My teacher took off marks because he said the reaction was SN1. Before I go confront him, please tell me if my reasoning is correct.

So, basically, I thought: we have two possibilities: 1. SN1: a 2°benzylic has the same stability of carbocation as a tertiary carbocation so, SN1 could work. 2. SN2: we have a secondary carbocation and a weak base cyanide: SN2 could work.

So I just randomly put the rate law for SN2 since the problem did not specify SN1 or SN2 and there is no way I can know which mechanism is prominent.

  • 1
    $\begingroup$ The fact that a benzene ring is attached to the potential carbocation center pushes the mechanism to the SN1 side. $\endgroup$
    – ron
    Oct 25, 2014 at 21:21
  • $\begingroup$ Ron, why? How can you qualitatively assess that? It is still a secondary carbocation and Sn2 will happen. How can you guess having Sn1 will be larger Sn2? $\endgroup$
    – yolo123
    Oct 25, 2014 at 21:30
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    $\begingroup$ The phenyl group makes a big difference, it is now a secondary benzylic carbocation. $\endgroup$
    – ron
    Oct 25, 2014 at 21:33
  • $\begingroup$ When I consult Solomons Organic Chemistry textbook, it says that secondary benzylic carbocations can react by SN2 or SN1. It does not say which one prevails. $\endgroup$
    – yolo123
    Oct 25, 2014 at 21:35
  • $\begingroup$ @ron any idea why? $\endgroup$
    – yolo123
    Oct 25, 2014 at 21:55

1 Answer 1


My inclination is that you are correct and the reaction proceeds by an SN2 reaction. Secondary alkyl halides can react by either an SN1 or SN2 mechanism, so the other reaction conditions are important to make a decision of the predominant pathway. Since cyanide is a strong nucleophile (and often utilized as the prototypical SN2 nucleophile), that suggests that the reaction should go through an SN2 pathway. One other piece of information that we don't have is the solvent. If this is run in a protic solvent, that could tip the pathway toward SN1, while polar aprotic will favor SN2. I also would not be surprised if both pathways are operating.

It is true that the aromatic ring will stabilize a carbocation at the benzylic position, but it will also stabilize the transition state of the SN2 reaction.

A brief literature search has revealed less insight than I expected. The closest result I have been able to find is that an optically active benzylic bromide treated with trimethylsilyl cyanide in acetonitrile (polar aprotic solvent) does give an optically active benzylic cyanide, suggesting SN2.

DOI: 10.1055/s-0030-1260187


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