Chapter 6: Haloalkanes and Haloarenes

The longest chain containing the $$\mathrm{C=C}$$ double bond has five carbons, so the parent hydrocarbon is pentene:

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}H={\overset{3}{C}}H{-}\overset{4}{C}HBr{-}\overset{5}{C}H_3}$$

The double bond is a suffix feature and must receive the lowest possible locant. Numbering from the left places the double bond at C-2 and $$\mathrm{Br}$$ at C-4; numbering from the right would place the double bond at C-3. So we number from the left, and the bromo group is cited as a prefix. The constitutional IUPAC name is 4-bromopent-2-ene.

Stereochemistry. Since the spatial arrangement is shown in the structure, the stereochemical descriptors are added:

• Double bond, C-2=C-3: C-2 carries $$\mathrm{CH_3}$$ and $$\mathrm{H}$$, and C-3 carries $$\mathrm{H}$$ and the $$\mathrm{{-}CHBrCH_3}$$ group, so each doubly-bonded carbon bears two different groups and the alkene shows geometrical (E/Z) isomerism. The higher-priority groups — $$\mathrm{CH_3}$$ on C-2 and $$\mathrm{{-}CHBrCH_3}$$ on C-3 — lie on opposite sides of the double bond as drawn, so the configuration is E.
• Chiral centre, C-4: C-4 is bonded to four different groups — $$\mathrm{Br}$$, $$\mathrm{H}$$, $$\mathrm{CH_3}$$ and $$\mathrm{{-}CH=CHCH_3}$$ — so it is a stereocentre and the molecule is chiral, existing as a pair of (R) and (S) enantiomers.

Including the descriptors, the full IUPAC name is (E)-4-bromopent-2-ene, with C-4 designated (R) or (S) according to its actual three-dimensional arrangement.

The longest chain containing the double bond runs through four carbons, so the parent is butene:

$$\mathrm{\overset{1}{C}H_2={\overset{2}{C}}(CH_3){-}\overset{3}{C}HBr{-}\overset{4}{C}H_3}$$

Numbering from the terminal $$\mathrm{CH_2}$$ end gives the double bond the lowest locant (C-1); a methyl group sits on C-2 and a bromo group on C-3. The constitutional IUPAC name is 3-bromo-2-methylbut-1-ene.

Stereochemistry.

• Double bond, C-1=C-2: C-1 is a terminal $$\mathrm{=CH_2}$$ carbon bearing two identical $$\mathrm{H}$$ atoms, so the double bond is not stereogenic — no E/Z isomerism is possible here.
• Chiral centre, C-3: C-3 is bonded to four different groups — $$\mathrm{Br}$$, $$\mathrm{H}$$, $$\mathrm{CH_3}$$ and the $$\mathrm{{-}C(CH_3)=CH_2}$$ group — so it is a stereocentre and the molecule is chiral, existing as a pair of (R) and (S) enantiomers.

Hence the IUPAC name is 3-bromo-2-methylbut-1-ene, with C-3 designated (R) or (S) according to its actual three-dimensional arrangement.

The longest chain containing the double bond has five carbons, so the parent is pentene:

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}H={\overset{3}{C}}(CH_3){-}\overset{4}{C}HBr{-}\overset{5}{C}H_3}$$

Numbering from the left places the double bond at C-2 (the lowest possible locant), a methyl substituent on C-3 and a bromo substituent on C-4. The constitutional IUPAC name is 4-bromo-3-methylpent-2-ene.

Stereochemistry.

• Double bond, C-2=C-3: C-2 carries $$\mathrm{CH_3}$$ and $$\mathrm{H}$$; C-3 carries a $$\mathrm{CH_3}$$ group and the $$\mathrm{{-}CHBrCH_3}$$ group. Each doubly-bonded carbon bears two different groups, so the alkene shows geometrical (E/Z) isomerism. The higher-priority groups are $$\mathrm{CH_3}$$ on C-2 and, on C-3, $$\mathrm{{-}CHBrCH_3}$$ (its first carbon carries $$\mathrm{Br}$$, which outranks the plain $$\mathrm{CH_3}$$). These lie on opposite sides of the double bond as drawn, so the configuration is E.
• Chiral centre, C-4: C-4 is bonded to four different groups — $$\mathrm{Br}$$, $$\mathrm{H}$$, $$\mathrm{CH_3}$$ and the $$\mathrm{{-}C(CH_3)=CHCH_3}$$ group — so it is a stereocentre and the molecule is chiral.

Including the descriptors, the full IUPAC name is (E)-4-bromo-3-methylpent-2-ene, with C-4 designated (R) or (S) according to its actual three-dimensional arrangement.

The longest chain containing the double bond has four carbons, so the parent is butene:

$$\mathrm{\overset{1}{C}H_2Br{-}{\overset{2}{C}}(CH_3)=\overset{3}{C}H{-}\overset{4}{C}H_3}$$

Numbering from the $$\mathrm{CH_2Br}$$ end places the double bond at C-2 and gives the bromo group the lowest locant, C-1; a methyl group is on C-2. The constitutional IUPAC name is 1-bromo-2-methylbut-2-ene.

Stereochemistry.

• Double bond, C-2=C-3: C-2 carries a $$\mathrm{{-}CH_2Br}$$ group and a $$\mathrm{CH_3}$$ group; C-3 carries $$\mathrm{CH_3}$$ and $$\mathrm{H}$$. Each doubly-bonded carbon bears two different groups, so the alkene shows geometrical (E/Z) isomerism. The higher-priority groups are $$\mathrm{{-}CH_2Br}$$ on C-2 (its carbon carries $$\mathrm{Br}$$, outranking $$\mathrm{CH_3}$$) and $$\mathrm{CH_3}$$ on C-3. These lie on opposite sides of the double bond as drawn, so the configuration is E.
• Chiral centre: no carbon carries four different groups (C-1 is a $$\mathrm{CH_2Br}$$ group with two $$\mathrm{H}$$ atoms), so the molecule has no stereocentre and is achiral.

Including the descriptor, the full IUPAC name is (E)-1-bromo-2-methylbut-2-ene.

The chain has four carbons with a double bond, so the parent is butene:

$$\mathrm{\overset{1}{C}H_2Br{-}\overset{2}{C}H=\overset{3}{C}H{-}\overset{4}{C}H_3}$$

Numbering from the $$\mathrm{CH_2Br}$$ end gives the bromo group the lowest locant (C-1) and the double bond at C-2. The constitutional IUPAC name is 1-bromobut-2-ene.

Stereochemistry.

• Double bond, C-2=C-3: C-2 carries a $$\mathrm{{-}CH_2Br}$$ group and $$\mathrm{H}$$; C-3 carries $$\mathrm{CH_3}$$ and $$\mathrm{H}$$. Each doubly-bonded carbon bears two different groups, so the alkene shows geometrical (E/Z) isomerism. The higher-priority groups are $$\mathrm{{-}CH_2Br}$$ on C-2 and $$\mathrm{CH_3}$$ on C-3; they lie on opposite sides of the double bond as drawn, so the configuration is E.
• Chiral centre: no carbon carries four different groups, so the molecule has no stereocentre and is achiral.

Including the descriptor, the full IUPAC name is (E)-1-bromobut-2-ene.

The longest chain containing the double bond has three carbons, so the parent is propene:

$$\mathrm{\overset{1}{C}H_2={\overset{2}{C}}(CH_3){-}\overset{3}{C}H_2Br}$$

Numbering from the $$\mathrm{CH_2}$$ end places the double bond at C-1. A methyl group sits on C-2 and a bromo group on C-3.

Hence the IUPAC name is 3-bromo-2-methylprop-1-ene (also written 3-bromo-2-methylpropene).

This is an electrophilic addition of $$\mathrm{HBr}$$ to an alkene, which follows Markovnikov's rule: $$\mathrm{H}$$ adds to the doubly-bonded carbon carrying more hydrogens and $$\mathrm{Br}$$ to the carbon carrying fewer.

Protonation can give two carbocations. Addition of $$\mathrm{H^+}$$ to the terminal $$\mathrm{=CH_2}$$ carbon gives the benzylic secondary cation $$\mathrm{C_6H_5\overset{+}{C}H{-}CH_3}$$, which is strongly stabilised by resonance with the ring. The alternative cation $$\mathrm{C_6H_5CH_2{-}\overset{+}{C}H_2}$$ is a primary, non-resonance-stabilised cation.

The more stable benzylic cation forms preferentially, and $$\mathrm{Br^-}$$ then attacks it:

$$\mathrm{C_6H_5CH=CH_2 + HBr \longrightarrow C_6H_5CHBrCH_3}$$

The product is (1-bromoethyl)benzene, $$\mathrm{C_6H_5CHBrCH_3}$$.

$$\mathrm{HCl}$$ adds to but-1-ene by electrophilic addition following Markovnikov's rule.

Protonation of the terminal $$\mathrm{=CH_2}$$ carbon gives the secondary carbocation $$\mathrm{CH_3CH_2\overset{+}{C}HCH_3}$$, which is more stable than the primary cation $$\mathrm{CH_3CH_2CH_2\overset{+}{C}H_2}$$ that would result from protonation at the internal carbon.

Thus $$\mathrm{H}$$ adds to the terminal carbon and $$\mathrm{Cl}$$ to the internal (more substituted) carbon:

$$\mathrm{CH_3CH_2CH=CH_2 + HCl \longrightarrow CH_3CH_2CHClCH_3}$$

The major product is 2-chlorobutane.

In the presence of a peroxide, $$\mathrm{HBr}$$ adds to a terminal alkene by a free-radical chain mechanism. This gives anti-Markovnikov (peroxide / Kharasch) addition — the opposite orientation to ionic addition.

A bromine atom adds first; it attaches to the terminal $$\mathrm{=CH_2}$$ carbon so as to generate the more stable secondary carbon radical $$\mathrm{C_6H_5CH_2\overset{\bullet}{C}HCH_2Br}$$. This radical then abstracts $$\mathrm{H}$$ from $$\mathrm{HBr}$$.

