Chloroethane can be used to make sodium propanoate.

$$\ce{chloroethane -> Q -> sodium propanoate}$$

The Intermediate, $\ce{Q}$, is hydrolysed with boiling aqueous sodium hydroxide, to give sodium propanoate.

Which reagent would produce the Intermediate, $\ce{Q}$, from chloroethane?

A concentrated ammonia solution
B dilute sulfuric acid
C hydrogen cyanide
D potassium cyanide

What is the difference between using hydrogen cyanide or potassium cyanide? Why would the answer be D?


2 Answers 2


Because hydrogen cyanide is not a very strong acid.

Potassium cyanide exists as potassium and cyanide ions, so here the cyanide ions can act as strong nucleophiles. Hydrogen cyanide is only a weak acid and so it will not dissociate to produce a significant amount of the cyanide ion nucleophiles. So it is a better idea to use potassium cyanide.

  • 2
    $\begingroup$ There's also the safety issue. Hydrogen cyanide is a toxic gas that can be lethal even at really small doses. $\endgroup$
    – Variax
    Aug 5, 2016 at 10:32

For cyanide to act as a nucleophile, we need the carbon-centred lone pair to be accessible as this orbital corresponds to the nucleophilic HOMO. In hydrogen cyanide, this is not a lone pair but occupied by a hydrogen atom forming a covalent bond. See the structures of hydrogen cyanide and the cyanide anion below.

$$\begin{array}{cc} \ce{H-C#N:} & \ce{:\overset{-}{C}#N:}\\ \text{hydrogen cyanide} & \text{cyanide anion}\end{array}$$

This immediately tilts the balance in favour of the cyanide anion. But hydrogen cyanide is also an acid (e.g. compare its German name Blausäure or blue acid). Can it not deprotonate to give cyanide? Well, it can, but it is a very weak acid. The Evans $\mathrm{p}K_\mathrm a$ table gives its $\mathrm pK_\mathrm a$ value as $9.4$ in water and $12.9$ in $\ce{DMSO}$. Therefore, relatively strong basic conditions are required to generate substantial amounts of the anion. The reaction conditions, however, do not mention any added base. It is therefore reasonable to assume neutral conditions, which means that the vast majority of cyanide species when using hydrogen cyanide will indeed be the free acid $\ce{HCN}$ which is not able to attack.

This is avoided by using the already deprotonated ionic compound potassium cyanide.


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