Hydride ion is a strong base...phenyl group will migrate more easily.
Q) When a methyl as well as phenyl grps are attached to same carbon atom, which shift takes place - 1,2-methyl or 1,2-phenyl shift and why?
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6 Answers
phenyl shift because. phenyl anion is more stable than methyl anion (sp2 C are more electronegative)
But the major doubt is what happens when H and Phenyl is there..
Many books say Phenyl and many say H- will be major
Ideally I think the order should be in the order of increasing energy. We want the migrating group and the carbocation left behind to be stable.
Ph > H > Methyl > Primary > Sec > Tert is what I think should be the order..
But what I studied at FIITJEE says that tertiary groups have the highest migratory aptitude among alkyl substituents. I don't know what to follow either.
Wait....these shifts occur by means of a ringed-transition state. So whichever group stabilises the transition state the best should be a good migrator, isn't it?
In the transition state, both the acceptor and receptor carbon atoms have partial positive charge. They will be quite electronegative, i.e., they will try to suck electrons from the group which is migrating. Whichever migrating group is stable enough to withstand this will have strong migratory aptitude.
Then according to this,
H > Ph > Tert > Sec > Primary > Methyl
And among substituted benzenes, those having activating groups will be better migrators, such as aniline, methoxybenzene, etc. than nitrobenzene, etc.
Ignore my previous comment about H losing out against Ph. H has to be the best migrator, it is smaller than Ph so moves faster, and hydride anion is highly basic whereas Ph anion may be a delocalised source of electrons, H still overrides it.
As I have Studied in FIITJEE that the Migratory aptitude is as follows:
H- > Ph- > tert > sec > primary > CH3-
H- is writeen first due to small size, not very stable and is more basic than Ph- as Ph- which can neutralize its -ve charge by resonance and bulky also.