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I was doing an exam question where benzyl chloride was undergoing a nucleophilic substitution with cyanide ion to form benzyl cyanide.

Please can someone enlighten me as I have never seen both conditions before just one or the other. I have Googled it but no sites talk about the conditions of nucleophilic substitution with a cyanide ion in a benzene context. Does the benzene alter the conditions?

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  • $\begingroup$ @orthocresol: Was benzene lost in the editing? Unclear where it comes into play. $\endgroup$
    – user55119
    Commented Apr 26, 2019 at 0:04
  • $\begingroup$ @user55119 I’m pretty sure “benzene” here refers to the electrophile being benzylic. $\endgroup$
    – orthocresol
    Commented Apr 26, 2019 at 6:51

2 Answers 2

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Since $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is 9.21 (Wikipedia), the cyanide ion in water is pretty alkaline and contains significant amounts of hydroxide ions, according to the following equilibrium: $$\ce{NC- + H2O <=> OH- + HCN}$$ That's why you smell strong $\ce{HCN}$ in aqueous $\ce{KCN}$ or $\ce{NaCN}$ solution. Thus, when you perform the reaction of $\ce{KCN}$ or $\ce{NaCN}$ with the haloalkane (including allyl and benzyl) in water you tend to get the substitution by $\ce{^-OH}$ instead of $\ce{^-CN}$. To get major product as cyanoalkane, you may need to do the reaction in absolute alcohol.

Not relevant, but note worthy that $\ce{^-CN}$ is an ambident nucleophile, meaning a nucleophile which can attack acitve site of a molecule through more than one sites (here either by $\ce{N}$ or $\ce{C}$). Thus, a reaction of alkyl halide ($\ce{R-X}$) with $\ce{KCN}$ gives alkyl nitrile ($\ce{R-CN}$) while a reaction of alkyl halide with $\ce{AgCN}$ gives an alkyl isonitrile ($\ce{R-NC}$).

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  • $\begingroup$ Not that I am willing to find out, but does KCN in ethanol smell of HCN? Certainly, the pKa's of water and ethanol are similar. Although the solubility of KCN in ethanol is less than water, I suspect that the greater solubility of benzyl chloride in ethanol is the determining factor. If benzyl alcohol is formed in aqueous KCN, might a solvolysis (SN1) be competing? Food for thought. $\endgroup$
    – user55119
    Commented Apr 26, 2019 at 14:26
  • $\begingroup$ Followup: Organic Synthesis: Benzyl chloride in ethanol added to aqueous NaCN. 80-90% yield distilled of phenylacetonitrile on a kilogram scale. orgsyn.org/demo.aspx?prep=CV1P0107 $\endgroup$
    – user55119
    Commented Apr 26, 2019 at 14:54
  • $\begingroup$ @user55119: I have seen this reference before I answered. Yet, I believe my argument has a sense. During the large scale synthesis, they have used about 35% water in ethanol and concentration of the solute is about $\pu{5 M}$. Thus, $\ce{^-CN}$ ion predominate over hydroxide and ethoxide. That's why in these syntheses nowadays, phase transfer catalyst is popular. One more point about water Vs ethanol: If you use 50% aq. ethanol for solvolysis of 1-chloro-1-phenylethane, you always get 1-phenylethanol (no ethoxy product). $\endgroup$
    – Mathew Mahindaratne
    Commented Apr 26, 2019 at 15:34
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The cyanide ion is a very good nucleophile and reacts readily with benzyl chloride. The problem is that the cyanide salts are soluble in water, while the benzyl chloride is not. Therefore, ethanol is added so that both reagents are slightly soluble in the medium and the reaction proceeds.

An alternative to this system is to use phase transfer catalysts like tetrabutylammonium salts. This cation can then transport the cyanide to the organic phase, for instance toluene, in which the benzyl cyanide is dissolved, and the reaction occurs there.

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  • $\begingroup$ More broadly speaking do reactions involving haloalkanes or halogen functional groups require alcoohlic conditions as the h-x bond is polar? $\endgroup$
    – daniel
    Commented Apr 26, 2019 at 21:33
  • $\begingroup$ No, it is not required. Apolar solvents are fine. In this case it is just a point of solubility. Note that the most polar reagent is the cyanide anion (and it's counter ion). The reaction itself is an Sn2, so not much polar stabilization is required. What is needed is to join the two reagents, either by choosing a solvent in which both are soluble or getting the cyanide in the solvent with a phase transfer catalyst $\endgroup$
    – Raoul Kessels
    Commented Apr 27, 2019 at 10:23

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