Chapter 10: Halogen Derivatives
In the late 1920s, a refrigerator was a dangerous thing to own. The machines were cooled by gases such as ammonia, sulphur dioxide and methyl chloride, and a cracked pipe in the night could poison an entire household. Thomas Midgley Jr., a chemist working with General Motors, was given a clear brief: design a cooling gas that will not burn and will not poison. He turned to the halogens. Starting from methane and replacing its hydrogen atoms with chlorine and fluorine, he arrived at dichlorodifluoromethane — the compound the world soon knew as Freon-12. It was odourless, non-flammable and remarkably unreactive, and Midgley is said to have demonstrated its safety before an audience of chemists by breathing it in. Within a few decades, chlorofluorocarbons were cooling homes, pushing paint out of spray cans and blowing bubbles into plastic foam.
That same unreactivity turned out to be the trouble. Nothing at ground level destroyed these molecules, so they drifted slowly upward into the stratosphere. In 1974, Mario Molina and F. Sherwood Rowland argued that ultraviolet radiation up there snaps the carbon–chlorine bond and sets free chlorine atoms, and that a single chlorine atom can go on to destroy an enormous number of ozone molecules. Their work led to the Montreal Protocol and, in 1995, to a Nobel Prize in Chemistry.
One bond, and one choice of halogen, altered the atmosphere of a planet. The picture in Fig. 10.0 shows how a molecule designed for a kitchen ended up in the sky.
Diagram
Fig. 10.0 – A CFC molecule (CCl₂F₂) rising from a refrigerator/aerosol can at ground level into the stratosphere, where a UV photon breaks the C–Cl bond to release a Cl atom that attacks an O₃ molecule; label C–Cl bond, UV radiation, ozone layer.
Keep these questions with you as you read — this chapter will give you everything you need to reason out each one.
10.1 Classification of Halogen Derivatives
The structures of these five compounds are drawn together in Fig. 10.7; look carefully at which carbon atom carries the halogen in each one, because that single detail is what this section is about.
Diagram
Fig. 10.7 – Structures of benzyl bromide (benzene ring with –CH₂Br), Freon-12 (CCl₂F₂), hexachlorobenzene (benzene ring with Cl on all six carbons), Westrosol (Cl–CH=CCl₂) and hexachlorocyclohexane (cyclohexane ring with one Cl on each carbon); label the halogen atoms in each.
The parent family of all organic compounds is the hydrocarbons — molecules built from carbon and hydrogen alone. Take any hydrocarbon, whether it is aliphatic (an open chain, such as ethane) or aromatic (a benzene-type ring), and replace one or more of its hydrogen atoms by halogen atoms. What you obtain is a new family of compounds: the halogen derivatives of hydrocarbons.
A halogen is any member of Group 17 of the periodic table — fluorine, chlorine, bromine or iodine. Since all four behave in a broadly similar way here, chemists use the general symbol \(\mathrm{X}\) to stand for any one of them and write a halogen derivative compactly as \(\mathrm{R\!-\!X}\), where \(\mathrm{R}\) is the hydrocarbon part of the molecule.
This replacement does far more than swap one small atom for another. A halogen is much more electronegative than carbon, so the new \(\mathrm{C\!-\!X}\) bond is polar, and that polarity turns an otherwise unreactive hydrocarbon into a molecule with a definite reactive site. This is why the halogen derivatives are so useful as starting materials for making alcohols, ethers, amines, nitriles and a great many other compounds. In this chapter we will study them in a systematic way.
Halogen derivatives are classified mainly in two ways.
Classification by the Carbon Skeleton
The first scheme looks at the kind of hydrocarbon skeleton to which the halogen atom is bonded. On this basis, halogen derivatives are classified as haloalkanes, haloalkenes, haloalkynes and haloarenes, as shown in Fig. 10.8.
\[\mathrm{CH_3\!-\!CH_2\!-\!X}\ \text{(haloalkane)} \qquad \mathrm{CH_2{=}CH\!-\!X}\ \text{(haloalkene)}\]
\[\mathrm{HC{\equiv}C\!-\!X}\ \text{(haloalkyne)} \qquad \mathrm{C_6H_5\!-\!X}\ \text{(haloarene)}\]
Diagram
Fig. 10.8 – Four general structures side by side: CH₃–CH₂–X labelled haloalkane, CH₂=CH–X labelled haloalkene, HC≡C–X labelled haloalkyne, and a benzene ring bearing X labelled haloarene.
The naming pattern is easy to read: the prefix halo- is simply attached to the name of the parent skeleton — alkane, alkene, alkyne or arene.
Classification by the Number of Halogen Atoms
The second scheme simply counts how many halogen atoms the molecule carries. On this basis, halogen derivatives are classified as mono, di, tri or poly halogen compounds. Representative structures are given in Fig. 10.9.
A monohalogen compound carries exactly one halogen atom, as in \(\mathrm{CH_3\!-\!CH_2\!-\!X}\) or \(\mathrm{C_6H_5\!-\!X}\). A dihalogen compound carries two, as in \(\mathrm{CH_3\!-\!CHX_2}\), in \(\mathrm{CH_2X\!-\!CH_2X}\), or in an ortho-dihalobenzene. A trihalogen compound carries three, as in \(\mathrm{CH_2X\!-\!CHX\!-\!CH_2X}\) or in a trihalobenzene.
Diagram
Fig. 10.9 – Three rows of examples: (row 1, monohalogen) CH₃–CH₂–X and benzene ring with one X; (row 2, dihalogen) CH₃–CHX₂, CH₂X–CH₂X and benzene ring with two X at adjacent positions; (row 3, trihalogen) CH₂X–CHX–CH₂X and benzene ring with three X at alternate positions.
Two dihalogen compounds may carry their halogens very differently. When both halogen atoms sit on the same carbon, as in \(\mathrm{CH_3\!-\!CHX_2}\), the compound is called a geminal dihalide; when they sit on adjacent carbons, as in \(\mathrm{CH_2X\!-\!CH_2X}\), it is called a vicinal dihalide. You will meet the vicinal dihalides again when alkenes add chlorine or bromine.
We will now consider the classification of monohalogen derivatives in more detail. The complete classification scheme is collected in FC 10.1.
Flowchart
FC 10.1 – Classification of halogen derivatives: halogen derivatives of hydrocarbons → branch A (basis: carbon skeleton) → haloalkanes / haloalkenes / haloalkynes / haloarenes; → branch B (basis: number of halogen atoms) → mono / di / tri / poly halogen compounds; mono halogen compounds → (basis: position of X and hybridization of C) → alkyl halide → further branch: primary (1°) / secondary (2°) / tertiary (3°); allylic halide; benzylic halide; vinylic halide; haloalkyne; aryl halide.
10.1.1 Classification of Monohalogen Compounds
Monohalogen compounds are classified further on the basis of two things: the position of the halogen atom in the molecule, and the type of hybridization of the carbon atom to which the halogen is attached.
Alkyl Halides or Haloalkanes
In alkyl halides, also called haloalkanes, the halogen atom is bonded to an \(\mathrm{sp^3}\) hybridized carbon which is part of a saturated carbon skeleton.
Alkyl halides are further described as primary, secondary or tertiary, depending on the substitution state of the carbon that carries the halogen — that is, on how many other carbon atoms are directly attached to it (refer to Std. XI Chemistry Textbook, section 14.3). If that carbon is joined to one carbon it is primary, to two carbons secondary, and to three carbons tertiary. These three types are drawn in Fig. 10.10.
\[\mathrm{R\!-\!CH_2\!-\!X}\ \text{(primary, 1}^\circ\text{)} \qquad \mathrm{R\!-\!CHX\!-\!R}\ \text{(secondary, 2}^\circ\text{)} \qquad \mathrm{R_3C\!-\!X}\ \text{(tertiary, 3}^\circ\text{)}\]
Diagram
Fig. 10.10 – Three structures with the halogen-bearing carbon highlighted: R–CH₂–X labelled primary halide (1° halide), R–CHX–R labelled secondary halide (2° halide), and R–C(R)(R)–X labelled tertiary halide (3° halide).
This 1°/2°/3° label is worth learning properly now. It will decide, later in this chapter, which mechanism an alkyl halide follows when a nucleophile attacks it.
Allylic Halides
In allylic halides, the halogen atom is bonded to an \(\mathrm{sp^3}\) hybridized carbon atom that lies next to a carbon–carbon double bond.
\[\mathrm{CH_2{=}CH\!-\!CH_2\!-\!X}\]
A cyclic example is 3-halocyclohexene, in which the halogen sits on the saturated carbon immediately next to the ring double bond.
Benzylic Halides
In benzylic halides, the halogen atom is bonded to an \(\mathrm{sp^3}\) hybridized carbon atom which is itself further bonded to an aromatic ring.
\[\mathrm{C_6H_5\!-\!CH_2\!-\!X} \qquad \mathrm{C_6H_5\!-\!C(CH_3)_2\!-\!X}\]
Notice the difference from an aryl halide: here the halogen is one carbon away from the ring, not on it.
Vinylic Halides
In vinylic halides, the halogen atom is bonded to an \(\mathrm{sp^2}\) hybridized carbon atom of an aliphatic chain. A vinylic halide is therefore a haloalkene.
\[\mathrm{CH_2{=}CH\!-\!X}\]
A halogen placed directly on the doubly bonded carbon of a cyclohexene ring gives a cyclic vinylic halide.
Haloalkynes
When a halogen atom is bonded to an \(\mathrm{sp}\) hybridized carbon atom, the compound is a haloalkyne.
\[\mathrm{CH{\equiv}C\!-\!X}\]
Aryl Halides or Haloarenes
In aryl halides, the halogen atom is directly bonded to an \(\mathrm{sp^2}\) hybridized carbon atom of an aromatic ring.
\[\mathrm{C_6H_5\!-\!X}\]
An example carrying a substituent as well is o-halotoluene, in which a methyl group occupies the position next to the halogen. All six classes are drawn together in Fig. 10.11.
Diagram
Fig. 10.11 – Six labelled structures in one panel with the hybridization of the halogen-bearing carbon marked below each: alkyl halide (R–CH₂–X, sp³), allylic halide (CH₂=CH–CH₂–X, sp³), benzylic halide (C₆H₅–CH₂–X, sp³), vinylic halide (CH₂=CH–X, sp²), haloalkyne (CH≡C–X, sp), aryl halide (benzene ring with X, sp²).
How This Is Tested
Classification rarely appears as a question by itself. In board and entrance papers it usually turns up as a match-the-following item pairing a structure with its class, or hidden inside a reasoning question — “which of these will not react with aqueous \(\mathrm{KOH}\)?” or “which one undergoes racemization?” — where the first step of the answer is simply to recognise whether the halogen sits on an \(\mathrm{sp^3}\), \(\mathrm{sp^2}\) or \(\mathrm{sp}\) carbon.
10.2 Nomenclature of Halogen Derivatives
Every halogen derivative can be referred to in two ways — by an older common (trivial) name that chemists have used for generations, and by a systematic IUPAC name that describes the structure precisely. You need both: the common name because bottles, catalogues and older books still carry it, and the IUPAC name because it is the one that can be read backwards to draw the structure.
Common Names
The common names of alkyl halides are built in two words. Name the alkyl group first, then name the halogen as a halide. So \(\mathrm{CH_3\!-\!I}\) is methyl iodide and \(\mathrm{(CH_3)_3C\!-\!Cl}\) is tert-butyl chloride.
IUPAC Names
According to the IUPAC system of nomenclature (Std. XI Chemistry Textbook, Chapter 14, section 14.4.7), alkyl halides are named as haloalkanes. Here the halogen is not treated as a principal functional group at all; it is cited as a substituent prefix — fluoro, chloro, bromo or iodo — placed before the name of the parent alkane. This is why the name never ends in a halogen suffix the way an alcohol ends in -ol.
Aryl halides are named as haloarenes in the common as well as the IUPAC system, so a benzene ring carrying a bromine atom is bromobenzene in both.
For a dihalogen derivative of an arene, the two systems differ in how they state the positions. The common system uses the prefixes o- (ortho), m- (meta) and p- (para), while the IUPAC system uses the numerals 1,2 ; 1,3 and 1,4 respectively. The three arrangements are drawn in Fig. 10.12.
Diagram
Fig. 10.12 – Three benzene rings each bearing two X substituents: adjacent positions labelled ortho (1,2), alternate positions labelled meta (1,3), opposite positions labelled para (1,4); number the ring carbons 1 to 6 on the first ring.
Building an IUPAC Name
Because the halogen is only a prefix, naming a halogen derivative comes down to naming the parent chain correctly and then attaching the prefix with the right locant. The working sequence is set out in FC 10.2.
Flowchart
FC 10.2 – Writing the IUPAC name of a halogen derivative: identify the longest continuous carbon chain (it must include any C=C or C≡C present) → number the chain from the end that gives the lowest locant to the multiple bond, if there is one; otherwise from the end nearer the first substituent → name the halogen as fluoro/chloro/bromo/iodo with its locant → if the same halogen occurs more than once, use di-, tri-, tetra- → arrange different substituent prefixes in alphabetical order (ignore di-, tri- while alphabetising) → assemble: locant-prefix + parent alkane/alkene/alkyne name.
One point in that sequence is easy to trip over. When the molecule also contains a double or triple bond, the multiple bond is the senior feature and takes the lower number, even if that forces the halogen to a higher number. That is why \(\mathrm{CH_2{=}CH\!-\!CH_2Br}\) is numbered from the double-bond end and is named 3-bromopropene, not 1-bromopropene.
Common and IUPAC names of some halogen derivatives are given in Table 10.1. The last three entries of the table are ring compounds; their structures are drawn in Fig. 10.13.
