Thus, the cyanide ion is a solid base.
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Also, the cyanide ion is a good nucleophile.
So in the reaction the alkyl halides v $\ceKCN$, a mixture of assets must be formed relying on the solvent and alkyl group.
However mine text suggests that the reaction proceeds only via $\mathrmS_N2$.
Shouldn"t the product created depend on the alkyl group, say because that a tertiary alkyl group, the reaction deserve to go via $\mathrmS_N1$ system for polar protic solvents and via $\mathrmE2$ for polar aprotic solvents and as the cyanide ion is strong base, the reaction is unlikely to go via $\mathrmE1$ and also due to steric components $\mathrmS_N2$ is additionally ruled out.
Also is there any type of difference if we use $\ceKCN$ or alcoholic $\ceKCN$?
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A general method for estimating $\mathrmpK_\mathrmb$ worths is to usage the equation $\mathrmpK_\mathrmb + \mathrmpK_\mathrma = 14$ wherein $\mathrmpK_\mathrmb$ is that of the conjugate acid; in this case, using it offers us $\mathrmpK_\mathrmb(\ceCN-) \approx 4.79$. Us can usually assume that ‘strong’ and also ‘weak’ in $\mathrmpK_\mathrmb$ state are an in similar way to be identified as for $\mathrmpK_\mathrma$ values. Therefore, cyanide is a weak base and also the premise of this concern is incorrect. This monitoring is constant with the truth that neutral cyanide remedies are relatively safe as lengthy as no mountain is introduced to protonate $\ceCN-$.
Note that the basic statement ‘the conjugate base of a weak acid is a solid base’ is incorrect. The opposite is frequently true, namely:
The conjugate base of a strong acid is generally a weak base.
And vice-versa if friend swap all occurences that acid and base. The word generally is forced here, because compounds such together $\ceH2Cl+$ are very strong acids whose conjugate bases are still very solid — however these compounds room not generally easily accessable and certainly no in aquaeous solution. A weak acid requirements to have actually $\mathrmpK_\mathrma > 15$ for the conjugate basic to be a solid one yet those space seldomly still considered ‘acidic’. An instance would it is in water ($\mathrmpK_\mathrma = 15.7$ thus $\mathrmpK_\mathrmb(\ceOH-) \approx -1.7$) i m sorry is a weak acid through a solid conjugate base.
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Since that is together a weak base, it has actually no access to $\mathrmE2$ pathways and $\mathrmE1$ pathways room disfavoured versus $\mathrmS_N1$ if one of two people is applicable. We don’t even need to think about $\mathrmE1_cb$. Thus, cyanide will typically react via one $\mathrmS_N2$ pathway unless the problems for $\mathrmS_N1$ are much better.