Net result: $$\mathrm{Br}$$ ends up on the terminal carbon and $$\mathrm{H}$$ on the internal carbon:

$$\mathrm{C_6H_5CH_2CH=CH_2 + HBr \xrightarrow{peroxide} C_6H_5CH_2CH_2CH_2Br}$$

The product is 1-bromo-3-phenylpropane (only $$\mathrm{HBr}$$ shows the peroxide effect; $$\mathrm{HCl}$$ and $$\mathrm{HI}$$ do not).

An $$\mathrm{S_N2}$$ reaction needs the nucleophile to approach the carbon bearing the halogen from the side opposite the leaving group. Bulky substituents around that carbon hinder this approach, so the rate falls in the order primary > secondary > tertiary.

In (chloromethyl)cyclohexane, $$\mathrm{C_6H_{11}CH_2Cl}$$, the $$\mathrm{Cl}$$ sits on a $$\mathrm{CH_2}$$ group — a primary carbon. In chlorocyclohexane the $$\mathrm{Cl}$$ is on a ring carbon attached to two other ring carbons — a secondary carbon, which is more crowded.

Hence the primary halide, (chloromethyl)cyclohexane, undergoes the $$\mathrm{S_N2}$$ reaction faster.

The two halides have the same alkyl group, so steric factors are identical; the difference is the leaving group.

A good leaving group is one that departs readily as a stable anion. Iodide ion is large, so the $$\mathrm{C{-}I}$$ bond is long and weak (low bond enthalpy) and the charge on $$\mathrm{I^-}$$ is well dispersed. The $$\mathrm{C{-}Cl}$$ bond is shorter and stronger. Hence $$\mathrm{I^-}$$ is a much better leaving group than $$\mathrm{Cl^-}$$.

Therefore the primary iodide $$\mathrm{R{-}I}$$ undergoes the $$\mathrm{S_N2}$$ reaction faster than the primary chloride $$\mathrm{R{-}Cl}$$.

The four isomeric bromobutanes are:

• $$\mathrm{CH_3CH_2CH_2CH_2Br}$$ — 1-bromobutane (primary, 1°)
• $$\mathrm{(CH_3)_2CHCH_2Br}$$ — 1-bromo-2-methylpropane (primary, 1°)
• $$\mathrm{CH_3CH_2CHBrCH_3}$$ — 2-bromobutane (secondary, 2°)
• $$\mathrm{(CH_3)_3CBr}$$ — 2-bromo-2-methylpropane (tertiary, 3°)

$$\mathrm{S_N1}$$ reactivity. The slow, rate-determining step is ionisation of the C–Br bond to give a carbocation, so the reaction is faster the more stable that carbocation is. Carbocation stability falls in the order tertiary > secondary > primary. Both primary halides give primary cations; the one from $$\mathrm{(CH_3)_2CHCH_2Br}$$ is slightly better stabilised (more alkyl groups close to the cationic carbon) than the one from $$\mathrm{CH_3CH_2CH_2CH_2Br}$$. Hence the order of $$\mathrm{S_N1}$$ reactivity, fastest first, is:

$$\mathrm{(CH_3)_3CBr > CH_3CH_2CHBrCH_3 > (CH_3)_2CHCH_2Br > CH_3CH_2CH_2CH_2Br}$$

that is, 3° > 2° > 1°.

$$\mathrm{S_N2}$$ reactivity. The nucleophile attacks the carbon bearing $$\mathrm{Br}$$ from the side opposite the leaving group, so the rate falls as steric crowding around that carbon increases. The order is exactly the reverse, fastest first:

$$\mathrm{CH_3CH_2CH_2CH_2Br > (CH_3)_2CHCH_2Br > CH_3CH_2CHBrCH_3 > (CH_3)_3CBr}$$

that is, 1° > 2° > 3°.

All four are benzylic bromides; the carbocations they form are stabilised by resonance with the phenyl ring(s).

$$\mathrm{S_N1}$$: The more phenyl groups attached to the cationic carbon, the greater the resonance stabilisation, hence the faster the $$\mathrm{S_N1}$$ reaction. The cation from $$\mathrm{C_6H_5C(CH_3)(C_6H_5)Br}$$ has two phenyl groups; that from $$\mathrm{C_6H_5CH(C_6H_5)Br}$$ also has two phenyls but one less alkyl; $$\mathrm{C_6H_5CH(CH_3)Br}$$ and $$\mathrm{C_6H_5CH_2Br}$$ have one phenyl. The order is

$$\mathrm{C_6H_5C(CH_3)(C_6H_5)Br > C_6H_5CH(C_6H_5)Br > C_6H_5CH(CH_3)Br > C_6H_5CH_2Br}$$

$$\mathrm{S_N2}$$: Reactivity is governed by steric crowding — more and bulkier groups slow the reaction. A phenyl group is bulkier than a methyl group. The order is the reverse:

$$\mathrm{C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br}$$

A molecule is chiral (non-superimposable on its mirror image) if it contains a carbon attached to four different groups; if any two groups are identical, the molecule is achiral.

First molecule ($$\mathrm{CH_3CHBrOH}$$): the central carbon carries four different groups — $$\mathrm{CH_3}$$, $$\mathrm{H}$$, $$\mathrm{Br}$$ and $$\mathrm{OH}$$. It is an asymmetric carbon, so this molecule is chiral.

Second molecule ($$\mathrm{CH_3CHBr_2}$$): the central carbon carries two identical $$\mathrm{Br}$$ atoms ($$\mathrm{CH_3}$$, $$\mathrm{H}$$, $$\mathrm{Br}$$, $$\mathrm{Br}$$). With two like groups it is not an asymmetric carbon, so this molecule is achiral.

First molecule (propan-2-ol, $$\mathrm{(CH_3)_2CHOH}$$): the carbinol carbon is bonded to $$\mathrm{OH}$$, $$\mathrm{H}$$ and two identical $$\mathrm{CH_3}$$ groups. Since two of the four groups are the same, it is not an asymmetric carbon — the molecule is achiral (its mirror image is superimposable on itself).

Second molecule (butan-2-ol, $$\mathrm{CH_3CH(OH)CH_2CH_3}$$): the carbinol carbon (C-2) is bonded to four different groups — $$\mathrm{OH}$$, $$\mathrm{H}$$, $$\mathrm{CH_3}$$ and $$\mathrm{C_2H_5}$$. This is an asymmetric carbon, so the molecule is chiral.

2-Bromobutane ($$\mathrm{CH_3CHBrCH_2CH_3}$$): the C-2 carbon bearing $$\mathrm{Br}$$ is attached to four different groups — $$\mathrm{CH_3}$$, $$\mathrm{H}$$, $$\mathrm{Br}$$ and $$\mathrm{C_2H_5}$$. It is an asymmetric carbon, so the molecule is chiral.

1-Bromobutane ($$\mathrm{CH_3CH_2CH_2CH_2Br}$$): the C-1 carbon bearing $$\mathrm{Br}$$ is a $$\mathrm{CH_2}$$ group — it carries two identical $$\mathrm{H}$$ atoms. No carbon in the molecule has four different groups, so the molecule is achiral.

The parent chain is pentane (5 carbons). Number the chain and place the substituents: a chloro group at C-2 and a methyl group at C-3.

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}HCl{-}\overset{3}{C}H(CH_3){-}\overset{4}{C}H_2{-}\overset{5}{C}H_3}$$

Hence the structure is $$\mathrm{CH_3CHClCH(CH_3)CH_2CH_3}$$.

The parent ring is cyclohexane. A chloro group is at ring carbon C-1 and an ethyl group ($$\mathrm{{-}C_2H_5}$$) at C-4, i.e. the two substituents are on opposite (1,4) carbons of the ring.

Draw a hexagonal ring; attach $$\mathrm{Cl}$$ to one carbon and $$\mathrm{CH_2CH_3}$$ to the carbon directly across the ring.

The parent chain is heptane (7 carbons). Place an iodo group at C-3 and a tert-butyl group $$\mathrm{{-}C(CH_3)_3}$$ at C-4.

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}H_2{-}\overset{3}{C}HI{-}\overset{4}{C}H[C(CH_3)_3]{-}\overset{5}{C}H_2{-}\overset{6}{C}H_2{-}\overset{7}{C}H_3}$$

Hence the structure is $$\mathrm{CH_3CH_2CHICH[C(CH_3)_3]CH_2CH_2CH_3}$$.

The parent is but-2-ene — a four-carbon chain with the double bond between C-2 and C-3. Bromo groups are placed at C-1 and C-4.

$$\mathrm{\overset{1}{B}rCH_2{-}\overset{2}{C}H=\overset{3}{C}H{-}\overset{4}{C}H_2Br}$$

Hence the structure is $$\mathrm{BrCH_2CH=CHCH_2Br}$$.

The parent is benzene. Number the ring so the substituents get: bromo at C-1, methyl at C-2 and a sec-butyl group at C-4.

The sec-butyl group is $$\mathrm{{-}CH(CH_3)CH_2CH_3}$$.

Draw a benzene ring carrying $$\mathrm{Br}$$ at C-1, $$\mathrm{CH_3}$$ at C-2 (ortho to Br) and $$\mathrm{{-}CH(CH_3)CH_2CH_3}$$ at C-4 (para to Br).

The three isomeric alkanes of formula $$\mathrm{C_5H_{12}}$$ are n-pentane, 2-methylbutane (isopentane) and 2,2-dimethylpropane (neopentane). The number of monochlorides equals the number of different types of hydrogen atoms.

A single monochloride is formed only if all the hydrogen atoms are equivalent. In 2,2-dimethylpropane (neopentane), $$\mathrm{C(CH_3)_4}$$, all 12 hydrogen atoms belong to four equivalent $$\mathrm{CH_3}$$ groups, so they are all of one type.

