathrm{R\!-\!OH_2^+}\), in a fast step. Protonation converts –OH, a poor leaving group, into \(\mathrm{-OH_2^+}\), which leaves as neutral water.
- Its slow dissociation into a carbocation. Being the slow step, it decides the rate of the whole reaction.
- Fast removal of a hydrogen ion from the adjacent carbon to form the alkene.
Diagram
Fig. 11.28 – Three-step dehydration mechanism with curly arrows: alcohol + H⁺ (fast) → protonated alcohol; protonated alcohol → (slow, –H₂O) → carbocation; carbocation → (fast, –H⁺) → alkene; label the four species “alcohol”, “protonated alcohol”, “carbocation”, “alkene”.
Flowchart
FC 11.7 – Dehydration of an alcohol: R–OH + H⁺ → protonated alcohol (fast) → loss of H₂O (slow, rate-determining) → carbocation → loss of H⁺ from an adjacent carbon (fast) → alkene; branch at the last step: if two different adjacent carbons can lose H⁺ → the more substituted alkene is the major product (Saytzeff rule) and the less substituted alkene is minor.
On reaction with an oxidising agent, primary and secondary alcohols undergo dehydrogenation to form carbonyl compounds, namely aldehydes and ketones respectively. Removing two hydrogen atoms — one from the –OH group and one from the carbon carrying it — creates the C=O double bond.
A primary alcohol on oxidation with \(\mathrm{CrO_3}\) forms an aldehyde. However, a better reagent to bring about this oxidation is PCC (pyridinium chlorochromate).
\[\mathrm{R\!-\!CH_2\!-\!OH} \xrightarrow{\ \mathrm{PCC},\ [\mathrm{O}]\ } \mathrm{R\!-\!CHO}\] \[(1^0\ \text{alcohol}) \qquad\qquad \text{(aldehyde)}\]
A secondary alcohol on oxidation with chromic anhydride (\(\mathrm{CrO_3}\)) forms a ketone.
\[\mathrm{R\!-\!CH(OH)\!-\!R'} \xrightarrow{\ \mathrm{CrO_3},\ [\mathrm{O}]\ } \mathrm{R\!-\!CO\!-\!R'}\] \[(2^0\ \text{alcohol}) \qquad\qquad \text{(ketone)}\]
When common oxidizing agents like nitric acid, potassium permanganate or potassium dichromate are used to oxidise a primary alcohol, the oxidation does not stop at the aldehyde stage, but the aldehyde formed is further oxidized to a carboxylic acid containing the same number of carbon atoms.
\[\mathrm{R\!-\!CH_2\!-\!OH} \xrightarrow{\ \mathrm{KMnO_4/K_2Cr_2O_7/HNO_3},\ [\mathrm{O}]\ } [\mathrm{R\!-\!CHO}] \xrightarrow{\ [\mathrm{O}]\ } \mathrm{R\!-\!CO\!-\!OH}\]
Tertiary alcohols are difficult to oxidise, since the carbon carrying the –OH group has no hydrogen atom left to remove. On oxidation with strong oxidising agents at high temperature, a tertiary alcohol undergoes breaking of C–C bonds and gives a mixture of carboxylic acids containing a smaller number of carbon atoms than the starting \(3^\circ\) alcohol.
Heating with Cu. When vapours of the various types of alcohols are passed over hot copper, the following reactions are observed.
\[\mathrm{R\!-\!CH_2\!-\!OH} \xrightarrow[\text{Oxidation}]{\ \mathrm{Cu}/573\ \text{K}\ } \mathrm{R\!-\!CHO} \qquad (1^0\ \text{alcohol} \rightarrow \text{aldehyde})\]
\[\mathrm{R\!-\!CH(OH)\!-\!R'} \xrightarrow[\text{Oxidation}]{\ \mathrm{Cu}/573\ \text{K}\ } \mathrm{R\!-\!CO\!-\!R'} \qquad (2^0\ \text{alcohol} \rightarrow \text{ketone})\]
\[\mathrm{H_3C\!-\!C(OH)(CH_3)\!-\!CH_3} \xrightarrow[\text{dehydration}]{\ \mathrm{Cu}/573\ \text{K}\ } \mathrm{H_3C\!-\!C(CH_3){=}CH_2} \qquad (3^0\ \text{alcohol} \rightarrow \text{alkene})\]
Note that hot copper oxidises the first two but merely dehydrates the third — once again because a tertiary alcohol has no hydrogen on the carbon bearing the –OH group.
Flowchart
FC 11.8 – Oxidation of alcohols by class: alcohol → is there a hydrogen on the carbon carrying –OH? → if no (3⁰) → resists oxidation; with strong oxidant at high temperature → C–C cleavage → mixture of carboxylic acids with fewer carbons; over Cu/573 K → dehydration → alkene. If yes → how many such hydrogens? → two (1⁰) → PCC or Cu/573 K → aldehyde → with KMnO₄/K₂Cr₂O₇/HNO₃ → further oxidation → carboxylic acid (same number of carbons). → one (2⁰) → CrO₃ or Cu/573 K → ketone.
d. Reactions of phenols
Phenol undergoes electrophilic substitution reactions more readily as compared to benzene. The –OH group in phenol is ring activating and an ortho-/para-directing group. The reason is the same resonance seen in the phenoxide ion: a lone pair on the oxygen is delocalised into the ring, raising the electron density at the ortho and para positions, and an electrophile — an electron-seeking species — attacks wherever electrons are richest.
i. Halogenation of phenol. Phenol reacts with an aqueous solution of bromine to give 2,4,6-tribromophenol (chlorine reacts in the same way).
\[\mathrm{C_6H_5OH} + 3\,\mathrm{Br_2} \xrightarrow{\ \mathrm{H_2O}\ } \mathrm{C_6H_2Br_3OH} + 3\,\mathrm{HBr}\] \[\text{(Phenol)} \qquad\qquad \text{(2,4,6-tribromophenol)}\]
If the reaction is carried out in a solvent of lower polarity than water, such as \(\mathrm{CHCl_3}\), \(\mathrm{CCl_4}\) or \(\mathrm{CS_2}\), a mixture of ortho- and para-bromophenol is formed, with the para compound as the major product.
\[\mathrm{C_6H_5OH} + \mathrm{Br_2} \xrightarrow[\text{Low temperature}]{\ \mathrm{CS_2}\ } \text{o-bromophenol} + \text{p-bromophenol (major)} + \mathrm{HBr}\]
Diagram
Fig. 11.29 – Halogenation of phenol: upper route, phenol + 3Br₂ in H₂O → 2,4,6-tribromophenol + 3HBr with the three Br atoms marked at positions 2, 4 and 6; lower route, phenol + Br₂ in CS₂ at low temperature → o-bromophenol + p-bromophenol (marked “major product”) + HBr.
ii. Nitration of phenol. Phenol reacts with dilute nitric acid at low temperature to give a mixture of ortho- and para-nitrophenol.
\[\mathrm{C_6H_5OH} + \mathrm{HO\!-\!NO_2} \rightarrow \text{o-nitrophenol} + \text{p-nitrophenol} + \mathrm{H_2O}\]
Phenol reacts with concentrated nitric acid to form 2,4,6-trinitrophenol (picric acid).
