Haloalkanes and Haloarenes

Rate of $S_N 2$ reaction of is faster than

Because iodine is a good leaving group due to large size of iodine. Which stabilises the $\mathrm{I}^{-}$ion.

$\bullet$ Aryl chlorides and bromides can easily be prepared by electrophilic substitution of arenes (toluene) with Cl$_2$ and Br$_2$ respectively in the presence of Lewis acid catalyst (Fe in dark).

$\bullet$ Reaction with I$_2$ is reversible and requires the presence of oxidising agent.

$\bullet$ Corresponding fluoroarene is not prepared by this method due to high reactivity of fluorine. Hence, ‘X’ is F.

Reactivity towards ,$\mathrm{S}_{\mathrm{N}} 1$ depends upon stability of carbocation.

Order of stability is

$\bullet$ Step-l is $\mathrm{S}_{\mathrm{N}}$ reaction with $\stackrel{\ominus}{\mathrm{C}} \mathrm{N}$ nucleophile.

$\bullet$ Step-II will give amide.

$\bullet$ Step-III is Hoffmann bromamide degradation reaction.

$\mathrm{[B] R-X+Ag-C\equiv N\to R-NC}$ Isonitrile

When alcoholic KOH react with haloalkanes, the reaction (rate) depends on the following factor :

(i) Size of halogen

(ii) Degree of C - atom, to which the halogen atom is bonded, i.e. more be the degree of haloalkane, more is its reactivity on the above basis.

As, bromine atom is large in size, thus

Hence, correct order is A > C > B > D.

Thus, option (b) is the correct.

Electron-donating groups (EDGs) bonded to a benzene ring activate the ring by the $+I$ effect, enhancing the electrophilic substitution reaction. Conversely, electron-withdrawing groups (EWGs) show a $-I$ effect, reducing the ring's tendency for electrophilic substitution.

Alkyl vs. Alkoxy Groups :

An alkyl group such as $ -\mathrm{CH_3} $ exhibits a $+I$ effect but less so than a methoxy group ($-\mathrm{OCH_3}$) due to the presence of the oxygen atom. Therefore, $\mathrm{C_6H_5OCH_3}$ shows better electrophilic substitution than $\mathrm{C_6H_5CH_3}$.

Chlorine vs. Formyl Groups :

Chlorine ($\mathrm{Cl}$) exhibits both $-I$ and $+R$ effects due to its lone pair of electrons, while the formyl group ($-\mathrm{CHO}$) has a $-R$ nature. Consequently, $\mathrm{C_6H_5Cl}$ shows more electrophilic substitution than $\mathrm{C_6H_5CHO}$.

Groups exhibiting a $+I$ and $+R$ effect show a higher tendency for electrophilic substitution than those showing either or $-I$ and $+R$ effects.

Therefore, the correct order is :

$ B > A > C > D $

and option (c) is the correct choice.

The Wurtz-Fittig reaction is the reaction of an aryl halide with alkyl halide and sodium metal to give substituted aromatic compound.

Thus, Wurtz Fittig reaction is

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}+2 \mathrm{Na}+\mathrm{CH}_3 \mathrm{I} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+2 \mathrm{NaI}$

Assertion and reason both are correct statements but reason is not the correct explanation of assertion. For the same hydrocarbon part, boiling point depends upon the atomic mass of halogen atom. Higher the mass of the halogen atom, higher will be the boiling point. So, we can say that boiling point decreases with decrease in atomic mass of halogen atom. Thus, the decreasing order of boiling point is

$R \mathrm{I}>R \mathrm{Br}>R \mathrm{Cl}>R \mathrm{~F}$