Question

When $C{H}_{3}C{H}_{2}Br$ is treated with aqueous NaCN, $C{H}_{3}C{H}_{2}CN$ is formed while $C{H}_{3}C{H}_{2}NC$ is formed when AgCN is used in the similar reaction. It is due to

• A. Ionic nature of $C-N$ bond in $NaCN$
• B. Covalent nature of $Ag-C$ bond in $AgCN$
• C. Covalent nature of $C-N$ bond in $NaCN$
• D. Ionic nature of $Ag-C$ bond in $AgCN$

Answer 1

Answer: (A), (B)

Covalent nature of $Ag-C$ bond in $AgCN$

$C{N}^{-}$ is the ambident nucleophile and can form different product with same substrate.
$C{H}_3-C{H}_2-Br+NaCN\to C{H}_3-C{H}_2-CN+C{H}_3-C{H}_2\stackrel{+}{N}\equiv C^{-}+NaBr\text{Ethyl cyanide}\text{Ethyl isocyanide}\text{Major}\text{Minor}$
Here, $NaCN$ is predominantly ionic and exist as free ions in solution. C site is more nucleophilic than N site and available to attack the ethyl bromide​. Hence, the major product will be ethyl cyanide​.
$AgCN$ is predominantly covalent. Although C site is more nucleophilic than N site but not available in case of $AgCN$. Hence, lone pair of N attacks the ethyl chloride and isocyanide is formed as the major product.​ Hence, the major product will be ethylisocyanide.
$C{H}_{3}C{H}_2-Br+AgCN\to C{H}_{3}C{H}_{2}CN+C{H}_{3}C{H}_{2}NC+AgBr\text{Ethyl cyanide}\text{Ethyl isocyanide}\text{Minor}\text{Major}$