Table 10.1 Names of some halogen derivatives
| Formula | Common name | IUPAC name |
|---|---|---|
| \(\mathrm{CH_2Cl_2}\) | Methylene chloride | Dichloromethane |
| \(\mathrm{CH_3CH_2Br}\) | Ethyl bromide | Bromoethane |
| \(\mathrm{CH_3CH(Cl)CH_3}\) | Isopropyl chloride | 2-Chloropropane |
| \(\mathrm{(CH_3)_2\,CH\!-\!CH_2Br}\) | Isobutyl bromide | 1-Bromo-2-methylpropane |
| \(\mathrm{(CH_3)_3\,C\,Br}\) | Tert-butyl bromide | 2-Bromo-2-methylpropane |
| \(\mathrm{(CH_3)_3\,C\,CH_2Cl}\) | Neopentyl chloride | 1-Chloro-2, 2-dimethyl propane |
| \(\mathrm{CH_2{=}CH\!-\!Cl}\) | Vinyl chloride | 1-Chloroethene |
| \(\mathrm{CH_2{=}CH\!-\!CH_2Br}\) | Allyl bromide | 3-Bromopropene |
| \(\mathrm{CH{\equiv}C\!-\!Cl}\) | Chloro acetylene | Chloroethyne |
| \(\mathrm{C_6H_5CH_2I}\) | Benzyl iodide | Iodophenylmethane |
| 4-iodo-substituted toluene ring | p-Iodotoluene | 1-Iodo-4-methyl benzene or 4-Iodotoluene |
| benzene ring with Cl at 1,3-positions | m-dichlorobenzene | 1, 3-dichlorobenzene |
Diagram
Fig. 10.13 – Three structures matching the last three rows of Table 10.1: benzene ring with –CH₂I (benzyl iodide); benzene ring with CH₃ at position 1 and I at position 4 (p-iodotoluene); benzene ring with Cl at positions 1 and 3 (m-dichlorobenzene).
How This Is Tested
Nomenclature is examined in both directions. A board paper will ask you to write the IUPAC name of a drawn structure, or to draw the structure of a named compound; entrance papers often hide the same skill inside a reaction question, where the product is described by name and you must convert it into a structure before you can go further. Match-the-following items pairing a common name with its IUPAC counterpart — vinyl chloride with 1-chloroethene, allyl bromide with 3-bromopropene — are also common.
10.3 Methods of Preparation of Alkyl Halides
An alkyl halide can be built in several ways, and the route a chemist chooses depends on what starting material is at hand and how pure the product has to be. The four families of methods covered in this section are collected in FC 10.3.
Flowchart
FC 10.3 – Routes to halogen derivatives: starting material → branch 1: alcohol → (a) halogen acid HX / (b) phosphorus halide PX₃ or PCl₅ / (c) thionyl chloride SOCl₂ → alkyl halide; branch 2: hydrocarbon → alkane + X₂ (light) gives a mixture / alkene + HX gives alkyl halide / alkene + X₂ gives vicinal dihalide → alkyl or dihalogen product; branch 3: another alkyl halide → NaI in acetone (Finkelstein) gives alkyl iodide, metal fluoride (Swartz) gives alkyl fluoride; branch 4: arene → electrophilic halogenation with Lewis acid, or diazonium salt (Sandmeyer) → aryl halide.
10.3.1 From Alcohol
The most widely used method of preparing an alkyl halide is the replacement of the hydroxyl group of an alcohol by a halogen atom. The reason is practical: alcohols are available in a wide variety, so almost any carbon skeleton you need can be bought or made as an alcohol first.
The hydroxyl group may be replaced by a halogen atom using (a) a halogen acid, (b) a phosphorus halide, or (c) thionyl chloride.
By Using a Halogen Acid or Hydrogen Halide (HX)
The conditions for the reaction of an alcohol with a halogen acid (HX) depend on the structure of the alcohol and on the particular halogen acid used. The order of reactivity of alcohols with a given haloacid is \(3^\circ > 2^\circ > 1^\circ\) (refer to section 11.2.1 a).
\[\mathrm{R\!-\!OH} + \mathrm{HX} \xrightarrow{\ \text{suitable condition}\ } \mathrm{R\!-\!X} + \mathrm{H_2O}\]
\[\text{(Alcohol)} \qquad\qquad \text{(Alkyl halide)}\]
Hydrogen chloride is used with zinc chloride (Grooves’ process) for primary and secondary alcohols, but tertiary alcohols readily react with concentrated hydrochloric acid in the absence of zinc chloride.
\[\mathrm{R\!-\!OH} + \mathrm{HCl} \xrightarrow{\ \text{anhydrous }\mathrm{ZnCl_2}\ } \mathrm{R\!-\!Cl} + \mathrm{H_2O}\]
Zinc chloride is a Lewis acid, that is, an electron-pair acceptor, and can therefore coordinate with the oxygen atom of the alcohol. Pulling electron density away from oxygen in this way weakens the \(\mathrm{R\!-\!O}\) bond and makes it easier to break. A mixture of concentrated \(\mathrm{HCl}\) and anhydrous \(\mathrm{ZnCl_2}\) is called Lucas reagent.
Constant boiling hydrobromic acid (48%) is used for preparing alkyl bromides. Primary alkyl bromides can also be prepared by reaction with \(\mathrm{NaBr}\) and \(\mathrm{H_2SO_4}\). Here \(\mathrm{HBr}\) is generated in situ, that is, produced inside the reaction mixture itself rather than being added from a bottle.
\[\mathrm{R\!-\!CH_2\!-\!OH} + \mathrm{HBr} \xrightarrow{\ \mathrm{NaBr},\ \mathrm{H_2SO_4},\ \text{heat}\ } \mathrm{R\!-\!CH_2\!-\!Br} + \mathrm{H_2O}\]
A good yield of alkyl iodides may be obtained by heating alcohols with sodium or potassium iodide in 95 % phosphoric acid. Here \(\mathrm{HI}\) is generated in situ.
\[\mathrm{R\!-\!OH} + \mathrm{HI} \xrightarrow{\ \mathrm{NaI}/\mathrm{H_3PO_4}\ } \mathrm{R\!-\!I} + \mathrm{H_2O}\]
By Using a Phosphorus Halide
An alkyl halide may be prepared by the action of a phosphorus halide on an alcohol. Phosphorus tribromide and triiodide are usually generated in situ by the action of red phosphorus on bromine and on iodine respectively. Phosphorus pentachloride reacts with an alcohol to give an alkyl chloride.
\[3\,\mathrm{R\!-\!OH} + \mathrm{PX_3} \rightarrow 3\,\mathrm{R\!-\!X} + \mathrm{H_3PO_3}\]
\[\mathrm{R\!-\!OH} + \mathrm{PCl_5} \rightarrow \mathrm{R\!-\!Cl} + \mathrm{HCl} + \mathrm{POCl_3}\]
Notice the stoichiometry of the first equation: one molecule of \(\mathrm{PX_3}\) can hand over all three of its halogen atoms, so three molecules of alcohol are consumed and phosphorous acid, \(\mathrm{H_3PO_3}\), is left behind.
Sometimes, during the replacement of \(\mathrm{-OH}\) by \(\mathrm{-X}\), alcohols tend to undergo rearrangement — the carbon skeleton quietly reorganises itself and the halogen ends up on the wrong carbon. This tendency can be minimised by the use of phosphorus halides. Straight chain primary alcohols react with a phosphorus trihalide to give unrearranged alkyl halides.
By Using Thionyl Chloride
Thionyl chloride reacts with straight chain primary alcohols to give unrearranged alkyl chloride. The byproducts obtained are gases. There is no need to put extra efforts for its separation, since both simply escape from the reaction vessel. This method is therefore preferred for the preparation of alkyl chlorides.
\[\mathrm{R\!-\!OH} + \mathrm{SOCl_2} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!Cl} + \mathrm{SO_2}\uparrow + \mathrm{HCl}\uparrow\]
10.3.2 From Hydrocarbon
Alkyl halides are formed from saturated as well as unsaturated hydrocarbons by various reactions.
Halogenation of Alkanes
Halogenation of alkanes is not suitable for the preparation of alkyl halides, as a mixture of mono and poly halogen compounds is formed. Once one hydrogen has been replaced, the remaining hydrogens of the same molecule are still available for attack, so the reaction does not stop neatly at the monohalogen stage. Separating the resulting mixture is tedious and the yield of any one compound is poor.
Addition of Hydrogen Halide to an Alkene
Alkyl halides are formed on addition of a hydrogen halide to alkenes. Refer to Std. XI Chemistry Textbook, Chapter 15, section 15.2.4 for all the details, including the order of reactivity of HX, Markownikov rule and the peroxide effect.
Three points from that treatment are worth recalling before you attempt the questions below. First, the order of reactivity of the hydrogen halides in this addition is \(\mathrm{HI} > \mathrm{HBr} > \mathrm{HCl}\). Second, when the alkene is unsymmetrical, the two possible products are not formed in equal amounts, and Markownikov’s rule tells you which one dominates. Third, hydrogen bromide alone reverses this preference when peroxides are present.
Diagram
Fig. 10.14 – Two reactions drawn one below the other: (a) 1-methylcyclohexene (cyclohexene ring with CH₃ on one doubly bonded carbon) + HBr → cyclohexane ring bearing CH₃ and Br on the same carbon; (b) methylenecyclohexane (cyclohexane ring with =CH₂ outside the ring) + HBr → the same product; label the product 1-bromo-1-methylcyclohexane in both.
Addition of Halogen to an Alkene
Alkenes form an addition product, a vicinal dihalide, with chlorine or bromine, usually in an inert solvent like \(\mathrm{CCl_4}\) at room temperature. The two halogen atoms end up on adjacent carbons, which is exactly what the word vicinal means.
\[\mathrm{R_2C{=}CR_2} + \mathrm{X_2} \xrightarrow{\ \mathrm{CCl_4}\ } \mathrm{R_2CX\!-\!CR_2X} \qquad (\mathrm{X = Cl,\ Br})\]
Allylic Substitution at High Temperature
When alkenes are heated with \(\mathrm{Br_2}\) or \(\mathrm{Cl_2}\) at high temperature, the hydrogen atom of the allylic carbon is substituted with a halogen atom, giving an allyl halide. Note the change of behaviour: at room temperature in \(\mathrm{CCl_4}\) the halogen adds across the double bond, but at high temperature the double bond survives and a hydrogen is replaced instead.
\[\mathrm{CH_2{=}CH\!-\!CH_3} + \mathrm{Cl_2} \rightarrow \mathrm{CH_2{=}CH\!-\!CH_2Cl} + \mathrm{HCl}\]
10.3.3 Halogen Exchange
Alkyl iodides are prepared conveniently by treating alkyl chlorides or bromides with sodium iodide in methanol or acetone solution. The sodium bromide or sodium chloride precipitates from the solution and can be separated by filtration.
\[\mathrm{R\!-\!Cl} + \mathrm{NaI} \xrightarrow{\ \text{acetone}\ } \mathrm{R\!-\!I} + \mathrm{NaCl}\downarrow\]
The reaction is known as the Finkelstein reaction.
Alkyl fluorides are prepared by heating alkyl chlorides or bromides with metal fluorides such as \(\mathrm{AgF}\), \(\mathrm{Hg_2F_2}\), \(\mathrm{AsF_3}\), \(\mathrm{SbF_3}\) etc.
\[\mathrm{R\!-\!Cl} + \mathrm{AgF} \rightarrow \mathrm{R\!-\!F} + \mathrm{AgCl}\downarrow\]
The reaction is known as the Swartz reaction. Here too an insoluble silver halide precipitates and drives the exchange forward.
10.3.4 Electrophilic Substitution
Aryl chlorides and bromides can be prepared by direct halogenation of benzene and its derivatives through electrophilic substitution. It may be conveniently carried out in the dark at ordinary temperature in the presence of a suitable Lewis acid catalyst like \(\mathrm{Fe}\), \(\mathrm{FeCl_3}\) or anhydrous \(\mathrm{AlCl_3}\). Darkness matters here: light would start a free-radical reaction at the side chain instead of the ring.
When toluene is brominated in the presence of iron, a mixture of ortho and para bromo toluene is obtained, as shown in Fig. 10.15.
\[\mathrm{C_6H_5CH_3} + \mathrm{Br_2} \xrightarrow{\ \mathrm{Fe},\ \text{dark}\ } \mathrm{CH_3C_6H_4Br}\ \text{(ortho and para)} + \mathrm{HBr}\]
Diagram
Fig. 10.15 – Toluene + Br₂ over an arrow marked Fe, dark, giving two products side by side: benzene ring with CH₃ at position 1 and Br at position 2, labelled o-bromotoluene, and benzene ring with CH₃ at position 1 and Br at position 4, labelled p-bromotoluene, plus HBr.
Aromatic electrophilic substitution with iodine is reversible: the hydrogen iodide released can attack the product and undo the reaction. In this case the use of \(\mathrm{HNO_3}/\mathrm{HIO_4}\) removes \(\mathrm{HI}\) by oxidising it to \(\mathrm{I_2}\); with one of the products being continually taken away, the equilibrium is shifted to the right and the iodo product is formed. Fluorine, being highly reactive, is not used this way, so fluoro compounds are not prepared by this method.
10.3.5 Sandmeyer’s Reaction
Aryl halides are most commonly prepared by replacement of the nitrogen of a diazonium salt. A diazonium salt is made from an aromatic amine and carries an \(\mathrm{-N_2^+}\) group, which is an outstandingly good leaving group because it departs as stable nitrogen gas. That is what makes this the standard laboratory route to aryl halides, in contrast with the sluggish direct halogenation described above. (For details refer to Chapter 13, section 13.6.)
