Answer

(i)

2-Bromobutane is a $2^{\circ}$ (secondary) alkylhalide whereas 1-Bromobutane is a $1^{\circ}$ (primary) alkyl halide. The approaching of nucleophile is more hindered in 2-Bromobutane than in 1-Bromobutane. Therefore, 1-Bromobutane reacts more rapidly than 2-Bromobutane by an ${S_{N}} 2$ mechanism.

(ii)

2-Bromobutane is $2^{\circ}$ alkylhalide whereas 2-Bromo-2-methylpropane is $3^{\circ}$ alkyl halide. Therefore, greater numbers of substituents are present in $3^{\circ}$ alkyl halide than in $2^{\circ}$ alkyl halide to hinder the approaching nucleophile. Hence, 2-Bromobutane reacts more rapidly than 2-Bromo-2-methylpropane by an ${S_{N}} 2$ mechanism.

(iii)

Both the alkyl halides are primary. However, the substituent $-{CH_3}$ is at a greater distance to the carbon atom linked to ${Br}$ in 1-Bromo-3-methylbutane than in 1-Bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by ${S_{N}} 2$ mechanism.

Answer

(i)

${S_{N}} 1$ reaction proceeds via the formation of carbocation. The alkyl halide (I) is $3^{\circ}$ while (II) is $2^{\circ}$. Therefore, (I) forms $3^{\circ}$ carbocation while (II) forms $2^{\circ}$ carbocation. Greater the stability of the carbocation, faster is the rate of ${S_{N}} 1$ reaction. Since $3^{\circ}$ carbocation is more stable than $2^{\circ}$ carbocation. (I), i.e. 2-Chloro-2-methylpropane, undergoes faster ${S_{N}} 1$ reaction than (II) i.e., 3-Chloropentane.

The alkyl halide (I) is $2^{\circ}$ while (II) is $1^{\circ} .2^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation. Therefore, $({I}), 2$ chloroheptane, undergoes faster ${S_{N}} 1$ reaction than (II), 1-Chlorohexane.

Answer