The product (C) in the below mentioned reaction is:
${CH}_{3} - {CH}_{2} - {CH}_{2} - Br\mathop{\to}_{∆}^{{KOH}_{\left(alc\right)}}A\overset{HBr}{\to}B\mathop{\to}_{{KOH}_{\left(aq\right)}}^{∆}C$

Question

The product (C) in the below mentioned reaction is: CH _ 3 - CH _ 2 - CH _ 2 - Br → _ ∆ ^ KOH _ (alc) A HBr → B → _ KOH _ (aq) ^ ∆ C

Quantrex Academy · Accepted Answer

Correct answer: D. Propan-2-ol

When haloalkane or alkyl halide with a β hydrogen is heated with alcoholic solution of $KOH$ , elimination of a hydrogen atom from $\beta$ carbon and halogen atom from $\alpha -$ carbon occurs as a result, alkene is formed as product. Since $\beta$ hydrogen atom is involved in the elimination reaction, it is often called $\beta$ elimination. An alkene is formed. This alkene then on reaction with $HBr$ undergoes addition reaction to give bromopropane, which on reaction with aqueous $KOH$ gives alcohol, propan $- 2 -$ ol.

The product (C) in the below mentioned reaction is: CH _ 3 - CH _ 2 - CH _ 2 - Br → _ ∆ ^ KOH _ (alc) A HBr → B → _ KOH

The product (C) in the below mentioned reaction is: CH _ 3 - CH _ 2...

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