How This Is Tested
Preparation methods are examined mainly as conversions and reagent-naming questions: “convert propan-1-ol into 1-iodopropane”, or “name the reagent used to bring about the following conversion”. A second favourite is the multi-step sequence in which compound A becomes B and then C, where each arrow carries only a reagent and you must supply the structures. Keep the reagent and its speciality paired in your mind — \(\mathrm{SOCl_2}\) with clean chlorides, \(\mathrm{NaI}\)/acetone with iodides, metal fluorides with fluorides, diazonium salts with aryl halides — because these are exactly the pairings the questions test.
10.4 Physical Properties
The physical properties of alkyl halides are considerably different from those of the corresponding alkanes. Replacing a single hydrogen atom by a halogen changes how strongly one molecule is pulled towards its neighbours, and that is what physical properties such as boiling point and solubility ultimately measure. The boiling points of alkyl halides are determined by the polarity of the \(\mathrm{C\!-\!X}\) bond as well as by the size of the halogen atom.
10.4.1 Nature of Intermolecular Forces
The halogens (\(\mathrm{X = F,\ Cl,\ Br}\) and \(\mathrm{I}\)) are more electronegative than carbon — that is, they attract the shared pair of a covalent bond more strongly than carbon does.
As a result, the carbon atom that carries the halogen develops a partial positive charge, while the halogen carries a partial negative charge. The carbon–halogen bond in an alkyl halide is therefore a polar covalent bond, and alkyl halides are moderately polar compounds. The separation of charge is drawn in Fig. 10.16.
\[\overset{\delta+}{\mathrm{C}}\!-\!\overset{\delta-}{\mathrm{X}}\]
Diagram
Fig. 10.16 – A carbon atom of a chain bonded to X, with δ⊕ marked above the carbon and δ⊖ above the halogen, and an arrow along the bond pointing from C towards X to show the direction of the dipole.
The size of the halogen atom increases from fluorine to iodine. Hence the \(\mathrm{C\!-\!X}\) bond length increases down the group. The \(\mathrm{C\!-\!X}\) bond strength decreases as the size of the halogen increases. The reason lies in orbital overlap: as the p-orbital of the halogen grows larger it becomes more diffused, its electron density is spread over a bigger region, and the extent of its overlap with the orbital of carbon decreases. A bond built on poorer overlap is a weaker bond.
Some typical bond lengths, bond enthalpies and dipole moments of the \(\mathrm{C\!-\!X}\) bond are given in Table 10.2.
Table 10.2 : Bond parameters of C-X bond
| Bond | Bond length (pm) | Bond enthalpy (kJ mol\(^{-1}\)) | Dipole moment (debye) |
|---|---|---|---|
| \(\mathrm{CH_3\!-\!F}\) | 139 | 452 | 1.847 |
| \(\mathrm{CH_3\!-\!Cl}\) | 178 | 351 | 1.860 |
| \(\mathrm{CH_3\!-\!Br}\) | 193 | 293 | 1.830 |
| \(\mathrm{CH_3\!-\!I}\) | 214 | 234 | 1.636 |
10.4.2 Boiling Point
The boiling points of alkyl halides are considerably higher than those of the corresponding alkanes, due to their higher polarity and higher molecular mass. Two molecules of an alkyl halide attract each other both through their permanent dipoles and through the ordinary van der Waals forces that all molecules feel, so more heat is needed to pull them apart.
Within the alkyl halides, for a given alkyl group, the boiling point increases with increasing atomic mass of the halogen, because the magnitude of the van der Waals force increases with increase in the size and mass of the halogen. A bigger halogen has a larger, looser electron cloud that is more easily distorted, and such a cloud produces stronger momentary attractions between neighbouring molecules.
Thus the boiling point of an alkyl halide decreases in the order
\[\mathrm{RI} > \mathrm{RBr} > \mathrm{RCl} > \mathrm{RF}\]
For example:
| Haloalkane | \(\mathrm{CH_3F}\) | \(\mathrm{CH_3Cl}\) | \(\mathrm{CH_3Br}\) | \(\mathrm{CH_3I}\) |
|---|---|---|---|---|
| Boiling point (K) | 194.6 | 248.8 | 276.6 | 315.4 |
For a given halogen, the boiling point rises with increasing carbon number, since a longer chain means a larger surface over which neighbouring molecules can touch and attract one another.
For example,
| Haloalkane | Boiling point (K) |
|---|---|
| \(\mathrm{CH_3Cl}\) | 248.8 |
| \(\mathrm{CH_3CH_2Cl}\) | 285.5 |
| \(\mathrm{CH_3CH_2CH_2Cl}\) | 320.0 |
| \(\mathrm{CH_3CH_2CH_2CH_2Cl}\) | 351.5 |
For isomeric alkyl halides, the boiling point decreases with increased branching, as the surface area decreases on branching and the van der Waals forces decrease. A branched molecule is closer to a compact sphere, and spheres touch each other over a smaller area than rods do. For example:
| Haloalkane | Boiling point (K) |
|---|---|
| \(\mathrm{CH_3CH_2CH_2CH_2Br}\) | 375 |
| \(\mathrm{CH_3\!-\!CHBr\!-\!CH_2\!-\!CH_3}\) | 364 |
| \(\mathrm{(CH_3)_3C\!-\!Br}\) | 346 |
Notice that all three compounds in this last table have exactly the same molecular formula and the same molecular mass; only their shape differs. That is what makes them such a clean demonstration of the effect of branching. The relationship between halogen size, bond strength and boiling point is plotted in Fig. 10.17.
Diagram
Fig. 10.17 – A dual plot for CH₃F, CH₃Cl, CH₃Br, CH₃I along the x-axis: left y-axis shows boiling point in K (194.6, 248.8, 276.6, 315.4) as a rising line with points; right y-axis shows C–X bond enthalpy in kJ mol⁻¹ (452, 351, 293, 234) as a falling line with points; label both curves.
10.4.3 Solubility
Though alkyl halides are moderately polar, they are insoluble in water. This is due to the inability of alkyl halides to form hydrogen bonds with water. The attraction between alkyl halide molecules is stronger than the attraction between an alkyl halide molecule and a water molecule, so the two liquids stay apart as separate layers rather than mixing. Alkyl halides are soluble in non-polar organic solvents, where the forces they must break and the forces they can form are of comparable strength.
Aryl Halides
Aryl halides are also insoluble in water but soluble in organic solvents. If aryl halides are not modified by the presence of any other functional group, they show properties similar to the corresponding alkyl halides.
The isomeric dihalobenzenes have nearly the same boiling points, but the melting points of these isomers show variation. The melting point of the para isomer is quite high compared to that of the ortho or meta isomer. This is because of its symmetrical structure, which can easily pack closely in the crystal lattice. As a result, the intermolecular forces of attraction are stronger and therefore greater energy is required to overcome its lattice energy.
| Dichlorobenzene | ortho | meta | para |
|---|---|---|---|
| b.p./K | 453 | 446 | 448 |
| m.p./K | 256 | 249 | 323 |
The contrast in this small table is worth pausing on. Melting is the breaking down of an ordered crystal, so how neatly the molecules stack matters enormously; boiling happens from a liquid in which that order has already been lost, so packing no longer helps. This is exactly why the three isomers differ by nearly \(70\ \text{K}\) in melting point but by only a few kelvin in boiling point.
How This Is Tested
Physical properties are almost always tested as ordering questions — arrange these four halides in increasing order of boiling point — or as short reasoning questions of the “give reasons” type: why are alkyl halides immiscible with water although they are polar; why does the para isomer melt so much higher. In assertion–reason items, the usual trap is to accept a correct statement paired with a wrong explanation, for instance attributing the insolubility of alkyl halides to their being non-polar, which they are not. The safe habit is to name the actual force involved each time — hydrogen bonding, dipole–dipole attraction, van der Waals forces or lattice energy.
10.5 Optical Isomerism in Halogen Derivatives
Isomers having the same bond connectivities, that is, the same structural formula, are called stereoisomers. They are built from the same atoms joined to the same partners; what differs is only the way those atoms are arranged in space.
Knowledge of optical isomerism, which is a kind of stereoisomerism, will be useful to understand the nucleophilic substitution reactions of alkyl halides (see section 10.6.3). There is a good reason to study it here rather than later: the spatial arrangement around a carbon atom turns out to be the most sensitive evidence we have about how a nucleophile attacks that carbon.
10.5.1 Chiral Atom and Molecular Chirality
Let us now jot down the atoms and groups attached to each carbon in 2-chlorobutane, numbering the chain from left to right.
\[\overset{1}{\mathrm{C}}\mathrm{H_3}\!-\!\overset{2}{\mathrm{C}}\mathrm{HCl}\!-\!\overset{3}{\mathrm{C}}\mathrm{H_2}\!-\!\overset{4}{\mathrm{C}}\mathrm{H_3}\]
| Carbon | The four atoms/groups attached to it |
|---|---|
| C-1 | \(\mathrm{-H}\), \(\mathrm{-H}\), \(\mathrm{-H}\), \(\mathrm{-CHCl\!-\!CH_2\!-\!CH_3}\) |
| C-2 | \(\mathrm{-H}\), \(\mathrm{-Cl}\), \(\mathrm{-CH_3}\), \(\mathrm{-CH_2\!-\!CH_3}\) |
| C-3 | \(\mathrm{-H}\), \(\mathrm{-H}\), \(\mathrm{-CH_3}\), \(\mathrm{-CHCl\!-\!CH_3}\) |
| C-4 | \(\mathrm{-H}\), \(\mathrm{-H}\), \(\mathrm{-H}\), \(\mathrm{-CH_2\!-\!CHCl\!-\!CH_3}\) |
Run your eye down the table. C-1, C-3 and C-4 each carry at least two identical atoms — two or three hydrogens. It can be seen that the four groups bonded to C-2 are all different from each other.
Thus, the C-2 in 2-chlorobutane is a chiral carbon. A chiral atom in a molecule is marked with an asterisk (*). For example, \(\mathrm{CH_3\!-\!{}^{*}CHCl\!-\!CH_2\!-\!CH_3}\).
When a molecule contains one chiral atom, it acquires a unique property. Such a molecule cannot superimpose perfectly on its mirror image. It is called a chiral molecule. A chiral molecule and its mirror image are not identical, as shown in Fig. 10.1.
Diagram
Fig. 10.1 : Nonsuperimposable mirror images – A tetrahedral carbon marked with an asterisk bearing CH₃, Cl, C₂H₅ and H, drawn on the left of a vertical mirror plane, with its mirror image drawn on the right; below, the two structures shown overlapped to demonstrate that they do not coincide; label the mirror plane.
A chiral molecule and its mirror image both have the same structural formula and, of course, the same molecular formula. The spatial arrangement of the four different groups around the chiral atom, however, is different. In other words, a chiral molecule and its mirror image are stereoisomers of each other (refer to Std. XI Chemistry Textbook, Chapter 14).
The relationship between a chiral molecule and its mirror image is similar to the relationship between left and right hands. Your two hands have the same parts joined in the same order, and each is the mirror image of the other, yet no amount of turning will make the left hand lie exactly over the right. This is why the property is called handedness or chirality. (Origin : Greek word cheir means hand.)
Figure 10.2 shows a few objects from day to day life which exhibit superimposable and non-superimposable mirror image relationships. A plain glass tumbler and its reflection can be superimposed; a hand or a foot and its reflection cannot.
Diagram
Fig. 10.2 : Superimposable and nonsuperimposable mirror image – Two groups of everyday objects: on the left, a pair of hands and a pair of feet with their mirror images, labelled Non superimposable; on the right, a chair and a drinking glass with their mirror images, labelled Superimposable.
The optical isomers differ from each other in terms of a measurable property called optical activity. To understand optical activity, we must first know what plane polarized light is.
An important point about the order of discovery is worth remembering here: the phenomenon of optical isomerism in organic compounds was observed first, and its origin in molecular chirality was recognised only later. Chemists could measure the effect for decades before they could explain it.
10.5.2 Plane Polarized Light
Ordinary light consists of electromagnetic waves having oscillations of the electric and magnetic field in all possible planes perpendicular to the direction of propagation of light. Picture the beam coming straight towards you: the vibrations are spread evenly in every direction around the line of travel, like the spokes of a wheel.
When ordinary light is passed through Nicol’s prism, oscillations in only one plane emerge out. The prism behaves like a narrow slit for vibrations, allowing through only those aligned in one direction and blocking the rest.
The difference between the two kinds of light is drawn in Fig. 10.18.
Diagram
Fig. 10.18 – Cross-section of a beam of ordinary light shown as arrows radiating in all directions within a circle, an arrow passing through a block labelled Nicol Prism, and on the right the cross-section of the emergent beam shown as arrows in a single vertical plane; label the two circles “cross section of ordinary light” and “cross section of plane polarized light”.
A Nicol prism is a special type of prism made from pieces of calcite, a crystalline form of \(\mathrm{CaCO_3}\), arranged in a specific manner. A Nicol prism is also called a polarizer.
10.5.3 Optical Activity
When an aqueous solution of certain organic compounds like sugar or lactic acid is placed in the path of plane polarized light, the transmitted light has oscillations in a different plane than the original. In other words, the incident light undergoes rotation of its plane of polarization. The plane of polarization rotates either to the right (clockwise) or to the left (anticlockwise).
The compounds which rotate the plane of plane polarized light are called optically active compounds, and those which do not rotate it are optically inactive compounds.
