2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields

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Solution

The correct option is C
Sodium methoxide in methanol is a strong base, it will undergo E2 elimination to give more substituted alkene product (Saytzeff product).

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2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:

(a) $C H_{3} | C_{2} H_{5} C H_{2} C - O C H_{3} | C H_{3}$

(b) $C_{2} H_{5} C H_{2} C = C H_{2} | C H_{3}$

R_{2} C H X

(C l_{3} C C O_{2} M e)

N a O M e

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