Draw the product of the following reaction and provide an explanation for this reaction knowing that $\mathrm{ROH}$ or $\mathrm{RNH_2}$ + $\mathrm{PhCl}$ do not react at all.

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I have no doubt that the answer is hydroxychloroquine (HCQ), but how would it be formed if the reaction: $\mathrm{RNH_2}$ + $\mathrm{PhCl}$ does not happen? Or, is the question hinting that the reaction takes the $\mathrm{S_N1}$ pathway where the phenylcation gets formed? If so, why does the Cl atom para to the nitrogen leave more readily than the other Chloride?


1 Answer 1


The 2 and 4 positions of quinolines are similar to the 2 and 4 positions of pyridine in reactivity. 2 and 4 haloquinolines/pyridines are highly reactive to nucleophiles by SNAr. Indeed 4-Cl-pyridine has to be handled as the hydrochloride because of its propensity to self condense.

The 7 position, by contrast, is unactivated and similar in reactivity to benzenes, so the 7-Cl is not going to undergo SNAr without specialist catalysis.


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