1. Tardigrade
  2. Question
  3. Chemistry
  4. In the following sequence of reactions, CH3CH2CH2Br xrightarrowKOH(alc.)(A) xrightarrowHBr(B) xrightarrowKOH(aq.) (C) The product C is:

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

JIPMERJIPMER 2001

Solution:

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}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}}\,\xrightarrow[KOH]{Alc.}\underset{\text{Propene}}{\mathop{C{{H}_{3}}CH==C{{H}_{2}}}}\, $ $ \xrightarrow{HBr}\overset{Br}{\mathop{\overset{|}{\mathop{C{{H}_{3}}CH\cdot C{{H}_{3}}}}\,}}\,\xrightarrow{KOH(aq.)} $ $ \underset{(C)\,\text{Propan-2-ol}}{\mathop{C{{H}_{3}}CH(OH)\cdot C{{H}_{3}}}}\, $