9
$\begingroup$

When a diazo compound is formed, and then a nucleophile is added to the mixture. Will nitrogen gas leave first or will the nucleophile have to "do" a substitution reaction? What I mean is will the carbocation be formed due to nitrogen leaving or will it be an SN2 reaction? Or will it depend on the carbon that the diazo group is attached to, (like primary, secondary, etc.) depending on the stability of carbocation formed? In short, I mean to ask if it's SN1 or SN2.

Is it a thumb rule that when an unstable diazo compound is formed, that the nitrogen leaves without any nucleophile intervention?

Improve this question
$\endgroup$
4
  • $\begingroup$ Because when I asked it to my professor, he said its a MUST Carbocation forming reaction (he didn't specify sn1... even though i asked him that if it is sn1...), however he is not someone i can trust THAT much.. Thanks:D $\endgroup$
    – Cinnamaldehyde
    Commented Oct 22, 2017 at 19:15
  • $\begingroup$ Probably depends a bit on the reaction. A real chemist would probably check kinetics of the reaction to see if it's first order in the nucleophile or not. $\endgroup$
    – Zhe
    Commented Oct 22, 2017 at 20:46
  • $\begingroup$ @Zhe Maybe a real physical chemist would. A real organic chemist would run with whatever they get out.] $\endgroup$
    – Jan
    Commented Oct 23, 2017 at 6:46
  • $\begingroup$ @Jan fair point and important missing qualifier.. $\endgroup$
    – Zhe
    Commented Oct 23, 2017 at 12:37

1 Answer 1

Reset to default
3
$\begingroup$

The reaction is $\ce{S_N1}$

enter image description here

Among the evidence for the $\ce{S_N1}$ mechanism with aryl cations as intermediates, is the following:

  1. The reaction rate is first order in diazonium salt and independent of the concentration of Y.
  2. When high concentrations of halide salts are added, the product is an aryl halide but the rate is independent of the concentration of the added salts.

  3. The effects of ring substituents on the rate are consistent with a unimolecular rate-determining cleavage.

  4. When reactions were run with substrate deuterated in the ortho position, isotope effects of $\sim 1.22$ were obtained. It is difficult to account for such high secondary isotope effects in any other way except that an incipient phenyl cation is stabilized by hyperconjugation, which is reduced when hydrogen is replaced by deuterium.

enter image description here

  1. That the first step is reversible cleavage was demonstrated by the observation that when $\ce{Ar^{15}N+ #N}$ was the reaction species, recovered starting material contained not only $\ce{Ar^{15}N+ #N}$ but also $\ce{ArN^{+} #N}$This could arise only If the nitrogen breaks away from the ring and then returns. Additional evidence was obtained by treating $\ce{PhN+ #N^{15}}$ with unlabeled $\ce{N2}$ at various pressures. At $\pu{300 atm}$, the recovered product had lost $3 \%$ of the labeled nitrogen, indicating that $\ce{PhN2+}$ was exchanging with atmospheric $\ce{N2}$

Source: March's Advanced Organic Chemistry: Aromatic Substitution: Nucleophilic and Organometallic

Improve this answer
$\endgroup$
2
  • $\begingroup$ What if it was a aliphatic diazonium compound ( i am aware that its unstable but assume it to be kept at a certain temp where its stable). Will it still remain SN1 or will it depend on the substrate? $\endgroup$
    – Cinnamaldehyde
    Commented Apr 29, 2018 at 4:27
  • $\begingroup$ @Cinnamaldehyde yes, you are right they are unstable hence, immediately nitrogen gas is given off and a planar carbocation is formed which reacts via $\ce{E1}$ or $\ce{SN1}$ $\endgroup$
    – Archer
    Commented Jun 22, 2018 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.