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Reaction of 2,2,5-trimethylcyclopentanol with conc H2SO4

Here I thought doing the following mechanisms

Reaction mechanism of 2,2,5-trimethylcyclopentanol dehydration

I thought the product option through the first pathway (left) would be the major product because the intermediate cation is more stabilised due to more alpha-H. I was told rate determining step is formation of carbo-cation so the pathway where a more stable cation is formed would be faster be major product.

The answer given was obtained from 2nd pathway. Is there any specific factor behind it? My professor told that in this case TC product dominated KC product

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    $\begingroup$ The most substituted alkene is the more thermodynamically favoured. This is an equilibrium situation, the product alkenes can both protonate under the reaction conditions. $\endgroup$
    – Waylander
    Commented Dec 29, 2023 at 12:28
  • $\begingroup$ The name of the compound is 2,3,5-trimethylcyclopentanol. Not 2,2-dimethyl-4-methylcyclopentanol as you stated. I have corrected it for readers' benefit. $\endgroup$
    – Mathew Mahindaratne
    Commented Dec 31, 2023 at 23:52

1 Answer 1

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The major rearrangement of carbocation resulted from the acid-catalyzed dehydration of original 2,3,5-trimethylcyclopentanol is correctly tracked in your scheme. Keep in mind that the dehydration is done in thermodynamically control manner (in higher temperature) so that all rearrangements are in reversible condition (See Waylander's comment).

The initial secondary carbocation at C-1 (the top structure in your scheme) can rearrange either by a 1,2-hydride ion shift from C-5 to give a tertiary carbocation (structure 1.1) or by 1,2-methide ion shift from C-2 to give a different tertiary carbocation (structure 2.1). Both hydride and methide groups are available so that both rearrangements would undergo in different rates under given conditions. These two new tertiary carbocations together with initial secondary carbocation would account for most of the alkene products in corresponding percentages according to their thermodynamic stabilities. The structure 2.2 (1,2,3-trimethylcyclopentene) would be the major product since it contains the most substituted double bond.

For example, dehydration of 2,2,4-trimethyl-3-pentanol with acid has given in Ref.1 as a complex mixture of the alkenes with following percentages in parentheses: I) 2,3,4-trimethyl-2-pentene (29%); II) 2,4,4-trimethyl-1-pentene (24%); III) 3,3,4-trimethyl-1-pentene (2%); IV) 2,4,4-trimethyl-2-pentene (24%); V) 2,3,4-trimethyl-1-pentene (18%); and VI) 2-isopropyl-3-methyl-1-butene (3%).

Suggested alkene producing mechanism for given example

References:

  1. Robert J. Ouellette and J. David Rawn, In Organic Chemistry Structure, Mechanism, & Synthesis; Second Edition, Academic Press: Cambridge, MA, 2018 (ISBN: 978-0-12-812838-1).
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  • $\begingroup$ Thanks a lot. Just one more thing, is more substituted alkene, in general, major product of dehydration? $\endgroup$
    – Sudarshan Kulkarni
    Commented Jan 2 at 15:21
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    $\begingroup$ As a rule of thumb, more substituted alkene (Zaitsev’s Rule) would generally produced under thermodynamic conditions. $\endgroup$
    – Mathew Mahindaratne
    Commented Jan 3 at 5:59

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