A) Ethyl bromide\[+\]alcoholic\[KCN\]
B) Propyl bromide\[+\]alcoholic\[KCN\]
C) Propyl bromide\[+\]alcoholic\[AgCN\]
D) Ethyl bromide\[+\]alcoholic\[AgCN\]
Correct Answer: A
Solution :
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.
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