Optical activity of a substance is expressed numerically in terms of optical rotation. The angle through which a substance rotates the plane of plane polarized light on passing through it is called optical rotation. The instrument used to measure it is a polarimeter, whose arrangement is shown in Fig. 10.19: light is first polarized, then sent through a tube containing the solution, and a second prism called the analyser is turned until the light passes through it fully once again. The angle through which the analyser had to be turned is the optical rotation.
Diagram
Fig. 10.19 – Schematic of a polarimeter arranged left to right: light source → polarizer (Nicol prism) → plane polarized light → sample tube containing the solution → rotated plane of polarization → analyser (second Nicol prism) marked with a rotation angle α on a circular scale → eyepiece.
In accordance with the direction of optical rotation, an optically active substance is either dextrorotatory or laevorotatory. A compound which rotates the plane of plane polarized light towards the right is called dextrorotatory and is designated by the symbol d- or by a (+) sign. A compound which rotates the plane of plane polarized light towards the left is called laevorotatory and is designated by the symbol l- or by a (−) sign.
Isomerism in which isomeric compounds have different optical activity is known as optical isomerism. The French scientist Louis Pasteur first recognised that optical activity is associated with a certain type of three-dimensional structure of molecules. Pasteur introduced the term enantiomers for the optical isomers having equal and opposite optical rotation.
Three ideas from this subsection should be kept clearly apart in your mind. Optical activity is an experimentally observable property of a compound — you measure it. Chirality is a description of molecular structure — you deduce it from the formula. Optical activity is the consequence of chirality. Molecules which contain one chiral atom are chiral, that is, they are nonsuperimposable on their mirror image; the two non-superimposable mirror image structures are called a pair of enantiomers; and enantiomers have equal and opposite optical rotation, which makes them a kind of optical isomer.
10.5.4 Enantiomers
For example, 2-chlorobutane exists as a pair of enantiomers (Fig. 10.1).
Enantiomers have identical physical properties — such as melting point, boiling point, density and refractive index — except the sign of optical rotation. The magnitude of their optical rotation is equal but the sign of optical rotation is opposite. They also have identical chemical properties, except towards an optically active reagent.
An equimolar mixture of enantiomers, that is, of the dextrorotatory and the laevorotatory forms in equal amounts, is called a racemic modification or racemic mixture.
A racemic modification is optically inactive because the optical rotation due to the molecules of one enantiomer is cancelled by the equal and opposite optical rotation due to the molecules of the other enantiomer. For every molecule turning the plane a little to the right there is another turning it by exactly as much to the left, so the net rotation observed is zero. A racemic modification is designated as (dl) or by the (±) sign.
10.5.5 Representation of Configuration of Molecules
Two representations are used to represent the configuration of a chiral carbon and the three-dimensional structure of optical isomers on plane paper. These are (a) the wedge formula and (b) the Fischer projection formula, also called the cross formula (Std. XI Chemistry Textbook, Chapter 14, section 14.2.3). Both exist to solve one problem: a tetrahedral carbon is a three-dimensional object, and the page is flat.
a. Fischer Projection Formula (Cross Formula)
In a Fischer projection the chiral carbon is placed at the point where two lines cross, and it is usually not written at all. The convention that gives the picture its meaning is fixed and must never be broken: the two horizontal lines represent bonds coming above the plane of the paper, towards the viewer, and the two vertical lines represent bonds going below the plane of the paper, away from the viewer. Fig. 10.3 shows both the convention and the tetrahedral structure it stands for.
Diagram
Fig. 10.3 Fischer projection formula – On the left, a cross with Cl at the top, H at the bottom, I on the left arm and Br on the right arm, labelled Fischer projection and with the crossing point marked “chiral carbon”; on the right, the same molecule drawn as a tetrahedral carbon with I and Br on horizontal bonds labelled “Bonds above the plane” and Cl and H on vertical bonds labelled “Bonds below the plane”; caption the pair “Convention of vertical and horizontal lines”.
b. Wedge Formula
When a tetrahedral carbon is imagined to be present in the plane of the paper, all four bonds at this carbon cannot lie in the same plane. Two of them must project out of it, one on each side.
In the wedge formula, the bonds in the plane of the paper are represented by normal lines, the bonds projecting above the plane of the paper are represented by solid wedges (or simply by bold lines), while the bonds going below the plane of the paper are represented by broken wedges (or simply by broken lines). The three kinds of line are shown in Fig. 10.20.
Diagram
Fig. 10.20 – A central carbon with four bonds: Br drawn upward as a normal line labelled “In the plane”, I drawn to the left as a broken wedge labelled “below the plane”, Cl drawn to the right as a normal line, and H drawn downward as a solid wedge labelled “Above the plane”.
How This Is Tested
The commonest question types are: identify the chiral molecule from a list of four; mark the chiral carbon in a given structure; draw the enantiomers of a named compound using Fischer or wedge formulae; and explain optical isomerism in a specific compound such as 2-chlorobutane. Assertion–reason items often pair a true statement about enantiomers having identical physical properties with a wrong reason, so check both halves separately. In the next section, chirality becomes a diagnostic tool rather than a topic in itself — whether a product turns out optically active or racemic will tell you which mechanism the reaction followed.
10.6 Chemical Properties
10.6.1 Laboratory Test of Haloalkanes
Haloalkanes are of neutral type in aqueous medium — they neither turn litmus red nor blue, so acidity or basicity gives no clue to their presence. Their halogen is locked into a covalent bond and therefore does not exist as a free ion in solution, which is why a fresh sample of an alkyl halide gives no precipitate with silver nitrate directly.
The test therefore begins by breaking that covalent bond. On warming with aqueous sodium or potassium hydroxide, the covalently bonded halogen in the haloalkane is converted to halide ion.
\[\mathrm{R\!-\!X} + \mathrm{OH^-} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!OH} + \mathrm{X^-}\]
When this reaction mixture is acidified by adding dilute nitric acid, and silver nitrate solution is added, a precipitate of silver halide is formed, which confirms the presence of halogen in the original organic compound.
\[\mathrm{Ag^+}\text{(aq)} + \mathrm{X^-}\text{(aq)} \rightarrow \mathrm{AgX}\downarrow\text{(s)}\]
The dilute nitric acid is added first for a practical reason: it neutralises the excess alkali, which would otherwise precipitate silver oxide and spoil the test.
10.6.2 Nucleophilic Substitution Reactions of Haloalkanes
The \(\mathrm{C\!-\!X}\) bond in alkyl halides is a polar covalent bond, and the carbon in the \(\mathrm{C\!-\!X}\) bond is positively polarized. In other words, the \(\mathrm{C\!-\!X}\) carbon is an electrophilic centre — a site short of electron density, hungry for a pair of electrons. It has, therefore, a tendency to react with a nucleophile (refer to Std. XI Chemistry Textbook, Chapter 14).
Alkyl halides react with a variety of nucleophiles to give nucleophilic substitution reactions, written in short as \(\mathrm{S_N}\). The reaction is represented in general form as shown below.
\[\mathrm{Nu^-} + \overset{\delta+}{\mathrm{C}}\!-\!\overset{\delta-}{\mathrm{X}} \longrightarrow \mathrm{C\!-\!Nu} + \mathrm{X^-}\]
When a substrate reacts fast it is said to be reactive. The reactivity of alkyl halides in an \(\mathrm{S_N}\) reaction depends upon two factors, namely, the substitution state (\(1^\circ\), \(2^\circ\) or \(3^\circ\)) of the carbon and the nature of the halogen. The order of reactivity influenced by these two factors is as shown below.
\[\text{tertiary }(3^\circ) > \text{secondary }(2^\circ) > \text{primary }(1^\circ)\]
\[\mathrm{R\!-\!I} > \mathrm{R\!-\!Br} > \mathrm{R\!-\!Cl}\]
The second of these orders should feel familiar from section 10.4.1: the \(\mathrm{C\!-\!I}\) bond is the longest and weakest of the three, so it breaks most readily and the iodide leaves most easily.
Examples of some important nucleophilic substitution reactions of alkyl halides are shown in Table 10.3. Read the table as a toolkit: each row tells you which reagent converts an alkyl halide into which class of compound, and together they explain why alkyl halides are such valuable starting materials.
Table 10.3 Nucleophilic substitution reactions of alkyl halides
| Sr. No. | Alkyl halide | Reagent | Substitution product |
|---|---|---|---|
| 1. | \(\mathrm{R\!-\!X}\) | \(\mathrm{NaOH}\)(aq) (or \(\mathrm{KOH}\)), \(\Delta\) | \(\mathrm{R\!-\!OH}\) (alcohol) \(+\ \mathrm{NaX}\) (or \(\mathrm{KX}\)) |
| 2. | \(\mathrm{R\!-\!X}\) | \(\mathrm{NaOR'}\) (sodium alkoxide), \(\Delta\) | \(\mathrm{R\!-\!O\!-\!R'}\) (ether) \(+\ \mathrm{NaX}\) |
| 3. | \(\mathrm{R\!-\!X}\) | \(\mathrm{R'COOAg}\) (silver carboxylate), \(\Delta\) | \(\mathrm{R'COOR}\) (ester) \(+\ \mathrm{AgX}\downarrow\) |
| 4. | \(\mathrm{R\!-\!X}\) | \(\mathrm{NH_3}\)(alc.) in excess, \(\Delta\), pressure | \(\mathrm{R\!-\!NH_2}\) (primary amine) \(+\ \mathrm{HX}\) |
| 5. | \(\mathrm{R\!-\!X}\) | \(\mathrm{KCN}\) (alc.), \(\Delta\) | \(\mathrm{R\!-\!CN}\) (nitrile / alkyl cyanide) \(+\ \mathrm{KX}\) |
| 6. | \(\mathrm{R\!-\!X}\) | \(\mathrm{AgCN}\) (alc.), \(\Delta\) | \(\mathrm{R\!-\!N{\equiv}C}\) (isocyanide) \(+\ \mathrm{AgX}\downarrow\) |
| 7. | \(\mathrm{R\!-\!X}\) | \(\mathrm{KO\!-\!N{=}O}\) (potassium nitrite) | \(\mathrm{R\!-\!O\!-\!N{=}O}\) (alkyl nitrite) \(+\ \mathrm{KX}\) |
| 8. | \(\mathrm{R\!-\!X}\) | \(\mathrm{Ag\!-\!O\!-\!N{=}O}\) (silver nitrite) | \(\mathrm{R\!-\!NO_2}\) (nitroalkane) \(+\ \mathrm{AgX}\downarrow\) |
Ambident Nucleophiles
Look carefully at rows 5 and 6, and again at rows 7 and 8. In each pair the same element of the nucleophile is present, yet the products are quite different compounds. The explanation lies in a special class of nucleophile.
The cyanide ion is capable of attacking through more than one site (atom), because its negative charge is shared between carbon and nitrogen, as the resonance structures in Fig. 10.21 show.
Diagram
Fig. 10.21 – Resonance structures of the two ambident nucleophiles: (a) ⁻:C≡N: ↔︎ :C=N:⁻ with the lone pairs and negative charge marked on carbon and on nitrogen; (b) nitrite ion ⁻O–N=O ↔︎ O=N–O⁻ showing the negative charge shared between the two oxygen atoms and the lone pair on nitrogen.
Such nucleophiles are called ambident nucleophiles. \(\mathrm{KCN}\) is predominantly ionic (\(\mathrm{K^+ \ C{\equiv}N^-}\)) and provides cyanide ions. Both carbon and nitrogen are capable of donating an electron pair. The \(\mathrm{C\!-\!C}\) bond being stronger than the \(\mathrm{C\!-\!N}\) bond, attack occurs through the carbon atom of the cyanide group, forming alkyl cyanides as the major product.
However, \(\mathrm{AgCN}\) (\(\mathrm{Ag\!-\!C{\equiv}N}\)) is mainly a covalent compound. Its carbon is already tied up in the bond to silver, and the nitrogen is free to donate a pair of electrons. Hence attack occurs through nitrogen, resulting in the formation of isocyanide.
Another ambident nucleophile is the nitrite ion, which can attack through ‘O’ or ‘N’.
10.6.3 Mechanism of \(\mathrm{S_N}\) Reaction
It can be seen from Table 10.3 that in nucleophilic substitution reactions of alkyl halides the halogen atom gets detached from the carbon, and a new bond is formed between that electrophilic carbon and the nucleophile. The covalently bonded halogen is converted into a halide ion (\(\mathrm{X^-}\)). It means that the two electrons constituting the original covalent bond are carried away by the halogen along with it. The halogen atom of the alkyl halide is therefore called the leaving group in the context of this reaction.
The substrate undergoes two changes during an \(\mathrm{S_N}\) reaction. The original \(\mathrm{C\!-\!X}\) bond undergoes heterolysis — an unequal splitting in which one atom takes both electrons — and a new bond is formed between the carbon and the nucleophile using two electrons of the nucleophile. These changes may occur in one or more steps.
The mechanism is deduced from the results of the study of the kinetics of \(\mathrm{S_N}\) reactions. This is an important point of method: nobody has ever watched a single molecule react, so the evidence is indirect. What chemists measure is how the rate responds when the concentration of each reactant is changed, and from that they reason backwards to the sequence of events. Two mechanisms are observed in various \(\mathrm{S_N}\) reactions. These are denoted as \(\mathrm{S_N}1\) and \(\mathrm{S_N}2\) mechanisms.
a. \(\mathrm{S_N}2\) Mechanism
The reaction between methyl bromide and hydroxide ion to give methanol follows second order kinetics, that is, the rate of this reaction depends on the concentration of two reacting species, namely, methyl bromide and hydroxide. Hence it is called substitution nucleophilic bimolecular, \(\mathrm{S_N}2\).
\[\mathrm{CH_3Br} + \mathrm{OH^-} \longrightarrow \mathrm{CH_3OH} + \mathrm{Br^-}\]
\[\text{rate} = k\,[\mathrm{CH_3Br}]\,[\mathrm{OH^-}]\]
The rate of a chemical reaction is influenced by the chemical species taking part in the slowest step of its mechanism. In the above reaction only two reactants are present and both are found to influence the rate of the reaction. This means that the reaction is a single step reaction, which can also be called the slow step. This further implies that the two changes, namely, bond breaking and bond forming at the carbon, take place simultaneously. This \(\mathrm{S_N}2\) mechanism is represented as shown in Fig. 10.4.