\[\mathrm{C_6H_5OH} + 3\,\mathrm{HO\!-\!NO_2} \xrightarrow{\ \text{conc. } \mathrm{H_2SO_4}\ } \mathrm{C_6H_2(NO_2)_3OH} + 3\,\mathrm{H_2O}\] \[\text{(picric acid)}\]
Diagram
Fig. 11.30 – Nitration of phenol: phenol + dilute HNO₂-free HNO₃ at low temperature → o-nitrophenol and p-nitrophenol drawn as ring structures; below, phenol + 3 HNO₃ with conc. H₂SO₄ → picric acid with NO₂ groups marked at positions 2, 4 and 6, + 3H₂O.
iii. Sulfonation of phenol. Phenol reacts with concentrated sulfuric acid at room temperature to give o-phenolsulfonic acid, and at 373 K, p-phenolsulfonic acid. The same reagent therefore gives two different products depending only on the temperature.
\[\mathrm{C_6H_5OH} + \mathrm{HO\!-\!SO_3H} \xrightarrow{\ 293\ \text{K}\ } \text{o-phenolsulfonic acid} + \mathrm{H_2O}\]
\[\mathrm{C_6H_5OH} + \mathrm{HO\!-\!SO_3H} \xrightarrow{\ 373\ \text{K}\ } \text{p-phenolsulfonic acid} + \mathrm{H_2O}\]
Diagram
Fig. 11.31 – Sulfonation of phenol: phenol with two branching arrows, one labelled “HO–SO₃H, 293 K” leading to o-phenolsulfonic acid and one labelled “HO–SO₃H, 373 K” leading to p-phenolsulfonic acid; draw both ring products with the –SO₃H group in place.
iv. Reimer–Tiemann Reaction. When phenol is treated with chloroform in aqueous sodium hydroxide solution, followed by hydrolysis with acid, salicylaldehyde is formed. This reaction is known as the Reimer–Tiemann reaction. The intermediate formed carries a \(\mathrm{-CHCl_2}\) group at the ortho position, which NaOH then converts into the –CHO group.
Diagram
Fig. 11.32 – Reimer–Tiemann reaction: phenol → (CHCl₃, aq. NaOH) → intermediate drawn as the sodium salt with a –CHCl₂ group ortho to –ONa → (NaOH) → sodium salt of salicylaldehyde → (H₃O⁺) → salicylaldehyde (ring with –OH and an adjacent –CHO); label the intermediate and the product.
Flowchart
FC 11.9 – Reimer–Tiemann reaction: phenol + CHCl₃ in aq. NaOH → ortho-substituted intermediate bearing –CHCl₂ → treatment with NaOH → sodium salt of the aldehyde → acid hydrolysis (H₃O⁺) → salicylaldehyde; branch: if CCl₄ is used in place of CHCl₃ → salicylic acid is obtained instead.
If carbon tetrachloride is used in place of chloroform, salicylic acid is formed.
v. Kolbe reaction. The treatment of sodium phenoxide with carbon dioxide at 398 K under a pressure of 6 atm, followed by acid hydrolysis, gives salicylic acid (o-hydroxybenzoic acid). This reaction is known as Kolbe’s reaction.
Diagram
Fig. 11.33 – Kolbe reaction: sodium phenoxide + CO₂ → (398 K, 6 atm) → sodium salicylate (ring with –OH and an adjacent –COONa) → (H₃O⁺) → salicylic acid (ring with –OH and an adjacent –COOH); label each structure.
Sodium phenoxide is more reactive than phenol towards electrophilic substitution, because the ring in the phenoxide ion carries a full negative charge spread over it rather than only a partial one. Hence it is able to react with a weak electrophile like \(\mathrm{CO_2}\) at high temperature and pressure in the Kolbe reaction.
vi. Oxidation of phenol. Phenol on oxidation with chromic anhydride or sodium dichromate in presence of \(\mathrm{H_2SO_4}\) gives p-benzoquinone.
\[\mathrm{C_6H_5OH} \xrightarrow{\ \mathrm{CrO_3}\ \text{or}\ \mathrm{Na_2Cr_2O_7/H_2SO_4},\ [\mathrm{O}]\ } \text{p-benzoquinone}\]
Phenol oxidizes slowly, giving a dark coloured mixture, in the presence of air. This is why a bottle of phenol that has stood for a long time is pink or brown rather than colourless.
vii. Catalytic hydrogenation of phenol. Phenol on catalytic hydrogenation gives cyclohexanol. In this reaction a mixture of vapours of phenol and hydrogen is passed over a nickel catalyst at 433 K. Three molecules of hydrogen are needed because the aromatic ring contains three double bonds in its Kekulé representation.
\[\mathrm{C_6H_5OH} + 3\,\mathrm{H_2} \xrightarrow{\ \mathrm{Ni},\ 433\ \text{K}\ } \mathrm{C_6H_{11}OH}\] \[\text{(Phenol)} \qquad\qquad \text{(Cyclohexanol)}\]
viii. Reduction of phenol. Phenol is reduced to benzene on heating with zinc dust. Here the whole –OH group is removed, the zinc taking the oxygen away as zinc oxide.
\[\mathrm{C_6H_5OH} + \mathrm{Zn} \rightarrow \mathrm{C_6H_6} + \mathrm{ZnO}\] \[\text{(Phenol)} \qquad \text{(Benzene)}\]
Diagram
Fig. 11.34 – Three reactions of phenol drawn with ring structures: (a) phenol → (CrO₃ or Na₂Cr₂O₇/H₂SO₄) → p-benzoquinone; (b) phenol + 3H₂ → (Ni, 433 K) → cyclohexanol; (c) phenol + Zn → benzene + ZnO.
11.5 Ethers
11.5.1 Preparation of ethers
a. Dehydration of alcohols
When an alcohol is heated with a dehydrating agent like concentrated \(\mathrm{H_2SO_4}\) or \(\mathrm{H_3PO_4}\), two products — either an ether or an alkene — can form, depending upon the temperature. Both products come from the loss of water; what differs is which two molecules lose it. At the lower temperature, two molecules of alcohol lose one molecule of water between them and are joined through oxygen, giving an ether. At the higher temperature, a single molecule loses water within itself and gives an alkene.
For example, dehydration of ethanol by \(\mathrm{H_2SO_4}\) gives ethoxyethane at 413 K, while ethene is formed at 443 K.
\[2\,\mathrm{C_2H_5OH} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \mathrm{H_2SO_4}/413\ \text{K}\ } \mathrm{C_2H_5\!-\!O\!-\!C_2H_5}\]
\[(\text{ethanol}) \qquad\qquad (\text{ethoxyethane})\]
\[\mathrm{C_2H_5OH} \xrightarrow[\ -\mathrm{H_2O}\ ]{\ \mathrm{H_2SO_4}/443\ \text{K}\ } \mathrm{CH_2{=}CH_2}\]
\[(\text{ethanol}) \qquad\qquad (\text{ethene})\]
Symmetrical ethers can be obtained from primary alcohols by this method. Use of a higher temperature, or of \(2^{\circ}\)/\(3^{\circ}\) alcohols, gives the alkene as the major product. The choice is set out in FC 11.10.
Flowchart
FC 11.10 – Dehydration of an alcohol with conc. H₂SO₄: primary alcohol at the lower temperature (413 K for ethanol) → two molecules combine → symmetrical ether; branch: same alcohol at the higher temperature (443 K for ethanol) → alkene; branch: secondary or tertiary alcohol at any temperature → alkene as the major product.
