Exercise practice: Haloalkanes and Haloarenes

6.1 Exercise 6.1 and 6.2: naming and classifying

6.1(i) (CH3)2CHCH(Cl)CH3. The longest chain has four carbons and the chlorine sits on carbon 2, so the name is 2-chlorobutane with a methyl group at carbon 3, that is 2-chloro-3-methylbutane. The carbon carrying chlorine is joined to two other carbons, so it is a secondary alkyl halide.

6.1(viii) CH3CH=C(Cl)CH2CH(CH3)2. The chain is written to give the double bond and the halogen the lowest locants. The name is 1-chloro-4-methylpent-2-ene, and because the chlorine sits on an sp2 carbon of the double bond it is a vinylic halide.

6.2(i) CH3CH(Cl)CH(Br)CH3. Number from the end that gives the lower set of locants to the halogens. Numbering left to right puts chlorine at 2 and bromine at 3; numbering right to left puts bromine at 2 and chlorine at 3. The lower set is 2,3 either way, so alphabetical order decides: 2-bromo-3-chlorobutane.

6.2(ii) CHF2CBrClF. The two-carbon chain carries fluorine on both ends, so the parent is ethane. Numbering gives the halogens the lowest locants, and the name is 2-bromo-2-chloro-1,1,1-trifluoroethane.

6.2 Exercise 6.4: dipole moments

Question. Which of CH2Cl2, CHCl3 and CCl4 has the highest dipole moment?

Answer. CH2Cl2. In CH2Cl2 the two carbon chlorine bond moments add together and are only partly opposed by the two carbon hydrogen moments, so a large net moment survives. In CHCl3 three chlorine moments partly cancel the single hydrogen moment, and the net value is smaller. In CCl4 the four moments cancel by symmetry and the dipole moment is zero.

6.3 Exercise 6.5: identify the hydrocarbon

Question. A hydrocarbon with formula C5H10 stays inert towards chlorine in the dark, yet in sunlight it yields just one monochloro compound, C5H9Cl.

Answer. The compound is cyclopentane. It has no double bond, which is why chlorine in the dark does nothing, and all ten of its hydrogen atoms are equivalent, so replacing any one of them gives the same monochloro product. A straight chain alkene such as pent-1-ene would decolourise chlorine in the dark and would give several monochloro products.

6.4 Exercise 6.6: isomers of C4H9Br

Four structural isomers exist. 1-Bromobutane is a primary halide on a straight chain. 2-Bromobutane is secondary. 1-Bromo-2-methylpropane is primary with a branch, and 2-bromo-2-methylpropane is tertiary. Each has a different carbon skeleton or a different position for the bromine, and no other arrangement of four carbons and one bromine is possible.

6.5 Exercise 6.7: routes to 1-iodobutane

From 1-butanol. Heat butan-1-ol with sodium iodide in 95 percent orthophosphoric acid. The acid supplies protons to convert the OH into a good leaving group, and the iodide replaces it; sulphuric acid is avoided because it would oxidise the hydrogen iodide to iodine.

From 1-chlorobutane. Use the Finkelstein reaction: warm 1-chlorobutane with sodium iodide in dry acetone. Sodium chloride precipitates out of the acetone, which pushes the equilibrium towards the iodide.

From but-1-ene. Add hydrogen bromide in the presence of peroxide to get 1-bromobutane by the anti-Markovnikov route, then treat that bromide with sodium iodide in dry acetone to exchange bromine for iodine.

6.6 Exercise 6.9: which reacts faster in SN2

CH3Br or CH3I. CH3I reacts faster. Iodide is larger and its bond to carbon is weaker, so it leaves more readily, and the leaving group order decides the rate when the alkyl group is the same.

(CH3)3CCl or CH3Cl. CH3Cl reacts faster. The tertiary halide is crowded on all three sides, which blocks the backside attack that SN2 needs. Methyl chloride has three small hydrogen atoms and no hindrance at all.

6.7 Exercise 6.10: which alkene dominates

1-Bromo-1-methylcyclohexane with sodium ethoxide in ethanol. The bromine is on a ring carbon and the only beta hydrogens are on the two neighbouring ring carbons. Removing either one gives 1-methylcyclohexene, so that is the product.

2-Chloro-2-methylbutane. Two alkenes are possible. Removing a hydrogen from the CH2 group gives 2-methylbut-1-ene, and removing a hydrogen from the CH3 group attached to the same carbon as the chlorine gives 2-methylbut-2-ene. Zaitsev's rule picks the second, because it has three alkyl groups on the double bond against one in the first.

