\(\mathrm{BrCH_2{-}CHO}\) \(\xrightarrow[\ \text{dry HCl gas}\ ]{\ \mathrm{EtOH\ excess}\ }\) "A" (major product) \(\xrightarrow{\ \mathrm{{}^tBuO^{-}K^{+}}\ }\) "B" (major product)
[where \(\mathrm{Et} \Rightarrow {-}\mathrm{C_2H_5}\), \(\mathrm{{}^tBu} \Rightarrow (\mathrm{CH_3})_3\mathrm{C{-}}\)]
Consider the above reaction sequence, Product "A" and Product "B" formed respectively are :
[where \(\mathrm{Et} \Rightarrow {-}\mathrm{C_2H_5}\), \(\mathrm{{}^tBu} \Rightarrow (\mathrm{CH_3})_3\mathrm{C{-}}\)]
Consider the above reaction sequence, Product "A" and Product "B" formed respectively are :
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The Correct Option is A
Solution and Explanation
Step 2: Potassium tert-butoxide is a strong, bulky base, so it removes the \(\mathrm{H}\) on the carbon bearing the two OEt groups (E2) with loss of \(\mathrm{Br^-}\): \(\mathrm{CH_2{=}C(OEt)_2}\) ("B", a ketene diethyl acetal).
Substitution by the bulky \(\mathrm{{}^tBuO^-}\) is not favoured, so the tert-butyl ether options are rejected.
Answer: (1) \(\mathrm{BrCH_2CH(OEt)_2}\) and \(\mathrm{CH_2{=}C(OEt)_2}\)
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