Question

Which one of the following forms propane nitrile as the major product?

• A. Ethyl bromide $+$ alcoholic $KCN$
• B. Propyl bromide $+$ alcoholic $KCN$
• C. Propyl bromide $+$ alcoholic $AgCN$
• D. Ethyl bromide $+$ alcoholic $AgCN$

Answer 1

Answer: (A)

When ethyl bromide reacts with alcoholic KCN, propane nitrile is obtained as main product. $C_2{H}_{5}Br+Alc.KCN\to \underset{propanenitrile}{C_2{H}_{5}CN}$ precipitation, tonic product > solubility product $(K_{sp})$ For $A{g}_{2}Cr{O}_4,$ Ionic product $={[A{g}^{+}]}^2[Cr{O}_4^{-}]$ $={({10}^{-4})}^2({10}^{-5})={10}^{-13}$ $K_{sp}$ of $A{g}_{2}Cr{O}_4=4\times {10}^{-12}$ Here, $K_{sp}>IP$ Thus, no precipitate is obtained. For $AgCl,$ Ionic product $=[A{g}^{+}][C{l}^{-}]$ $=[{10}^{-4}][{10}^{-5}]$ $={10}^{-9}$ $K_{sp}(AgCl)=1\times {10}^{-10}$ Here, $IP>K_{sp}$ So, precipitate will form. Thus, silver chloride gets precipitated first.