Dehydration of alcohols to form an ether is an \(\mathrm{S_N2}\) reaction. The protonated alcohol species undergoes a backside attack by a second molecule of alcohol in a slow step. Subsequent fast deprotonation results in the formation of the ether. The three stages are drawn in Fig. 11.35.
i. Protonation.
\[\mathrm{C_2H_5\!-\!O\!-\!H} + \mathrm{H^+} \rightarrow \mathrm{C_2H_5\!-\!\overset{+}{O}H_2}\]
ii. \(\mathrm{S_N2}\) attack. A second molecule of ethanol uses a lone pair on its oxygen to attack the carbon from the side opposite the leaving group, and water departs.
\[\mathrm{C_2H_5\!-\!OH} + \mathrm{CH_3CH_2\!-\!\overset{+}{O}H_2} \xrightarrow[\ -\mathrm{H_2O}\ ]{} \mathrm{C_2H_5\!-\!\overset{+}{O}(H)\!-\!CH_2\!-\!CH_3}\]
iii. Deprotonation.
\[\mathrm{C_2H_5\!-\!\overset{+}{O}(H)\!-\!C_2H_5} \xrightarrow{\ -\mathrm{H^+}\ } \mathrm{C_2H_5\!-\!O\!-\!C_2H_5}\]
Diagram
Fig. 11.35 – Three-stage SN2 mechanism of ether formation: (i) C₂H₅–O–H + H⊕ → C₂H₅–OH₂⊕; (ii) curly arrow from the lone pair of a second C₂H₅–OH oxygen to the carbon of the protonated alcohol, with water leaving from the opposite side, giving the protonated ether; (iii) loss of H⊕ giving C₂H₅–O–C₂H₅. Mark the backside approach clearly in step (ii).
Because the attack must come from behind the leaving group, the carbon under attack has to be reasonably open. That is exactly why this method works well for primary alcohols and fails for the bulkier ones.
b. Williamson Synthesis
Simple as well as mixed ethers can be prepared in the laboratory by Williamson Synthesis. In this method an alkyl halide is treated with sodium alkoxide or sodium phenoxide to give dialkyl ethers or alkyl aryl ethers.
\[\mathrm{R\!-\!X} + \mathrm{Na^+O^-\!-\!R} \rightarrow \mathrm{R\!-\!O\!-\!R} + \mathrm{NaX}\]
\[\mathrm{R\!-\!X} + \mathrm{Na^+O^-\!-\!Ar} \rightarrow \mathrm{R\!-\!O\!-\!Ar} + \mathrm{NaX}\]
This reaction follows the \(\mathrm{S_N2}\) mechanism. The ether is formed as a result of a backside attack by the alkoxide or phenoxide ion — a nucleophile — on the alkyl halide, as shown in Fig. 11.36. The alkyl halide used in this reaction must be primary. For example, t-butyl methyl ether can be synthesised by the reaction of methyl bromide with sodium t-butoxide.
\[\mathrm{(CH_3)_3C\!-\!O^-Na^+} + \mathrm{CH_3\!-\!Br} \rightarrow \mathrm{(CH_3)_3C\!-\!O\!-\!CH_3} + \mathrm{NaBr}\]
\[(\text{sodium t-butoxide}) \quad (\text{methyl bromide}) \qquad (\text{t-butyl methyl ether})\]
Diagram
Fig. 11.36 – Williamson synthesis as an SN2 process: alkoxide ion R–O⊖ approaching the carbon of R′–X from the side directly opposite the C–X bond, curly arrow from the oxygen lone pair to carbon and a second curly arrow from the C–X bond to the halide; show the halide leaving and the ether R–O–R′ forming.
If secondary or tertiary alkyl halides are used, the reaction leads mainly to alkene formation (an elimination reaction). For example:
\[\mathrm{(CH_3)_3C\!-\!Cl} + \mathrm{Na^+O^-\!-\!C_2H_5} \rightarrow \mathrm{(CH_3)_2C{=}CH_2} + \mathrm{C_2H_5OH} + \mathrm{NaCl}\]
\[(\text{t-butyl chloride}) \quad (\text{sodium ethoxide}) \qquad (\text{isobutene}) \qquad (\text{ethanol})\]
Aryl halides do not give Williamson’s synthesis, because the carbon–halogen bond of an aryl halide has partial double bond character and is not opened by a nucleophile under these conditions.
Williamson synthesis is an effective method for the preparation of ethers from hydroxy compounds, because both the substrates it needs — the nucleophile and the alkyl halide — are themselves obtained from hydroxy compounds, as shown in FC 11.11.
Flowchart
FC 11.11 – Both Williamson substrates come from hydroxy compounds: phenol + NaOH → sodium phenoxide (the nucleophile); branch: C₂H₅–OH + Na → C₂H₅ONa (alkoxide, the nucleophile); branch: C₂H₅–OH + HX → C₂H₅X (the alkyl halide); the alkoxide/phenoxide and the alkyl halide then combine → ether.
11.5.2 Physical properties
a. Physical state and boiling points
Dimethyl ether and ethyl methyl ether are gases. Other ethers are colourless liquids with a pleasant odour.
Lower ethers are highly volatile and highly inflammable substances.
The boiling points of ethers show a gradual increase with the increase in molecular mass.
| Ether | B.P. / K |
|---|---|
| \(\mathrm{CH_3\!-\!O\!-\!CH_3}\) | 248 |
| \(\mathrm{C_2H_5\!-\!O\!-\!CH_3}\) | 284 |
| \(\mathrm{C_2H_5\!-\!O\!-\!C_2H_5}\) | 308 |
b. Polarity and solubility
Since the \(\mathrm{-C\!-\!O\!-\!C\!-}\) bond angle is \(110^{\circ}\) and not \(180^{\circ}\), the bond dipole moments of the two C–O bonds do not cancel each other; therefore ethers possess a small net dipole moment. For example, the dipole moment of diethyl ether is 1.18 D. The geometry is drawn in Fig. 11.37.
Diagram
Fig. 11.37 – A bent C–O–C linkage with the two R groups drawn at an angle of 110° at the oxygen, the two C–O bond dipoles drawn as small arrows pointing towards oxygen, and their resultant drawn as a single arrow labelled “net dipole moment”.
Weak polarity of ethers does not affect their boiling points, which are about the same as those of alkanes having comparable molecular mass (see Table 11.6).
Table 11.6 Comparative boiling points of alkane, ether and alcohol
| Name | n-Heptane | Methyl n-pentyl ether | n-Hexyl alcohol |
|---|---|---|---|
| Molecular mass | 100 | 102 | 102 |
| Boiling point / K | 371 | 373 | 430 |
The intermolecular hydrogen bonding that holds alcohol molecules together strongly is not present in ethers and alkanes. An ether has no O–H bond at all, so although its oxygen can accept a hydrogen bond, it has no hydrogen of its own to donate — and hydrogen bonding between two ether molecules is therefore impossible.
However, the solubility or miscibility of ethers in water is similar to that of alcohols of comparable molecular mass. This is because ethers can form hydrogen bonds with water through the ethereal oxygen — water supplies the hydrogen that the ether lacks, as drawn in Fig. 11.38.
Diagram
Fig. 11.38 – An ether molecule R–O–R with δ⊖ marked on the oxygen, and a water molecule H–O–H with δ⊕ marked on one hydrogen, joined by a dotted hydrogen bond from the water hydrogen to the ether oxygen.
For example, diethyl ether and n-butyl alcohol have respective miscibilities of 7.5 and 9 g per 100 g of water.
