Haloalkanes and Haloarenes

The correct match is A-III, B-I, C-II.

Sandmeyer reaction is a reaction used to synthesis aryl halides from aryl diazonium salt.

Finkelstien reaction is a halogen $(X)$ exchange method. It is used to prepare alkyl iodide by reaction of alkyl chlorides/bromides with NaI in dry acetone.

$ \underset{(x=\mathrm{Cl}, \mathrm{Br})}{R-X}+\mathrm{NaI} \xrightarrow{\text { Acetone }} R \mathrm{I}+\mathrm{NaX} $

Swartz reaction is used to get alkyl fluorides from alkyl chlorides.

$ R-\mathrm{Br}+\mathrm{AgF} \longrightarrow R-\mathrm{F}+\mathrm{AgBr} $

The ammonia gas released will turn red litmus to blue. Thus, the compound $X$ and $Y$ are $\mathrm{KCN} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$.

Since, benzyl halides are more reactive than aryl halides, therefore reaction occur at the benzyl chloride part.

The reaction is as follow,

The rate of $ \mathrm{S}_{\mathrm{N}} 2 $ reaction is the maximum with $ 1^{\circ} $ alkyl halide having least steric hindrance. Out of

$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} $ and $ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} $ has the least steric hindrance and hence has the highest relative rate.

Tertiary alkyl halides readily undergo dehydrohalogenation to form alkenes.

The correct option from the pairs having both compounds with a net non-zero dipole moment is:

Option A: $\text{cis-butene, trans-butene}$ cis-butene has a net dipole moment because the two alkyl groups are on the same side of the molecule, causing an imbalance in the distribution of electrons. However, trans-butene is symmetric, so the dipole moments of the two alkyl groups cancel each other out, giving it a net dipole moment of zero. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

Option B: $\text{CH}_2\text{Cl}_2, \text{CHCl}_3$ Both $\text{CH}_2\text{Cl}_2$ (dichloromethane) and $\text{CHCl}_3$ (chloroform) have a net non-zero dipole moment due to the difference in electronegativity between the hydrogen and chlorine atoms. Hence, this pair has both compounds with a net non-zero dipole moment.

Option C: $\text{Benzene, Anisidine}$ Benzene has a symmetrical structure, leading to a net zero dipole moment. Anisidine, however, does have a net dipole moment due to the asymmetry in its structure. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

Option D: $\text{1,4-Dichlorobenzene, 1,3-Dichlorobenzene}$ $\text{1,4-Dichlorobenzene}$ has a symmetrical structure, leading to a net zero dipole moment. However, $\text{1,3-Dichlorobenzene}$ has a non-zero dipole moment due to the asymmetry in its structure. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

Rate of $\mathrm{S_N 1}$ $\propto$ stability of carbocation produced

(D) $\mathrm{C_2H_5Cl+Nal\longrightarrow C_2H_5l+NaCl}$ (Finkelstein reaction)

The isomeric deuterated bromide with molecular formula $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{DBr}$ having two chiral carbon atoms is

2-Bromo-3-deuterobutane

Swarts reaction is a halogen exchange reaction. It is used to get alkyl fluorides from alkyl chlorides or bromides.

$ R-\mathrm{Cl}+\mathrm{AgF} \longrightarrow R-\mathrm{F}+\mathrm{AgCl} $

Among the given options, chlorobenzene is least reactive towards nucleophilic substitution reaction due to the delocalisation of the electrons of $\mathrm{C}-\mathrm{Cl}$ bond and a partial double bond character developed in the bond which makes it difficult for nucleophile to deave $\mathrm{C}-\mathrm{Cl}$ bond.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[\mathrm{Mg} / \text { Dry ether }]{\mathrm{D}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{D}$

For $\mathrm{S}_{\mathrm{N}} 1$ reaction, carbocation intermediate is formed and stability of carbocation defines feasibility of reaction. The order of stability of carbocation is tertiary > secondary > primary

Order of reactivity of alkyl halide is tertiary halide $>$ secondary halide $>$ primary halide.

Swart's reaction is used for synthesis of alkyl fluorides from alkyl bromides.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow{\mathrm{CoF}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~F} $

The order of reactivity towards $\mathrm{S}_{\mathrm{N}} 2$ reaction for alkyl halide is methyl $>$ primary halide $\left(1^{\circ}\right)>$ secondary halide $\left(2^{\circ}\right)>$ tertiary halide $\left(3^{\circ}\right)$.

Thus, option (d) is least reactive towards $\mathrm{S}_{\mathrm{N}} 2$ reaction as it is a $3^{\circ}$ halide.