Chlorination therefore gives only $$\mathrm{ClCH_2C(CH_3)_3}$$ (1-chloro-2,2-dimethylpropane).

n-Pentane, $$\mathrm{CH_3CH_2CH_2CH_2CH_3}$$, has three different types of hydrogen atoms:

• the C-1 / C-5 $$\mathrm{CH_3}$$ hydrogens,
• the C-2 / C-4 $$\mathrm{CH_2}$$ hydrogens,
• the C-3 $$\mathrm{CH_2}$$ hydrogens.

Replacing each type by chlorine gives three isomeric monochlorides: 1-chloropentane, 2-chloropentane and 3-chloropentane.

2-Methylbutane (isopentane), $$\mathrm{(CH_3)_2CHCH_2CH_3}$$, has four different types of hydrogen atoms:

• the six $$\mathrm{H}$$ of the two $$\mathrm{CH_3}$$ groups on C-2,
• the single $$\mathrm{H}$$ on the tertiary C-2,
• the two $$\mathrm{H}$$ of the $$\mathrm{CH_2}$$ (C-3),
• the three $$\mathrm{H}$$ of the terminal $$\mathrm{CH_3}$$ (C-4).

Replacing each type by chlorine gives four isomeric monochlorides: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane.

Thionyl chloride replaces the $$\mathrm{{-}OH}$$ of an alcohol by $$\mathrm{{-}Cl}$$, the other products being gases $$\mathrm{SO_2}$$ and $$\mathrm{HCl}$$ which escape (Darzen's process):

$$\mathrm{C_6H_{11}{-}OH + SOCl_2 \longrightarrow C_6H_{11}{-}Cl + SO_2\uparrow + HCl\uparrow}$$

Hence the major monohalo product is chlorocyclohexane (cyclohexyl chloride).

In the presence of heat or UV light (and absence of a halogen carrier), bromine reacts by a free-radical mechanism that substitutes a hydrogen on the side chain, not on the ring.

Substitution occurs at the benzylic carbon — the $$\mathrm{CH_2}$$ directly attached to the ring — because the benzylic radical formed there is stabilised by resonance with the ring.

$$\mathrm{p\text{-}O_2N{-}C_6H_4{-}CH_2CH_3 \xrightarrow{Br_2,\ \Delta/h\nu} p\text{-}O_2N{-}C_6H_4{-}CHBrCH_3 + HBr}$$

The major product is 1-(1-bromoethyl)-4-nitrobenzene, $$\mathrm{p\text{-}O_2N{-}C_6H_4{-}CHBrCH_3}$$.

The molecule has two $$\mathrm{{-}OH}$$ groups: a phenolic $$\mathrm{{-}OH}$$ attached directly to the ring, and a benzylic alcohol $$\mathrm{{-}CH_2OH}$$.

The phenolic $$\mathrm{C{-}OH}$$ bond has partial double-bond character (the lone pair on O is in resonance with the ring) and is very strong; it is not cleaved by $$\mathrm{HCl}$$. The benzylic $$\mathrm{{-}CH_2OH}$$, however, is on an $$sp^3$$ carbon and is readily replaced by $$\mathrm{Cl}$$.

$$\mathrm{p\text{-}HO{-}C_6H_4{-}CH_2OH + HCl \xrightarrow{\Delta} p\text{-}HO{-}C_6H_4{-}CH_2Cl + H_2O}$$

The major monohalo product is 4-(chloromethyl)phenol, $$\mathrm{p\text{-}HO{-}C_6H_4{-}CH_2Cl}$$.

$$\mathrm{HI}$$ adds across the ring double bond by electrophilic addition, following Markovnikov's rule.

In 1-methylcyclohexene the double bond joins C-1 (which bears the methyl group) and C-2. Protonation at C-2 gives the more stable tertiary carbocation at C-1 (stabilised by the methyl group and the two ring carbons).

Iodide then attacks C-1, so $$\mathrm{H}$$ adds to C-2 and $$\mathrm{I}$$ to C-1:

The major product is 1-iodo-1-methylcyclohexane — a cyclohexane ring bearing both $$\mathrm{I}$$ and $$\mathrm{CH_3}$$ on the same (C-1) carbon.

This is the Finkelstein reaction — a halogen-exchange carried out in dry acetone:

$$\mathrm{CH_3CH_2Br + NaI \xrightarrow{dry\ acetone} CH_3CH_2I + NaBr\downarrow}$$

$$\mathrm{NaI}$$ is soluble in acetone whereas $$\mathrm{NaBr}$$ is not; $$\mathrm{NaBr}$$ precipitates out, and by Le Chatelier's principle this drives the reaction forward.

The major monohalo product is iodoethane (ethyl iodide), $$\mathrm{CH_3CH_2I}$$.

Bromine adds across a $$\mathrm{C=C}$$ double bond at low temperature in the dark. But under heat or UV light the reaction takes a free-radical course and gives allylic substitution instead of addition — a hydrogen on the carbon adjacent to the double bond (the allylic carbon) is replaced by bromine.

The allylic radical/product is favoured because it is resonance-stabilised. For cyclohexene the allylic carbon is C-3.

$$\mathrm{C_6H_{10} + Br_2 \xrightarrow{\Delta/h\nu} 3\text{-}bromocyclohexene + HBr}$$

The major monohalo product is 3-bromocyclohex-1-ene.

For these halomethanes the boiling point rises with increasing molecular mass (and number of electrons), because the van der Waals forces of attraction get stronger.

So the boiling points increase in the order:

$$\text{Chloromethane} < \text{Bromomethane} < \text{Dibromomethane} < \text{Bromoform}$$

1-Chlorobutane ($$\mathrm{C_4H_9Cl}$$) has the highest molecular mass, so its van der Waals forces are strongest — it has the highest boiling point.

1-Chloropropane and isopropyl chloride (2-chloropropane) are isomers with the same molecular mass. Isopropyl chloride is more branched (more spherical, smaller surface area), so its molecules touch over a smaller area, the van der Waals forces are weaker, and its boiling point is lower than that of the straight-chain 1-chloropropane.

Hence the boiling points increase in the order:

$$\text{Isopropyl chloride} < \text{1-Chloropropane} < \text{1-Chlorobutane}$$

In an $$\mathrm{S_N2}$$ reaction the nucleophile attacks the carbon bearing the halogen from the back side; crowding around that carbon slows the reaction. Reactivity therefore falls in the order $$1^\circ > 2^\circ > 3^\circ$$.

$$\mathrm{CH_3CH_2CH_2CH_2Br}$$ (1-bromobutane) is a primary halide; $$\mathrm{CH_3CH_2CHBrCH_3}$$ (2-bromobutane) is a secondary halide and so is more hindered.

Hence $$\mathrm{CH_3CH_2CH_2CH_2Br}$$ reacts faster by the $$\mathrm{S_N2}$$ mechanism.

$$\mathrm{S_N2}$$ reactivity decreases with increasing steric crowding at the carbon bearing $$\mathrm{Br}$$: $$1^\circ > 2^\circ > 3^\circ$$.

$$\mathrm{CH_3CH_2CHBrCH_3}$$ (2-bromobutane) is a secondary halide; $$\mathrm{(CH_3)_3CBr}$$ (2-bromo-2-methylpropane) is a tertiary halide whose three bulky methyl groups strongly block back-side attack.

Hence the secondary halide $$\mathrm{CH_3CH_2CHBrCH_3}$$ reacts faster by the $$\mathrm{S_N2}$$ mechanism.

Both compounds are primary bromides, so we compare the steric crowding near the reaction centre.

In $$\mathrm{(CH_3)_2CHCH_2CH_2Br}$$ (1-bromo-3-methylbutane) the methyl branching is on C-3 — the $$\gamma$$-carbon, two carbons away from the carbon bearing $$\mathrm{Br}$$.

In $$\mathrm{CH_3CH_2CH(CH_3)CH_2Br}$$ (1-bromo-2-methylbutane) the methyl branch is on C-2 — the $$\beta$$-carbon, adjacent to the carbon bearing $$\mathrm{Br}$$. Branching at the $$\beta$$-carbon hinders the back-side approach of the nucleophile much more strongly.

Hence $$\mathrm{(CH_3)_2CHCH_2CH_2Br}$$, with its branch further from the reaction centre, reacts faster by the $$\mathrm{S_N2}$$ mechanism.

An $$\mathrm{S_N1}$$ reaction goes through a carbocation intermediate formed in the slow, rate-determining step. The more stable that carbocation, the faster the $$\mathrm{S_N1}$$ reaction. Stability of carbocations follows $$3^\circ > 2^\circ > 1^\circ$$, because alkyl groups release electron density (+I effect) and provide hyperconjugation.

$$\mathrm{(CH_3)_3C{-}Cl}$$ on ionisation gives a tertiary carbocation $$\mathrm{(CH_3)_3C^+}$$, whereas the secondary chloride gives only a less stable secondary carbocation.

Hence the tertiary chloride $$\mathrm{(CH_3)_3C{-}Cl}$$ undergoes the $$\mathrm{S_N1}$$ reaction faster.

$$\mathrm{S_N1}$$ rate depends on the stability of the carbocation formed in the slow step ($$3^\circ > 2^\circ > 1^\circ$$).

2-Chlorohexane is a secondary chloride; on ionisation it gives a secondary carbocation. 1-Chlorohexane is a primary chloride and would have to form a much less stable primary carbocation.

Hence the secondary chloride, 2-chlorohexane, undergoes the $$\mathrm{S_N1}$$ reaction faster.

Expanded structure: $$\mathrm{CH_3{-}CH(CH_3){-}CHCl{-}CH_3}$$. The longest chain has four carbons (butane). Numbering to give the lowest locants puts $$\mathrm{Cl}$$ at C-2 and a methyl group at C-3.

IUPAC name: 2-chloro-3-methylbutane.

The carbon bearing $$\mathrm{Cl}$$ is attached to two other carbon atoms, so it is a secondary (2°) alkyl halide.

