Why can't we use $\ce{HCN (g)}$ and $\ce{HCl (aq)}$ to convert 1,2-dibromoethane into butanedioic acid?

I found out that the answer is $\ce{KCN (aq/alcoholic)}$ and $\ce{H2SO4 (aq)}$, I don't understand the differences between these reagents and their reactions.


1 Answer 1


In the first step — aside from the fact that nobody wants to work with gaseous $\ce{HCN}$ due to its toxicity — using $\ce{HCN}$ wouldn’t work. The reaction in question is a bimolecular nucleophilic substitution. For that, you need to have a nucleophilic lone pair that is capable of attacking and it has to be on carbon (otherwise we would arrive at a diisonitrile). However, in hydrogen cyanide there is no free lone pair on carbon; it would need to get deprotonated before it can attack. Rather than using the toxic, gaseous $\ce{HCN}$ and deprotonating it, it is much quicker to just immediately start from an ionic compound such as $\ce{KCN}$ wherein the cyanide is already present as $\ce{^-C#N}$ anions ready for nucleophilic attack.

$$\ce{Br-CH2-CH2-Br + 2 CN- -> NC-CH2-CH2-CN + 2 Br-}\tag{1}$$

In your second step, it really doesn’t matter which acid you use. In fact, many problem sets wouldn’t even specify the acid, they would just write what I did in equation $(2)$. Thus, you should get the same points for using $\ce{HCl}$ as you would for using $\ce{H2SO4}$. You should use a strong acid, though; acetic acid will not do the trick.

$$\ce{NC-CH2-CH2-CN ->[H+/H2O][-2 NH3] HOOC-CH2-CH2-COOH}\tag{2}$$

  • $\begingroup$ Would $\ce{NAOH(aq)}$ and $\ce{K2Cr2O7/H2SO4(aq)}$ work because of the $\ce{OH}$ having a lone pair? $\endgroup$
    – rudraksh
    Commented Jan 22, 2017 at 18:10
  • $\begingroup$ @rudraksh Hydroxide does not work. Cyanide provides two extra carbons to give the 4-carbon chain. $\endgroup$
    – Zhe
    Commented Jan 22, 2017 at 18:17

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