Deciding E1/E2/SN1/SN2 for cyanide ion

Question

The $K_\mathrm{a}$ for $\ce{HCN}$ is $6.17 \cdot 10^{-10}$ with a $\mathrm{p}K_\mathrm{a}$ of $9.21$

Thus, the cyanide ion is a strong base.

Also, the cyanide ion is a good nucleophile.

So in the reaction of alkyl halides with $\ce{KCN}$, a mixture of products must be formed depending on the solvent and alkyl group.

However my text suggests that the reaction proceeds only via $\mathrm{S_N2}$.

Shouldn't the product formed depend on the alkyl group, say for a tertiary alkyl group, the reaction can go via $\mathrm{S_N1}$ mechanism for polar protic solvents and via $\mathrm{E2}$ for polar aprotic solvents and as the cyanide ion is strong base, the reaction is unlikely to go via $\mathrm{E1}$ and due to steric factors $\mathrm{S_N2}$ is also ruled out.

Also is there any difference if we use $\ce{KCN}$ or alcoholic $\ce{KCN}$?

Answer 1

A general method for estimating $\mathrm{p}K_\mathrm{b}$ values is to use the equation $\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} = 14$ where $\mathrm{p}K_\mathrm{b}$ is that of the conjugate acid; in this case, applying it gives us $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. We can generally assume that ‘strong’ and ‘weak’ in $\mathrm{p}K_\mathrm{b}$ terms are similarly to be defined as for $\mathrm{p}K_\mathrm{a}$ values. Therefore, cyanide is a weak base and the premise of this question is incorrect. This observation is consistent with the fact that neutral cyanide solutions are relatively safe as long as no acid is introduced to protonate $\ce{CN-}$.

Note that the general statement ‘the conjugate base of a weak acid is a strong base’ is incorrect. The opposite is often true, namely:

The conjugate base of a strong acid is generally a weak base.

And vice-versa if you swap all occurences of acid and base. The word generally is required here, since compounds such as $\ce{H2Cl+}$ are very strong acids whose conjugate bases are still very strong — but these compounds are not typically easily accessable and certainly not in aquaeous solution. A weak acid needs to have $\mathrm{p}K_\mathrm{a} > 15$ for the conjugate base to be a strong one but those are seldomly still considered ‘acidic’. An example would be water ($\mathrm{p}K_\mathrm{a} = 15.7$ therefore $\mathrm{p}K_\mathrm{b}(\ce{OH-}) \approx -1.7$) which is a weak acid with a strong conjugate base.

Since it is such a weak base, it has no access to $\mathrm{E2}$ pathways and $\mathrm{E1}$ pathways are disfavoured against $\mathrm{S_N1}$ if either is applicable. We don’t even need to consider $\mathrm{E1_{cb}}$. Thus, cyanide will generally react via an $\mathrm{S_N2}$ pathway unless the conditions for $\mathrm{S_N1}$ are much better.

Answer 2

By definition, a weak base or weak acid is something that does not completely ionize in solution. Consider acetic acid, which is weak acid, and has a $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH}) \approx 4.756$. Now consider that $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. This means that acetic acid ionizes even more than the cyanide ion does, though it is still a weak acid. Clearly then, the cyanide ion is a weak base, and cannot react with haloalkanes via $\mathrm{E_1}$ or $\mathrm{E_2}$ pathways.

What I assume you mean when you ask about the differences between $\ce{KCN}$ and alcoholic $\ce{KCN}$ is the use of alcohol as a solvent. A polar aprotic solvent (like DMSO or acetone) will greatly increase the the nucleophilicity of $\ce{CN-}$, because they effectively solvate $\ce{K+}$. They do not solvate anions like $\ce{CN-}$ as well though, meaning it will be relatively free to react with the haloalkane. A polar protic solvent (like an alcohol or water) will decrease the nucleophilicity of $\ce{CN-}$ because it solvates it, and thereby reduces its ability to participate in a reaction.

For your reference, the relative rates of reaction of 1-bromobutane with azide happens 1300 times faster in DMSO than in methanol$^1$. Although azide is not the same molecule as cyanide and this is only data for a primary haloalkane, the trend should be clear.

1 - Brown, W., & Iverson, B. (2014). Organic chemistry (7th ed., p. 359)