Question

(f) Which of the following alkenes is the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol?

A) (E)-4-methylpent-2-ene
B) (Z)-4-methylpent-2-ene
C) 2-methylpent-1-ene
D) 2-methylpent-2-ene
E) 4-methylpent-1-ene

Answer 1

D) 2-methylpent-2-ene

Step-by-step explanation:

This is an elimination reaction of Halogenoalkane. 2-bromo-2-methylpentane when is heated with NaOH or NaOC2O5( sodium ethoxide) in ethanol will form alkene rather than alcohol.

2-methylpent-1-ene is minor product since double bond form with secondary Carbon rather than primary Carbon.

Answer 2

The major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is: 2-methylpent-2-ene

Step-by-step explanation:

The elimination reaction generates the most substituted alkene (Zaitsev's rule)

Hence, the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is 2-methylpent-2-ene.

In the attached picture, you can see that the 2-methylpent-1-ene is less substituted