What happens if a primary alcohol that also bears a halide substituent on the alcoholic carbon (the carbon to which the hydroxyl group is attached) when it is reacted with copper at 300 °C ? I know primary alcohols, on treatment with copper at 300 °C react to give aldehydes. But I am confused what happens if halides are present on the same alcoholic carbon.


1 Answer 1

  1. You are correct that copper metal is a catalyst for alcohol dissociation to aldehyde + $\ce{H2}$. The reaction of ethanol, for example, can be written as:

$$ \ce{CH3CH2OH ->[\ce{Cu}~\textrm{metal}][300~^{\circ} \textrm{C}] CH3CHO + H2} $$

  1. Your question about halo-substituted alcohols is very interesting, because those species are very unstable. If I understand you correctly, you are not referring to halohydrins such as 2-chloroethanol, which are species that have a halo substituent beta to the alcoholic carbon, but rather to species such as 1-chloroethanol where the halo and hydroxyl substituents are on the same carbon atom.

  2. alpha halo alcohols are very unstable, and tend to eliminate halo-acid and form the corresponding carbonyl compound. For example, gaseous chloromethanol spontaneously eliminates hydrogen chloride to form formaldehyde at 296 K (~23 °C) with a half-life of about 300 seconds.

  3. Thus I think the best answer to your question is that:

    • Halo alcohols like 1-chloroethanol or chloromethanol are already oxidized to the same degree as carbonyl compounds.
    • Even before reacting them with copper metal at 300 °C, they are likely to decompose to the carbonyl compound via elimination of hydrogen halide.
    • Unlike the reaction of simple alcohols on hot copper metal, this reaction is not a redox dissociation, but is a simple elimination reaction.$%edit$
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    $\begingroup$ I wouldn’t call that a disproportionation, because different atom kinds get reduced and oxidised. In my understanding, disproportionations only include reactions where atoms of the same element get simultaneously reduced and oxidised such as in $\ce{H2O2 -> H2O + 1/2 O2}$. $\endgroup$
    – Jan
    Nov 29, 2015 at 15:51
  • $\begingroup$ Jan, I can see your point. What would be an appropriate term? Dissociation? $\endgroup$
    – Curt F.
    Nov 29, 2015 at 19:06
  • $\begingroup$ I just edited to make it say dissociation instead of disproportionation. Thanks for pointing out my non-standard use of the term! $\endgroup$
    – Curt F.
    Nov 29, 2015 at 19:36

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