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By James Ashenhurst

Rearrangement Reactions (1) – Hydride Shifts

Last updated: March 26th, 2019 |

For nucleophilic substitution, the pattern of bonds that form and break is pretty straightforward. You break C-(leaving group) and you form C-(nucleophile). A straight swap. But every once in awhile you might see a “weird” substitution reaction. If you look closely at the pattern of bonds formed and bonds broken in the second reaction below, there’s an extra set!

In other words it’s a substitution reaction where the hydrogen has moved. We call these movements “rearrangements”, for reasons that will become clear shortly.

The big question is, what’s going on? How did this happen? 

As it turns out, reactions that go through carbocations can sometimes undergo rearrangements. And looking back at substitution reactions, recall that theSN1 reaction goes through a carbocation intermediate. In this post we’ll go through when you’ll expect to see a rearrangement reaction.

Let’s think back to carbocations. They’re carbon atoms with six electrons bearing a positive charge. In other words, they’re electron deficient – 2 electrons short of a full octet. So it would make sense that carbocations become more stable as you increase the number of electron donating groups attached to them. Alkyl groups are a perfect example. That’s why carbocation stability increases as you go from primary to secondary to tertiary.

(It’s also worth pointing out that carbocations are also stabilized by resonance, which allows the positive charge to be delocalized or “spread out” over a greater area on the molecule.)

So what does this have to do with rearrangements? As it turns out,  if  a situation exists where an unstable carbocation can be transformed into  a more stable carbocation, a rearrangement is possible. 

One rearrangement pathway where an unstable carbocation can be transformed into a more stable carbocation is called a hydride shift. Look at the diagram below.

In this reaction we have a secondary carbocation on the left hand side. In this rearrangement reaction, the pair of electrons in the C-H bond is transferred to the empty p orbital on the carbocation. In the transition state of this reaction, there’s a partial C-H bond on C3 and a partial C-H bond on C2. The transition state here is kind of like that split second in a relay race where one sprinter is passing the baton to another sprinter and they both have their hands on it. Then, as the C2-H bond shortens and the C3-H bond weakens, we end up with a carbocation on C3 (a tertiary carbocation) in the product which is more stable.

Note that we only need one arrow to show this occurring!

Here are some examples of “allowed” rearrangement reactions. Notice how we’re always going from a less substituted carbocation to a more substituted carbocation. One exception is at the very bottom; the rearrangement is favorable because the new carbocation is resonance stabilized.

Now we’re ready to show how the rearrangement reaction occurs with the SN1. Recall that the first step in the SN1 is that the leaving group leaves to give a carbocation. In the case below, the carbocation that is formed is secondary, and there’s a tertiary carbon next door. Therefore, a rearrangement can occur to give the more stable tertiary carbocation, which is then attacked by the nucleophile (water in this case). Finally, the water is deprotonated to give the neutral alcohol. So this is an example of an SN1 reaction with rearrangement.

I’ve given some more examples of SN1 reactions with rearrangements below. See if you can draw the mechanisms! In the next post we’ll talk about a slightly different rearrangement pathway with substitution reactions.

Next Post: Rearrangement Reactions (2) – Alkyl Shifts

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Comment section

44 thoughts on “Rearrangement Reactions (1) – Hydride Shifts

  1. can we do hydride and alkyl shift multiple times in order to get a stable carbocation or is it allowed only once?

  2. Sir,
    What happens if a chiral centre is generated?
    For eg,
    (C2H5)(CH3)CH—–C((Cl)(CH3)(C3H8)———–> (C2H5)(CH3)CH—–C+(CH3)(C3H8)
    ——————>(1,2-hydride shift) (C2H5)(CH3)C+—–CH(CH3)(C3H8)
    What is R/S configuration of rearranged carbocation(inversion or retention or both)?

    1. Since a carbocation is planer the resulting stereocenter following reaction with a nucleophile would be racemic or an even mixture of (R) and (S) at the position of the carbocation. This can become more complex with the influence of additional stereochemical considerations from other portions of the molecule that don’t involve the cation or molecules with limited degrees of freedom.

  3. Might be a dumb question, but are multiple hydride and or methyl shifts allowed? If so what is the justification/if not why as well?

    Been thinking as I was solving some tricky sn1 with ring openings and stuff (how do I even see that?!)

    1. Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.

      For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.

