[lrw1793]

This is one of the questions on an organic past paper I'm going through (no answer schemes)

http://imgur.com/v1KuKg9

The first Q was to draw the Sn1 mechanism to form F
The next Q:
(ii) With reference to the reactive intermediate drawn in your mechanism in
part (c)(i), explain why nitrile G is formed in preference to F.

Is this because the carbocation intermediate has 3 resonance forms where the positive charge is on the ortho and para positions, but the para is preferred for some reason? Maybe due to sterics?

(iii) Draw an arrow-pushing mechanism for the formation of 1,2-
dimethylbenzene H and explain why it is the major product in the reaction.

I really have no idea on this and I'm not sure what I should be looking up in my textbooks etc. so if someone could point me in the right direction I'd be grateful!

[critzz]

You are in the right direction with the resonance structures, but it has nothing to do with sterics. 

Try to draw each resonance structure and consult your book for the relative stability of carbocations (or you already might know the stability order).

[lrw1793]

Thanks! Okay think I've got it... When the positive charge is in the para position, it is more stable because it is inbetween two sp2 hybridised carbons and so can share the charge between the p orbitals. Whereas in the ortho position it is inbetween an sp2 and an sp3 hybridised carbon. The sp3 orbitals can only partially stabilise the charge as they are not parallel or the same shape.
Anyone got any clues about the second part?

[critzz]

Now for the second part, what kind of carbocation will you get if a methyl group shifts to the positive charge (former OH-group)?

And what happends next under sn1/e1-conditions?

[lrw1793]

You'd get a tertiary carbocation, so that's preferred because it's the most stable. Thanks so much  sometimes my brain doesn't seem to work properly! Bit stumped on how it forms a double bond but I'll bet that's really simple as well

[critzz]

Haha, at least you find it logical now.

The double bond is formed when the hydrogen attached to the carbon (the carbon where the methyl is shifted to) eliminates.

This means that the electrons in the C-H bond "fill up" the positive charge and form a double bond, and a really stable benzene ring is formed!

The remaining proton (H⁺) is then picked up by. a CN^--ion.

Hope it helps. 

[lrw1793]

Ahh I see! Thanks for all your help, I actually understand now!