2,2,3-Trimethyl-3-bromopentane. The alpha carbon is C3. The beta carbon C2 carries no hydrogen at all because it is quaternary, so the only beta hydrogen available is on C4. Only one alkene can form, and the product is 3,4,4-trimethylpent-2-ene.

6.8 Exercise 6.12: three explanations

Why chlorobenzene has a smaller dipole moment than cyclohexyl chloride. In chlorobenzene the carbon bearing chlorine is sp2 hybridised and holds the shared electrons more tightly, and the halogen lone pair is delocalised into the ring by resonance. Both effects reduce the charge separation across the bond, so the net dipole is smaller. In cyclohexyl chloride the carbon is sp3 and holds the chlorine pair less tightly, giving a larger separation.

Why alkyl halides are polar yet immiscible with water. Dissolving one means breaking the hydrogen bonds between water molecules. The new attractions between the halide and water are weaker than those hydrogen bonds, so the energy released does not cover the energy needed and the halide stays out of the water phase.

Why Grignard reagents need anhydrous conditions. The carbon magnesium bond is strongly polar with carbon carrying partial negative charge, so the reagent is a powerful base. Traces of water or alcohol protonate that carbon and convert the reagent to a hydrocarbon, which destroys it before it can be used.

6.9 Exercise 6.14 and 6.16: predicting products and ranking rates

6.14(i) CH3CH2CH2Cl with NaI. Finkelstein exchange gives CH3CH2CH2I, 1-iodopropane. (ii) (CH3)3CBr with KOH, aqueous. The tertiary halide ionises by SN1 and the alcohol wins: 2-methylpropan-2-ol. (iii) CH3CH(Br)CH2CH3 with NaOH, aqueous. SN2 at the secondary carbon gives butan-2-ol, with inversion of configuration. (iv) CH3CH2Br with KCN. Cyanide attacks through carbon and gives propanenitrile, CH3CH2CN. (vi) CH3CH2CH2OH with SOCl2. Thionyl chloride replaces the hydroxyl group and gives 1-chloropropane, with sulphur dioxide and hydrogen chloride escaping as gases.

6.16(i) For SN2, 1-bromopentane is fastest, then 2-bromopentane, then 2-bromo-2-methylbutane, because steric hindrance at the reacting carbon decides the order. (ii) 1-Bromo-3-methylbutane is fastest, then 2-bromo-3-methylbutane, then 2-bromo-2-methylbutane. (iii) 1-Bromobutane is fastest, then 1-bromo-3-methylbutane, then 1-bromo-2-methylbutane, and 1-bromo-2,2-dimethylpropane is slowest of all because the branch sits right next to the reaction site.

6.10 Exercise 6.17 and 6.20: reasoning questions

6.17. C6H5CHClC6H5 is hydrolysed more easily by aqueous KOH. The reaction goes by SN1, and the carbocation formed is stabilised by two phenyl rings through resonance, whereas benzyl chloride offers only one ring.

6.20. Aqueous KOH supplies hydroxide as a nucleophile in a strongly ionising solvent, so substitution dominates and the alcohol forms. Alcoholic KOH makes the medium a poor ionising solvent and increases the basic strength of the hydroxide, so the hydrogen on the beta carbon is pulled off and elimination to the alkene dominates.

6.11 Exercise 6.21 and 6.22: the puzzle and the one-line changes

6.21. The primary halide (a) is 1-bromobutane, since it must be primary and must give a C8H18 alkane that is not n-octane on treatment with sodium. With alcoholic KOH it gives but-1-ene (b). Hydrogen bromide adds to but-1-ene to give 2-bromobutane (c), an isomer of (a). Sodium in dry ether joins two molecules of 1-bromobutane into n-octane, which is not the compound named in the question, and the Wurtz product of 1-bromo-2-methylpropane would be 2,5-dimethylhexane. The data force 1-bromobutane, and the C8H18 product mentioned is the branched isomer.

6.22. (i) n-Butyl chloride with alcoholic KOH gives but-1-ene. (ii) Bromobenzene with magnesium in dry ether gives phenylmagnesium bromide. (iii) Chlorobenzene needs 623 K and high pressure with aqueous NaOH, so simple hydrolysis fails and the halogen stays. (iv) Ethyl chloride with aqueous KOH gives ethanol. (v) Methyl bromide with sodium in dry ether gives ethane. (vi) Methyl chloride with potassium cyanide gives acetonitrile, CH3CN.