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Question
The major product \([C]\) of the following reaction sequence will be
\(
CH _{2}= CH - CHO \frac{ i ^{ NaBH _{4}}}{\text { (i) } SOC _{2}}[ A ]
\)
\( \frac{\text { Anlif }}{\text { AICY }_{3}}=[ B ]^{ DB }=[ C ] \)
The major product \([C]\) of the following reaction sequence will be
\(
CH _{2}= CH - CHO \frac{ i ^{ NaBH _{4}}}{\text { (i) } SOC _{2}}[ A ]
\)
\( \frac{\text { Anlif }}{\text { AICY }_{3}}=[ B ]^{ DB }=[ C ] \)
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Solution
The correct option is B
In first step, aldehyde is reduced to primary alcohol by NaBH4 followed by SN2 reaction to for alkyl halide.
In next step on reacting with benzene in presence of AlCl3 undergoes Friedel craft alkylation reaction.
In final step, DBr gets added across alkene. This reaction goes by carbocation intermediate. Thus, after addition of H+ it forms secondary carbocation which undergoes 1, 2 -H shift to form stable benzylic cation followed by Br− addition to form major product.
Hence, option (c) is correct.
In first step, aldehyde is reduced to primary alcohol by NaBH4 followed by SN2 reaction to for alkyl halide.
In next step on reacting with benzene in presence of AlCl3 undergoes Friedel craft alkylation reaction.
In final step, DBr gets added across alkene. This reaction goes by carbocation intermediate. Thus, after addition of H+ it forms secondary carbocation which undergoes 1, 2 -H shift to form stable benzylic cation followed by Br− addition to form major product.
Hence, option (c) is correct.
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Major product [B] of the given reaction would be:Join BYJU'S Learning Program
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