11.5.3 Chemical properties of ethers
a. Laboratory test for ethers
Ethers are neutral compounds in aqueous medium. Ethers do not react with bases, cold dilute acids, reducing agents, oxidizing agents and active metals. However, ethers dissolve in cold concentrated \(\mathrm{H_2SO_4}\) due to the formation of oxonium salts. A lone pair on the ethereal oxygen accepts a proton, and the resulting ion is soluble.
\[\mathrm{R\!-\!O\!-\!R'} + \mathrm{H_2SO_4} \rightarrow [\mathrm{R\!-\!\overset{+}{O}(H)\!-\!R'}]\ \mathrm{HSO_4^-}\]
This property distinguishes ethers from hydrocarbons, which stay as a separate layer under the same treatment.
b. Reaction involving alkyl group of ether
i. Formation of peroxide. Ethers combine with atmospheric oxygen to form a peroxide. The oxygen attacks the carbon next to the ethereal oxygen, inserting an \(\mathrm{-O\!-\!O\!-\!H}\) group there.
\[\mathrm{CH_3\!-\!CH_2\!-\!O\!-\!C_2H_5} + \mathrm{O_2} \xrightarrow{\ \text{Long contact with air}\ } \mathrm{CH_3\!-\!CH(O\!-\!OH)\!-\!O\!-\!C_2H_5}\]
\[(\text{diethyl ether}) \quad (\text{oxygen}) \qquad\qquad (\text{peroxide of diethyl ether})\]
All ethers which have been exposed to the atmosphere contain peroxide. This is a very undesirable reaction. Peroxides are hazardous because they decompose violently at high temperature.
c. Reactions involving the C–O bond
i. Reaction with hot dilute sulphuric acid (Hydrolysis). Ethers, when heated with dilute sulfuric acid under pressure, undergo hydrolysis to give alcohols or phenols.
\[\mathrm{R\!-\!O\!-\!R} + \mathrm{H\!-\!O\!-\!H} \xrightarrow[\ \Delta\ ]{\ \mathrm{H_3O^+}\ } 2\,\mathrm{R\!-\!OH}\]
\[\mathrm{R\!-\!O\!-\!R'} + \mathrm{H\!-\!O\!-\!H} \xrightarrow[\ \Delta\ ]{\ \mathrm{H^+}\ } \mathrm{R\!-\!OH} + \mathrm{R'\!-\!OH}\]
\[\mathrm{Ar\!-\!O\!-\!R} + \mathrm{H\!-\!O\!-\!H} \xrightarrow[\ \Delta\ ]{\ \mathrm{H^+}\ } \mathrm{Ar\!-\!OH} + \mathrm{R\!-\!OH}\]
ii. Reaction with \(\mathrm{PCl_5}\). Ethers react with \(\mathrm{PCl_5}\) to give alkyl chlorides.
\[\mathrm{R\!-\!O\!-\!R'} + \mathrm{PCl_5} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!Cl} + \mathrm{R'\!-\!Cl} + \mathrm{POCl_3}\]
Both C–O bonds are broken here, so a single ether gives two alkyl chlorides.
iii. Reaction with hot concentrated acid. Alkyl ethers react with hot and concentrated HI and HBr to give an alcohol and an alkyl halide.
\[\mathrm{R\!-\!O\!-\!R} + \mathrm{HX} \rightarrow \mathrm{R\!-\!X} + \mathrm{R\!-\!OH}\]
\[\mathrm{R\!-\!OH} \xrightarrow{\ \mathrm{HX}\ } \mathrm{R\!-\!X} + \mathrm{H_2O}\]
The order of reactivity of HX is \(\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\).
The reaction of an ether with hot concentrated HI involves the formation of an oxonium ion by protonation in the first step, and subsequent nucleophilic substitution brought about by the powerful nucleophile \(\mathrm{I^-}\). The least substituted carbon in the oxonium ion is attacked by \(\mathrm{I^-}\), following the \(\mathrm{S_N2}\) mechanism.
\[\mathrm{CH_3\!-\!O\!-\!CH_2\!-\!CH_3} + \mathrm{H\!-\!I} \overset{\Delta}{\rightleftharpoons} \mathrm{CH_3\!-\!\overset{+}{O}(H)\!-\!CH_2\!-\!CH_3} + \mathrm{I^-}\]
\[\mathrm{I^-} + \mathrm{CH_3\!-\!\overset{+}{O}(H)\!-\!CH_2\!-\!CH_3} \rightarrow \mathrm{CH_3\!-\!I} + \mathrm{CH_3\!-\!CH_2\!-\!OH}\]
Diagram
Fig. 11.39 – Cleavage of ethyl methyl ether by hot concentrated HI: protonation of the oxygen to give the oxonium ion; then I⊖ approaching the less substituted (methyl) carbon from the side opposite the oxygen, with a transition state in square brackets showing partial I···CH₃ and CH₃···O bonds carrying half-charges; products CH₃–I and CH₃CH₂–OH.
Two further points follow from this mechanism.
- Use of excess HI converts the alcohol formed into the alkyl iodide as well, so that both fragments end up as iodides.
- In the case of an ether having one tertiary alkyl group, the reaction with hot HI follows the \(\mathrm{S_N1}\) mechanism, and the tertiary iodide is formed rather than the tertiary alcohol. Here the crowded tertiary carbon blocks a backside attack, but it forms a comparatively stable carbocation instead.
\[\text{Step 1:} \quad \mathrm{(CH_3)_3C\!-\!\overset{+}{O}(H)\!-\!CH_3} \xrightarrow{\ \text{slow}\ } \mathrm{(CH_3)_3C^+} + \mathrm{CH_3OH}\]
\[\text{Step 2:} \quad \mathrm{(CH_3)_3C^+} + \mathrm{I^-} \xrightarrow{\ \text{fast}\ } \mathrm{(CH_3)_3C\!-\!I}\]
Flowchart
FC 11.12 – Cleavage of an ether by hot concentrated HI: ether + HI → protonation of the ethereal oxygen → oxonium ion → branch A (both groups primary/secondary): I⊖ attacks the least substituted carbon by SN2 → smaller alkyl iodide + alcohol from the larger group; branch B (one tertiary group): slow heterolysis → tertiary carbocation + alcohol from the smaller group → fast attack by I⊖ → tertiary iodide; branch C (excess HI): the alcohol produced reacts further → second alkyl iodide + H₂O.
Aryl alkyl ethers. Aryl alkyl ethers have a stronger and shorter bond between the oxygen and the aromatic ring, because the lone pair of the oxygen is delocalised into the ring and gives that bond partial double bond character. Hence an aryl alkyl ether undergoes cleavage of the oxygen–alkyl bond and yields a phenol and an alkyl halide on reaction with HI.
\[\mathrm{C_6H_5\!-\!O\!-\!CH_3} + \mathrm{HI} \xrightarrow{\ 398\ \text{K}\ } \mathrm{C_6H_5\!-\!OH} + \mathrm{CH_3I}\]
\[(\text{anisole}) \qquad\qquad (\text{phenol}) \quad (\text{methyl iodide})\]
d. Electrophilic substitution in aromatic ethers
The alkoxy group in an aromatic ether is a ring activating and ortho-, para-directing group towards electrophilic aromatic substitution. This is evident from the resonance structures drawn in Fig. 11.40.
Diagram
Fig. 11.40 – Five resonance structures, labelled (I) to (V), of an aryl alkyl ether Ar–O–R: structure I with both lone pairs on oxygen; structures II, III and IV with a positive charge on oxygen and a negative charge at the ortho, para and other ortho ring carbon respectively; structure V back to the neutral form; draw curly arrows between successive structures.
