# Question #47fbe

Jan 2, 2018

Here's what I get.

#### Explanation:

(a) 1-Chloro-3-methylcyclopentane with aqueous $\text{NaOH}$

This is an ${\text{S}}_{\textrm{N}} 2$ substitution. The product is 3-methylcyclopentanol.

(b) 2-Bromo-2-methylpentane with ethanolic $\text{NaOH}$

This is an $\text{E2}$ elimination. The major product is 2-methylpent-2-ene, and the minor product is 2-methylpent-1-ene.

(c) Mechanism

The hydroxide ion reacts with ethanol to produce ethoxide ion in an equilibrium reaction.

$\text{CH"_3"CH"_2"OH" + "OH"^"-" ⇌ "CH"_3"CH"_2"O"^"-" + "H"_2"O}$

The ethoxide ion is a stronger base than $\text{OH"^"-}$.

(d) "We leave it as an exercise for the student" to compare and contrast the ${\text{S}}_{\textrm{N}} 2$ and $\text{E2}$ mechanisms.