Diagram
Fig. 10.4 : S_N2 mechanism – HO⁻ approaching the carbon of CH₃Br from the side opposite to Br, with a curly arrow from the lone pair of OH⁻ to carbon and another curly arrow from the C–Br bond to Br; the transition state drawn in square brackets with HO···C···Br collinear, each of HO and Br carrying ½⊖, and the three C–H bonds in one plane at 120°; the product HO–CH₃ drawn with the three hydrogens swung over to the opposite side, plus Br⁻; label the transition state (T.S.) and mark the umbrella-like inversion.
Salient features of the \(\mathrm{S_N}2\) mechanism:
Single step mechanism with simultaneous bond breaking and bond forming.
Backside attack of the nucleophile: The nucleophile attacks the carbon undergoing substitution from the side opposite to that of the leaving group. This is to avoid steric repulsion (repulsion due to bulkiness of the groups) and electrostatic repulsion between the incoming nucleophile and the leaving group.
In the transition state (T.S.) the nucleophile and the leaving group are bonded to the carbon with partial bonds and carry partial negative charge, so the total negative charge is diffused, that is, spread out rather than concentrated at one point. A spread-out charge means a lower-energy arrangement, which is why such a crowded structure can exist at all.
The T.S. contains pentacoordinate carbon having three \(\sigma\) (sigma) bonds in one plane making bond angles of \(120^\circ\) with each other, and two partial covalent bonds along a line perpendicular to this plane. Note that this carbon is not forming five full bonds; two of the five are half-made and half-broken, which is exactly what makes the transition state a fleeting arrangement rather than a compound you could bottle.
When an \(\mathrm{S_N}2\) reaction is brought about at a chiral carbon (in an optically active substrate), the product is found to have the opposite configuration compared to that of the substrate. In other words, the \(\mathrm{S_N}2\) reaction is found to proceed with inversion of configuration. This is like the flipping of an umbrella in a strong wind (see Fig. 10.4). It is known as Walden inversion. The inversion in configuration is the result of the backside attack of the nucleophile.
Flowchart
FC 10.4 – S_N2 mechanism: R–X + Nu⁻ → nucleophile approaches from the side opposite the leaving group → single transition state with pentacoordinate carbon, partial Nu···C and C···X bonds, ½⊖ on each → simultaneous bond formation and bond breaking → R–Nu + X⁻; branch: if the carbon is chiral → product has inverted configuration (Walden inversion).
Diagram
Fig. 10.22 – Two graphs side by side, each with concentration of nucleophile [Nu⁻] on the x-axis and reaction rate on the y-axis: graph (a) labelled reaction of CH₃Br with OH⁻ shows a straight line rising through the origin; graph (b) labelled reaction of (CH₃)₃CBr with OH⁻ shows a horizontal line independent of [Nu⁻].
b. \(\mathrm{S_N}1\) Mechanism
The reaction between tert-butyl bromide and hydroxide ion to give tert-butyl alcohol follows first-order kinetics, that is, the rate of this reaction depends on the concentration of only one species, which is the substrate molecule, tert-butyl bromide. Hence it is called substitution nucleophilic unimolecular, \(\mathrm{S_N}1\).
\[\mathrm{(CH_3)_3C\!-\!Br} + \mathrm{OH^-} \longrightarrow \mathrm{(CH_3)_3C\!-\!OH} + \mathrm{Br^-}\]
\[\text{rate} = k\,[\mathrm{(CH_3)_3CBr}]\]
It can be seen in this reaction that the concentration of only the substrate appears in the rate equation; the concentration of the nucleophile does not influence the reaction rate. In other words, tert-butyl bromide reacts with hydroxide by a two step mechanism. In the slow step the \(\mathrm{C\!-\!X}\) bond in the substrate undergoes heterolysis, and in the subsequent fast step the nucleophile uses its electron pair to form a new bond with the carbon undergoing change. This \(\mathrm{S_N}1\) mechanism is represented as shown in Fig. 10.5.
Step I
\[\mathrm{(CH_3)_3C\!-\!Br} \xrightarrow{\ \text{slow}\ } \mathrm{(CH_3)_3C^+} + \mathrm{Br^-}\]
Step II
\[\mathrm{(CH_3)_3C^+} + \mathrm{OH^-} \xrightarrow{\ \text{fast}\ } \mathrm{(CH_3)_3C\!-\!OH}\]
Diagram
Fig. 10.5 : S_N1 mechanism – Step I: (CH₃)₃C–Br with a curly arrow from the C–Br bond to Br, over a slow, reversible double arrow, giving the planar carbocation (CH₃)₃C⁺ drawn with three methyl groups at 120° in one plane, plus Br⁻; label “carbocation intermediate”. Step II: the planar carbocation with OH⁻ approaching, giving (CH₃)₃C–OH; show that OH⁻ can approach from either face of the plane.
Salient features of the \(\mathrm{S_N}1\) mechanism:
Two step mechanism.
Heterolysis of the \(\mathrm{C\!-\!X}\) bond in the slow and reversible first step to form a planar carbocation intermediate.
Attack of the nucleophile on the carbocation intermediate in the fast second step to form the product.
When an \(\mathrm{S_N}1\) reaction is carried out at a chiral carbon in an optically active substrate, the product formed is nearly racemic. This indicates that the \(\mathrm{S_N}1\) reaction proceeds mainly with racemization. This means both the enantiomers of the product are formed in almost equal amount. Racemization in an \(\mathrm{S_N}1\) reaction is the result of the formation of the planar carbocation intermediate (Fig. 10.5). The nucleophile can attack the planar carbocation from either side, which results in the formation of both the enantiomers of the product.
Flowchart
FC 10.5 – S_N1 mechanism: R–X → slow, reversible heterolysis of the C–X bond → planar carbocation R⁺ + X⁻ → fast attack by Nu⁻ → R–Nu; branch at the carbocation: attack from the top face gives one enantiomer, attack from the bottom face gives the other → nearly racemic product when the carbon is chiral.
10.6.4 Factors Influencing \(\mathrm{S_N}1\) and \(\mathrm{S_N}2\) Mechanism
a. Nature of Substrate
\(\mathrm{S_N}2\) : The transition state of the \(\mathrm{S_N}2\) mechanism is pentacoordinate and thus crowded (see Fig. 10.4). Every extra alkyl group on the reacting carbon adds bulk to an already congested arrangement, raising its energy and slowing the reaction. As a result, the \(\mathrm{S_N}2\) mechanism is favoured in primary halides and least favoured in tertiary halides.
\(\mathrm{S_N}1\) : A planar carbocation intermediate is formed in the \(\mathrm{S_N}1\) reaction. It has no steric crowding, since only three groups surround the carbon and they lie flat at \(120^\circ\) to one another. Bulky alkyl groups can be easily accommodated in a planar carbocation (see Fig. 10.5). As a result, the \(\mathrm{S_N}1\) mechanism is most favoured in tertiary halides and least favoured in primary halides. Formation of the planar carbocation intermediate in fact results in a relief from the steric crowding present in the tertiary halide substrate.
Secondly, the carbocation intermediate is stabilized by the \(+\mathrm{I}\) effect of alkyl substituents — their tendency to push electron density towards the electron-deficient carbon — and also by the hyperconjugation effect of alkyl substituents containing \(\alpha\)-hydrogens. The more alkyl groups there are, the more help of both kinds the positive carbon receives. As a result, the \(\mathrm{S_N}1\) mechanism is most favoured in tertiary halides and least favoured in primary halides. Both effects are drawn out in Fig. 10.6, and the resulting opposite trends are summarised in Fig. 10.23.
Diagram
Fig. 10.6 : Influence of substrate in S_N1 and S_N2 – Row (a) Transition states in S_N2: four pentacoordinate transition states drawn in a row for CH₃X, primary, secondary and tertiary substrates, with Nu···C···X collinear and ½⊖ on each end, the number of alkyl groups increasing left to right, under a long arrow labelled “crowding and destabilization increases”. Row (b) Carbocation intermediates in S_N1: four planar carbocations for the same four substrates, α-hydrogens marked, under a long arrow labelled “steric relief, stabilization by +I and hyperconjugation of α-hydrogens increases”.
Diagram
Fig. 10.23 – A horizontal line of substrates CH₃–X, 1°, 2°, 3° with an arrow above pointing right labelled “S_N1 rate increases” and an arrow below pointing left labelled “S_N2 rate increases”.
Tertiary halides undergo nucleophilic substitution by the \(\mathrm{S_N}1\) mechanism, while primary halides follow the \(\mathrm{S_N}2\) mechanism. Secondary halides react by either of the mechanisms, or by a mixed mechanism, depending upon the exact conditions.
Diagram
Fig. 10.24 – Two panels: (a) Resonance stabilization of the allylic carbocation, showing CH₂=CH–CH₂⁺ ↔︎ ⁺CH₂–CH=CH₂ with curly arrows for the shift of the π electrons; (b) Resonance stabilization of the benzylic carbocation, showing C₆H₅–CH₂⁺ and its four resonance structures in which the positive charge appears at the ortho and para ring carbons, with curly arrows drawn on each.
b. Nucleophilicity of the Reagent
A nucleophile is a species that uses its electron pair to form a bond with carbon. The nucleophilic character of any species is expressed in its electron releasing tendency, which can be correlated to its strength as a Lewis base.
A more powerful nucleophile attacks the substrate faster and favours the \(\mathrm{S_N}2\) mechanism. The rate of the \(\mathrm{S_N}1\) mechanism is independent of the nature of the nucleophile. The nucleophile does not react in the slow step of \(\mathrm{S_N}1\); it waits till the carbocation intermediate is formed, and then reacts fast with it. A stronger nucleophile therefore cannot speed up something it is not part of.
Three working rules help you compare nucleophiles. First, a negatively charged nucleophile is more powerful than its conjugate acid; for example, \(\mathrm{R\!-\!O^-}\) is a better nucleophile than \(\mathrm{R\!-\!OH}\), because the full negative charge holds a more available electron pair. Second, when the donor atoms are from the same period of the periodic table, nucleophilicity decreases from left to right in a period; for example, \(\mathrm{H_2O}\) is a less powerful nucleophile than \(\mathrm{NH_3}\), since the more electronegative oxygen holds its lone pairs more tightly. Third, when the donor atoms are from the same group of the periodic table, nucleophilicity increases down the group; for example, \(\mathrm{I^-}\) is a better nucleophile than \(\mathrm{Cl^-}\), because the larger, more loosely held electron cloud of iodide is more easily offered to carbon.
c. Solvent Polarity
The \(\mathrm{S_N}1\) mechanism proceeds via the formation of a carbocation intermediate. A good ionizing solvent, that is, a polar solvent, stabilizes the ions by solvation — surrounding each ion with a shell of solvent molecules oriented towards its charge. Solvation of the carbocation is relatively poor, and solvation of the anion is particularly important. Anions are solvated by hydrogen bonding solvents, that is, protic solvents such as water and alcohols. Thus the \(\mathrm{S_N}1\) reaction proceeds more rapidly in polar protic solvents than in aprotic solvents.
Polar protic solvents usually decrease the rate of the \(\mathrm{S_N}2\) reaction. In the rate determining step of the \(\mathrm{S_N}2\) mechanism, the substrate as well as the nucleophile is involved. A polar solvent stabilizes the nucleophile, one of the reactants, by solvation. Thus the solvent deactivates the nucleophile by stabilizing it — wrapping it in a cage of solvent molecules that must be stripped away before it can attack. Hence aprotic solvents, or solvents of low polarity, will favour the \(\mathrm{S_N}2\) mechanism.
10.6.5 Elimination Reaction : Dehydrohalogenation
When an alkyl halide having at least one \(\beta\)-hydrogen is boiled with an alcoholic solution of potassium hydroxide, it undergoes elimination of a hydrogen atom from the \(\beta\)-carbon and a halogen atom from the \(\alpha\)-carbon, resulting in the formation of an alkene.
Before going further, fix the labels firmly. The carbon bearing the halogen is commonly called the \(\alpha\)-carbon (alpha carbon), and any carbon attached to the \(\alpha\)-carbon is a \(\beta\)-carbon (beta carbon). Hydrogens attached to a \(\beta\)-carbon are \(\beta\)-hydrogens. An alkyl halide with no \(\beta\)-hydrogen at all simply cannot undergo this reaction.
This reaction is called a \(\beta\)-elimination (or 1,2-elimination) reaction, as it involves elimination of the halogen and a \(\beta\)-hydrogen atom from adjacent carbons, as drawn in Fig. 10.25.
Diagram
Fig. 10.25 – General β-elimination: a base B: with its lone pair shown attacking a β-hydrogen of the fragment H–Cβ–Cα(X)–, with curly arrows for the removal of the β-hydrogen, formation of the C=C π bond and departure of X⁻; over the arrow write alc. KOH, Δ; products drawn as the alkene C=C, B⁺H and X⁻; label base, alkyl halide and alkene.
As hydrogen and halogen are removed in this reaction, it is also known as a dehydrohalogenation reaction.
