Acetaldehyde on treatment with a few drops of concentrated $\ce{H2SO4}$ gives?


First, I formed an enol intermediate in acidic medium $\ce{CH2=CHOH}$. Then, I performed electrophilic addition of another acetaldehyde molecule on this enol intermediate to get $\ce{CH3CH(OH)CH2CHO}$.

But answer given in my book is:

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Please let me know the error in my attempt.

  • $\begingroup$ You have it kinda backwards, the enol O acts as nucleophile on a second protonated acetaldehyde molecule. $\endgroup$ – Waylander Aug 7 '18 at 20:20

You're not wrong.

Your mechanism describes the classic acid catalyzed synthesis of 3-hydroxybutanal by dimerisation of acetaldehyde (Borodin 1869, Wurtz 1872).

enter image description here Image: Whoopie23, Wikipedia

However, under different conditions acetaldehyde forms different addion and condensation products. E.g. with concentrated sulphuric acid at low temperatures metaldehyde is formed whereas at room temperature paraldehyde (the answer given in your book) is formed.

enter image description here

  • $\begingroup$ Under what conditions is the dimerisation product formed? $\endgroup$ – Archer Aug 9 '18 at 3:18
  • $\begingroup$ Suitable conditions can be either acidic or basic. Wurtz (see link above) describes the dimerisation with hydrochloric acid at -10°C. $\endgroup$ – aventurin Aug 9 '18 at 13:54

I'm only giving what I believe to be a reasonable explanation here, but I think it begins with protonation of the oxygen resulting in breaking of the $\ce{C=O}$ double bond and formation of a carbocationic alcohol. The next best nucleophile in solution would be another aldehyde molecule, so you get nucleophillic attack at the $\ce{C+}$ centre resulting in another $\ce{C=O}$ cleavage and the process repeating once more resulting in the 6-membered cyclic ring above. Enol formation would be unfavourable in this case as pKa of the alpha carbon on the aldehyde is much higher than anything else in the solution.

Formation of the 6 membered ring is both kinetic and thermodynamic in nature; 6 membered rings are simply favourable due to low torsional strain (thermodynamic) and the relevant atoms being relatively close in space (kinetic).

  • $\begingroup$ The way I've explained is confusing as the process would not be Sn1. Cannot form carbocation at primary carbon. But also Sn2 should not be possible at sp2 centre. Thus I can only assume it is a pseudo-sn2 where build up of charge on the carbon occurs and then attack occurs causing C=O cleavage. $\endgroup$ – Ishy Aug 7 '18 at 20:30

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