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I should put these carbocations in decreasing order of stability.

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Structure 1 is a primary carbocation.

Structure 2 is a secondary carbocation.

Structure 3 is a tertiary carbocation.

Structure 4 (a primary carbocation) and structure 5 (a primary carbocation) are less substituted, but one could form resonance structures.

Therefore I would say 5 > 4 > 3 > 2 > 1.

But I'm not sure why 5 should be more stable than 4. In both cases the positive charge is stabilized by the + I-effect through the sigma bond of the two methyl groups and both can be stabilized through the free electron pair of the heteroatoms.

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  • $\begingroup$ I'm correcting myself: the right order should be 4 > 5 > 3 > 2 > 1. Nitrogen is less willing to give its free electron pair to the positive charge because it's more electronegative than bromine. So structure 4 should come before 5, would that be the right explanation? $\endgroup$ – carl192 Sep 29 '20 at 12:27
  • $\begingroup$ 5 will get electron donation from the N and the positive charge will locate on the nitrogen - so as drawn the cation is highly unstable $\endgroup$ – Waylander Sep 29 '20 at 12:42
  • $\begingroup$ So, the right order should be: 3 > 2 > 1 > 4 > 5? Structure 4 will be then unstable too. If it donates its free electron pair to the positive carbocation, the positive charge will be located on the bromine atom. I'm just a little bit confused by 4 and 5. $\endgroup$ – carl192 Sep 29 '20 at 12:44
  • $\begingroup$ Br is nowhere near as much of an electron donor as N $\endgroup$ – Waylander Sep 29 '20 at 13:33
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    $\begingroup$ @Waylander Don't confuse the OP! carl, he just meant that the mesomeric structure written there is less important then the other one, with double bond. As far as expected order for such question is concerned, your original order is correct. $\endgroup$ – Mithoron Sep 29 '20 at 13:33
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In case of 5


there is a lone pair with nitrogen atom which can help to reduce to positive charge and make double bond with a positive charge on nitrogen so we say this carbocation is extra stable due to its involvement in resonance . Though there is a -I of nh2 but resonance overpowers it


In case of 4


Br has a -I effect and also shows -M effect I.e helps to neutralize the positive charge but the point to note here is that order of mesomeric effect is greater for N than Br


So 5>4>3>2>1

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