If there are two or more non-equivalent \(\beta\)-hydrogen atoms in a halide, then this reaction gives a mixture of products. Thus, 2-bromobutane on heating with alcoholic \(\mathrm{KOH}\) gives a mixture of but-1-ene and but-2-ene. Writing the substrate as \(\mathrm{CH_3\!-\!CH_2\!-\!CHBr\!-\!CH_3}\), the \(\alpha\)-carbon is the one carrying bromine; the \(\mathrm{CH_2}\) on its left is one \(\beta\)-carbon, labelled \(\beta^1\), and the terminal \(\mathrm{CH_3}\) on its right is the other, labelled \(\beta^2\). The two possible eliminations are shown in Fig. 10.26.
Loss of a \(\beta^2\)-hydrogen, from the terminal methyl group, places the double bond at the end of the chain and gives but-1-ene, \(\mathrm{CH_3\!-\!CH_2\!-\!CH{=}CH_2}\). Loss of a \(\beta^1\)-hydrogen, from the \(\mathrm{CH_2}\) group, places the double bond in the middle of the chain and gives but-2-ene, \(\mathrm{CH_3\!-\!CH{=}CH\!-\!CH_3}\).
Flowchart
FC 10.6 – Dehydrohalogenation of 2-bromobutane: CH₃–CH₂–CHBr–CH₃ + alc. KOH, Δ → branch 1: loss of β²-hydrogen (from the terminal CH₃) → CH₃–CH₂–CH=CH₂, but-1-ene (minor); branch 2: loss of β¹-hydrogen (from the CH₂) → CH₃–CH=CH–CH₃, but-2-ene (major, by Saytzeff rule).
Diagram
Fig. 10.26 – 2-Bromobutane drawn as CH₃–CH₂–CH(Br)–CH₃ with the α-carbon and the two β-carbons labelled β¹ and β², and two downward arrows from it marked alc. KOH, Δ: the left arrow labelled “loss of β²-hydrogen” leading to CH₃–CH₂–CH=CH₂ (But-1-ene) and the right arrow labelled “loss of β¹-hydrogen” leading to CH₃–CH=CH–CH₃ (But-2-ene).
The different products of elimination do not form in equal proportion. After studying a number of elimination reactions, the Russian chemist Saytzeff formulated an empirical rule, given below.
Therefore, in the above reaction but-2-ene is the preferred product, and is formed as the major product. Count the alkyl groups to check this: but-2-ene has a methyl group on each of the two doubly bonded carbons, while but-1-ene has one ethyl group on one of them and nothing on the other.
It turned out that more highly substituted alkenes are also more stable alkenes. Hence Saytzeff elimination is the preferred formation of the more highly stabilized alkene during an elimination reaction. The stability order of alkyl substituted alkenes is:
\[\mathrm{R_2C{=}CR_2} > \mathrm{R_2C{=}CHR} > \mathrm{R_2C{=}CH_2},\ \mathrm{RCH{=}CHR} > \mathrm{RCH{=}CH_2}\]
Elimination versus Substitution
Alkyl halides undergo substitution as well as elimination reactions. Both reactions are brought about by a basic reagent, hence there is always a competition between these two reactions. The same species can pull a proton off the \(\beta\)-carbon, giving elimination, or attack the \(\alpha\)-carbon, giving substitution. The reaction which actually predominates depends upon the following factors.
a. Nature of alkyl halides : Tertiary alkyl halides prefer to undergo elimination reaction, whereas primary alkyl halides prefer to undergo substitution reaction.
b. Strength and size of nucleophile : A bulkier electron rich species prefers to act as a base by abstracting a proton, and thus favours elimination — a large species finds a \(\beta\)-hydrogen on the outside of the molecule easier to reach than the crowded carbon within. Substitution is favoured in the case of comparatively weaker bases, which prefer to act as nucleophiles.
c. Reaction conditions : A less polar solvent and a high temperature favour elimination, whereas a low temperature and a polar solvent favour substitution reaction.
How This Is Tested
This section supplies more examination questions than any other in the chapter. Expect: distinguish between \(\mathrm{S_N}1\) and \(\mathrm{S_N}2\) mechanisms (a standard long-answer question); choose the compound that reacts fastest by a stated mechanism; identify which compound undergoes racemization during alkaline hydrolysis; predict the major product of dehydrohalogenation using Saytzeff’s rule; and complete a conversion in which the reagent must be chosen to give substitution rather than elimination. Two habits protect you here. First, always classify the substrate as \(1^\circ\), \(2^\circ\) or \(3^\circ\) before anything else. Second, read the solvent and the temperature, because they, and not the alkali alone, often decide the answer.
10.7 Uses and Environmental Effects of Some Polyhalogen Compounds
Compounds carrying more than one halogen atom occupy an uncomfortable place in chemistry. The very properties that make them useful — chemical stability, low flammability, an ability to dissolve oils and greases — are the properties that make them persist in the body and in the environment long after their work is done. The six compounds in this section are studied together for exactly this reason: each has a genuine use, and each carries a cost. Their structures are drawn in Fig. 10.36.
Diagram
Fig. 10.36 – Structural formulae of five polyhalogen compounds drawn side by side with names below each: CH₂Cl₂ (dichloromethane), CHCl₃ (chloroform), CCl₄ (carbon tetrachloride), CHI₃ (iodoform) and CCl₂F₂ (Freon-12); show all bonds from the central carbon.
10.7.1 Dichloromethane / Methylene Chloride (\(\mathrm{CH_2Cl_2}\))
It is a colourless volatile liquid with a moderately sweet aroma. It is used as a solvent, and used as a propellant in aerosols.
Over exposure to dichloromethane causes dizziness, fatigue, nausea, headaches, numbness and weakness. It is highly dangerous if it comes in direct contact with the eyes, as it damages the cornea.
Its volatility is what makes it useful and dangerous at the same time: a liquid that evaporates readily leaves a work surface quickly, but it also fills the air of a closed room quickly.
10.7.2 Chloroform / Trichloromethane (\(\mathrm{CHCl_3}\))
It is a colourless liquid with a peculiar sweet smell. It is used to prepare chlorofluoromethane, a refrigerant R-22. It is used as a solvent for the extraction of natural products like gums, fats and resins. It is also used as a source of dichlorocarbene.
Chloroform causes depression of the central nervous system. Inhaling chloroform for a short time causes fatigue, dizziness and headache. Long exposure to chloroform may affect the liver.
Chloroform, when exposed to air and light, forms a poisonous compound phosgene, so it is stored in dark coloured air tight bottles. The dark glass keeps out the light and the tight stopper keeps out the air; remove either precaution and the bottle slowly manufactures its own poison.
10.7.3 Carbon Tetrachloride / Tetrachloromethane (\(\mathrm{CCl_4}\))
It is a colourless liquid with a sweet smell. It is a very useful solvent for oils, fats and resins. It serves as a source of chlorine. It is used as a cleaning agent.
It is highly toxic to the liver. Exposure to a high concentration of \(\mathrm{CCl_4}\) can affect the central nervous system, and it is suspected to be carcinogenic. Prolonged exposure may cause death. It is also a greenhouse gas.
10.7.4 Iodoform or Triiodomethane (\(\mathrm{CHI_3}\))
It is a yellow crystalline substance with a disagreeable smell. It is used in medicine as a healing agent and antiseptic in the dressing of wounds; however, its use is limited.
It causes irritation to the skin and eyes. It may cause respiratory irritation or breathing difficulty, dizziness, nausea, depression of the central nervous system and visual disturbance.
10.7.5 Freons
These are organic compounds of chlorine and fluorine — chlorofluorocarbons, CFCs — commonly used as refrigerants. The most common representative is dichlorodifluoromethane (Freon-12); others include chlorodifluoromethane (R-22) and trichlorofluoromethane (R-11), and so on.
They are used as refrigerants in fridges and air conditioning, as propellants in aerosols, and as solvents. They are used as blowing agents in making foams and packing materials.
Chlorofluorocarbons are responsible for ozone depletion in the stratosphere. Regular large inhalation of freons results in breathing problems, organ damage and loss of consciousness.
How CFCs Destroy the Ozone Layer
When ultraviolet radiation (UV) strikes a CFC molecule such as \(\mathrm{CFCl_3}\) in the upper atmosphere, the carbon–chlorine bond breaks and produces a highly reactive chlorine atom.
\[\mathrm{CFCl_3} \rightarrow \mathrm{CFCl_2} + \mathrm{Cl}\]
This reactive chlorine atom decomposes an ozone molecule (\(\mathrm{O_3}\)) into an oxygen molecule (\(\mathrm{O_2}\)).
\[\mathrm{O_3} + \mathrm{Cl} \rightarrow \mathrm{O_2} + \mathrm{ClO}\]
\[\mathrm{ClO} + \mathrm{O} \rightarrow \mathrm{O_2} + \mathrm{Cl}\]
One atom of chlorine can destroy up to 100,000 ozone molecules. The sequence is set out as a cycle in FC 10.9.
Flowchart
FC 10.9 – Catalytic destruction of ozone by chlorine: CFCl₃ + UV radiation → CFCl₂ + Cl atom → Cl attacks O₃ → O₂ + ClO → ClO meets an O atom → O₂ + Cl atom regenerated → the regenerated Cl atom returns to attack another O₃ molecule, closing the cycle; note beside the cycle that one Cl atom can destroy up to 100,000 O₃ molecules.
You met the origin of this problem in the story that opened this chapter, where a refrigerant designed to be harmless proved harmless only at ground level. The chemistry above is the reason why: the same unreactivity that keeps a CFC safe in a kitchen also keeps it intact all the way up to the stratosphere, where ultraviolet radiation strong enough to break the C–Cl bond is waiting.
Diagram
Fig. 10.37 – A schematic dual-trend graph with time in decades along the x-axis: one curve showing atmospheric CFC concentration rising steeply from the mid-twentieth century, flattening near the end of the 1980s and slowly declining afterwards; a second curve showing stratospheric ozone over Antarctica falling as the first curve rises and beginning to recover late; label both axes and mark the point where international restrictions on CFC production began.
10.7.6 Dichlorodiphenyltrichloroethane (DDT)
It is a colourless, tasteless and odourless crystalline compound having insecticidal property. Its structure is shown in Fig. 10.38 — two chlorine-substituted benzene rings joined to a carbon that also carries a trichloromethyl group.
It kills insects such as houseflies, mosquitoes and body lice. It was used for controlling malaria and typhus.
Diagram
Fig. 10.38 – Structure of DDT: a central carbon bonded to a CCl₃ group above it and to two benzene rings, each ring carrying a Cl atom at the para position; label the CCl₃ group and the two para chlorine atoms.
Exposure to high doses of DDT may cause vomiting, tremors or shakiness. Laboratory animal studies showed adverse effects of DDT on the liver and reproduction.
DDT is a persistent organic pollutant, readily absorbed in soils, and it tends to accumulate in the ecosystem. When dissolved in oil or another liquid, it is readily absorbed by the skin. It is resistant to metabolism. It accumulates in fatty tissues. There is a ban on the use of DDT due to all these adverse effects.
How This Is Tested
Questions on this section are short and factual, and they nearly always pair a use with a hazard: “give two uses of chloroform and state why it is stored in dark bottles”, “which compound is used as a source of dichlorocarbene”, “why is DDT called a persistent organic pollutant”. The CFC–ozone equations are asked as a three-mark item requiring both steps and the conclusion that chlorine is regenerated. Assertion–reason questions often use the pairing “CFCs are chemically unreactive” with “CFCs are harmless to the environment”, where the assertion is true but the reason is false — and the explanation for that mismatch is the whole content of section 10.7.5.
10.8 Summary
Replacing one or more hydrogen atoms of an aliphatic or aromatic hydrocarbon by halogen atoms gives a halogen derivative, written in general as \(\mathrm{R\!-\!X}\) where \(\mathrm{X} = \mathrm{F}\), \(\mathrm{Cl}\), \(\mathrm{Br}\) or \(\mathrm{I}\).
On the basis of the carbon skeleton they are classified as haloalkanes, haloalkenes, haloalkynes and haloarenes; on the basis of the number of halogen atoms as mono, di, tri or poly halogen compounds.
Monohalogen compounds are classified further by the position of the halogen and the hybridization of the carbon carrying it: alkyl (sp³), allylic (sp³, next to C=C), benzylic (sp³, attached to a ring), vinylic (sp²), haloalkyne (sp) and aryl (sp², on the ring).
Alkyl halides are primary, secondary or tertiary according to the number of carbon atoms attached to the carbon that bears the halogen. This label decides much of the chemistry that follows.
Common names take the alkyl group followed by the halide: ethyl bromide, tert-butyl chloride. IUPAC names treat the halogen only as a prefix — fluoro, chloro, bromo, iodo — before the parent alkane name, so the same compound is bromoethane.
For dihalogen derivatives of an arene, the common system uses o-, m-, p- while the IUPAC system uses 1,2- ; 1,3- and 1,4-.
When a multiple bond is present it takes the lower locant, even if that pushes the halogen to a higher number.
From alcohols, using a halogen acid: \(\mathrm{R\!-\!OH} + \mathrm{HX} \rightarrow \mathrm{R\!-\!X} + \mathrm{H_2O}\). The reactivity order of alcohols is \(3^\circ > 2^\circ > 1^\circ\). Primary and secondary alcohols need anhydrous \(\mathrm{ZnCl_2}\) with \(\mathrm{HCl}\) (Grooves’ process); concentrated \(\mathrm{HCl}\) with anhydrous \(\mathrm{ZnCl_2}\) is Lucas reagent.
