0
$\begingroup$

At which position will electrophilic substitution on p-cyanonitrobenzene occur?

As both are -M groups, there are two possible sites of reaction, meta to each of the groups.
The answer is meta to $\ce{CN-}$. Why is this so? According to me it should be meta to $\ce{NO2}$, because $\ce{NO2}$ is the stronger -M group?

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. $\endgroup$ – user15489 Aug 27 '15 at 9:11
  • 2
    $\begingroup$ For such compound most eletrophilic subst. wouldn't work $\endgroup$ – Mithoron Aug 27 '15 at 12:15
3
$\begingroup$

As Mithoron suggests, the correct answer is probably "no reaction." There are no examples of electrophilic aromatic substitution of p-cyanonitrobenzene in SciFinder, by the introductory organic chemistry mechanism at least.

There is one report of halogenation (with N-halosuccinimide) of this compound with palladium catalysis. The cyano group serves as a directing group to place a halogen ortho to the cyano. Chlorination and bromination are both reported.

Reference: Palladium-Catalyzed Highly Selective ortho-Halogenation (I, Br, Cl) of Arylnitriles via sp2 C–H Bond Activation Using Cyano as Directing Group

$\endgroup$
  • $\begingroup$ Do you mean that there is no such reaction at all ? $\endgroup$ – mathemather Aug 28 '15 at 6:05
  • $\begingroup$ @arutoregni using the typical conditions. For example, reaction of p-cyanonitrobenzene with bromine and iron tribromide has never been reported in the literature. Halogenation has been accomplished by a different mechanism, which is what is described in the answer. $\endgroup$ – jerepierre Aug 28 '15 at 15:54

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.