JEE Main Chemistry question
A cyclic alcohol precursor is treated with (i) then (ii) to give the rearranged alcohol (major product, via a ring-expansion hydride shift), which is then converted to the bromide with HBr, and finally treated with the bulky base and heat to give the elimination product (major product). Compound is:
Answer: (C)
Solution
Protonation of the cyclic alcohol generates a carbocation that undergoes a 1,2-hydride shift with ring expansion to a more stable, larger-ring secondary/tertiary carbocation, which is captured by water to give the rearranged alcohol . Treatment of with HBr converts the alcohol to the corresponding
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