Haloalkanes and Haloarenes

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.

Primary halides undergo $\mathrm{S}_{\mathrm{N}} 2$ reaction with $\mathrm{aq} . \mathrm{KOH}$. The reaction forms a transition state where the attacking nucleophile and leaving group both are partially bonded to carbon.

$3^{\circ}$ alkyl halide undergoes nucleophilic substitution reaction via formation of a stable $3^{\circ}$ carbocation. Thus, it shows $\mathrm{S}_{\mathrm{N}} 1$ reaction where rate of reaction depends only on concentration of substrate.

$ \text { (C) } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I} \xrightarrow{\text { Alc. } \mathrm{KOH}}\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 $

Primary halides prefer to undergo E2 elimination reaction.

The reaction path followed is E1 reaction.

∴ The correct option is $\mathrm{A} \rightarrow \mathrm{IV}, \mathrm{B} \rightarrow \mathrm{I}, \mathrm{C} \rightarrow \mathrm{II}$ and $\mathrm{D} \rightarrow \mathrm{III}$.

$\therefore X$ is alc. KOH $Y$ is NaOI

Due to presence of peroxide, addition of HBr takes place according to anti-Markownikoff rule in which negative part of addendum goes to carbon which has more number of H -atom attached to it.

Hydrolysis of cyano group will yield or produce.

(i) Firstly toluene reacts with bromine in the presence of light to give benzyl bromide. This reaction is free radical substitution reaction.

(ii) Benzyl bromide reacts with magnesium to give benzyl magnesium bromide (Grignard reagent).

(iii), (iv) In next step, benzyl magnesium bromide reacts with formaldehyde $\left(\mathrm{CH}_2 \mathrm{O}\right)$ followed by acidic hydrolysis to give 2-phenylethanol.

(v) In last step, alcohol is converted to alkyl halide in presence of $\mathrm{PBr}_3$.

Mechanism involved

Step (i) Bromination reaction Electrophilic substitution takes place.

Step (ii) Friedel crafts alkylation Electrophilic substitution takes place. As, Br is ortho-para directing group, so, alkylation occurs at para-position.

$ \text { Step (iii) Grignard reagent formation } $

Step (iv) Nucleophilic addition reaction of Grignard reagent

$ \text { Step (v) Hydrolysis } $

The $\mathrm{S}_{\mathrm{N}} 1$ reaction proceed by the formation of carbonium ion. The order of stability of carbonium ion is as follows

$ 3^{\circ}>2^{\circ}>1^{\circ} $

Also, more the number of carbon atoms, more will be the stability of carbonium ion. Therefore, the correct order of reactivity of $\mathrm{S}_{\mathrm{N}} 1$ reaction is as follows

$ \text { I }<\text { III }<\text { II }<\text { IV } $

Cl has $-I$ and $+R$-effect. Due to electron donating resonance effect it is ortho/para-directing and donates the electron density to benzene ring. Thus, the incoming electrophile -Br gets attached on para-position and para disubstituted product is obtained as major product and ortho as minor product.