-1
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.