Answer

(i)

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

(iii)

Both the alkyl halides are primary. However, the substituent $-\mathrm{CH_3}$ is at a greater distance to the carbon atom linked to $\mathrm{Br}$ 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 $\mathrm{S_\mathrm{N}} 2$ mechanism.

Answer

(i)

$\mathrm{S_\mathrm{N}} 1$ 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 $\mathrm{S_\mathrm{N}} 1$ reaction. Since $3^{\circ}$ carbocation is more stable than $2^{\circ}$ carbocation. (I), i.e. 2 -chloro-2-methylpropane, undergoes faster $\mathrm{S_\mathrm{N}} 1$ 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, $(\mathrm{I}), 2$ chloroheptane, undergoes faster $\mathrm{S_\mathrm{N}} 1$ reaction than (II), 1-chlorohexane.

Answer

Since $\mathrm{D}$ of $\mathrm{D_2} \mathrm{O}$ gets attached to the carbon atom to which $\mathrm{MgBr}$ is attached, $\mathrm{C}$ is

Therefore, the compound $\mathrm{R}$ - $\mathrm{Br}$ is

2-Bromopropane

When an alkyl halide is treated with $\mathrm{Na}$ in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, $\mathrm{R}^{1}-\mathrm{X}$, is

Therefore, compound D is

And, compound $\mathrm{E}$ is