# Are carbocation intermediates in equilibrium?

So for the reaction of $\ce{H_2O}$ with an alkyl ethene, in the presence of $\ce{D^+}$ ions, forms two possible carbocations:

$\ce{ R-CHD-CH2^+}$ and $\ce{R-CH^+-CH2D}$

The major product formed is the secondary carbocation as it is stabler than the primary carbocation.

So my questions are:

1. Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?

2. If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back to being a less stable primary carbocation?

3. Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a more stable carbocation can be formed (depending on the attached alkyl group)?

## 2 Answers

1. Are the two carbocations in an equilibrium or are they formed in specific, fixed quantities?

$\ce{R-CH=CH2 + H^+ <=> R-CH2-CH2^+ + RCH^+-CH3 <=> alcohol~products}$

Yes, the two carbocations formed are in equilibrium. Initially the two carbocations will be formed in a ratio dependent upon the difference in activation energies leading to them (kinetic control). However, since all of the steps in this reaction involve equilibria, after a while when equilibrium is established several things can happen

• many carbocations will revert back to starting material which can then be reprotonated and reform the carbocations
• or alcoholic product can be reprotonated, eliminate water and reform the carbocations
• or the carbocations themselves can equilibrate directly by hydrogen shift

Once the equilibrium is established, the ratio of carbocations formed will be dependent upon the energy difference between the carbocations themselves (thermodynamic control). Often, but not always, the relative stabilities of the transition states leading to the carbocations matches the relative stabilities of the carbocation intermediates themselves.

1. If they are in equilibrium, i.e., the primary and secondary carbocation, why and how would the secondary carbocation revert back to being a less stable primary carbocation?

why - that's what happens in equibria; even though phenol is much, much more stable than its enol counterpart, there will be an exceedingly small amount of the enol present at equilibrium

how - by any of the 3 routes mentioned in the bullets above

1. Also, is the hydride shift that converts the primary to a secondary carbocation the only shift that can occur; even if after the shift, a more stable carbocation can be formed (depending on the attached alkyl group)?

Other groups like alkyl or phenyl can shift too, but hydrogen shift is usually faster than the others.

• Say the alkyl group was a third degree carbon, would a possible hydride shift also be included in the equilibria; or would the reaction already have moved on to the next step? – tkhanna42 Apr 24 '15 at 0:04
• Not sure I fully understand your question. If you're asking if a tertiary alkyl group can compete with a hydride shift, usually the hydride shift will be faster. The shift(s) will occur before the intermediate carbocation moves on to product. – ron Apr 24 '15 at 0:14
• I mean like if there are two or more successive shifts possible, would all the possible carbocations exist in an equilibrium? – tkhanna42 Apr 24 '15 at 0:17
• Yes, in relative amounts that correspond to their relative stabilities. – ron Apr 24 '15 at 0:19

Actually four possible intermediates are there, two from autolysis of water and two from added deutrium.

1) $\rm 1>2\gg3>4$
Since autolysis of water is very minute and a secondary carbocation is always stable than a primary carbocation since it has more hyperconjugations.

2) They would be in equilibrium since every reaction is reversible, though it's rate might be minute for a specific direction. There could also be one in a gazillion hydride shift case.

3) This depends on the migratory aptitude which is generalised as $\rm H>Ph>1^\circ>2^\circ>3^\circ$