The +R effect of the \(\mathrm{-\ddot{O}R}\) group results in increased electron density at the para- and the two ortho-positions (see resonance structures II, III and IV). An electrophile therefore attacks at these positions, and the para product usually dominates because the two ortho positions are crowded by the alkoxy group itself.
i. Halogenation. Anisole undergoes bromination with bromine in acetic acid even in the absence of a ferric bromide catalyst. It is due to the activation of the benzene ring by the methoxy group.
\[\mathrm{C_6H_5\!-\!OCH_3} + \mathrm{Br_2} \xrightarrow{\ \mathrm{CH_3COOH}\ } \text{p-bromoanisole (major)} + \text{o-bromoanisole (minor)}\]
Diagram
Fig. 11.41 – Anisole + Br₂ in CH₃COOH giving two products: a ring with –OCH₃ at C-1 and Br at C-4, labelled “(major) (p-bromo anisole)”; and a ring with –OCH₃ at C-1 and Br at C-2, labelled “(minor) (o-bromo anisole)”.
ii. Friedel Crafts reaction. Anisole reacts with an alkyl halide and with an acyl chloride in the presence of anhydrous \(\mathrm{AlCl_3}\), a Lewis acid, as catalyst.
\[\mathrm{C_6H_5\!-\!OCH_3} + \mathrm{CH_3Cl} \xrightarrow{\ \text{Anhydrous } \mathrm{AlCl_3}\ } \text{4-methoxytoluene (major)} + \text{2-methoxytoluene (minor)}\]
\[\mathrm{C_6H_5\!-\!OCH_3} + \mathrm{CH_3COCl} \xrightarrow{\ \text{Anhydrous } \mathrm{AlCl_3}\ } \text{4-methoxyacetophenone (major)} + \text{2-methoxyacetophenone (minor)}\]
Diagram
Fig. 11.42 – Two Friedel–Crafts reactions of anisole drawn with ring structures: with CH₃Cl/anhydrous AlCl₃ giving the ring with –OCH₃ at C-1 and –CH₃ at C-4 labelled “(4-Methoxytoluene) (major)” and the ring with –CH₃ at C-2 labelled “(2-Methoxytoluene) (minor)”; with CH₃COCl/anhydrous AlCl₃ giving the corresponding –COCH₃ products labelled “(4-Methoxyacetophenone) (major)” and “(2-Methoxyacetophenone) (minor)”.
A Lewis acid is an electron pair acceptor; anhydrous \(\mathrm{AlCl_3}\) pulls the chloride away from \(\mathrm{CH_3Cl}\) or \(\mathrm{CH_3COCl}\) and so generates the electrophile that the ring then attacks. It must be anhydrous, because water would destroy it.
iii. Nitration. Anisole reacts with concentrated nitric acid in the presence of concentrated sulfuric acid (nitrating mixture) to give a mixture of o-nitroanisole and p-nitroanisole.
\[\mathrm{C_6H_5\!-\!OCH_3} \xrightarrow{\ \text{conc. } \mathrm{HNO_3}\ /\ \text{conc. } \mathrm{H_2SO_4}\ } \text{4-nitroanisole (major)} + \text{2-nitroanisole (minor)}\]
Diagram
Fig. 11.43 – Anisole with conc. HNO₃/conc. H₂SO₄ giving the ring with –OCH₃ at C-1 and –NO₂ at C-4, labelled “(4-Nitroanisole) (major)”, plus the ring with –NO₂ at C-2, labelled “(2-Nitroanisole) (minor)”.
11.6 Uses of alcohols, phenols and ethers
Everything learnt so far about these three families — their polarity, their hydrogen bonding, their acidity and their reactions — shows up in the jobs they are put to. Each use below is worth reading twice: once for the fact, and once for the property that makes the fact possible.
Alcohols
Methyl alcohol is used as a solvent for paints and varnishes. Its molecule has a short hydrocarbon part and a polar –OH group, so it dissolves both the resin and the pigment, and its low boiling point lets it evaporate quickly once the coat has been applied.
Ethyl alcohol is used as an antifreeze agent in automobile radiators. It is also used as a solvent.
Phenols
- Phenol is used in the preparation of phenol formaldehyde resin, for example, bakelite.
Diagram
Fig. 11.44 – Phenol reacting with formaldehyde (HCHO) to give a cross-linked phenol–formaldehyde network: draw two or three phenol rings joined through –CH₂– bridges at the ortho and para positions, with the –OH groups retained; caption “Phenol–formaldehyde resin (bakelite)”.
The reason phenol can build such a network is one we have already established: the –OH group activates the ring at three positions, two ortho and one para, so each phenol molecule can be linked to three neighbours through \(\mathrm{-CH_2-}\) bridges. A rigid, three-dimensional solid results.
- Phenols are used as antiseptics in common products like air fresheners, deodorants, mouthwash, calamine lotions, floor cleaners, etc. They act by attacking the proteins of micro-organisms, which is also the reason concentrated phenol burns skin and must never be handled without gloves.
Ethers
Earlier, diethyl ether was used as a general anaesthetic in surgical operations.
Diethyl ether is used as a solvent for Grignard reagents, fats, waxes, oil, etc.
An ether makes a good solvent for exactly the reason it makes a poor liquid to store: its oxygen has no hydrogen of its own, so ether molecules do not cling to one another, and non-polar substances such as fats, waxes and oils dissolve in them readily. At the same time the lone pairs on that oxygen can coordinate to the magnesium of a Grignard reagent and keep it in solution.
11.7 Summary
An alcohol has a hydroxyl group (–OH) attached to a saturated (\(sp^3\)) carbon atom. A phenol has the –OH group attached directly to a carbon of a benzene ring, while in an aromatic alcohol the –OH sits on a side-chain carbon. An ether has an oxygen atom bonded to two alkyl or aryl groups, \(\mathrm{R\!-\!O\!-\!R'}\), \(\mathrm{R\!-\!O\!-\!Ar}\) or \(\mathrm{Ar\!-\!O\!-\!Ar'}\); an epoxide is a cyclic ether whose oxygen forms part of a three-membered ring.
Alcohols and phenols are classified as monohydric, dihydric, trihydric or polyhydric according to the number of –OH groups in one molecule.
Monohydric alcohols with an \(sp^3\mathrm{C\!-\!OH}\) bond are primary (\(1^0\)), secondary (\(2^0\)) or tertiary (\(3^0\)), according as the carbon carrying –OH is bonded to one, two or three other carbon atoms. Each type may in addition be allylic (that carbon is bonded to a C=C carbon) or benzylic (bonded to an aromatic ring carbon). When the –OH is fixed on an \(sp^2\) carbon of a double bond, the alcohol is vinylic.
Ethers are symmetrical (simple) when the two groups on oxygen are identical, and unsymmetrical (mixed) when they differ.
Alcohols are named in three systems: common names (alkyl group + “alcohol”), the carbinol system (alkyl groups in alphabetical order + “carbinol”), and the IUPAC system, in which the final ‘e’ of the parent alkane is replaced by ‘ol’. For polyhydric alcohols the ‘e’ is retained and di, tri, etc. are inserted before ‘ol’, as in ethane-1,2-diol.
The IUPAC name of phenol is benzenol, though phenol is also accepted; ortho, meta and para in common names correspond to the locants 2-, 3- and 4-. Ethers are named as alkoxyalkanes, the larger group being the parent alkane and the smaller becoming the alkoxy prefix.
