JEE Main 2022 · 29 Jul · Shift 1 · Chemistry Q74

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ChemistrySingle correct+4 / -1Question 74 of 90

Ethyl chloride is treated with AgCN in $\mathrm{C_2H_5OH-H_2O}$ to give 'A' (Major Product), and with NaCN in $\mathrm{C_2H_5OH-H_2O}$ to give 'B' (Major Product), as shown.

Considering the above reactions, the compound 'A' and compound 'B' respectively are :

Ethyl chloride in the centre: with AgCN (C2H5OH-H2O) to the left it gives 'A', with NaCN (C2H5OH-H2O) to the right it gives 'B'.

\mathrm{CH_3CH_2-\overset{+}{N}{\equiv}\overset{-}{C}}

\mathrm{CH_3CH_2-\overset{+}{N}{\equiv}\overset{-}{C}}

Option A: ethyl isocyanide, ethyl isocyanide.

\mathrm{CH_3CH_2-C{\equiv}N}

\mathrm{CH_3CH_2-C{\equiv}N}

Option B: propanenitrile (ethyl cyanide), propanenitrile.

\mathrm{CH_3CH_2-\overset{+}{N}{\equiv}\overset{-}{C}}

\mathrm{CH_3CH_2-C{\equiv}N}

Option C: ethyl isocyanide, propanenitrile.

\mathrm{CH_3CH_2-C{\equiv}N}

\mathrm{CH_3CH_2-\overset{+}{N}{\equiv}\overset{-}{C}}

Option D: propanenitrile, ethyl isocyanide.

Answer and worked solution

Official answer: C

Ambident cyanide. AgCN is largely covalent, so attack is through N and gives ethyl isocyanide. NaCN is ionic, and the more nucleophilic C end attacks to give propanenitrile.

(C)

Topics:Cyanides and isocyanides SN1 and SN2

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JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 74