I was given the following question on a sample exam for my organic chemistry course:


I am quite certain that it would happen in a SN2-type reaction considering that

  1. the nucleophile is $\ce{CN-};$
  2. the solvent is DMF.

However, I'm not sure what electrophilic center would be attacked first. I'd appreciate any explanation about why the nucleophile would choose one carbon over the other (if that were the case). The question also states that there are 2 product formed, so could it be possible that an elimination reaction is happening as well?

  • 1
    $\begingroup$ What do you think would happen if it attacked the carbonyl? Ok, you add into the carbonyl group, then what happens? $\endgroup$ – orthocresol Jan 11 at 10:50
  • $\begingroup$ The $\pi$ electrons would go up to the carbonyl oxygen, giving it a negative charge? $\endgroup$ – w_p_d_01 Jan 11 at 11:11

If it attacks the carbonyl carbon, then you form an unstable tetrahedral intermediate with a negative formal charge on the oxygen. Unless you treat the compound with water so that the oxygen can remove it's formal charge, what will happen is that the reaction goes backwards where the CN group is kicked off.

However, if the CN attacks the Cl group, then you can have a substitution where you get a stable compound finally.

  • $\begingroup$ Can the elimination reaction also occur? The question on the exam also stated that there are two products (A & B) formed $\endgroup$ – w_p_d_01 Jan 11 at 11:43
  • 2
    $\begingroup$ If the elimination occurs, the product is a Michael acceptor and cyanide will add giving the same product as the direct nucleophilic displacement of chloride. The other product is likely the cyanohydrin $\endgroup$ – Waylander Jan 11 at 12:28

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