Question

A tertiary alkyl halide (X), C\(_4\)H\(_9\)Br, reacted with alc.KOH to give compound (Y). Compound (Y) reacted with HBr in presence of peroxide to give compound (Z). The compounds (Y) and (Z) are respectively:

• A. Propene and tert-butylbromide
• B. 2-methyl-1-propene and 1-bromo-2-methylpropane
• C. but-1-ene and 2-bromopropane
• D. but-2-ene and 2-methylpropane
• E. but-2-ene and 3-methylpropane

Hint:

In the presence of peroxides, HBr follows the anti-Markovnikov rule, where bromine adds to the less substituted carbon.

Answer 1

The Correct Option is B

The reaction of a tertiary alkyl halide with alcoholic KOH leads to elimination (E2 mechanism), forming an alkene (Y). The addition of HBr in the presence of peroxides follows the anti-Markovnikov rule, leading to (Z). For C\(_4\)H\(_9\)Br, the correct intermediate (Y) is 2-methyl-1-propene, which reacts with HBr (in peroxides) to form 1-bromo-2-methylpropane (Z).