  4. Although hydride shifts seem like I good method to offer shortcuts in synthesis, I’ve often wondered why you don’t see them that often and also why I learned about them relatively late in my chemical education. I think it’s because they often occur from carbocations which only form when they are stabilized. And hydride shifts occur to increase the stability of the carbocations, so there’s not an awful lot of scope for them. For aliphatic compounds this only really permits the carbocation to change from secondary to tertiary.

    1. They’re also pretty unselective, so if there’s multiple neighbours where a hydride can be donated from, it’s difficult to control for which one will happen. Plus there are side reactions (like elimination)

  5. Hii!! Is it possible that if a primary carbocation is generated and to its side we have a 2 degree carbon and to that 2 degree carbon’s side we have a 4 degree carbon somewhat like (ch3)3- c -ch2-ch2+ .. Then is it possible that directly a 3 degree carbocation is generated ( somewhat like simultaneous shift ) ??? Thankyou

    1. Benzyl carbo cation is always more stable than alkyl due to resonance ..
      So (Ph)-CH2+. Is more stable ….but at the same time remember that (Ph)+ is most unstable also..

  6. Do shifts only occur between adjacent carbons? For example, if a shift between carbons 1 and 3 created a more stable carbocation, but a shift between 1 and 2 created a less stable carbocation, would the shift from 1 to 3 still occur? Or the rearrangement only occurs if a shift to an adjacent carbon produces a more stable carbocation? Thanks

    1. The shift between 1 and 3 or 1 and 4 is prohibited becos for shift to take place the orbital overlap has to take place between the c-h and the empty orbital on C+.

      Which would be difficult with 3rd and 4th..

      But as the chain gets longer then it can bend and C+ comes near C-H. So, 1,5 shifts take place…

    1. It can happen anytime a carbocation is generated. That can happen via loss of a leaving group as the beginning step of E1/SN1 or it can also happen during addition of acids to alkenes (if you’ve covered that).

  7. As aPHD organic chemist

    I personally in the lab was able to remove hydride from activated carbon at 1100C.
    The issue here with this PHD Chemist is the carbon originally was in a two electron covalent bond, Therefore removing a hydride leaves an electron on the carbon.
    1 That means the carbon has 7 electrons.
    2 Chemist are weird! They call this a carbocation.


    Wayne Harlan PHD

  8. Say if we are having an intermediate Br-CH2-CH (+)-CH3. Will the positive charge shift towards CH2 as it will be stabilised by mesomeric effect of bromine? (There are many questions in my book in which this is not done!!)

    1. There is no way that exists as a free carbocation. The Br will form a 3-membered bromonium ion and will not be prone to rearrangement.

      In some cases – such as if you replaced the CH3 in your example with Ph – then the resulting carbocation would be quite stable, and it is during such situations that breakdown of the typically observed ‘anti’ selectivity for bromination is observed.

  9. Hi,
    Please help me with following questions
    A carbon and hydrogen bond energy is around 413kJ/mol. Is it feasible for such hydride shifts to occur? Because I feel more energy is spent on breaking C-H bond than the energy released by making carbocation stable.
    Secondly, how we could justify shifting of hydride from one molecule to other as in Meervein-pondroff-verley reaction?

    1. You have to balance out the energetic “cost” of breaking the bond with the energetic “gain” of forming a new bond. The bond strength of the C-H that’s being broken is usually within 10 kcal/mol of the bond strength of the C-H that’s being formed.

      What we call “instability” of various carbocations (e.g. secondary, aryl, etc.) is really just another word for high electron affinity, which means that the more “unstable” the neighboring carbocation, the more energy will be released when it completes its octet. If the carbocation resulting from the shift has a relatively low electron affinity (e.g. tertiary or allylic) then the net energy gain can be quite considerable. For instance going from a typical secondary carbocation to a typical tertiary carbocation can be worth about 18 kcal/mol which is a LOT. (Based on gas phase R-H bond dissociation energies. See March’s Advanced Organic Chemistry 5th ed p. 224 table 5.2. )

  10. Why double bond of cyclohexene won’t break on addition of Br₂, UV light and heat? The product I expected was bromocyclohexane but it was 3-bromocyclohex-1-ene!

  11. If we’re adding H-Br to an alkene and on both sides of the alkene carbons we have carbons that can potentially form more stable carbocations but one involves a methyl shift and the other involves a hydride shift, which pathway would we follow when doing the mechanism and adding the proton in the first step?
    Thank you

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