Question

In the following sequence of reactions, $C{H}_{3}C{H}_{2}C{H}_{2}Br\stackrel{KOH(alc.)}{\to }(A)\stackrel{HBr}{\to }(B)$ $\stackrel{KOH(aq.)}{\to }(C)$ The product C is:

• A. propan-2-ol
• B. propan-1-ol
• C. propyne
• D. propene

Answer 1

Answer: (A)

Alkyl hallides dehydrohalogenates in presence of alcohalic alkali to give alkenes when added HBr. Hence, the reactions proceed as follows: $\underset{\text{Propyl}\text{bromide}}{C{H}_{3}C{H}_{2}C{H}_{2}Br}\underset{KOH}{\overset{Alc.}{\to }}\underset{\text{Propene}}{C{H}_{3}CH==C{H}_2}$ $\stackrel{HBr}{\to }\stackrel{Br}{\stackrel{|}{C{H}_{3}CH\cdot C{H}_3}}\stackrel{KOH(aq.)}{\to }$ $\underset{(C)\text{Propan-2-ol}}{C{H}_{3}CH(OH)\cdot C{H}_3}$