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73
Subjective

Elimination reactions (especially $\beta$-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Specify the reagents used in both cases.

Explanation

Elimination reactions are as common as the nucleophilic substitution reaction in case of alkyl halides as two reactions occur simultaneously. Generally, at lower temperature and by using weaker base nucleophilic substitution reaction occur while at higher temperature and by using a stronger base elimination reactions (especially $\beta$ - elimination) take place. e.g., If ethyl bromide is treated with aq. KOH , at low temperature it gives ethanol while if it is treated with alc, KOH at high temperature then it gives ethene.

$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[373 \mathrm{~K}]{\text { Aq. } \mathrm{KOH}} \underset{\substack{\text { Ethanol }}}{\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}}$ (Nucleophilic substitution reaction)

$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[473-523 \mathrm{~K}]{\mathrm{Alc} \mathrm{KOH}} \underset{\text { Ethene }}{\mathrm{CH}_2=\mathrm{CH}_2}$ (Elimination reaction)

74
Subjective

How will you obtain monobromobenzene from aniline?

Explanation

When aniline, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. This diazo salt on treatment with cuprous bromide gives monobromobenzene.

This reaction is named as Sandmeyer's reaction. If benzene diazonium chloride is treated with copper in HBr then the product formed is bromobenzene. This reaction is known as Gattermann reaction.

75
Subjective

Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution.

Explanation

Aryl halides are less reactive towards nucleophilic substitution reaction. Presence of electron withdrawing group at ortho and para position increases the stability of intermediates and hence increases the reactivity of aryl halides towards nucleophilic substitution reaction.

Now, more the number of EWG at ortho and para position, higher will be the reactivity of aryl halide. Compound (III) has three EWG so, it is most reactive and compound (I) has only one EWG so, it is least reactive. So, the order of reactivity is (I)<(II)<(III)

76
Subjective

tert-Butylbromide reacts with aq. NaOH by $\mathrm{S}_{\mathrm{N}} 1$ mechanism while n -butylbromide reacts by $\mathrm{S}_{\mathrm{N}} 1$ mechanism. Why?

Explanation

Tert. butyl bromide reacts with aq. NaOH as follows

tert. butyl bromide when treated with aq. NaOH , it forms tert. corbocation which is more stable intermediate. This intermediate is further attacked by ${ }^{-} \mathrm{OH}$ ion. As tert. carbocation is highly stable so tert butylbromide follow $\mathrm{S}_{\mathrm{N}} 1$ mechanism. In case of $n$-nutylbromide, primary carbocation is formed which is least stable so, it does not follow $\mathrm{S}_{\mathrm{N}} 1$ mechanism. Here, stearic hindrance is very less so, it follow $\mathrm{S}_{\mathrm{N}} 2$ mechanism. In $\mathrm{S}_{\mathrm{N}} 2$ mechanism, ${ }^{-} \mathrm{OH}$ will attack from backside and a transition state is formed. The leaving group is then pushed off the eopposite side and the product is formed.

77
Subjective

Predict the major product formed when HCl is added to isobutylene, Explain the mechanism involved.

Explanation

Reaction between the isobutylene added to HCl

Electrophilic addition reaction takes place in accordance with Markownikoff's rule.

We know that $3^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation because in further step $3^{\circ}$ carbocation is further attacked by $\mathrm{Cl}^{-}$ion.