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Question:

Predict the product of: ($\ce{C^*}$ is a $\ce{C^14}$ carbon)

enter image description here

My attempt:

I simply assumed the super-leaving group $\ce{-OTs^-}$ would leave, forming a primary ethylbenzene carbocation. That would rearrange via a hydride shift to form a secondary carbocation on $\ce{C^*}$ atom. The nucleophilic $\ce{-OH}$ group of the $\ce{-COOH}$ group will then attack and lose a proton to form 1-phenylethylethanoate (with the ester linkage at the $\ce{C^*}$ carbon):

enter image description here

However, my book has instead taken a different route. In the very first step, after $\ce{OTs-}$ leaves, they have taken the NGP mechanism and formed this transition state:

enter image description here

and said that now the $\ce{OAc-}$ will attack either of the two carbons (in that trigonal ring). Hence, according to their logic, both types of 1-phenylethylethanoate will be formed - one with the ester linkage at the $\ce{C^*}$ carbon, and the other at the $\ce{C}$ carbon.

My question:

I have confirmed from Clayden (Chapter 37 Rearrangements page 976) that the product ratio is indeed 50:50. So:

  • Why is the NGP mechanism being favored here over the other mechanism I described, even though the NGP mechanism involves a loss of aromaticity?
  • Are there other cases where the NGP mechanism dominates over the simple carbocation mechanisms, or is this the only such case?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q16 in Alkyl Halides (Substitution)

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    $\begingroup$ At some level, the mechanism must be similar to that described (losing aromaticity etc) as that predicts the observed results. Not the other way around, surely. $\endgroup$
    – sjb-2812
    Mar 13, 2018 at 9:36
  • $\begingroup$ Your second question doesn't make much sense. If there was only a unique case there wouldn't be even such a name as NGP. Every time when neighbouring group can delocalise carbocation it does that. $\endgroup$
    – Mithoron
    Mar 13, 2018 at 17:37
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    $\begingroup$ The experimental results dictate the choice of mechanism. The spiro cation is a stabilized cation, even if aromaticity is interrupted and it is less stable than its precursor. Is there an electrophilic aromatic substitution that doesn't interrupt aromaticity? $\endgroup$
    – user55119
    Mar 13, 2018 at 22:39
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    $\begingroup$ Hydride shift? Do you really want to form a primary carbocation? $\endgroup$
    – Apoorv Potnis
    Mar 14, 2018 at 13:29
  • $\begingroup$ @ApoorvPotnis I mentioned that "That would rearrange via a hydride shift to form a secondary carbocation on C* atom", so I am not really sure why you think I'm forming a primary carbocation instead :/ $\endgroup$
    – Gaurang Tandon
    Mar 14, 2018 at 14:23

1 Answer 1

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From Clayden (Chapter 37 Rearrangements page 976):

Intramolecular reactions (including participation of a neighbouring group) that give three-, five-, or six-membered rings are usually faster than intermolecular reactions.

It also gives the exact same example as your question:

enter image description here

Notice that both products are given to be exactly 50% each.

NGP is very often faster than bimolecular reactions as for the latter to take place two molecules have to collide first.

1:1 ratio can be attributed to the fact that in NGP the transition state involves loss of aromaticity BUT NGP supports the ionization of the starting material.

There are a lot of examples this one reaction isn't a black swan. Interested readers may see chapter 36: Participation, rearrangement and fragmentation in Clayden to find more details and examples.

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  • $\begingroup$ I added a few things. You may wanna delete your 2 bullet points. The data is exactly 50:50 given in book. $\endgroup$
    – Gaurang Tandon
    Mar 13, 2018 at 8:12
  • $\begingroup$ I don't understand why NGP is favored even over losing aromaticity. And both of your reasons don't make sense. Can you please give a page number where I can read about "favored by high entropy changes"? And what exactly do you mean by "The ionization of the starting material? $\endgroup$
    – Gaurang Tandon
    Mar 13, 2018 at 8:19
  • $\begingroup$ @Gaurang See clayden CH31 pg 806-807 for the ring thing . Ionization of the starting material means ngp doesn't depend upon a collision so it ionizes faster. $\endgroup$
    – Avyansh Katiyar
    Mar 13, 2018 at 8:34
  • $\begingroup$ Pg 806-807 Ch31 doesn't mention "entropy" anywhere. And $\ce{OTs-}$ is a good leaving group, I doubt if it even needs any collisions to leave. $\endgroup$
    – Gaurang Tandon
    Mar 13, 2018 at 8:38
  • $\begingroup$ @Gaurang my south asian edition does $\endgroup$
    – Avyansh Katiyar
    Mar 13, 2018 at 8:40

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