Consider the conversion, CH33C-Br→alc.KOHA→peroxideHBrB ; The compound ‘B’ in the process is
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a
Iso butyl bromide
b
Tertiary butyl bromide
c
n-butyl bromide
d
Secondary butyl bromide
answer is A.
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Detailed Solution
CH33C-Br→alc.KOHCH32C=CH2→peroxideHBrCH32CH -CH2Br; Step I…Dehydrohalogenation to give isobutene;Step II… Addition of HBr by anti markonikov's rule(free radical addition) without rearrangement to give iso butyl bromide;