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A question taken from a test series (Vibrant, India):

Amongst butan-2-ol and butane-2,3-diol, which has higher rate of dehydration?

The carbocation formed in butane-2,3-diol is more stable as it will rearrange to get +M stability from alcohol group. However, the product will be aldehyde (due to tautomerism) for butane-2,3-diol and alkene for butan-2-ol.

As rate is determined by stability of carbocation, butane-2,3-diol should have faster rate. Also, the products don't give any extra hint. Right? But the answer given for higher rate is butan-2-ol. Is there any point that I have missed out?

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    $\begingroup$ I would have refused to answer this question on the grounds that "rate of dehydration" is ambiguous when comparing a mono-ol with a diol. $\endgroup$
    – Zhe
    Commented Jul 21, 2020 at 18:18
  • $\begingroup$ Not to say about lack of any info about circumstances, like temperature, solvent, catalyst, or even phase! $\endgroup$
    – Mithoron
    Commented Jul 21, 2020 at 22:18
  • $\begingroup$ Sorry, but I was given this question in a test and devoid of any such details, hence I didn't any info over here. $\endgroup$
    – Khushi Ladha
    Commented Jul 22, 2020 at 4:06

1 Answer 1

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For this question, given the lack of initial information, I'm making the following assumptions:

  • Butane-2,3-diol dehydrates to give butan-2-one
  • 2-butanol dehydrates to give but-2-ene

Butane-2,3-diol dehydration is nicely described in this paper from Emerson1:

Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of sulfuric acid to form methyl ethyl ketone. The reaction could be described by a pinacol rearrangement mechanism and had an activation energy of 36 kcal/mol.

Finding the data for dehydration of 2-butanol proved more elusive. One study2 conducted dehydration rate experiments for secondary alcohols at 250 °C and 40 bar pressure and obtained a rate constant for 3-hexanol to be $k = \pu{(0.067 \pm 0.021) h^{-1}}$. Assuming this rate would be similar to 2-butanol, we can find the activation energy via the arrhenius equation, which comes out to be around 48 kJ/mol, which is around 11.52 kcal/mol. Another source3 conducted the same experiment over a copper-chromite catalyst, and obtained an activation energy of 13 kcal/mol. This means that 2-butanol will dehydrate faster than butane-2,3-diol.

From a conceptual standpoint, this makes sense: The pinacol rearrangement mechanism would require a hydride shift, which would be the RDS and would take more time and energy than a simple carbocation formation. However, if you're comparing absolute reaction rates, a large number of factors come into play, and given the right conditions, butane-2,3-diol can dehydrate faster than butan-2-ol.

References:

  1. Emerson, Richard R., et al. “Kinetics of Dehydration of Aqueous 2,3-Butanediol to Methyl Ethyl Ketone.” Industrial & Engineering Chemistry Product Research and Development, vol. 21, no. 3, 1982, pp. 473–77. doi:10.1021/i300007a025
  2. Bockisch, Christiana, et al. “Kinetics and Mechanisms of Dehydration of Secondary Alcohols Under Hydrothermal Conditions.” ACS Earth and Space Chemistry, vol. 2, no. 8, 2018, pp. 821–32. doi:10.1021/acsearthspacechem.8b00030.
  3. Joseph Sires Cantrell, "Rate and activation energy of the first order dehydration of 2-butanol over a copper chromite catalyst", 1954, source
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    $\begingroup$ Why do you think hydride shift is RDS? All literature I have seen says it’s the cation formation that is rate-determining. It would be slower than 2-butanol because of electron withdrawing effect of second OH which destabilizes cation, apparently enough to overcome the statistical advantage of two OH groups. $\endgroup$
    – Andrew
    Commented Jul 21, 2020 at 18:40
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    $\begingroup$ I just looked up the Emerson reference that you cited, and it also says the departure of H2O is generally believed to be rate-limiting, although it may be concerted with and aided by the hydride shift. Certainly, the hydride shift alone is not the RDS and would not "take more time and energy than a simple carbocation formation". In fact, concerted hydride shift would lower the reaction barrier by assisting the H2O elimination and leading directly to a more stable carbocation. $\endgroup$
    – Andrew
    Commented Jul 21, 2020 at 20:18
  • $\begingroup$ @Andrew: The mechanism is similar to neopentyl alcohol elimination. Primary carbocation formation is assisted by concomitant shift of methide group. Nevertheless, the reaction is slow. $\endgroup$
    – Mathew Mahindaratne
    Commented Jul 21, 2020 at 20:26

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