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

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?

My friend got full credit for this:

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:

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.

• Jerepierre, How could I have known SN1 would have won? Solvent effet and branching at beta makes the situation a bit ambiguous and when we see CN-, we are tempted to do SN2? – yolo123 Nov 1 '14 at 15:05
• You make a hydride shift from the tertiary carbon. – yolo123 Nov 1 '14 at 16:05

Murpy, TJ, J. Chem. Ed. 2009, 86, 519-524 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.

• Perhaps you could give some more details about the paper in question and ideally provide a link to it. At the moment this seems more like a comment than an answer. – bon Nov 17 '15 at 20:19
• 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 $\ce{CN^-}$

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

• Although HCN is a weak acid, in water, $\ce CN^-$ will stay as is because if it takes $\ce H^+$ from water, the net equilibrium constant of the reaction will be $\ce {K_w}/{K_a}$ = $\ce 10^{10} * 10^{-16}$ = $\ce 10^{-6}$ – Ayushmaan Oct 6 '17 at 17:59