5
$\begingroup$

I was reading about nucleophilic substitutions of aryl halides, and there are three different mechanisms discussed, namely the SN1, SNAr (addition-elimination), and benzyne (elimination-addition) mechanisms.

However, I am confused as to how to identify which mechanism is operative. How can I tell this from the reaction conditions and reagents?

$\endgroup$
8
$\begingroup$

Before comparing the three main mechanisms, I will first explain why aryl halides typically do not undergo SN2-type reactions. Take bromobenzene as a typical example of an aryl halide. A nucleophile, like hydroxide ion, cannot simply displace the bromide from bromobenzene. Quoting from Organic Chemistry (1st ed., p 589) by Clayden et al.,

In the aromatic compound, the C–Br bond is in the plane of the ring as the carbon atom is trigonal. To attack from the back, the nucleophile would have to appear inside the benzene ring and invert the carbon atom in an absurd way. This reaction is not possible!


In rare cases, SN1 reactions can take place; these characteristically occur when there is an excellent leaving group on the arene. One of the few examples is in the case of an aryldiazonium ion, where the corresponding leaving group is nitrogen gas.

Sample SN1 reaction

Note that the presence of a diazonium ion does not necessarily imply a SN1 mechanism; there are also other possible mechanistic pathways, mainly involving radical chemistry.


The next possibility is that of an addition-elimination reaction, commonly termed SNAr. This is made possible by electron-withdrawing groups (e.g. nitro or carbonyl groups) when they are ortho and/or para to the halogen, as the negative charge in the intermediate can be delocalised into these electron-withdrawing groups:

Sample SNAr mechanism


Finally, the benzyne mechanism. Consider bromobenzene: it cannot undergo SN2 reactions (as explained above), SN1 reactions (bromide is not a sufficiently good leaving group), or SNAr reactions (no electron-withdrawing group ortho or para to bromine). However, under strongly basic conditions, it is possible to eliminate HBr by abstracting a proton ortho to the halide. Examples of such strong bases include sodium amide (NaNH2), LDA/lithium diisopropylamide (LiNi-Pr2), or potassium tert-butoxide (KOt-Bu).

Sample benzyne mechanism


To summarise the three categories:

  1. If you have a very good leaving group - primarily nitrogen gas - on the aromatic compound, then a SN1 mechanism is possible.
  2. If you have good electron withdrawing groups ortho or para to the halide, then it is likely an addition-elimination or SNAr reaction.
  3. If the above two are not possible, and if you have an extremely strong base such as NaNH2 or KOt-Bu, then the benzyne mechanism may be applicable.
$\endgroup$

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.