Why Is This Considered an SN2 Reaction?

[phEight]

Homework Statement

CH3CH2-Br + (CH3)3C-O¯+K ----> (CH3)3C-O-CH2CH3
ethylbromide + potassium t-butoxide ---> ethyl t-butyl either

can someone please tell me why this is a Sn2 reaction?

Homework Equations

The Attempt at a Solution

I figured it should be Sn1 because the nucleophile is tertiary, but that is not the case. Is it because O is primary? Even then the nucleophile seems very crowded for an Sn2 reaction. Thanks in advanced.

 

[Kalirren]

The whole reason why tertiary/secondary/primary substitution level matters is that it stabilizes the intermediate carbocation, i.e., substitution level matters on the electrophile, not the nucleophile. It doesn't matter whether the nucleophile is tertiary or primary.

CH3CH2-Br is synthetically equivalent to CH3CH2(+) - hence, it is the electrophile. Since the electrophile is primary, and there is no resonance stabilization of the carbocation, the reaction is likely to proceed by an SN2 mechanism.

 

[Blueshift5]

The reaction shown is indeed an Sn2 reaction. Sn2 reactions are characterized by a single step mechanism where the nucleophile attacks the substrate and simultaneously displaces the leaving group. This reaction mechanism is favored when the substrate is primary or secondary and the nucleophile is strong and unhindered. In this case, the substrate (ethylbromide) is primary and the nucleophile (t-butoxide) is strong and unhindered, making it a good candidate for an Sn2 reaction. The fact that the t-butoxide is tertiary does not hinder the reaction because it is not involved in the rate-determining step. The crowded nature of the t-butoxide does not affect the reaction because the bulky tert-butyl group is able to stabilize the intermediate carbocation, making it a favorable Sn2 reaction.