enter image description here As you can see in the image diazonium can attack at 2 different positions to form different products then why only the first product is correct? As there are two ortho positions available why does it attack only one? resonance

I tried to draw the intermediates see if they are useful and correct

is it because there are 7 intermediates in the first case? or steric factors play some role here?

  • $\begingroup$ Not sure if it's alright to ask another question in the comments, but it's closely related to this question here... What would happen if instead of $\ce{\beta-napthol}$, $\ce{\alpha-napthol}$ were to be used? Would the attack take place at the ortho or para position? $\endgroup$ May 17 '18 at 17:07
  • $\begingroup$ it takes at para but idk why $\endgroup$
    – amish dua
    May 17 '18 at 17:13
  • 1
    $\begingroup$ Perhaps it could be solved using quantum mechanics? Like a computational calculation of the LUMO coefficients for the activated naphthol? Not sure if that helps though. $\endgroup$ May 24 '18 at 5:39
  • $\begingroup$ i just finished high school so i don't know what you are talking about @Justanotherchemist $\endgroup$
    – amish dua
    May 24 '18 at 6:56
  • $\begingroup$ HOMO has the biggest coefficient at that site. $\endgroup$
    – EJC
    May 24 '18 at 16:06

The most important thing to point out here is that your arrows are backwards. The diazonium is not the nucleophile. It is the electrophile. The diazonium ion has one lone pair. If it were a nucleophile, there's no way that you'd end up with the diazo compounds drawn, where each nitrogen has a lone pair.

This also means that drawing out all of your resonance structures doesn't mean anything because that is not the intermediate. The real question to ask here is which site on the naphthol ring is more nucleophilic.

Consider the naphthol as an enol for the purposes of the addition. Your two starting structures are exactly the right ones for focus one. The unsaturation in the enol in the top example is such that the nucleophilic site is benzylic while, in the other case, it is far from the ring.


The intermediates after attack by an electrophile are shown (technically, the carbonyls should be protonated still, but I'm too busy to fix that now). Only in the top case, where we substitute adjacent to the other ring, is aromaticity still preserved in the other ring. This is why the substitution is strongly preferred there as opposed to the other position, where we proceed via an intermediate that is not aromatic.


The formation of para-red is an example of electrophilic aromatic substitution. The electrophile reacts with the ring carbon that is ortho to both the activating group ($\ce{-OH}$) and the other ring because even the ring is weakly activating through $\text{+M}$ effect.

The reaction is similar to the sulfonation of napthalene. If the electrophile attacks the first ortho position, then for the resulting carbocation two resonance structures can be drawn which contain aromatic sextets. On the other hand if the electrophile attacks the second ortho position, only one resonance structure can be drawn which has an aromatic sextet.

Thus, by the Hammond postulate, the first product is major.

  • $\begingroup$ can you please elaborate the sextet part $\endgroup$
    – amish dua
    May 24 '18 at 6:57
  • $\begingroup$ @amishdua Look at the resonance structures you have drawn for both cases and spot the ones with an aromatic benzene ring. $\endgroup$
    – Archer
    May 24 '18 at 7:01
  • $\begingroup$ I don't think the Hammond postulate is particularly useful in predicting the major product here. Perhaps we can say that the TS will resemble the reactants as it is an endothermic reaction,but beyond that, I don't think you can comment on the stability of final product using this(In fact, your product is mostly kinetically controlled).. $\endgroup$ May 12 '19 at 5:07

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.