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\xrightarrow[{}]{{}}\underset{propane\text{ }nitrile}{\mathop{{{C}_{2}}{{H}_{5}}CN}}\, $ precipitation, tonic product > solubility product $ ({{K}_{sp}}) $ For $ A{{g}_{2}}Cr{{O}_{4}}, $ Ionic product $ ={{[A{{g}^{+}}]}^{2}}[CrO_{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.