Alcohols are prepared from alkyl halides by hydrolysis with aqueous alkali or moist silver oxide; by acid catalyzed hydration of alkenes (Markownikoff addition, through a carbocation); by hydroboration–oxidation of alkenes with \(\mathrm{B_2H_6}\) followed by \(\mathrm{H_2O_2}/\mathrm{OH^-}\) (antimarkownikoff); by reduction of aldehydes, ketones and carboxylic acids; and by addition of a Grignard reagent to a carbonyl compound.
Aldehydes give \(1^0\) alcohols and ketones give \(2^0\) alcohols on reduction. \(\mathrm{LiAlH_4}\) leaves an isolated C=C bond untouched, whereas \(\mathrm{H_2}/\mathrm{Ni}\) reduces it as well.
With a Grignard reagent, formaldehyde gives a \(1^0\) alcohol, any other aldehyde a \(2^0\) alcohol and a ketone a \(3^0\) alcohol; an epoxide gives a \(1^0\) alcohol two carbons longer than the reagent’s alkyl group.
Phenol is prepared from chlorobenzene (Dow process, NaOH at 623 K/300 atm), from cumene by air oxidation in the presence of Co-naphthenate (the commercial route, giving acetone as a valuable by-product), from benzene sulfonic acid by fusion with NaOH at 573 K, and from aniline through benzene diazonium chloride.
Flowchart
FC 11.13 – The four routes to phenol at a glance: chlorobenzene → NaOH, 623 K/300 atm → sodium phenoxide → dil. HCl → phenol; cumene → O₂, Co-naphthenate, 423 K → cumene hydroperoxide → dil. HCl → phenol + acetone; benzene sulfonic acid → NaOH → sodium benzene sulfonate → NaOH, 573 K → sodium phenoxide → dil. HCl → phenol; aniline → NaNO₂/HCl, 273 K → benzene diazonium chloride → H₂O, Δ → phenol + N₂ + HCl.
Alcohols and phenols are strongly polar and are held together by intermolecular hydrogen bonding. Their boiling points rise with molecular mass and fall with branching, and lower alcohols (up to three carbons) and phenols dissolve appreciably in water by hydrogen bonding with it. o-Nitrophenol is much less soluble than p-nitrophenol because its hydrogen bonding is intramolecular.
In the laboratory, alcohols are neutral to litmus while phenols turn blue litmus red. Phenols dissolve in NaOH but do not react with \(\mathrm{NaHCO_3}\); alcohols react with neither, but liberate hydrogen with sodium metal.
\[2\,\mathrm{R\!-\!OH} + 2\,\mathrm{Na} \rightarrow 2\,\mathrm{R\!-\!O^-Na^+} + \mathrm{H_2}\uparrow\]
- Phenols give a deep purple, violet or green colour with neutral ferric chloride, which is their characteristic test.
\[3\,\mathrm{Ar\!-\!OH} + \mathrm{FeCl_3} \rightarrow (\mathrm{Ar\!-\!O})_3\mathrm{Fe} + 3\,\mathrm{HCl}\]
The Lucas test (conc. HCl with \(\mathrm{ZnCl_2}\)) distinguishes the three classes of alcohol by the time taken for turbidity: instantaneous for \(3^0\), slow for \(2^0\), and only on heating for \(1^0\).
Phenol is weakly acidic because the phenoxide ion is resonance stabilized, the negative charge being shared with the ortho and para ring carbons. An alkoxide ion has no such stabilization and is further destabilized by the +I effect of alkyl groups, so alcohols are practically neutral. An electron withdrawing group such as \(\mathrm{-NO_2}\) raises acidity further, giving the order p-nitrophenol > phenol > ethanol > tert-butyl alcohol.
Alcohols and phenols form esters with carboxylic acids, acid anhydrides and acid chlorides; the reaction with a carboxylic acid is called esterification and is reversible, so water is removed to drive it forward. Acetylation introduces the \(\mathrm{CH_3\!-\!CO\!-}\) group, and the number of acetyl groups taken up equals the number of –OH groups present. Aspirin is the acetyl derivative of salicylic acid.
The C–O bond of an alcohol is broken by hydrogen halides (\(\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\), with \(3^0 > 2^0 > 1^0\); HCl needs anhydrous \(\mathrm{ZnCl_2}\)), and by \(\mathrm{PCl_5}\) or \(\mathrm{PX_3}\), giving alkyl halides.
\[\mathrm{R\!-\!OH} + \mathrm{PCl_5} \rightarrow \mathrm{R\!-\!Cl} + \mathrm{POCl_3} + \mathrm{HCl}\]
Dehydration with concentrated \(\mathrm{H_2SO_4}\), \(\mathrm{H_3PO_4}\) or alumina converts alcohols to alkenes through a carbocation, and the more substituted alkene is the major product in accordance with Saytzeff rule.
On oxidation, \(1^0\) alcohols give aldehydes with PCC and carboxylic acids with \(\mathrm{KMnO_4}\), \(\mathrm{K_2Cr_2O_7}\) or \(\mathrm{HNO_3}\); \(2^0\) alcohols give ketones with \(\mathrm{CrO_3}\); \(3^0\) alcohols resist oxidation and break at the C–C bonds only under drastic conditions. Passed over copper at 573 K, \(1^0\) alcohols give aldehydes, \(2^0\) give ketones and \(3^0\) undergo dehydration to alkenes.
The –OH group of phenol activates the ring and directs incoming electrophiles to the ortho and para positions. Phenol therefore gives 2,4,6-tribromophenol with bromine water (and only o- and p-bromophenol in a solvent of low polarity), o- and p-nitrophenol with dilute \(\mathrm{HNO_3}\), picric acid with concentrated \(\mathrm{HNO_3}\), and o- or p-phenolsulfonic acid on sulfonation at 293 K or 373 K respectively.
The Reimer–Tiemann reaction converts phenol into salicylaldehyde using chloroform and aqueous NaOH (carbon tetrachloride gives salicylic acid instead), and Kolbe’s reaction converts sodium phenoxide into salicylic acid with \(\mathrm{CO_2}\) at 398 K and 6 atm.
Phenol is oxidised to p-benzoquinone by \(\mathrm{CrO_3}\) or \(\mathrm{Na_2Cr_2O_7}/\mathrm{H_2SO_4}\), hydrogenated to cyclohexanol over nickel at 433 K, and reduced to benzene by zinc dust.
\[\mathrm{C_6H_5OH} + 3\,\mathrm{H_2} \xrightarrow[\ 433\ \text{K}\ ]{\ \mathrm{Ni}\ } \mathrm{C_6H_{11}OH}\]
- Ethers are prepared by dehydration of alcohols (conc. \(\mathrm{H_2SO_4}\) at the lower temperature gives the ether, the higher temperature gives the alkene; suitable only for symmetrical ethers from \(1^0\) alcohols) and by Williamson synthesis, in which a sodium alkoxide or phenoxide attacks a primary alkyl halide by an \(\mathrm{S_N2}\) mechanism. Secondary and tertiary halides give alkenes instead, and aryl halides do not react at all.
\[\mathrm{R\!-\!X} + \mathrm{Na^+O^-\!-\!R'} \rightarrow \mathrm{R\!-\!O\!-\!R'} + \mathrm{NaX}\]
Because the C–O–C angle is \(110^{\circ}\) and not \(180^{\circ}\), the two bond dipoles do not cancel and ethers have a small net dipole moment (1.18 D for diethyl ether). Having no O–H bond, ethers cannot hydrogen bond to one another, so they boil at about the same temperature as alkanes of comparable mass; but they dissolve in water much like alcohols do, because the ethereal oxygen can accept a hydrogen bond from water.
