Why does enantiopure sec-butyl alcohol retain its optical activity over aqueous base but forms a racemic mixture over dilute sulphuric acid ?

My reasoning is that, sec-butyl alcohol get dehydrated into a alkene because of sulphuric acid and then get hydrolysed to form alcohol again. Thus a equilibrium is formed between alkene and alcohol and between $R-$ and $S-$ sec butyl alcohol. Dehydration cannot occur with a base, so the alcohol remains optically active.

I feel that conversion of alcohol to alkene and then back to alcohol seems ridiculous and impossible. What do you think, is this reasoning correct or just plain non-sense ?

To be specific, I have some doubt whether alkene will convert back to alcohol or not in dil sulphuric acidic ?

  • 5
    $\begingroup$ Why is your speculated mechanism ridiculous? You have proposed a reasonable mechanism. The rate of the racemization isn't given. Even if you think the process will be slow, it explains the observation. $\endgroup$
    – jerepierre
    Jul 20, 2017 at 18:12
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    $\begingroup$ I don't think you need to go all the way to the alkene. The protonated $OH_2^+$ is a good leaving group and makes an $S_N1$ reaction with more water. I'm not sure trans-2,3-butene would hydrolyse in dilute acidic solution? $\endgroup$
    – Karl
    Jul 20, 2017 at 18:34
  • $\begingroup$ @jerepierre I have doubts on whether alkene will convert to alcohol or not just as Karl asked. $\endgroup$ Jul 22, 2017 at 15:47
  • 1
    $\begingroup$ @Mockingbird Sulfate is a lousy nucleophile, if I remember correctly. $\endgroup$
    – Karl
    Jul 22, 2017 at 17:46
  • 1
    $\begingroup$ Back to the alcohol by water addition (the medium is aq sulfuric acid) $\endgroup$
    – Waylander
    Jul 22, 2017 at 19:25

1 Answer 1


Although sulphuric acid is a notable dehydrating agent, it is concentrated rather than dilute sulphuric acid that would be typically used for such a purpose.

Here, it is due to the improvement of $\ce{H2O}$ as a leaving group compared to $\ce{OH-}$, due to the lack of any charge as a leaving group.

Under acidic conditions, the sec-butyl alcohol will be protonated in a small amount creating a much better leaving group. The protonation also makes the alcoholic carbon centre more electrophilic and vulnerable to $S_N2$ attack by other water molecules in solution, leading to inversion of the chiral centre.

This in not possible under neutral conditions as the unprotonated alcohol would have $\ce{OH-}$ as a much poorer leaving group, preventing elimination or substitution reactions.

Under acidic conditions, with neutral water as a nucleophile $S_N2$ substition will be the preferred path.

  • A secondary alcohol on a small, un-branched carbon chain represents a relatively accessible spot for nucleophilic attack.
  • $E_1$ eliminiation or $S_N1$ substituion would require the formation of a secondary carbocation with relatively poor stabilisation compared to a tertiary carbocation.
  • $E_2$ elimination would require the removal of a hydrogen of low acidity from the carbon chain by a very weak base, water.

This allows a equilibrium converting molecules back and forth between the $R$ and $S$ forms of the alcohol to be set up and leads to racemisation in solution.

  • $\begingroup$ While there is no clear cut line between the SN1 and SN2, I'd argue that the SN2 is much more pronounced. I doubt that there is the formation of an actual carbocation, and that one or two water molecules be loosely coordinated to the (alcoholic) carbon centre at all times. (Maybe you have switched the two, your explanation otherwise is very sound.) $\endgroup$ Jul 23, 2017 at 5:27
  • 1
    $\begingroup$ Dilute sulfuric acid is not dehydrating at all. And $S_N1$ does not lead to inversion, but is not stereospecific at all. $S_N2$ does that. Not that it would matter here. $\endgroup$
    – Karl
    Jul 23, 2017 at 10:51
  • $\begingroup$ Thanks for pointing that out, I got $S_N1$ and $S_N2$ labels mixed up. Edited. $\endgroup$
    – user213305
    Jul 23, 2017 at 12:04

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