question_answer

2-bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is [CBSE PMT 1998]

A) Pentene-1

B) cis pentene-2

C) trans pentene-2

D) 2-ethoxypentane

Correct Answer: C

Solution :

           \[C{{H}_{3}}-\underset{Br\,\,}{\mathop{\underset{|\,\,\,}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}+KOH\] \[\underset{\begin{smallmatrix}  \text{Elimination} \\  \,\,\,\,\,\text{reaction} \end{smallmatrix}}{\mathop{\xrightarrow{{{C}_{2}}{{H}_{5}}OH}}}\,\underset{\text{2-Pentene}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}}}\,\] When alkyl halide reacts with alc. KOH then it favours elimination reaction (Dehydrohalogenation). Since, Trans pentene-2 is more symmetrical than cis isomers. Hence, it is main product.