\(\mathrm{HBr}\) is generated in situ from \(\mathrm{NaBr}\) and \(\mathrm{H_2SO_4}\); \(\mathrm{HI}\) from \(\mathrm{NaI}\) in 95% phosphoric acid.
From alcohols, using a phosphorus halide: \(3\,\mathrm{R\!-\!OH} + \mathrm{PX_3} \rightarrow 3\,\mathrm{R\!-\!X} + \mathrm{H_3PO_3}\) and \(\mathrm{R\!-\!OH} + \mathrm{PCl_5} \rightarrow \mathrm{R\!-\!Cl} + \mathrm{HCl} + \mathrm{POCl_3}\). These minimise rearrangement of the carbon skeleton.
From alcohols, using thionyl chloride: \(\mathrm{R\!-\!OH} + \mathrm{SOCl_2} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!Cl} + \mathrm{SO_2}\uparrow + \mathrm{HCl}\uparrow\). Both by-products are gases, so no separation is needed and the method is preferred for alkyl chlorides.
From hydrocarbons: halogenation of alkanes is unsuitable because a mixture of mono and poly halogen compounds forms. Addition of \(\mathrm{HX}\) to an alkene follows Markownikov’s rule, and \(\mathrm{HBr}\) in the presence of peroxide adds the other way round (peroxide effect). Alkenes with \(\mathrm{Cl_2}\) or \(\mathrm{Br_2}\) in \(\mathrm{CCl_4}\) at room temperature give vicinal dihalides, while at high temperature the allylic hydrogen is substituted instead.
Halogen exchange: \(\mathrm{R\!-\!Cl} + \mathrm{NaI} \xrightarrow{\ \text{acetone}\ } \mathrm{R\!-\!I} + \mathrm{NaCl}\downarrow\) is the Finkelstein reaction; \(\mathrm{R\!-\!Cl} + \mathrm{AgF} \rightarrow \mathrm{R\!-\!F} + \mathrm{AgCl}\downarrow\) is the Swartz reaction.
Aryl halides are prepared by electrophilic halogenation of benzene and its derivatives, in the dark with a Lewis acid catalyst (\(\mathrm{Fe}\), \(\mathrm{FeCl_3}\), anhydrous \(\mathrm{AlCl_3}\)), or from diazonium salts by Sandmeyer’s reaction. Iodination is reversible and needs \(\mathrm{HNO_3/HIO_4}\) to remove \(\mathrm{HI}\); fluorination is too violent to use.
The \(\mathrm{C\!-\!X}\) bond is polar covalent: carbon carries \(\delta+\) and halogen \(\delta-\), which makes the carbon the point of attack for nucleophiles.
From \(\mathrm{F}\) to \(\mathrm{I}\) the bond length increases and the bond enthalpy decreases, because the halogen p-orbital becomes more diffuse and overlaps the carbon orbital less well.
Boiling points are higher than those of the corresponding alkanes, and follow \(\mathrm{RI} > \mathrm{RBr} > \mathrm{RCl} > \mathrm{RF}\). They rise with chain length and fall with branching, since branching reduces surface area and weakens van der Waals forces.
Alkyl halides and aryl halides are insoluble in water, because they cannot form hydrogen bonds with it, but dissolve in non-polar organic solvents.
Isomeric dihalobenzenes have nearly the same boiling points, but the para isomer melts much higher because its symmetrical structure packs closely in the crystal lattice.
A carbon carrying four different groups is a chiral carbon, marked with an asterisk. A molecule with one chiral carbon is chiral: it is not superimposable on its mirror image, just as a left hand is not superimposable on a right.
Plane polarized light has oscillations in only one plane and is obtained by passing ordinary light through a Nicol’s prism.
Optical activity is the rotation of the plane of polarization by a substance, measured as the optical rotation. Rotation to the right is dextrorotatory, d- or (+); to the left is laevorotatory, l- or (−).
Non-superimposable mirror images are enantiomers. They have identical physical properties except for the sign of optical rotation, and identical chemical properties except towards optically active reagents.
An equimolar mixture of two enantiomers is a racemic modification, (dl) or (±), and is optically inactive because the two rotations cancel.
Configuration is shown on paper by the wedge formula (bold wedge above the plane, broken wedge below) or by the Fischer projection, in which horizontal bonds lie above the plane and vertical bonds below.
Laboratory test: warming with aqueous alkali gives the halide ion, \(\mathrm{R\!-\!X} + \mathrm{OH^-} \xrightarrow{\ \Delta\ } \mathrm{R\!-\!OH} + \mathrm{X^-}\); acidifying with dilute \(\mathrm{HNO_3}\) and adding \(\mathrm{AgNO_3}\) then precipitates silver halide, \(\mathrm{Ag^+} + \mathrm{X^-} \rightarrow \mathrm{AgX}\downarrow\).
The C–X carbon is an electrophilic centre, so haloalkanes undergo nucleophilic substitution (\(\mathrm{S_N}\)) with a wide range of nucleophiles, giving alcohols, ethers, esters, amines, nitriles, isocyanides, alkyl nitrites and nitroalkanes.
Cyanide and nitrite are ambident nucleophiles: \(\mathrm{KCN}\) gives mainly the cyanide (nitrile) and \(\mathrm{AgCN}\) the isocyanide; \(\mathrm{KNO_2}\) gives the alkyl nitrite and \(\mathrm{AgNO_2}\) the nitroalkane.
Reactivity in \(\mathrm{S_N}\) reactions increases with substitution state, \(3^\circ > 2^\circ > 1^\circ\), and with the halogen in the order \(\mathrm{R\!-\!I} > \mathrm{R\!-\!Br} > \mathrm{R\!-\!Cl}\). The halogen that leaves with the bond pair is the leaving group.
\(\mathrm{S_N}2\) is a single step reaction in which bond breaking and bond forming happen together. The nucleophile attacks from the side opposite the leaving group, the transition state has a pentacoordinate carbon, and the product has the opposite configuration — inversion, known as Walden inversion.
\(\mathrm{S_N}1\) is a two step reaction: slow, reversible heterolysis of the C–X bond gives a planar carbocation, which is then attacked rapidly by the nucleophile. Because attack can occur from either face, the product is largely racemic.
\(\mathrm{S_N}2\) is favoured by primary halides and \(\mathrm{S_N}1\) by tertiary halides; secondary halides may follow either route. Allylic and benzylic halides react well by \(\mathrm{S_N}1\) because their carbocations are resonance stabilised.
A stronger nucleophile favours \(\mathrm{S_N}2\); the rate of \(\mathrm{S_N}1\) does not depend on the nucleophile at all. Polar protic solvents favour \(\mathrm{S_N}1\) by solvating the ions, while aprotic or less polar solvents favour \(\mathrm{S_N}2\) because they leave the nucleophile free.
Elimination: boiling an alkyl halide having at least one \(\beta\)-hydrogen with alcoholic \(\mathrm{KOH}\) removes a hydrogen from the \(\beta\)-carbon and the halogen from the \(\alpha\)-carbon to give an alkene. This \(\beta\)-elimination is also called dehydrohalogenation.
Where two or more non-equivalent \(\beta\)-hydrogens are available, a mixture forms and Saytzeff’s rule applies: the preferred product is the alkene with the greater number of alkyl groups on the doubly bonded carbons, which is also the more stable alkene. The stability order is \(\mathrm{R_2C{=}CR_2} > \mathrm{R_2C{=}CHR} > \mathrm{R_2C{=}CH_2}\), \(\mathrm{RCH{=}CHR} > \mathrm{RCH{=}CH_2}\).
Substitution and elimination always compete. Tertiary halides, bulky strong bases, less polar solvents and high temperature favour elimination; primary halides, weaker bases acting as nucleophiles, polar solvents and low temperature favour substitution.
Sodium, magnesium and cadmium give organometallic compounds, in which carbon is bonded directly to a metal and therefore carries a partial negative charge and behaves as a nucleophile.
\(\mathrm{R\!-\!X} + \mathrm{Mg} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!Mg\!-\!X}\) gives the Grignard reagent, which is destroyed by water, alcohols or ammonia and must therefore be handled under anhydrous conditions.
Wurtz reaction: \(2\,\mathrm{R\!-\!X} + 2\,\mathrm{Na} \xrightarrow{\ \text{dry ether}\ } \mathrm{R\!-\!R} + 2\,\mathrm{NaX}\). Two different alkyl halides together give all three possible alkanes.
Wurtz–Fittig reaction joins an aryl halide to an alkyl halide with sodium in dry ether, alkylating the ring; with only aryl halide the product is biphenyl, the Fittig reaction.
Haloarenes resist nucleophilic substitution for two reasons: a lone pair of the halogen is in conjugation with the ring \(\pi\)-electrons, giving the C–X bond partial double bond character (C–Cl is 169 pm in chlorobenzene against 178 pm in an alkyl chloride), and the halogen is held by an \(\mathrm{sp^2}\) carbon. \(\mathrm{S_N}1\) is ruled out because the phenyl cation is not resonance stabilised, and \(\mathrm{S_N}2\) because the ring blocks backside attack.
Chlorobenzene is converted to phenol only at about 623 K and 300 atm. Electron withdrawing groups at the ortho and/or para positions activate the halogen strongly by stabilising the negatively charged intermediate; the more such groups, the milder the conditions needed. The same group at the meta position has practically no effect.
Haloarenes undergo electrophilic substitution more slowly than benzene, because the halogen deactivates the ring through its −I effect; but its +R effect raises the electron density at the ortho and para positions, so the halogen is o-/p- directing. The para product usually predominates because the ortho position is crowded.
Halogenation, nitration, sulfonation and Friedel–Craft’s alkylation and acylation all follow this pattern, giving ortho and para products with para as the major one.
Dichloromethane is a solvent and aerosol propellant; over exposure causes dizziness and nausea and it damages the cornea.
Chloroform is used to make the refrigerant R-22, as a solvent for gums, fats and resins, and as a source of dichlorocarbene. It depresses the central nervous system, affects the liver, and forms poisonous phosgene in air and light, so it is stored in dark air-tight bottles.
Carbon tetrachloride dissolves oils, fats and resins, serves as a source of chlorine and as a cleaning agent. It is highly toxic to the liver, suspected of being carcinogenic, and is a greenhouse gas.
Iodoform is a yellow crystalline antiseptic used in dressings, though its use is limited; it irritates skin, eyes and the respiratory tract.
Freons are chlorofluorocarbons used as refrigerants, propellants, solvents and foam blowing agents. Ultraviolet radiation splits off a chlorine atom, \(\mathrm{CFCl_3} \rightarrow \mathrm{CFCl_2} + \mathrm{Cl}\), which then destroys ozone by \(\mathrm{O_3} + \mathrm{Cl} \rightarrow \mathrm{O_2} + \mathrm{ClO}\) followed by \(\mathrm{ClO} + \mathrm{O} \rightarrow \mathrm{O_2} + \mathrm{Cl}\). Because the chlorine atom is regenerated, one atom can destroy up to 100,000 ozone molecules.
DDT is a crystalline insecticide once used widely against mosquitoes and lice. It is a persistent organic pollutant: resistant to metabolism, readily absorbed through the skin, and stored in fatty tissue, so it accumulates in the ecosystem. Its use is now banned.
Flowchart
FC 10.10 – The chapter at a glance: alcohols / hydrocarbons / other halides / arenes → (preparation: HX, PX₃, PCl₅, SOCl₂; HX or X₂ addition; Finkelstein and Swartz exchange; electrophilic halogenation and Sandmeyer) → halogen derivative R–X or Ar–X → (physical properties: polar C–X bond, boiling point order RI > RBr > RCl > RF, insoluble in water) → branch A: nucleophilic substitution → S_N2 for primary halides, inversion of configuration; S_N1 for tertiary halides, racemization → alcohols, ethers, esters, amines, nitriles; branch B: elimination with alcoholic KOH → alkene, Saytzeff product major; branch C: active metals → Grignard reagent with Mg, alkane by Wurtz with Na; branch D: haloarenes → nucleophilic substitution only under forcing conditions or with o/p electron withdrawing groups, and slow electrophilic substitution giving mainly the para product; branch E: polyhalogen compounds → uses as solvents, refrigerants and insecticides, and their environmental effects.
10.9 Objective Questions
- The correct order of increasing reactivity of the C–X bond towards a nucleophile in the following compounds is: (I) a benzene ring bearing X; (II) a benzene ring bearing X with an \(\mathrm{{-}NO_2}\) group at the para position; (III) \(\mathrm{(CH_3)_3C\!-\!X}\); (IV) \(\mathrm{(CH_3)_2CH\!-\!X}\).
- I < II < III < IV
- II < I < III < IV
- III < IV < II < I
- IV < III < I < II
- The major product of the reaction \(\mathrm{CH_3\!-\!CH{=}CH_2} + \mathrm{HI} \xrightarrow{\ \text{peroxide}\ }\) is,
- \(\mathrm{I\!-\!CH_2\!-\!CH{=}CH_2}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2I}\)
- \(\mathrm{CH_3\!-\!CHI\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!CHI\!-\!CH_2OH}\)
- Which of the following is likely to undergo racemization during alkaline hydrolysis? (I) \(\mathrm{CH_3\!-\!CHCl\!-\!C_2H_5}\); (II) a benzene ring bearing \(\mathrm{{-}CH_2Cl}\); (III) a benzene ring bearing \(\mathrm{{-}Cl}\); (IV) \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2Cl}\).
- Only I
- Only II
- II and IV
- Only IV
- The best method for the preparation of alkyl fluorides is
- Finkelstein reaction
- Swartz reaction
- Free radical fluorination
- Sandmeyer’s reaction
- Identify the chiral molecule from the following.