Expanded structure: $$\mathrm{CH_3{-}CH_2{-}CH(CH_3){-}CHCl{-}CH_2{-}CH_3}$$. The longest chain contains six carbons (hexane). Numbering to give the lowest locants places $$\mathrm{Cl}$$ at C-3 and a methyl group at C-4.

IUPAC name: 3-chloro-4-methylhexane.

The carbon bearing $$\mathrm{Cl}$$ is attached to two other carbon atoms, so it is a secondary (2°) alkyl halide.

Expanded structure: $$\mathrm{CH_3{-}CH_2{-}C(CH_3)_2{-}CH_2I}$$. The longest chain has four carbons (butane). Numbering from the $$\mathrm{CH_2I}$$ end gives $$\mathrm{I}$$ at C-1 and two methyl groups at C-2.

IUPAC name: 1-iodo-2,2-dimethylbutane.

The carbon bearing $$\mathrm{I}$$ is attached to only one other carbon atom, so it is a primary (1°) alkyl halide.

Expanded structure: $$\mathrm{(CH_3)_3C{-}CH_2{-}CHBr{-}C_6H_5}$$. Taking the longest carbon chain as butane and the phenyl ring as a substituent, the carbon bearing $$\mathrm{Br}$$ is C-1; it also carries the phenyl group, and there are two methyl groups on C-3.

IUPAC name: 1-bromo-3,3-dimethyl-1-phenylbutane.

The $$\mathrm{Br}$$ is on an $$sp^3$$ carbon directly attached to the benzene ring, so it is a benzylic halide; that carbon also carries one other alkyl carbon, making it a secondary benzylic halide.

Expanded structure: $$\mathrm{(CH_3)_2CH{-}CHBr{-}CH_3}$$. The longest chain has four carbons (butane). Numbering to give the lowest locants puts $$\mathrm{Br}$$ at C-2 and a methyl group at C-3.

IUPAC name: 2-bromo-3-methylbutane.

The carbon bearing $$\mathrm{Br}$$ is attached to two other carbon atoms, so it is a secondary (2°) alkyl halide.

Expanded structure: $$\mathrm{CH_3{-}C(C_2H_5)_2{-}CH_2Br}$$. The longest chain (running through one of the ethyl groups) has four carbons (butane). Numbering from the $$\mathrm{CH_2Br}$$ end gives $$\mathrm{Br}$$ at C-1, with an ethyl group and a methyl group on C-2.

IUPAC name: 1-bromo-2-ethyl-2-methylbutane.

The carbon bearing $$\mathrm{Br}$$ is attached to only one other carbon atom, so it is a primary (1°) alkyl halide.

Expanded structure: $$\mathrm{CH_3{-}CCl(C_2H_5){-}CH_2{-}CH_3}$$. The central carbon carries two ethyl-type chains; the longest chain runs through both and has five carbons (pentane). $$\mathrm{Cl}$$ and a methyl group are both on the middle carbon, C-3.

IUPAC name: 3-chloro-3-methylpentane.

The carbon bearing $$\mathrm{Cl}$$ is attached to three other carbon atoms, so it is a tertiary (3°) alkyl halide.

Expanded structure: $$\mathrm{CH_3{-}CH=CCl{-}CH_2{-}CH(CH_3){-}CH_3}$$. The longest chain containing the double bond has six carbons (hexene). Numbering to give the double bond the lowest locant places it at C-2, with $$\mathrm{Cl}$$ at C-3 and a methyl group at C-5.

IUPAC name: 3-chloro-5-methylhex-2-ene.

The $$\mathrm{Cl}$$ is bonded directly to an $$sp^2$$ carbon of the $$\mathrm{C=C}$$ double bond, so it is a vinyl (vinylic) halide.

Expanded structure: $$\mathrm{CH_3{-}CH=CH{-}CBr(CH_3){-}CH_3}$$. The longest chain containing the double bond has five carbons (pentene). Numbering to give the double bond the lowest locant places it at C-2, with $$\mathrm{Br}$$ and a methyl group on C-4.

IUPAC name: 4-bromo-4-methylpent-2-ene.

The $$\mathrm{Br}$$ is on an $$sp^3$$ carbon next to the $$\mathrm{C=C}$$ double bond, i.e. an allylic carbon, and that carbon is bonded to three other carbons. So it is an allylic halide (tertiary).

The compound is a benzene ring carrying a chloro group and, at the para position, an isobutyl group $$\mathrm{{-}CH_2CH(CH_3)_2}$$ (the IUPAC name of this side chain is 2-methylpropyl).

IUPAC name: 1-chloro-4-(2-methylpropyl)benzene (i.e. 1-chloro-4-isobutylbenzene).

The $$\mathrm{Cl}$$ is bonded directly to an $$sp^2$$ carbon of the aromatic ring, so it is an aryl halide.

The compound is a benzene ring carrying two side chains at the meta (1,3) positions: a chloromethyl group $$\mathrm{{-}CH_2Cl}$$ and a neopentyl group $$\mathrm{{-}CH_2C(CH_3)_3}$$ (IUPAC: 2,2-dimethylpropyl).

IUPAC name: 1-(chloromethyl)-3-(2,2-dimethylpropyl)benzene.

The $$\mathrm{Cl}$$ is on an $$sp^3$$ carbon ($$\mathrm{CH_2}$$) attached to the aromatic ring, and that carbon carries no other alkyl carbon, so it is a primary benzylic halide.

The compound is a benzene ring carrying a bromo group and, at the ortho position, a sec-butyl group $$\mathrm{{-}CH(CH_3)CH_2CH_3}$$ (IUPAC: butan-2-yl).

IUPAC name: 1-bromo-2-(butan-2-yl)benzene (i.e. 1-bromo-2-sec-butylbenzene).

The $$\mathrm{Br}$$ is bonded directly to an $$sp^2$$ carbon of the aromatic ring, so it is an aryl halide.

The chain has four carbons (butane). The halogens are on C-2 and C-3; both numbering directions give the locant set $$\{2,3\}$$. When there is a tie, the lower locant goes to the substituent first in alphabetical order — bromo before chloro.

Hence $$\mathrm{Br}$$ is at C-2 and $$\mathrm{Cl}$$ at C-3.

IUPAC name: 2-bromo-3-chlorobutane.

The parent chain has two carbons (ethane). One carbon carries $$\mathrm{Br}$$, $$\mathrm{Cl}$$ and $$\mathrm{F}$$; the other carries two $$\mathrm{F}$$ and one $$\mathrm{H}$$.

Numbering for the lowest set of locants makes the $$\mathrm{CBrClF}$$ carbon C-1: then $$\mathrm{Br}$$, $$\mathrm{Cl}$$ and one $$\mathrm{F}$$ are at C-1 and the two remaining $$\mathrm{F}$$ at C-2. Substituents are cited alphabetically — bromo, chloro, fluoro.

IUPAC name: 1-bromo-1-chloro-1,2,2-trifluoroethane.

The chain has four carbons with a triple bond between C-2 and C-3 (but-2-yne). The halogens are on the terminal carbons; both directions give the locant set $$\{1,4\}$$ for the halogens.

The tie is broken alphabetically: bromo comes before chloro, so $$\mathrm{Br}$$ gets the lower locant, C-1.

IUPAC name: 1-bromo-4-chlorobut-2-yne.

The central carbon bears one $$\mathrm{Cl}$$ and three $$\mathrm{CCl_3}$$ groups. The longest chain runs through the central carbon and two of the $$\mathrm{CCl_3}$$ groups, giving a three-carbon parent (propane); the third $$\mathrm{CCl_3}$$ becomes a substituent — a trichloromethyl group.

Chlorine atoms: three on C-1, one on C-2, three on C-3 — seven in all (heptachloro). The trichloromethyl group is on C-2.

IUPAC name: 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane.

The parent carbon chain is butane: $$\mathrm{CH_3{-}C{-}CHBr{-}CH_3}$$. Two para-chlorophenyl (4-chlorophenyl) groups are on one inner carbon and $$\mathrm{Br}$$ on the other.

Numbering for the lowest set of locants: placing the two aryl groups at C-2 and $$\mathrm{Br}$$ at C-3 gives $$\{2,2,3\}$$, which is lower than the alternative $$\{2,3,3\}$$.

IUPAC name: 3-bromo-2,2-bis(4-chlorophenyl)butane.

The longest chain containing the double bond has four carbons (butene): $$\mathrm{CCl=CH{-}C(CH_3)_2{-}CH_3}$$. Numbering from the $$\mathrm{CCl}$$ end gives the double bond the lowest locant, C-1.

So C-1 carries $$\mathrm{Cl}$$ and a para-iodophenyl (4-iodophenyl) group; C-3 carries two methyl groups. Substituents are cited alphabetically — chloro, (4-iodophenyl), methyl.

IUPAC name: 1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene.

The parent chain is pentane (5 carbons). Place a chloro group at C-2 and a methyl group at C-3:

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}HCl{-}\overset{3}{C}H(CH_3){-}\overset{4}{C}H_2{-}\overset{5}{C}H_3}$$

Structure: $$\mathrm{CH_3CHClCH(CH_3)CH_2CH_3}$$.

The prefix p- (para) means the two substituents lie on opposite (1,4) carbons of the benzene ring. Here the ring carries one $$\mathrm{Br}$$ and one $$\mathrm{Cl}$$ para to each other.

Draw a benzene ring with $$\mathrm{Br}$$ on C-1 and $$\mathrm{Cl}$$ on C-4 — i.e. 1-bromo-4-chlorobenzene.

The parent ring is cyclohexane (six-membered ring of $$\mathrm{CH_2}$$ groups). A chloro group sits on ring carbon C-1, and an ethyl group $$\mathrm{{-}CH_2CH_3}$$ sits on C-4 — the carbon directly across the ring (the 1,4 or para-type positions).

