Question

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

Answer 1

First, let me preface by saying that no reaction is necessarily `100%` `S_N1` or `S_N2`. 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.
• `"Bu"` is shorthand for `"CH"_3"CH"_2"CH"_2"CH"_2-`, or butyl.
• When I wrote substitution, I included both `S_N1` and `S_N2` under one umbrella, and similarly, when I wrote elimination, I mean both `E1` and `E2` combined.

That aside, let's see.

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MAIN POINTS ABOUT SN1 AND SN2

See the following for an overview on these types of reactions:
http://socratic.org/scratchpad/0980681261bdd0bf3e98

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_N1` than `S_N2`.

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_N1` or `S_N2` predominates, depending on the solvent conditions...

If I had to guess, I would say in general, `S_N1` 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_N1` predominates, I wouldn't know. All we can say without definitive data to prove it is that `S_N1` is more prominent than `S_N2`.