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The question is to compare the SN1 reactivity for the following compounds. According to me it should be like R>S>Q>P.

But the correct answer is Q>P>R>S. How can a lone pair donating into a vacant p be more stable than allyl/benzyl carbocation?

These are the four compounds in the order PQRS respectively:
enter image description here

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    $\begingroup$ The most crucial pair to compare is the methoxymethyl (P) with benzyl (R) and then it's easier to devise Q>P and R>S (as you did). Why do you think (R) should give a more stable cation than (P) ? I suggest you to spend some more time pondering on that question - it's going to pay off later. btw this is a nice practical question. All these reagents would be routinely used in a med chem project for example to optimise an ether molecule. $\endgroup$
    – K_P
    Jul 25, 2015 at 16:18
  • $\begingroup$ @abhirikshma Yes, such cations are more stable, why you think they wouldn't be? $\endgroup$
    – Mithoron
    Jul 25, 2015 at 21:24
  • $\begingroup$ I dont know, but I tend to have a general notion that benzyl cations are very stable because of the extent of delocalisation of the positive charge. On the other hand, lone pair donating into a vacant p does not have any such attributes. $\endgroup$
    – Abhirikshma
    Jul 26, 2015 at 4:15
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    $\begingroup$ A cation derived from Q can be thought of as having a double bond $\ce{N=C}$ and plus on N. It is basically the same as ammonium cation with some organic substituents. These are stable, like real compounds; you may take some of them off the shelf. As for benzyl cation, it is indeed very stable... for a reaction intermediate. That's a wholly different degree of "very stable". Well, with suitable anion like $\mathrm{SbF_6^-}$ you may be able to isolate one as well, but that requires some effort. $\endgroup$
    – Ivan Neretin
    Aug 30, 2015 at 21:41

2 Answers 2

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Benzyl and allyl cations are stabilised by the neighbouring $\unicode[Times]{x3C0}$-systems. The electrons in the $\unicode[Times]{x3C0}$-system are actually required for (formally) form double bonds to ensure all-octet structures as well as possible. The cationic structures will contain an electron sextet somewhere (most often drawn on the benzylic carbon for the benzyl cation).

The $\unicode[Times]{x3B1}$-oxygen and $\unicode[Times]{x3B1}$-nitrogen systems on the other hand contain free lone pairs on the heteroatoms. These are not used for double bonds, so they can be shared with the cation to form a new double bond. The resulting mesomeric structure is an all-octet structure which are usually much more stable than non-octet structures (at least for main group elements).

The nitrogen-containing carbocation could even, after the double bond has been formed, loose its proton to create a neutral all-octet structure, the formal $\ce{HCl}$ elimination product.

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In carbocation stability, mesomeric effect has highest priority, then hyperconjugation and then resonance.

Mesomeric effect of NH in 'Q' is greater than O in 'P'. 'R' has resonance and in 'S' carbocation has least stability.

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  • $\begingroup$ What is the difference betwen mesomeric and resonence effect ? $\endgroup$
    – Abhirikshma
    Aug 1, 2015 at 4:43
  • $\begingroup$ There's no difference and they are all stabilized mesomerically, only more ore less strongly. $\endgroup$
    – Mithoron
    Sep 29, 2015 at 21:50

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