The structure can be written in skeletal form as:

$$\mathrm{\underset{\textstyle Cl}{\underset{|}{\overset{1}{C}H}}\langle\overset{2}{C}H_2{-}\overset{3}{C}H_2{-}\underset{\textstyle C_2H_5}{\underset{|}{\overset{4}{C}H}}{-}\overset{5}{C}H_2{-}\overset{6}{C}H_2\rangle}$$

where the angle brackets indicate that C-1 and C-6 are joined to close the six-membered ring. Equivalently, drawn as a hexagon: $$\mathrm{Cl}$$ on one vertex (C-1) and $$\mathrm{{-}CH_2CH_3}$$ on the vertex three bonds away (C-4):

$$\mathrm{Cl{-}\underset{1}{C}H{-}\underset{2}{C}H_2{-}\underset{3}{C}H_2{-}\underset{4}{C}H({-}CH_2CH_3){-}\underset{5}{C}H_2{-}\underset{6}{C}H_2{-}}\ \text{(ring: C-6 bonded back to C-1)}$$

The parent chain is octane (8 carbons). C-1 carries an iodo group, and C-2 carries a 2-chlorophenyl group — a benzene ring (attached through one of its carbons) bearing $$\mathrm{Cl}$$ at the adjacent ortho position.

Structure: $$\mathrm{ICH_2{-}CH(o\text{-}ClC_6H_4){-}CH_2CH_2CH_2CH_2CH_2CH_3}$$.

The parent chain is butane. Place a bromo group on C-2:

$$\mathrm{\overset{1}{C}H_3{-}\overset{2}{C}HBr{-}\overset{3}{C}H_2{-}\overset{4}{C}H_3}$$

Structure: $$\mathrm{CH_3CHBrCH_2CH_3}$$.

The parent chain is heptane (7 carbons). Place an iodo group at C-3 and a tert-butyl group $$\mathrm{{-}C(CH_3)_3}$$ at C-4:

$$\mathrm{CH_3CH_2CHICH[C(CH_3)_3]CH_2CH_2CH_3}$$.

The parent is benzene. Number the ring so the substituents are bromo at C-1, methyl at C-2 and a sec-butyl group $$\mathrm{{-}CH(CH_3)CH_2CH_3}$$ at C-4.

Draw a benzene ring carrying $$\mathrm{Br}$$ at C-1, $$\mathrm{CH_3}$$ at C-2 (ortho to $$\mathrm{Br}$$) and $$\mathrm{{-}CH(CH_3)CH_2CH_3}$$ at C-4 (para to $$\mathrm{Br}$$).

The parent is but-2-ene — a four-carbon chain with the double bond between C-2 and C-3. Place bromo groups at C-1 and C-4:

$$\mathrm{BrCH_2{-}CH=CH{-}CH_2Br}$$

The three chloromethanes given in this question — $$\mathrm{CH_2Cl_2}$$, $$\mathrm{CHCl_3}$$ and $$\mathrm{CCl_4}$$ — are all tetrahedral around carbon, so their net dipole moment is the vector sum of the individual $$\mathrm{C{-}H}$$ and $$\mathrm{C{-}Cl}$$ bond dipoles.

$$\mathrm{CH_2Cl_2}$$ has two $$\mathrm{C{-}Cl}$$ bonds (large bond moments, pointing toward $$\mathrm{Cl}$$) and two $$\mathrm{C{-}H}$$ bonds (small bond moments, with $$\mathrm{C}$$ the slightly negative end). The molecule has $$C_{2v}$$ symmetry: the resultant of the two $$\mathrm{C{-}Cl}$$ bond dipoles and the resultant of the two $$\mathrm{C{-}H}$$ bond dipoles both lie along the same $$C_2$$ axis and, because the negative pole is on the $$\mathrm{Cl}$$ side for both pairs, the two resultants point in the same direction and add together. The bond moments therefore do not cancel.

The measured dipole moment of $$\mathrm{CH_2Cl_2}$$ is about $$\mathbf{1.60\ D}$$ — the highest of the three molecules listed. (For comparison, $$\mathrm{CHCl_3}$$ has $$\mu \approx 1.04\ \mathrm{D}$$ and $$\mathrm{CCl_4}$$ has $$\mu = 0$$.)

$$\mathrm{CHCl_3}$$ (chloroform) is tetrahedral with three $$\mathrm{C{-}Cl}$$ bonds and one $$\mathrm{C{-}H}$$ bond ($$C_{3v}$$ symmetry). Pick the unique $$\mathrm{C{-}H}$$ direction as the molecular axis. The three $$\mathrm{C{-}Cl}$$ bond moments are symmetric about this axis; their components perpendicular to the axis cancel exactly, leaving only the components along the axis, which point away from $$\mathrm{H}$$ toward the $$\mathrm{Cl_3}$$ side. The small $$\mathrm{C{-}H}$$ bond moment points in the opposite direction along the same axis, partially opposing the $$\mathrm{C{-}Cl}$$ resultant.

Hence the net dipole moment of $$\mathrm{CHCl_3}$$ ($$\mu \approx 1.04\ \mathrm{D}$$) is smaller than that of $$\mathrm{CH_2Cl_2}$$ ($$\mu \approx 1.60\ \mathrm{D}$$), where the $$\mathrm{C{-}H}$$ and $$\mathrm{C{-}Cl}$$ resultants reinforce each other instead of opposing.

Therefore $$\mathrm{CHCl_3}$$ does not have the highest dipole moment among the three given molecules — $$\mathrm{CH_2Cl_2}$$ does.

$$\mathrm{CCl_4}$$ is a perfectly symmetric tetrahedron ($$T_d$$ symmetry) with four identical $$\mathrm{C{-}Cl}$$ bonds. The four equal bond dipoles point from $$\mathrm{C}$$ toward the corners of the tetrahedron and, because of this high symmetry, their vector sum is exactly zero.

Hence the dipole moment of $$\mathrm{CCl_4}$$ is $$\mu = 0$$ — it is a non-polar molecule.

Comparing the three given molecules:

The overall order is $$\mathrm{CH_2Cl_2 > CHCl_3 > CCl_4}$$, so $$\mathrm{CH_2Cl_2}$$ has the highest dipole moment.

Heat 1-butanol with sodium iodide (or potassium iodide) in the presence of phosphoric acid (a non-oxidising acid). $$\mathrm{NaI}$$ and $$\mathrm{H_3PO_4}$$ generate $$\mathrm{HI}$$ in situ, which then converts the alcohol into the alkyl iodide:

$$\mathrm{NaI + H_3PO_4 \longrightarrow HI + NaH_2PO_4}$$

$$\mathrm{CH_3CH_2CH_2CH_2OH + HI \xrightarrow{\Delta} CH_3CH_2CH_2CH_2I + H_2O}$$

Combined (net) equation:

$$\mathrm{CH_3CH_2CH_2CH_2OH + NaI + H_3PO_4 \xrightarrow{\Delta} CH_3CH_2CH_2CH_2I + NaH_2PO_4 + H_2O}$$

Sulphuric acid is deliberately avoided here because it would oxidise $$\mathrm{HI}$$ to $$\mathrm{I_2}$$; phosphoric acid does not.

This is the Finkelstein reaction — a halogen exchange carried out in dry acetone:

$$\mathrm{CH_3CH_2CH_2CH_2Cl + NaI \xrightarrow{dry\ acetone} CH_3CH_2CH_2CH_2I + NaCl\downarrow}$$

$$\mathrm{NaCl}$$ is insoluble in acetone and precipitates out; by Le Chatelier's principle this drives the equilibrium towards the iodide.

Direct addition of $$\mathrm{HI}$$ to but-1-ene would, by Markovnikov's rule, give 2-iodobutane — not 1-iodobutane — and $$\mathrm{HI}$$ does not show the peroxide (anti-Markovnikov) effect. So a two-step route is used.

Step 1 — anti-Markovnikov addition of $$\mathrm{HBr}$$ in the presence of peroxide gives the terminal (primary) bromide:

$$\mathrm{CH_3CH_2CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2CH_2Br}$$

Step 2 — Finkelstein halogen exchange converts it to the iodide:

$$\mathrm{CH_3CH_2CH_2CH_2Br + NaI \xrightarrow{dry\ acetone} CH_3CH_2CH_2CH_2I + NaBr}$$

$$\mathrm{CH_3Br}$$ and $$\mathrm{CH_3I}$$ have the same methyl group, so $$\mathrm{S_N2}$$ reactivity is decided entirely by the leaving group.

Iodide ion is larger than bromide, so the $$\mathrm{C{-}I}$$ bond is longer and weaker than the $$\mathrm{C{-}Br}$$ bond, and $$\mathrm{I^-}$$ is a more stable, better leaving group.

Hence $$\mathrm{CH_3I}$$ reacts faster with $$\mathrm{{}^-OH}$$ in an $$\mathrm{S_N2}$$ reaction.

$$\mathrm{S_N2}$$ reactivity decreases sharply with steric crowding at the carbon bearing the halogen: $$\mathrm{CH_3X > 1^\circ > 2^\circ > 3^\circ}$$.

$$\mathrm{CH_3Cl}$$ is a methyl halide — its carbon carries only three small hydrogen atoms, so back-side attack by $$\mathrm{{}^-OH}$$ is easy. $$\mathrm{(CH_3)_3CCl}$$ is a tertiary halide whose three bulky methyl groups block the back-side approach.

Hence $$\mathrm{CH_3Cl}$$ reacts faster in the $$\mathrm{S_N2}$$ reaction.

Sodium ethoxide removes $$\mathrm{HBr}$$ (dehydrohalogenation, an $$\mathrm{E2}$$ elimination). The $$\mathrm{Br}$$ is on C-1 of the ring, the carbon that also carries the methyl group. A $$\beta$$-hydrogen can be taken either from a ring carbon (C-2/C-6) or from the methyl group.

• Removing a $$\beta$$-H from a ring carbon gives 1-methylcyclohexene — the double bond lies in the ring and is trisubstituted.
• Removing a $$\beta$$-H from the methyl group gives methylenecyclohexane — an exocyclic $$\mathrm{=CH_2}$$, a disubstituted alkene.

