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.

  • $\begingroup$ 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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