# Would the following molecules react by Sn1 or Sn2 mechanism: 1-methyl-1-bromo-cyclohexane and 2-bromohexane?

Feb 18, 2016

First, let me preface by saying that no reaction is necessarily 100% ${S}_{N} 1$ or ${S}_{N} 2$. In many cases, at a suitable temperature, it is a mix of the two. Here is a set of reactions from a 2013 O-Chem worksheet that I did:

Some things I could define:

• The compound under each arrow is a solvent.
• $\text{Bu}$ is shorthand for ${\text{CH"_3"CH"_2"CH"_2"CH}}_{2} -$, or butyl.
• When I wrote substitution, I included both ${S}_{N} 1$ and ${S}_{N} 2$ under one umbrella, and similarly, when I wrote elimination, I mean both $E 1$ and $E 2$ combined.

That aside, let's see.

MAIN POINTS ABOUT SN1 AND SN2

See the following for an overview on these types of reactions:

1-BROMO-1-METHYLCYCLOHEXANE

1-bromo-1-methylcyclohexane is a tertiary cyclohaloalkane. It means it has steric hindrance on carbon-1. Therefore, it is more likely to commit to ${S}_{N} 1$ than ${S}_{N} 2$.

Why? Because the steric hindrance makes it more difficult to come up from behind and backside-attack. There's too much bulk and obstruction around the target site that it isn't easy for the reactant to get to where it needs to.

2-BROMOHEXANE

This is where it gets tricky. It is a secondary haloalkane, so it's anyone's guess as to whether ${S}_{N} 1$ or ${S}_{N} 2$ predominates, depending on the solvent conditions...

If I had to guess, I would say in general, ${S}_{N} 1$ is more prominent by a small margin because secondary haloalkanes have some steric obstruction, but not as much as a comparable tertiary haloalkane.

As to what extent ${S}_{N} 1$ predominates, I wouldn't know. All we can say without definitive data to prove it is that ${S}_{N} 1$ is more prominent than ${S}_{N} 2$.