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Question: 2-bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is....

2-bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is.

A

Pentene-1

B

cis pentene-2

C

trans pentene-2

D

2-ethoxypentane

Answer

trans pentene-2

Explanation

Solution

CH3−CH∣Br−CH2−CH2−CH3+KOHCH_{3} - \underset{Br}{\underset{|}{CH}} - CH_{2} - CH_{2} - CH_{3} + KOH

& \text{Elimination} \\ & \text{reaction} \end{aligned}}{\overset{\quad C_{2}H_{5}OH\quad}{\rightarrow}}\underset{\text{2-Pentene}}{CH_{3} - CH = CH - CH_{2} - CH_{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. <img src="https://cdn.pureessence.tech/canvas_139.png?top_left_x=900&top_left_y=1024&width=288&height=131" style="width:2.52083in;height:1.14583in" />