Rearrangement Reactions (2) – Alkyl Shifts

by James

in Organic Chemistry 1

In the last post we saw how certain carbocations can sometimes rearrange (through hydride shifts)to give more stable carbocations.

However, sometimes there are situations where a hydride shift would not lead to a more stable carbocation, such as in this case. If a hydride shift occurred, we’d be going to a less stable (primary) carbocation.

You might note something with this example, however: it is possible for a more stable tertiary carbocation to be formed if an alkyl  group migrates instead!

The most common situation where alkyl shifts can occur is when a quaternary carbon (that’s a carbon attached to 4 carbons) is adjacent to a secondary carbocation. How does this work? Well, the pair of electrons from the C-C bond can donate into the empty p orbital on the carbocation (side note: this means they have to be aligned in the same plane). In the transition state, there are partial bonds between the carbon being transferred and each of the two adjacent carbon atoms. Then, as one bond shortens and the other lengthens, we end up with a (more stable) tertiary carbocation.


Rearrangements can potentially occur any time a carbocation is formed. That includes SN1 reactions (and as we’ll later see, elimination and addition reactions).

 Here’s an example of an SN1 with an alkyl shift (note that the CH3 groups here are just shown as lines).

It doesn’t always have to be a methyl group that moves. One interesting example is when a carbocation is formed adjacent to a strained ring, such as a cyclobutane. Even though the CH3 could potentially migrate in this case, it’s favorable to shift one of the alkyl groups in the ring, which leads to ring expansion and the formation of a less strained, five-membered ring.

Here’s an example of an SN1 where an alkyl shift leads to ring expansion.

Having gone through two types of rearrangements in substitution reactions, the next series of posts will cover a different class of reactions: elimination reactions.

 Next Post: Walkthrough of Elimination Reactions (1) 

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{ 12 comments… read them below or add one }

dr.klbajaj

Why the attack of water molecule is at C-2 instead of C -3 carbocation.please explain.drklbajaj

Reply

james

It was an error. Thanks for correcting me Krishnan.

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satyajit chakraborty

I think in the very last reaction the attack by the water is shown at a wrong place.

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Wafa

no, i think it’s correct. When you expand you realize that the C that shifted is satisfied with 4 bonds C-H, C-CH2, C-CH3, C-C+. Whereas the C to which OH later gets attached has only 3 bonds C-CH2, C-CH3 and C-CH. Moreover if the OH had attached to the C you think it should’ve, it wouldn’t have enough free valence e- to another atom.
I hope I haven’t confused you.

Reply

james

Fixed, thank you Satyajit.

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chetan

Both reduction amd rearrangement occur in reaction…???

Reply

James

Replacing Cl with OH is not a reduction. Both are more electronegative than carbon.

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Raj

The 1,2-hydride shift occurs because of the hydrogen atom moving to the adjacent carbon with both of the electrons in the bond, according to what my teacher explained. He said that even alkyl groups can shift (though alkyls are much larger than hydrogen) since only the electron cloud is actually shifting.

Is it then possible to have any other shift that stabilizes the carbocation intermediate in a reaction, e.g. a -OH shift? You mentioned “allowed” rearrangement reactions, so what reactions are disallowed and why?

Thanks in advance!

Reply

Helena

Hi James,

Greetings.
How would you explain the methyl and ethyl carboxylate groups attached to the same carbon and only the ester group migrates as the 1,2 shift?
Is the migration via a radical or to the carbanion? I do not think that to the carbocation center. You said there are no 1,2-shifts in radicals.

Best,
Helena

Reply

James

It would help to see the specific example you’re thinking of, but it wouldn’t be a radical shift. There are two lone pairs on oxygen that could come down to help stabilize any positive charge buildup on the carbonyl carbon during the shift.

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ishtiyak qadir

Please explain why the reaction of neopentane yields 1- chloro- 2,2-dimethyl propane when treated with chlorine molecule during free radical mechanism ?
why not 1,2 methyl shift to give more stable tertiary free radical as compared to less stable primary free radicle????
please reply.

Reply

James

1,2-shifts do not occur with radical substrates. Only carbocations.

Reply

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