I was recently came to know as a fact that cyclohexane-1,4-dicarbaldehyde does not give the aldol condensation reaction in spite of possessing α-hydrogens. I was told that this is due to steric reasons, but I am trying to come up with a more specific reason, which does not require prior knowledge of this fact. Can anyone guide me towards a better explanation?

Edit 1: I am able to understand that there is absolutely no scope for an intramolecular aldol condensation. But what I am asking is whether even the aldol REACTION would take place, as basically the exact wording of the book is " cyclohexane-1,4-dicarbaldehyde gives zero aldol reaction products". So I am confused that what part is aldol reaction and/or condensation

  • $\begingroup$ I presume you mean intermolecularly, not intramolecularly. Any CONDENSATION reaction, by definition, must lose a small molecule, in this instance water. With only one alpha-hydrogen, the best you can do is an aldol REACTION (aldol = ALdehyde alcohOL), which is reversible. You need at least two alpha-hydrogens to effect elimination of water. $\endgroup$
    – user55119
    Oct 23, 2018 at 16:31
  • $\begingroup$ @user55119 But once it has undergone aldol reaction, it will have formed a beta-hydroxy aldehyde, from where it can surely undergo $\ce{E1CB}$ reaction to form the alpha-beta unsaturated ketone(I think). $\endgroup$ Oct 23, 2018 at 16:53
  • $\begingroup$ It seems that you are considering an intramolecular reaction. Forget it! Any aldol product will reverse to the dialdehyde owing to strain issues. n-Butyraldehyde can undergo an aldol CONDENSATION because it has two alpha-hydrogens vincinal to the aldehyde group. Removal of the first hydrogen leads to the aldol product. Removal of the second hydrogen effects elimination of hydroxide (net loss of water). $\endgroup$
    – user55119
    Oct 23, 2018 at 17:04
  • $\begingroup$ @user55119 So by that reasoning, isopropyl aldehyde will undergo intermolecular aldol reaction, but not condensation. Have I followed you properly? $\endgroup$ Oct 23, 2018 at 17:12
  • $\begingroup$ It's isobutyraldehyde. Depends on the conditions as to where the equilibrium lies. To obtain the aldol of acetone, acetone is boiled in a Soxhlet extractor (look it up, it's like a percolator coffee pot) with Ca(OH)2 in the cup. This way, 4-hydroxy-4-methyl-pentan-2-one is formed and returned to the base-free pot faster than elimination can occur. The dimer has a high enough bp that it doesn't come in contact with the base again. $\endgroup$
    – user55119
    Oct 23, 2018 at 17:27

1 Answer 1


Your claim is worded very poorly, and taken in the broadest sense, is not true. However, it's more complex than an outright yes or no.

If we're interested in fundamental reactivity, it suffices to look at cyclohexanecarbaldehyde, as there is no reason that the dialdehyde should be any different (apart from the likelihood that it gives very messy reactions in practice). The main issue is that "aldol reaction" is a very unspecific term, which is why you (and everybody else) is having trouble explaining it.

  1. Firstly, it can refer either to condensations where you eliminate a molecule of water after the initial addition of an enolate, or to additions where the product is a β-hydroxyketone.

    Formally, any condensation involving cyclohexanecarbaldehyde as a nucleophilic component is not possible, because the α-carbon is tertiary. This includes the self-condensation, where it acts as both the nucleophile and electrophile. After forming a new bond, the α-carbon will become a quaternary centre, which cannot take part in elimination of water.

    Impossibility of H2O loss after crossed aldol condensation

  2. However, if CyCHO is the electrophilic component (i.e. a different enolate reacts with CyCHO), condensations are still possible. The crossed aldol condensation with acetone is well-documented in the literature:

    Crossed aldol condensation between acetone and cyclohexanecarbaldehyde

  3. If we go back to simple aldol additions, which still count as aldol reactions, there are many examples of CyCHO acting as the electrophilic component in an aldol addition. Here is a nice example (Tetrahedron: Asymmetry 2000, 11 (15), 3211–3220):

    syn-selective aldol reaction with CyCHO

  4. On the other hand, aldol additions with cyclohexanecarbaldehyde acting as the nucleophilic component are fairly rare. This reflects a general trend that aldol reactions generating quaternary centres (i.e. aldol reactions involving α,α-disubstituted enolates, such as the enolate of CyCHO) are traditionally very difficult to accomplish. This is mainly because of steric hindrance – the reacting centre of the enolate (the α-carbon) is bulky.

  5. The above is a good rule of thumb, but there still are examples of CyCHO acting as the nucleophilic component in aldol reactions. Many of them involve secondary amine catalysis (example below taken from Org. Lett. 2003, 5 (23), 4369–4372), although there is one example of the lithium enolate of CyCHO attacking a carbonyl group.

    Organocatalysed aldol reaction with 4-nitrobenzaldehyde

    There are also plenty of examples where the silyl enol ether of CyCHO reacts with π-electrophiles. Silyl enol ethers are weaker nucleophiles than lithium enolates, so generally Lewis acid catalysis is required to form more powerful electrophiles. Still, these are considered to be aldol-type reactions (specifically Mukaiyama aldols).


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