I was quite puzzled by the answer for this question in my Carboxylic Acids and Derivatives tutorial question set. It is question 6 (a), on the synthesis of malic acid, as shown in the image below.

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The idea for the synthesis is to convert the $\ce {Br}$ to $\ce {CN}$ and the carbonyl group to a cyanohydrin, then convert the $\ce {CN}$ to $\ce {COOH}$. I have no disputes about the last step, which is to heat with aqueous hydrochloric acid, or some other mineral acid. However, what I proposed was to perform the two initial conversions in one step, which is to heat the compound with $\ce {KCN (aq)}$ and ethanol. Would that not allow for both the nucleophilic addition, as well as the substitution, to take place, all in one step?

The proposed answer for steps 1 and 2 was to first heat with reflux, with ethanolic $\ce {KCN}$, then react with $\ce {HCN (aq)}$, with trace $\ce {KCN}$, or $\ce {NaOH}$ as catalyst, under temperature of 10-20 degrees C. Is it really necessary to perform the initial two conversions in two separate steps?

  • $\begingroup$ Cyanohydrin formation is a net addition of HCN. KCN + EtOH doesn’t generate any HCN, crucially it doesn’t provide any protons so the anionic RR’(CN)(O-) will just collapse back to the aldehyde. $\endgroup$
    – orthocresol
    Jul 1, 2018 at 7:43
  • $\begingroup$ @orthocresol But isn't the attacking nucleophile cyanide? The protonation can occur as long as there is a proton source available. In this case, water can provide that so we don't really need to add HCN specifically right? $\endgroup$ Jul 1, 2018 at 8:52
  • $\begingroup$ There are two reasons why this is a bad question, which is obviously not your fault. (1) Chloroacetaldehyde exists as the hydrate, and likely, bromoacetaldehyde does the same. This is an important fact that the problem solver should know about. (2) Malic acid is a commodity chemical made from oxidation of maleic anhydride, which is produced from oxidation of n-butane. The suggestion that we should take this route is, therefore, a terrible idea but nevertheless useful as an exercise. $\endgroup$
    – Zhe
    Jul 9, 2018 at 17:10

1 Answer 1


I have clarified with my chemistry tutor and it appears that performing the substitution and the addition as two separate steps with two separate sets of conditions is absolutely necessary. The most important reason would be the temperature, which is also the most stark difference between the two different set of conditions.

Due to entropy considerations, addition should ideally be carried out at low temperatures. Thus, if we would like to have the addition of $\ce {HCN}$ to the substrate, it should be done at much lower temperatures as the reaction has a negative entropy change. However, the substition of $\ce {Br}$ by $\ce {CN}$ does not need such low temperlatures and should be done at higher temperatures to provide enough energy of activitation to break the carbon-bromide bond.

  • $\begingroup$ Did you count the number of carbon atoms in the product? Adding KCN only gives you 3 carbon atoms. $\endgroup$
    – ringo
    Jul 9, 2018 at 14:50
  • 3
    $\begingroup$ @ringo I think the point is that a cyanide equivalent would add into the aldehyde and another would substitute the bromide, increasing the length of the chain by 2. $\endgroup$
    – Zhe
    Jul 9, 2018 at 17:13

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