Will this beta-branched secondary alkyl halide undergo SN1 or SN2?

[![](https://i.stack.imgur.com/ovimO.png "source: imgur.com")](https://i.stack.imgur.com/yYuuh.jpg)


What is the correct mechanism here? Would SN1 occur here? I chose SN2 because we have a secondary alkyl halide with a weak base $\ce{CN-}$.


However, the solution says SN1 is major. Can you explain why?




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My friend got full credit for this:  

[![](https://i.stack.imgur.com/zcXPC.png "source: imgur.com")](https://i.stack.imgur.com/Q5yyO.jpg)




Secondary alkyl halides are on the borderline of $\ce{S\_{N}2}$ and $\ce{S\_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S\_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S\_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.


Full mechanism:  

![full mechanism](https://i.stack.imgur.com/19nO3.jpg)


I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.  

I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S\_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S\_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.


 15

[Murpy, TJ, *J. Chem. Ed.* **2009**, 86, 519-524](http://dx.doi.org/10.1021/ed086p519) shows that secondary alkyl halides do not undergo SN1 reactions.
This is counter to what we have been (and still are in many cases) teaching.


 1

* It is in $2^\circ$ position, so both, $\ce{S\_{N}1}$ and $\ce{S\_{N}2}$ can happen.
* $\ce{Br}$ is a good leaving group probably the third best in the groups very common(depends on opinion, but it is a good one anyways though):
$\ce{OTs>I>Br>Cl>...}$ but LG accelerates both reactions
* $\ce{CN-}$ is a very good nucleophile in the list of common ones(depends on opinion, but it is a good one anyways though): $\ce{PhS- > CN- > I- > ...}$, the second best so $\ce{S\_{N}2}$ is favourable.
* The solvent is water, which is polar protic which highly favours $\ce{S\_{N}1}$
* The temperature is normal so there's no elimination
* Finally fourth point overthrows thirds and moreover water reduces nucleophilicity of $\ce{CN^-}$


 0

I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required


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