By Saytzeff's rule the more substituted alkene predominates.

Major product: 1-methylcyclohexene; minor product: methylenecyclohexane.

2-Chloro-2-methylbutane is $$\mathrm{(CH_3)_2CCl{-}CH_2CH_3}$$. The $$\mathrm{Cl}$$ is on C-2; $$\beta$$-hydrogens are available on the two methyl carbons and on the C-3 $$\mathrm{CH_2}$$.

• Removing a $$\beta$$-H from C-3 gives 2-methylbut-2-ene, $$\mathrm{CH_3C(CH_3)=CHCH_3}$$ — a trisubstituted alkene.
• Removing a $$\beta$$-H from a methyl group gives 2-methylbut-1-ene, $$\mathrm{CH_2=C(CH_3)CH_2CH_3}$$ — a disubstituted alkene.

By Saytzeff's rule the more substituted alkene is the major product.

Major product: 2-methylbut-2-ene; minor product: 2-methylbut-1-ene.

2,2,3-Trimethyl-3-bromopentane is $$\mathrm{CH_3{-}C(CH_3)_2{-}CBr(CH_3){-}CH_2{-}CH_3}$$ (C-1 to C-5); the $$\mathrm{Br}$$ is on C-3.

Examine the $$\beta$$-carbons of C-3:

• C-2 is quaternary — it carries no hydrogen, so elimination towards C-2 is impossible.
• C-4 ($$\mathrm{CH_2}$$) has hydrogens; removing a $$\beta$$-H from C-4 gives 3,4,4-trimethylpent-2-ene — a trisubstituted alkene.
• The methyl group on C-3 has hydrogens; removing one gives 2-ethyl-3,3-dimethylbut-1-ene — a disubstituted (terminal) alkene.

By Saytzeff's rule the more substituted alkene predominates.

Major product: 3,4,4-trimethylpent-2-ene; minor product: 2-ethyl-3,3-dimethylbut-1-ene.

Step 1: Dehydrate ethanol to ethene, then add bromine across the double bond:

$$\mathrm{CH_3CH_2OH \xrightarrow{conc.\ H_2SO_4,\ 443\ K} CH_2=CH_2 \xrightarrow{Br_2} BrCH_2CH_2Br}$$

Step 2: Double dehydrohalogenation gives ethyne, which is converted to sodium acetylide:

$$\mathrm{BrCH_2CH_2Br \xrightarrow{2KOH(alc.)} HC{\equiv}CH \xrightarrow{Na} HC{\equiv}C^-Na^+}$$

Step 3: Prepare bromoethane separately and alkylate the acetylide:

$$\mathrm{CH_3CH_2OH \xrightarrow{HBr} CH_3CH_2Br}$$

$$\mathrm{HC{\equiv}C^-Na^+ + CH_3CH_2Br \longrightarrow HC{\equiv}C{-}CH_2CH_3 + NaBr}$$

The product is but-1-yne.

Step 1: Free-radical bromination of ethane gives bromoethane:

$$\mathrm{CH_3CH_3 + Br_2 \xrightarrow{UV\ light} CH_3CH_2Br + HBr}$$

Step 2: Dehydrohalogenation with alcoholic $$\mathrm{KOH}$$ gives ethene, which then adds bromine:

$$\mathrm{CH_3CH_2Br \xrightarrow{KOH(alc.)} CH_2=CH_2 \xrightarrow{Br_2} BrCH_2CH_2Br}$$

Step 3: One dehydrohalogenation with alcoholic $$\mathrm{KOH}$$ gives bromoethene:

$$\mathrm{BrCH_2CH_2Br \xrightarrow{KOH(alc.)} CH_2=CHBr + KBr + H_2O}$$

Step 1: Add $$\mathrm{HBr}$$ to propene in the presence of peroxide (anti-Markovnikov) to obtain the primary bromide:

$$\mathrm{CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br}$$

Step 2: Treat 1-bromopropane with silver nitrite; $$\mathrm{AgNO_2}$$ (largely covalent) gives the nitroalkane as the main product:

$$\mathrm{CH_3CH_2CH_2Br + AgNO_2 \longrightarrow CH_3CH_2CH_2NO_2 + AgBr}$$

The product is 1-nitropropane.

Step 1: Free-radical (side-chain) chlorination of toluene in sunlight gives benzyl chloride:

$$\mathrm{C_6H_5CH_3 + Cl_2 \xrightarrow{UV\ light} C_6H_5CH_2Cl + HCl}$$

Step 2: Hydrolysis of benzyl chloride with aqueous $$\mathrm{NaOH}$$ (or boiling water) gives benzyl alcohol:

$$\mathrm{C_6H_5CH_2Cl + NaOH(aq.) \longrightarrow C_6H_5CH_2OH + NaCl}$$

Step 1: Add bromine across the double bond to get the vicinal dibromide:

$$\mathrm{CH_3CH=CH_2 + Br_2 \longrightarrow CH_3CHBrCH_2Br}$$

Step 2: Double dehydrohalogenation. Alcoholic $$\mathrm{KOH}$$ removes one $$\mathrm{HBr}$$ to give a bromopropene; the second $$\mathrm{HBr}$$ is removed by the stronger base sodium amide:

$$\mathrm{CH_3CHBrCH_2Br \xrightarrow{KOH(alc.)} CH_3CBr=CH_2 \xrightarrow{NaNH_2} CH_3C{\equiv}CH}$$

The product is propyne.

Step 1: Convert ethanol to ethyl chloride:

$$\mathrm{CH_3CH_2OH + PCl_5 \longrightarrow CH_3CH_2Cl + POCl_3 + HCl}$$

Step 2: Heat the haloalkane with a metallic fluoride (Swarts reaction):

$$\mathrm{CH_3CH_2Cl + AgF \longrightarrow CH_3CH_2F + AgCl}$$

The product is ethyl fluoride (fluoroethane).

Step 1: Convert bromomethane to methyl cyanide (acetonitrile) using $$\mathrm{KCN}$$:

$$\mathrm{CH_3Br + KCN \longrightarrow CH_3CN + KBr}$$

Step 2: React the nitrile with the Grignard reagent $$\mathrm{CH_3MgBr}$$ (itself made from bromomethane) and hydrolyse the resulting ketimine salt:

$$\mathrm{CH_3CN + CH_3MgBr \longrightarrow CH_3{-}C(=NMgBr){-}CH_3 \xrightarrow{H_2O/H^+} CH_3COCH_3}$$

The product is propanone (acetone).

Step 1: Markovnikov addition of $$\mathrm{HBr}$$ (no peroxide) places $$\mathrm{Br}$$ on the more substituted carbon:

$$\mathrm{CH_3CH_2CH=CH_2 + HBr \longrightarrow CH_3CH_2CHBrCH_3}$$

Step 2: Dehydrohalogenation of 2-bromobutane with alcoholic $$\mathrm{KOH}$$; by Saytzeff's rule the more substituted alkene, but-2-ene, is the major product:

$$\mathrm{CH_3CH_2CHBrCH_3 \xrightarrow{KOH(alc.)} CH_3CH=CHCH_3}$$

This is the Wurtz reaction — two molecules of an alkyl halide couple when heated with sodium in dry ether:

$$\mathrm{2\,CH_3CH_2CH_2CH_2Cl + 2Na \xrightarrow{dry\ ether} CH_3(CH_2)_6CH_3 + 2NaCl}$$

The product is n-octane.

Step 1: Chlorinate benzene in the presence of a Lewis-acid catalyst:

$$\mathrm{C_6H_6 + Cl_2 \xrightarrow{anhyd.\ AlCl_3} C_6H_5Cl + HCl}$$

Step 2: Heat chlorobenzene with sodium in dry ether (Fittig reaction); two aryl halide molecules couple:

$$\mathrm{2\,C_6H_5Cl + 2Na \xrightarrow{dry\ ether} C_6H_5{-}C_6H_5 + 2NaCl}$$

The product is biphenyl (diphenyl).

The dipole moment depends on the polarity and length of the $$\mathrm{C{-}Cl}$$ bond and on any opposing effects.

In cyclohexyl chloride the $$\mathrm{Cl}$$ is attached to an $$sp^3$$ carbon; the $$\mathrm{C{-}Cl}$$ bond is a simple polar single bond.

In chlorobenzene two effects lower the dipole moment: (a) the $$\mathrm{Cl}$$ is attached to an $$sp^2$$ carbon, which has more s-character, is more electronegative and holds the bonding electrons more tightly, giving a shorter and less polar $$\mathrm{C{-}Cl}$$ bond; (b) a lone pair on $$\mathrm{Cl}$$ is in conjugation (resonance) with the ring, so the $$\mathrm{C{-}Cl}$$ bond acquires partial double-bond character and chlorine partly donates electron density back towards the ring, opposing the $$\mathrm{C{\rightarrow}Cl}$$ polarity.

Because of these effects chlorobenzene has a lower dipole moment than cyclohexyl chloride.

For a substance to dissolve in water, the new solute–water attractions must release roughly as much energy as is needed to break the existing attractions in the two pure liquids.

Water molecules are held together by strong hydrogen bonds. Alkyl halides cannot form hydrogen bonds with water; the only new forces possible (dipole–dipole and dispersion forces) are much weaker.

So the energy released when an alkyl halide mixes with water is far less than the energy needed to break the water–water hydrogen bonds. The process is energetically unfavourable, so alkyl halides remain immiscible with water even though they are polar.

A Grignard reagent, $$\mathrm{R{-}MgX}$$, has a highly polar and very reactive carbon–magnesium bond in which the carbon is strongly nucleophilic (almost carbanion-like).

It reacts at once with any compound containing active hydrogen — and water has active hydrogen:

$$\mathrm{R{-}MgX + H_2O \longrightarrow R{-}H + Mg(OH)X}$$

If even traces of moisture are present, the Grignard reagent is destroyed (decomposed to an alkane) instead of being available for the intended reaction. Therefore Grignard reagents must be prepared and used under strictly anhydrous (moisture-free) conditions.