Ethers are neutral and unreactive towards bases, cold dilute acids, oxidising and reducing agents and active metals, but dissolve in cold concentrated \(\mathrm{H_2SO_4}\) as oxonium salts. On long contact with air they form peroxides, which decompose violently on heating.
Ethers are hydrolysed by hot dilute \(\mathrm{H_2SO_4}\) to alcohols or phenols, give two alkyl chlorides with \(\mathrm{PCl_5}\), and are cleaved by hot concentrated HI or HBr to an alkyl halide and an alcohol. The iodide attacks the least substituted carbon by \(\mathrm{S_N2}\); with a tertiary group the reaction follows \(\mathrm{S_N1}\) and gives the tertiary iodide. An aryl alkyl ether always yields a phenol and an alkyl halide, never an aryl halide.
\[\mathrm{C_6H_5\!-\!O\!-\!CH_3} + \mathrm{HI} \xrightarrow{\ 398\ \text{K}\ } \mathrm{C_6H_5\!-\!OH} + \mathrm{CH_3I}\]
The alkoxy group of an aromatic ether is ring activating and ortho-, para-directing because of its +R effect. Anisole therefore brominates in acetic acid without a catalyst, undergoes Friedel–Crafts alkylation and acylation with anhydrous \(\mathrm{AlCl_3}\), and is nitrated by the nitrating mixture — the para product being the major one in each case.
Methyl alcohol is a solvent for paints and varnishes; ethyl alcohol is an antifreeze and a solvent; phenol makes phenol formaldehyde resin such as bakelite and is used in antiseptic products; diethyl ether was formerly a general anaesthetic and is a solvent for Grignard reagents, fats, waxes and oils.
11.8 Objective Questions
- Which of the following represents the increasing order of boiling points of (1), (2) and (3)?
- \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2\!-\!OH}\) (2) \(\mathrm{(CH_3)_2CH\!-\!O\!-\!CH_3}\) (3) \(\mathrm{(CH_3)_3COH}\)
- < (2) < (3)
- < (1) < (3)
- < (2) < (1)
- < (3) < (1)
- Which is the best reagent for carrying out the conversion of 1-methylcyclohexene into 2-methylcyclohexan-1-ol, shown in Fig. E11.1?
- \(\mathrm{LiAlH_4}\)
- Conc. \(\mathrm{H_2SO_4}\), \(\mathrm{H_2O}\)
- \(\mathrm{H_2}/\mathrm{Pd}\)
- \(\mathrm{B_2H_6}\), \(\mathrm{H_2O_2}\!-\!\mathrm{NaOH}\)
Diagram
Fig. E11.1 – A cyclohexene ring carrying a –CH₃ group on one of the doubly bonded carbons, an arrow, and a saturated cyclohexane ring carrying –CH₃ on one carbon and –OH on the adjacent carbon.
- Which of the following reactions will give an ionic organic product on reaction?
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH} + \mathrm{Na} \rightarrow\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH} + \mathrm{SOCl_2} \rightarrow\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH} + \mathrm{PCl_5} \rightarrow\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH} + \mathrm{H_2SO_4} \rightarrow\)
- Which is the most resistant alcohol towards oxidation reaction among the following?
- \(\mathrm{CH_3\!-\!CH_2\!-\!OH}\)
- \(\mathrm{(CH_3)_2CH\!-\!OH}\)
- \(\mathrm{(CH_3)_3C\!-\!OH}\)
- \(\mathrm{C_2H_5\!-\!CH(CH_3)\!-\!OH}\)
- Resorcinol on distillation with zinc dust gives
- Cyclohexane
- Benzene
- Toluene
- Benzene-1,3-diol
- Anisole on heating with concentrated HI gives
- Iodobenzene
- Phenol + Methanol
- Phenol + Iodomethane
- Iodobenzene + methanol
- Which of the following is the least acidic compound?
- Phenol
- 4-Methylphenol
- 3-Nitrophenol
- 4-Nitrophenol
- The compound incapable of hydrogen bonding with water is ……
- \(\mathrm{CH_3\!-\!CH_2\!-\!O\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_3}\)
- Phenol
- \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!OH}\)
- Ethers are kept in air tight brown bottles because
- Ethers absorb moisture
- Ethers evaporate readily
- Ethers oxidise to explosive peroxide
- Ethers are inert
- Ethers react with cold and concentrated \(\mathrm{H_2SO_4}\) to form
- oxonium salt
- alkene
- alkoxides
- alcohols
11.9 Check Your Concepts
Classify each of the following as an alcohol, a phenol, an aromatic alcohol or an ether, and where it is an alcohol state whether it is \(1^0\), \(2^0\) or \(3^0\): (i) \(\mathrm{C_6H_5\!-\!CH_2\!-\!CH_2\!-\!OH}\) (ii) \(\mathrm{C_6H_5\!-\!OH}\) (iii) \(\mathrm{(CH_3)_3C\!-\!OH}\) (iv) \(\mathrm{C_6H_5\!-\!O\!-\!C_2H_5}\).
Prop-2-en-1-ol and prop-1-en-2-ol both contain a double bond and an –OH group, yet only one of them is called a vinylic alcohol. Which one, and on what basis is the decision made?
How many structural isomers of molecular formula \(\mathrm{C_4H_{10}O}\) are ethers? Draw each and give its IUPAC name.
Name t-butyl alcohol by the carbinol system, and explain in one line how that system decides the name.
Assertion (A): Ethanol is named ethanol but ethylene glycol is named ethane-1,2-diol. Reason (R): The final ‘e’ of the parent alkane is dropped only when the suffix that follows begins with a vowel. State whether both A and R are true, and whether R is the correct explanation of A.
Propene is treated in two separate experiments: (a) with conc. \(\mathrm{H_2SO_4}\) followed by water, and (b) with \(\mathrm{B_2H_6}\) followed by \(\mathrm{H_2O_2}\) in alkaline medium. Name the alcohol obtained in each case and explain why the two differ.
But-2-enal, \(\mathrm{CH_3\!-\!CH{=}CH\!-\!CHO}\), is to be converted into but-2-en-1-ol without disturbing the double bond. Which reducing agent would you choose, and why would the other one be unsuitable?
Which carbonyl compound and which Grignard reagent would you use to prepare (i) a primary alcohol, (ii) a secondary alcohol and (iii) a tertiary alcohol? Illustrate each with a general equation.
An epoxide is treated with a Grignard reagent \(\mathrm{R\!-\!MgX}\) followed by acidic hydrolysis. By how many carbon atoms is the chain of the R group extended, and what class of alcohol results?
The cumene process is preferred industrially over the Dow process for making phenol. Give two reasons, one economic and one environmental.
Why does the preparation of phenol from chlorobenzene need a temperature of 623 K and a pressure of 300 atm, while the preparation from aniline is carried out near 273 K?
Arrange the following in decreasing order of boiling point and justify your order: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol.
Phenol and n-butyl alcohol have comparable molecular masses, yet phenol dissolves in water more than ten times as freely. Explain.
Assertion (A): o-Nitrophenol is less soluble in water than p-nitrophenol. Reason (R): o-Nitrophenol forms intramolecular hydrogen bonds. State whether both A and R are true, and whether R correctly explains A.