- 1-Bromobutane
- 1,1-Dibromobutane
- 2,3-Dibromobutane
- 2-Bromobutane
- An alkyl chloride on Wurtz reaction gives 2,2,5,5-tetramethylhexane. The same alkyl chloride on reduction with zinc–copper couple in alcohol gives a hydrocarbon with molecular formula \(\mathrm{C_5H_{12}}\). What is the structure of the alkyl chloride?
- \(\mathrm{(CH_3)_3C\!-\!CH_2Cl}\)
- \(\mathrm{CH_3\!-\!CCl(CH_3)\!-\!CH_2CH_3}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!CHCl\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH(CH_3)\!-\!CH_2Cl}\)
- Butanenitrile may be prepared by heating
- propanol with KCN
- butanol with KCN
- n-butyl chloride with KCN
- n-propyl chloride with KCN
- Choose the compound from the following that will react fastest by \(\mathrm{S_N}1\) mechanism.
- 1-iodobutane
- 1-iodopropane
- 2-iodo-2-methylbutane
- 2-iodo-3-methylbutane
- For the sequence (chlorocyclohexane) \(+ \mathrm{Mg} \xrightarrow{\ \text{dry ether}\ } \mathrm{A} \xrightarrow{\ \mathrm{H_2O}\ } \mathrm{B}\), the product ‘B’ in the above reaction sequence is,
- cyclohexyl–Mg
- cyclohexyl–Mg–Cl
- Cl–cyclohexyl–Mg
- cyclohexane
- Which of the following is used as a source of dichlorocarbene?
- tetrachloromethane
- chloroform
- iodoform
- DDT
10.10 Check Your Concepts
Classify each of the following, stating the hybridization of the carbon that carries the halogen: (i) \(\mathrm{CH_3\!-\!CH{=}CH\!-\!CH_2\!-\!Br}\), (ii) \(\mathrm{C_6H_5\!-\!CH(CH_3)\!-\!Cl}\), (iii) \(\mathrm{CH{\equiv}C\!-\!Br}\), (iv) a benzene ring bearing \(\mathrm{{-}I}\) at one position and \(\mathrm{{-}CH_3}\) at the next.
Assertion–Reason. Assertion (A): Vinyl chloride does not react with aqueous \(\mathrm{KOH}\) under ordinary conditions. Reason (R): In vinyl chloride the chlorine is attached to an \(\mathrm{sp^2}\) carbon and one of its lone pairs is in conjugation with the double bond. Choose the correct option: (a) both A and R are true and R is the correct explanation of A; (b) both A and R are true but R is not the correct explanation of A; (c) A is true, R is false; (d) A is false, R is true.
The \(\mathrm{C\!-\!I}\) bond enthalpy is much lower than the \(\mathrm{C\!-\!F}\) bond enthalpy. Explain the structural reason for this difference, and state one consequence it has for the reactivity of alkyl halides in nucleophilic substitution.
Arrange 1-chlorobutane, 2-chlorobutane and 2-chloro-2-methylpropane in increasing order of reactivity towards the \(\mathrm{S_N}1\) mechanism, and give the reason for the order you choose.
A student attempts to prepare a Grignard reagent by adding magnesium turnings to bromoethane dissolved in ethanol. Explain what will happen, write the equation for the reaction that actually takes place, and name the organic product.
Predict the major product formed when 2-bromo-2-methylbutane is boiled with alcoholic \(\mathrm{KOH}\). Name the rule you have used and state why the major product is the more stable alkene.
Complete the reaction and name both products, indicating which is the major one: chlorobenzene \(+\ \mathrm{Cl_2} \xrightarrow{\ \text{anhydrous}\ \mathrm{FeCl_3}\ }\). Explain why one of the two predominates.
Distinguish between a geminal dihalide and a vicinal dihalide, giving one example of each, and state which of the two is obtained when an alkene reacts with bromine in \(\mathrm{CCl_4}\) at room temperature.
The boiling points of the three isomeric dichlorobenzenes lie within a few kelvin of one another, but their melting points differ by nearly seventy kelvin. Explain this observation.
Match the reagent in Column I with the product it gives from an alkyl halide \(\mathrm{R\!-\!X}\) in Column II.
| Column I | Column II |
|---|---|
| a. \(\mathrm{NaOR'}\) | i. nitroalkane |
| b. alcoholic \(\mathrm{KCN}\) | ii. primary amine |
| c. alcoholic \(\mathrm{AgCN}\) | iii. ether |
| d. \(\mathrm{AgNO_2}\) | iv. alkyl cyanide |
| e. \(\mathrm{NH_3}\) (excess, under pressure) | v. isocyanide |
An optically active alkyl halide is hydrolysed and the product is found to have the configuration opposite to that of the starting material. Which mechanism has operated? Explain how the mechanism accounts for the change in configuration, and name the effect.
Chlorobenzene needs about 623 K and 300 atm to be converted into phenol, while 2,4,6-trinitrochlorobenzene gives the corresponding phenol with merely warm water. Explain the difference.
What happens when a mixture of methyl bromide and ethyl bromide is treated with sodium in dry ether? Write the products, and explain why this route is rarely used for preparing a particular alkane.
A chemist needs a sample of pure 1-chlorobutane from butan-1-ol. Explain why thionyl chloride would be preferred to phosphorus pentachloride for this preparation, and write the equation for the reaction chosen.
Assertion–Reason. Assertion (A): Chlorobenzene directs an incoming electrophile mainly to the para position. Reason (R): Chlorine activates the benzene ring towards electrophilic substitution. Choose the correct option: (a) both A and R are true and R is the correct explanation of A; (b) both A and R are true but R is not the correct explanation of A; (c) A is true, R is false; (d) A is false, R is true.
Draw the transition state formed when methyl bromide reacts with hydroxide ion. Mark the partial bonds and the partial charges, state the number of \(\sigma\) bonds lying in one plane and the angle between them, and explain why the nucleophile must approach from the side opposite the bromine.
Explain why an \(\mathrm{S_N}1\) reaction carried out on an optically active substrate gives a product that is very nearly optically inactive, while an \(\mathrm{S_N}2\) reaction on the same substrate does not.
Identify A and B, and name the type of reaction in each step: propan-2-ol \(\xrightarrow{\ \text{conc. HCl, anhydrous}\ \mathrm{ZnCl_2}\ } \mathrm{A} \xrightarrow{\ \text{alcoholic KOH},\ \Delta\ } \mathrm{B}\).
Which will react faster with sodium iodide in acetone — 1-chlorobutane or 2-chloro-2-methylpropane? Name the reaction, state the mechanism involved, and justify your choice.
Arrange chlorobenzene, m-nitrochlorobenzene, p-nitrochlorobenzene and 2,4-dinitrochlorobenzene in increasing order of reactivity towards nucleophilic substitution, and explain why the position of the nitro group matters as much as the number of such groups.
A compound with molecular formula \(\mathrm{C_4H_9Br}\) is optically active. Draw its structure, mark the chiral carbon with an asterisk and give its IUPAC name. Explain why the isomer 1-bromobutane is not optically active.
Bromination of benzene with \(\mathrm{Br_2}\) and \(\mathrm{FeBr_3}\) is carried out in the dark. Explain what would happen if the reaction were exposed to bright light instead, and name the type of reaction that would then compete.
Write the two equations by which a chlorine atom released from a CFC destroys ozone in the stratosphere. Explain why a single chlorine atom is able to destroy a very large number of ozone molecules, and identify the role the chlorine atom is playing.
Give reasons for each of the following: (i) DDT accumulates in the fatty tissue of fish rather than being excreted; (ii) chloroform is stored in dark coloured, air-tight bottles; (iii) carbon tetrachloride, though useful as a solvent, is no longer used as a cleaning agent in schools and homes.
What happens if an alkyl halide with at least one \(\beta\)-hydrogen is treated with (i) aqueous \(\mathrm{KOH}\) and (ii) alcoholic \(\mathrm{KOH}\)? Name the two competing reactions and list three factors that decide which one predominates.
Write the IUPAC name of the following compounds.
- \(\mathrm{CH_3\!-\!CH{=}C(CH_3)\!-\!CH(CH_3)\!-\!Br}\)
- \(\mathrm{CH_3\!-\!CHCl\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}\)
- a cyclohexane ring bearing \(\mathrm{{-}Cl}\) and \(\mathrm{{-}C_2H_5}\)
- a benzene ring bearing two \(\mathrm{{-}Cl}\) groups and one \(\mathrm{{-}CH_3}\) group
- Write the structure and IUPAC name of the major product in each of the following reactions.
- \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2Cl} + \mathrm{NaI} \xrightarrow{\ \text{acetone}\ }\)
- \(\mathrm{CH_3\!-\!CH_2Br} + \mathrm{SbF_3} \rightarrow\)
- \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH{=}CH_2} + \mathrm{HBr} \xrightarrow{\ \text{peroxide}\ }\)
- cyclohexanol \(+\ \mathrm{SOCl_2} \rightarrow\)
- toluene \(+\ \mathrm{Cl_2} \xrightarrow{\ \mathrm{Fe},\ \text{dark}\ }\)
- Identify the chiral molecule/s from the following.
- \(\mathrm{CH_3\!-\!CH(OH)\!-\!CH_2\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!CHBr\!-\!CH_2\!-\!CH_3}\)
- \(\mathrm{CH_3\!-\!CH_2\!-\!CH_2\!-\!CH_2Br}\)
- \(\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}\)
- Which one compound from the following pairs would undergo \(\mathrm{S_N}2\) faster?
- a cyclohexane ring bearing \(\mathrm{{-}CH_2Cl}\), and a cyclohexane ring bearing \(\mathrm{{-}Cl}\)
- \(\mathrm{CH_3CH_2CH_2I}\) and \(\mathrm{CH_3CH_2CH_2Cl}\)
- Complete the following reactions giving the major product.
- \(\mathrm{CH_3\!-\!CH{=}CH_2} \xrightarrow{\ \mathrm{HBr},\ \text{peroxide}\ } \mathrm{A} \xrightarrow{\ \text{alc. KOH}\ } \mathrm{B}\)
- \(\mathrm{CH_3\!-\!CH(OH)\!-\!CH_3} \xrightarrow{\ \text{Red P}/\mathrm{Br_2}\ } \mathrm{A} \xrightarrow{\ \mathrm{Ag_2O}/\mathrm{H_2O}\ } \mathrm{B}\)
- \(\mathrm{(CH_3)_3C\!-\!CH_2\!-\!Cl} \xrightarrow{\ \mathrm{Na}/\text{dry ether}\ } \mathrm{A}\)
- (chlorobenzene) \(\xrightarrow{\ \mathrm{Mg}/\text{dry ether}\ } \mathrm{A}\)
- Name the reagent used to bring about the following conversions.
- Bromoethane to ethoxyethane
- 1-Chloropropane to 1-nitropropane
- Ethyl bromide to ethyl isocyanide
- Chlorobenzene to biphenyl
- Arrange the following in the increasing order of boiling points.
- 1-Bromopropane
- 2-Bromopropane
- 1-Bromobutane
- 1-Bromo-2-methylpropane
- Match the pairs.
| Column I | Column II |
|---|---|
| a. \(\mathrm{CH_3\!-\!CHX\!-\!CH_3}\) | i. vinyl halide |
| b. \(\mathrm{CH_2{=}CH\!-\!CH_2X}\) | ii. alkyl halide |
| c. \(\mathrm{CH_2{=}CH\!-\!X}\) | iii. allyl halide |
| iv. benzyl halide | |
| v. aryl halide |
- Give reasons.
- Haloarenes are less reactive than haloalkanes.
- Alkyl halides, though polar, are immiscible with water.
- Reactions involving Grignard reagent must be carried out under anhydrous conditions.
- Alkyl halides are generally not prepared by free radical halogenation of alkanes.
Distinguish between \(\mathrm{S_N}1\) and \(\mathrm{S_N}2\) mechanisms of substitution reaction.
Explain optical isomerism in 2-chlorobutane.
Convert the following.
- Propene to propan-1-ol
- Benzyl alcohol to benzyl cyanide
- Ethanol to propane nitrile
- But-1-ene to n-butyl iodide
- 2-Chloropropane to propan-1-ol
- tert-Butyl bromide to isobutyl bromide
- p-Nitrochlorobenzene to p-nitrophenol
- Propene to 1-nitropropane
HCl is added to a hydrocarbon ‘A’ \((\mathrm{C_4H_8})\) to give a compound ‘B’, which on hydrolysis with aqueous alkali forms the tertiary alcohol ‘C’ \((\mathrm{C_4H_{10}O})\). Identify ‘A’, ‘B’ and ‘C’.
Complete the following reaction sequences by writing the structural formulae of the organic compounds ‘A’, ‘B’ and ‘C’.
- 2-Bromobutane \(\xrightarrow{\ \text{alc. KOH}\ } \mathrm{A} \xrightarrow{\ \mathrm{Br_2}\ } \mathrm{B} \xrightarrow{\ \mathrm{NaNH_2}\ } \mathrm{C}\)
- Isopropyl alcohol \(\xrightarrow{\ \mathrm{PBr_3},\ \Delta\ } \mathrm{A} \xrightarrow{\ \mathrm{NH_3}\ \text{excess}\ } \mathrm{B}\)
- Observe the following and answer the questions given below. \[\mathrm{CH_2{=}CH\!-\!\ddot{X}\!:} \longleftrightarrow {}^{\ominus}\mathrm{CH_2\!-\!CH{=}\overset{\oplus}{X}\!:}\]
- Name the type of halogen derivative.
- Comment on the bond length of the C–X bond in it.
- Can it react by \(\mathrm{S_N}1\) mechanism? Justify your answer.