This is the Finkelstein halogen-exchange reaction in acetone — chloride is replaced by iodide:

$$\mathrm{CH_3CH_2CH_2Cl + NaI \xrightarrow{acetone,\ \Delta} CH_3CH_2CH_2I + NaCl\downarrow}$$

Major product: 1-iodopropane.

$$\mathrm{(CH_3)_3CBr}$$ is a tertiary halide, and $$\mathrm{KOH}$$ in ethanol with heat favours elimination (dehydrohalogenation):

$$\mathrm{(CH_3)_3CBr + KOH \xrightarrow{ethanol,\ \Delta} (CH_3)_2C=CH_2 + KBr + H_2O}$$

Major product: 2-methylprop-1-ene (isobutylene).

$$\mathrm{NaOH}$$ in water favours nucleophilic substitution; $$\mathrm{{}^-OH}$$ replaces $$\mathrm{Br}$$:

$$\mathrm{CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{water} CH_3CH(OH)CH_2CH_3 + NaBr}$$

Major product: butan-2-ol.

With $$\mathrm{KCN}$$ (ionic, supplying free $$\mathrm{CN^-}$$) the attack takes place through carbon, giving a nitrile:

$$\mathrm{CH_3CH_2Br + KCN \xrightarrow{aq.\ ethanol} CH_3CH_2CN + KBr}$$

Major product: propanenitrile (ethyl cyanide), $$\mathrm{CH_3CH_2CN}$$.

This is a Williamson ether synthesis — the phenoxide ion displaces chloride from ethyl chloride:

$$\mathrm{C_6H_5ONa + C_2H_5Cl \longrightarrow C_6H_5OC_2H_5 + NaCl}$$

Major product: ethoxybenzene (phenetole), $$\mathrm{C_6H_5OC_2H_5}$$.

Thionyl chloride replaces the $$\mathrm{{-}OH}$$ group by $$\mathrm{{-}Cl}$$, the by-products escaping as gases:

$$\mathrm{CH_3CH_2CH_2OH + SOCl_2 \longrightarrow CH_3CH_2CH_2Cl + SO_2\uparrow + HCl\uparrow}$$

Major product: 1-chloropropane.

$$\mathrm{HBr}$$ in the presence of peroxide adds in the anti-Markovnikov sense — $$\mathrm{Br}$$ goes to the terminal carbon:

$$\mathrm{CH_3CH_2CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2CH_2Br}$$

Major product: 1-bromobutane.

$$\mathrm{HBr}$$ adds to $$\mathrm{CH_3CH=C(CH_3)_2}$$ (2-methylbut-2-ene) by Markovnikov's rule. Protonation gives the more stable tertiary carbocation on the $$\mathrm{C(CH_3)_2}$$ carbon, so $$\mathrm{Br}$$ ends up there:

$$\mathrm{CH_3CH=C(CH_3)_2 + HBr \longrightarrow CH_3CH_2{-}CBr(CH_3)_2}$$

Major product: 2-bromo-2-methylbutane.

$$\mathrm{S_N2}$$ reactivity decreases as steric crowding at the carbon bearing $$\mathrm{Br}$$ increases: $$1^\circ > 2^\circ > 3^\circ$$.

1-Bromopentane is primary, 2-bromopentane is secondary and 2-bromo-2-methylbutane is tertiary.

Order of $$\mathrm{S_N2}$$ reactivity:

$$\text{1-Bromopentane} > \text{2-Bromopentane} > \text{2-Bromo-2-methylbutane}$$

$$\mathrm{S_N2}$$ reactivity follows $$1^\circ > 2^\circ > 3^\circ$$ (less steric crowding reacts faster).

1-Bromo-3-methylbutane is primary, 2-bromo-3-methylbutane is secondary and 2-bromo-2-methylbutane is tertiary.

Order of $$\mathrm{S_N2}$$ reactivity:

$$\text{1-Bromo-3-methylbutane} > \text{2-Bromo-3-methylbutane} > \text{2-Bromo-2-methylbutane}$$

All four are primary bromides, so reactivity is decided by branching near the carbon bearing $$\mathrm{Br}$$ — branching at the $$\beta$$-carbon hinders the back-side attack the most.

• 1-Bromobutane — the $$\beta$$-carbon is an unbranched $$\mathrm{CH_2}$$: least hindered.
• 1-Bromo-3-methylbutane — branching is at the $$\gamma$$-carbon, well away from the reaction centre: little hindrance.
• 1-Bromo-2-methylbutane — one methyl branch on the $$\beta$$-carbon: more hindered.
• 1-Bromo-2,2-dimethylpropane (neopentyl bromide) — two methyl branches on the $$\beta$$-carbon: most hindered.

Order of $$\mathrm{S_N2}$$ reactivity:

$$\text{1-Bromobutane} > \text{1-Bromo-3-methylbutane} > \text{1-Bromo-2-methylbutane} > \text{1-Bromo-2,2-dimethylpropane}$$

Step 1: Anti-Markovnikov addition of $$\mathrm{HBr}$$ in the presence of peroxide gives the primary bromide:

$$\mathrm{CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br}$$

Step 2: Hydrolysis with aqueous $$\mathrm{KOH}$$ (or $$\mathrm{NaOH}$$) gives propan-1-ol:

$$\mathrm{CH_3CH_2CH_2Br + KOH(aq.) \longrightarrow CH_3CH_2CH_2OH + KBr}$$

The carbon skeleton must be extended from $$\mathrm{C_2}$$ to $$\mathrm{C_4}$$, so the route is to make ethyne (from ethanol) and bromoethane (also from ethanol) and couple them via the acetylide.

Step 1 — ethanol to ethyne. Dehydrate ethanol to ethene, brominate, then doubly dehydrohalogenate:

$$\mathrm{CH_3CH_2OH \xrightarrow{conc.\ H_2SO_4,\ 443\ K} CH_2=CH_2 \xrightarrow{Br_2} BrCH_2CH_2Br \xrightarrow{2\,KOH(alc.),\ \Delta} HC{\equiv}CH}$$

Step 2 — ethyne to its sodium acetylide. The terminal $$\equiv$$C–H of ethyne is weakly acidic ($$pK_a \approx 25$$), so a very strong base is needed. The standard reagent is sodamide ($$\mathrm{NaNH_2}$$) in liquid ammonia:

$$\mathrm{HC{\equiv}CH + NaNH_2 \xrightarrow{liq.\ NH_3} HC{\equiv}C^-Na^+ + NH_3}$$

Step 3 — bromoethane from ethanol (in parallel):

$$\mathrm{CH_3CH_2OH + HBr \longrightarrow CH_3CH_2Br + H_2O}$$

Step 4 — alkylation of sodium acetylide (an $$\mathrm{S_N2}$$ displacement; a primary halide such as bromoethane is essential):

$$\mathrm{HC{\equiv}C^-Na^+ + CH_3CH_2Br \xrightarrow{liq.\ NH_3} HC{\equiv}C{-}CH_2CH_3 + NaBr}$$

The product is but-1-yne.

Step 1: Dehydrohalogenate 1-bromopropane with alcoholic $$\mathrm{KOH}$$ to give propene:

$$\mathrm{CH_3CH_2CH_2Br \xrightarrow{KOH(alc.)} CH_3CH=CH_2}$$

Step 2: Add $$\mathrm{HBr}$$ in the absence of peroxide; by Markovnikov's rule $$\mathrm{Br}$$ goes to the middle carbon:

$$\mathrm{CH_3CH=CH_2 + HBr \longrightarrow CH_3CHBrCH_3}$$

The product is 2-bromopropane.

Step 1: Free-radical (side-chain) chlorination of toluene in sunlight gives benzyl chloride:

$$\mathrm{C_6H_5CH_3 + Cl_2 \xrightarrow{UV\ light} C_6H_5CH_2Cl + HCl}$$

Step 2: Hydrolysis of benzyl chloride with aqueous $$\mathrm{NaOH}$$ (or boiling water) gives benzyl alcohol:

$$\mathrm{C_6H_5CH_2Cl + NaOH(aq.) \longrightarrow C_6H_5CH_2OH + NaCl}$$

Step 1: Brominate benzene first (because bromine is an ortho/para-directing substituent):

$$\mathrm{C_6H_6 + Br_2 \xrightarrow{anhyd.\ FeBr_3} C_6H_5Br + HBr}$$

Step 2: Nitrate bromobenzene; the $$\mathrm{Br}$$ already present directs the incoming nitro group mainly to the para position:

$$\mathrm{C_6H_5Br + HNO_3 \xrightarrow{conc.\ H_2SO_4} p\text{-}O_2N{-}C_6H_4{-}Br + H_2O}$$

The major product is 4-bromonitrobenzene. (If benzene were nitrated first, the meta-directing nitro group would not give the para product.)

Step 1: Convert benzyl alcohol to benzyl chloride:

$$\mathrm{C_6H_5CH_2OH + SOCl_2 \longrightarrow C_6H_5CH_2Cl + SO_2 + HCl}$$

Step 2: Treat benzyl chloride with $$\mathrm{KCN}$$ to introduce one extra carbon as a nitrile:

$$\mathrm{C_6H_5CH_2Cl + KCN \longrightarrow C_6H_5CH_2CN + KCl}$$

Step 3: Acidic hydrolysis of the nitrile gives the carboxylic acid:

$$\mathrm{C_6H_5CH_2CN + 2H_2O \xrightarrow{H^+} C_6H_5CH_2COOH + NH_3}$$

The product is 2-phenylethanoic acid (phenylacetic acid).

Step 1: Convert ethanol to bromoethane:

$$\mathrm{CH_3CH_2OH + HBr \longrightarrow CH_3CH_2Br + H_2O}$$

Step 2: Treat bromoethane with $$\mathrm{KCN}$$; the cyanide adds through carbon:

$$\mathrm{CH_3CH_2Br + KCN \longrightarrow CH_3CH_2CN + KBr}$$

The product is propanenitrile, $$\mathrm{CH_3CH_2CN}$$.