Phenols react with NaOH but not with \(\mathrm{NaHCO_3}\), whereas HCl reacts with both. What does this tell you about the relative acid strengths involved?
Arrange the following in increasing order of acid strength and justify each placement: 2-methylpropan-2-ol, ethanol, phenol, 4-nitrophenol.
Predict, with reasons, whether 4-methoxyphenol is a stronger or a weaker acid than phenol, and do the same for 4-chlorophenol.
Explain how the Lucas test distinguishes primary, secondary and tertiary alcohols, and arrange the three in order of the speed with which turbidity appears.
A compound takes up three acetyl groups on complete acetylation with acetic anhydride. What can you conclude about the compound? Name one compound from this chapter that would behave in this way.
What is the role of pyridine when an alcohol is treated with an acid chloride? What would happen if it were left out?
Complete the following reactions and name the major product in each case:
- \(\mathrm{C_6H_5OH} + 3\,\mathrm{Br_2} \xrightarrow{\ \mathrm{H_2O}\ } ?\)
- \(\mathrm{C_6H_5\!-\!O^-Na^+} + \mathrm{CO_2} \xrightarrow[\ 6\ \text{atm}\ ]{\ 398\ \text{K}\ } ? \xrightarrow{\ \mathrm{H_3O^+}\ } ?\)
- \(\mathrm{C_6H_5OH} + 3\,\mathrm{HNO_3} \xrightarrow{\ \text{conc. } \mathrm{H_2SO_4}\ } ?\)
- \(\mathrm{C_6H_5OH} + \mathrm{Zn} \rightarrow ?\)
Butan-2-ol is heated with concentrated \(\mathrm{H_2SO_4}\). Write both alkenes that can form, state which is the major product, and name the rule that decides this.
Write the reactions needed to convert propan-1-ol into propan-2-ol, and explain why the –OH group cannot simply be shifted from one carbon to the next.
State what happens when: (a) ethanol vapour is passed over copper at 573 K, (b) 2-methylpropan-2-ol vapour is passed over copper at 573 K, (c) propan-2-ol is treated with \(\mathrm{CrO_3}\), (d) ethanol is oxidised with acidified \(\mathrm{KMnO_4}\).
Why does phenol undergo electrophilic substitution far more readily than benzene, and why do the incoming groups take the ortho and para positions?
Bromination of phenol in water gives a tribromo product, but in \(\mathrm{CS_2}\) at low temperature it gives mainly a monobromo product. Account for the difference.
Sodium phenoxide reacts with carbon dioxide in Kolbe’s reaction, although \(\mathrm{CO_2}\) is only a weak electrophile. Why is the phenoxide ion able to do what phenol itself cannot?
Match each reagent in Column I with the product it gives from phenol in Column II.
| Column I (reagent and conditions) | Column II (product) |
|---|---|
| a. \(\mathrm{CHCl_3}\) + aq. NaOH, then \(\mathrm{H_3O^+}\) | i. Cyclohexanol |
| b. \(\mathrm{CCl_4}\) + aq. NaOH, then \(\mathrm{H_3O^+}\) | ii. Benzene |
| c. \(\mathrm{H_2}\)/Ni at 433 K | iii. Salicylaldehyde |
| d. Zinc dust, heat | iv. p-Benzoquinone |
| e. \(\mathrm{CrO_3}\) or \(\mathrm{Na_2Cr_2O_7}/\mathrm{H_2SO_4}\) | v. Salicylic acid |
Ethanol is dehydrated with conc. \(\mathrm{H_2SO_4}\) at 413 K in one experiment and at 443 K in another. Name the product in each case and explain why the temperature decides the outcome.
Sodium ethoxide is treated with (i) ethyl chloride and (ii) t-butyl chloride. Predict the major product in each case and name the type of reaction that dominates.
Which alkoxide and which alkyl halide would you choose to prepare 2-methoxy-2-methylpropane by Williamson synthesis? Explain why the alternative combination fails.
Why do aryl halides fail to undergo Williamson synthesis?
Assertion (A): Ethers are stored in dark, airtight brown bottles. Reason (R): Ethers are highly volatile liquids. State whether both A and R are true, and whether R is the correct explanation of A.
Methyl n-pentyl ether and n-hexyl alcohol have almost the same molecular mass, but the ether boils at 373 K and the alcohol at 430 K. Explain the difference, and explain also why the ether nevertheless dissolves in water almost as well as the alcohol.
Why do ethers possess a small net dipole moment even though they contain two identical C–O bonds in many cases?
Give a simple chemical test to distinguish between (i) diethyl ether and n-hexane, (ii) ethanol and phenol, (iii) propan-1-ol and 2-methylpropan-2-ol.
Predict the products, and name the mechanism followed, when each of the following is heated with concentrated HI: (i) \(\mathrm{CH_3\!-\!O\!-\!CH_2CH_2CH_3}\), (ii) \(\mathrm{CH_3\!-\!O\!-\!C(CH_3)_3}\), (iii) \(\mathrm{C_6H_5\!-\!O\!-\!CH_3}\).
An organic compound of molecular formula \(\mathrm{C_7H_8O}\) does not react with sodium metal, and on heating with concentrated HI gives phenol and methyl iodide. Identify the compound and write the equation for its cleavage.
Arrange HI, HBr and HCl in decreasing order of reactivity towards ethers, and give the reason for the order.
Anisole undergoes bromination in acetic acid without any \(\mathrm{FeBr_3}\) catalyst, and the para product predominates. Explain both observations using the resonance structures of the aryl alkyl ether.
Phenol has three activated positions on its ring. Show how this fact explains the formation of a rigid, cross-linked network when phenol is heated with formaldehyde.
Give one use each of methyl alcohol, ethyl alcohol, phenol and diethyl ether, and in each case name the property of the compound that makes that use possible.
Hydroboration-oxidation of propene gives …..
Write the IUPAC name of the alcohol having molecular formula \(\mathrm{C_4H_{10}O}\) which is resistant towards oxidation.
Write the structure of the optically active alcohol having molecular formula \(\mathrm{C_4H_{10}O}\).
Write the name of the electrophile used in Kolbe’s Reaction.
Why is phenol more acidic than ethyl alcohol?
Why is p-nitrophenol a stronger acid than phenol?
Write two points of difference between the properties of phenol and ethyl alcohol.
Give the reagents and conditions necessary to prepare phenol from a. Chlorobenzene b. Benzene sulfonic acid.
Give the equations of the reactions for the preparation of phenol from isopropyl benzene.
Give a simple chemical test to distinguish between ethanol and ethyl bromide.
An ether (A), \(\mathrm{C_5H_{12}O}\), when heated with excess of hot HI produces two alkyl halides which on hydrolysis form compound (B) and (C). Oxidation of (B) gave an acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E).
Write structural formulae for a. 3-Methoxyhexane b. Methyl vinyl ether c. 1-Ethylcyclohexanol d. Pentane-1,4-diol e. Cyclohex-2-en-1-ol
Write IUPAC names of the compounds shown in Fig. E11.2.
Diagram
Fig. E11.2 – Four structures for naming: (i) a cyclopentane ring carrying –OH and –CH₃ on the same carbon; (ii) CH₃–CH(OH)–CH(CH₃)–CH₂–OH; (iii) a benzene ring with –OH on C-1 and –NO₂ on C-2; (iv) a cyclohexane ring carrying an –O–CH₃ group.