Step 1: Diazotise aniline with nitrous acid (generated from $$\mathrm{NaNO_2}$$ and $$\mathrm{HCl}$$) at 273–278 K:

$$\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273\text{-}278\ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O}$$

Step 2: Treat the benzenediazonium chloride with cuprous chloride (Sandmeyer reaction):

$$\mathrm{C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2}$$

The product is chlorobenzene.

This is the Wurtz reaction — two molecules of 2-chlorobutane couple at their $$\mathrm{Cl}$$-bearing carbons when heated with sodium in dry ether:

$$\mathrm{2\,CH_3CHClCH_2CH_3 + 2Na \xrightarrow{dry\ ether} CH_3CH_2CH(CH_3)CH(CH_3)CH_2CH_3 + 2NaCl}$$

Joining two sec-butyl groups gives 3,4-dimethylhexane.

Add $$\mathrm{HCl}$$ to 2-methylpropene by Markovnikov's rule. Protonation gives the stable tertiary carbocation $$\mathrm{(CH_3)_3C^+}$$, so $$\mathrm{Cl}$$ becomes attached to the central carbon:

$$\mathrm{(CH_3)_2C=CH_2 + HCl \longrightarrow (CH_3)_3CCl}$$

The product is 2-chloro-2-methylpropane (tert-butyl chloride).

Step 1: Treat ethyl chloride with $$\mathrm{KCN}$$ to introduce one extra carbon as a nitrile:

$$\mathrm{CH_3CH_2Cl + KCN \longrightarrow CH_3CH_2CN + KCl}$$

Step 2: Hydrolyse the nitrile with dilute acid:

$$\mathrm{CH_3CH_2CN + 2H_2O \xrightarrow{H^+} CH_3CH_2COOH + NH_3}$$

The product is propanoic acid.

Step 1: Anti-Markovnikov addition of $$\mathrm{HBr}$$ (peroxide) gives the terminal bromide:

$$\mathrm{CH_3CH_2CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2CH_2Br}$$

Step 2: Finkelstein halogen exchange with $$\mathrm{NaI}$$ in acetone:

$$\mathrm{CH_3CH_2CH_2CH_2Br + NaI \xrightarrow{acetone} CH_3CH_2CH_2CH_2I + NaBr}$$

The product is n-butyl iodide (1-iodobutane).

Step 1: Dehydrohalogenate 2-chloropropane with alcoholic $$\mathrm{KOH}$$ to give propene:

$$\mathrm{(CH_3)_2CHCl \xrightarrow{KOH(alc.)} CH_3CH=CH_2}$$

Step 2: Anti-Markovnikov addition of $$\mathrm{HBr}$$ (peroxide):

$$\mathrm{CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br}$$

Step 3: Hydrolysis with aqueous $$\mathrm{KOH}$$:

$$\mathrm{CH_3CH_2CH_2Br + KOH(aq.) \longrightarrow CH_3CH_2CH_2OH + KBr}$$

Isopropyl alcohol, $$\mathrm{(CH_3)_2CHOH}$$, contains the $$\mathrm{CH_3CH(OH){-}}$$ grouping, so it answers the iodoform test.

On warming with iodine and sodium hydroxide (i.e. $$\mathrm{NaOI}$$), it is first oxidised to acetone; the $$\mathrm{CH_3CO{-}}$$ group is then triiodinated and cleaved to give iodoform:

$$\mathrm{(CH_3)_2CHOH + 4I_2 + 6NaOH \longrightarrow CHI_3\downarrow + CH_3COONa + 5NaI + 5H_2O}$$

The yellow precipitate is iodoform, $$\mathrm{CHI_3}$$.

Step 1: Nitrate chlorobenzene; $$\mathrm{Cl}$$ is ortho/para-directing, so the para product predominates:

$$\mathrm{C_6H_5Cl + HNO_3 \xrightarrow{conc.\ H_2SO_4} p\text{-}O_2N{-}C_6H_4{-}Cl + H_2O}$$

Step 2: Hydrolyse with aqueous $$\mathrm{NaOH}$$ on heating; the para-nitro group activates the ring towards nucleophilic substitution, so $$\mathrm{Cl}$$ is replaced by $$\mathrm{{-}OH}$$. Acidification then liberates the phenol:

$$\mathrm{p\text{-}O_2N{-}C_6H_4{-}Cl \xrightarrow{NaOH(aq.),\ \Delta} p\text{-}O_2N{-}C_6H_4{-}ONa \xrightarrow{H^+} p\text{-}O_2N{-}C_6H_4{-}OH}$$

The product is p-nitrophenol.

Step 1: Dehydrohalogenate 2-bromopropane with alcoholic $$\mathrm{KOH}$$ to give propene:

$$\mathrm{(CH_3)_2CHBr \xrightarrow{KOH(alc.)} CH_3CH=CH_2}$$

Step 2: Add $$\mathrm{HBr}$$ in the presence of peroxide (anti-Markovnikov), so $$\mathrm{Br}$$ goes to the terminal carbon:

$$\mathrm{CH_3CH=CH_2 + HBr \xrightarrow{peroxide} CH_3CH_2CH_2Br}$$

The product is 1-bromopropane.

This is the Wurtz reaction — two molecules of chloroethane couple with sodium in dry ether:

$$\mathrm{2\,CH_3CH_2Cl + 2Na \xrightarrow{dry\ ether} CH_3CH_2CH_2CH_3 + 2NaCl}$$

The product is butane.

Step 1 — chlorination of benzene. Treat benzene with chlorine in the presence of a Lewis-acid catalyst (anhydrous $$\mathrm{FeCl_3}$$ or $$\mathrm{AlCl_3}$$). The Lewis acid polarises $$\mathrm{Cl_2}$$ to generate the electrophile $$\mathrm{Cl^+}$$, which substitutes a ring hydrogen:

$$\mathrm{C_6H_6 + Cl_2 \xrightarrow{anhyd.\ FeCl_3} C_6H_5Cl + HCl}$$

Step 2 — Fittig reaction. Heat chlorobenzene with sodium metal in dry ether; two molecules of the aryl halide couple, eliminating $$\mathrm{NaCl}$$ (this aryl–aryl coupling is called the Fittig reaction; the closely related coupling of an aryl halide with an alkyl halide using sodium is the Wurtz–Fittig reaction):

$$\mathrm{2\,C_6H_5Cl + 2Na \xrightarrow{dry\ ether} C_6H_5{-}C_6H_5 + 2NaCl}$$

The product is biphenyl (diphenyl), $$\mathrm{C_6H_5{-}C_6H_5}$$.

Step 1: Dehydrohalogenate tert-butyl bromide with alcoholic $$\mathrm{KOH}$$ to give 2-methylpropene:

$$\mathrm{(CH_3)_3CBr \xrightarrow{KOH(alc.)} (CH_3)_2C=CH_2}$$

Step 2: Add $$\mathrm{HBr}$$ in the presence of peroxide (anti-Markovnikov), so $$\mathrm{Br}$$ goes to the terminal $$\mathrm{CH_2}$$:

$$\mathrm{(CH_3)_2C=CH_2 + HBr \xrightarrow{peroxide} (CH_3)_2CHCH_2Br}$$

The product is isobutyl bromide (1-bromo-2-methylpropane).

Aniline, a primary amine, undergoes the carbylamine reaction when warmed with chloroform and alcoholic $$\mathrm{KOH}$$:

$$\mathrm{C_6H_5NH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O}$$

The product is phenyl isocyanide (phenyl carbylamine), $$\mathrm{C_6H_5NC}$$, recognised by its extremely offensive smell.

n-Butyl chloride treated with alcoholic $$\mathrm{KOH}$$ undergoes dehydrohalogenation — elimination of $$\mathrm{HCl}$$:

$$\mathrm{CH_3CH_2CH_2CH_2Cl + KOH(alc.) \xrightarrow{\Delta} CH_3CH_2CH=CH_2 + KCl + H_2O}$$

The product is but-1-ene.

Bromobenzene reacts with magnesium in dry ether to form a Grignard reagent:

$$\mathrm{C_6H_5Br + Mg \xrightarrow{dry\ ether} C_6H_5MgBr}$$

The product is phenylmagnesium bromide.

The $$\mathrm{C{-}Cl}$$ bond in chlorobenzene has partial double-bond character (due to resonance) and the ring is electron-rich, so chlorobenzene is very unreactive towards nucleophiles. Its hydrolysis therefore needs drastic conditions — heating with aqueous $$\mathrm{NaOH}$$ at about 623 K under a pressure of about 300 atm (Dow process):

$$\mathrm{C_6H_5Cl + NaOH \xrightarrow{623\ K,\ 300\ atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH}$$

Sodium phenoxide, on acidification, gives phenol.

Ethyl chloride treated with aqueous $$\mathrm{KOH}$$ undergoes nucleophilic substitution — $$\mathrm{{}^-OH}$$ replaces $$\mathrm{Cl}$$:

$$\mathrm{CH_3CH_2Cl + KOH(aq.) \longrightarrow CH_3CH_2OH + KCl}$$

The product is ethanol.

Methyl bromide reacts with sodium in dry ether by the Wurtz reaction; two methyl groups couple:

$$\mathrm{2\,CH_3Br + 2Na \xrightarrow{dry\ ether} CH_3{-}CH_3 + 2NaBr}$$

The product is ethane.

Methyl chloride reacts with $$\mathrm{KCN}$$. Since $$\mathrm{KCN}$$ is ionic it supplies free $$\mathrm{CN^-}$$ ions, which attack through carbon to give a nitrile:

$$\mathrm{CH_3Cl + KCN \longrightarrow CH_3CN + KCl}$$

The product is methyl cyanide (acetonitrile / ethanenitrile).