I came across this question recently,

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I tried to devise a mechanism but I'm struck here, My try -

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Moreover, I think my mechanism is somewhat inefficient because of formation of some byproducts, which we don't want.

In short in my mechanism,

First we convert succinic acid to its ester (by esterification using ethanol).

Then, we use this ester to react with acetaldehyde in basic medium (dilute NaOH), intramolecular magic occurs to yield an ESTER of the desired product.

Now, here's where i'm struck, how do I convert ester of the desired product into acid, Can I do Hydrolysis, but I think then the ring might get cleaved!

Moreover, I think during this reaction, some acetaldehyde may undergo self condensation to yield a byproduct, this may be true for ester of succinic acid.

Any help will be appreciated!


1 Answer 1


The mechanism looks OK. Though catalytic NaOEt would be a better base than NaOH as the EtO- is regenerated as the reaction proceeds.

The opening of the lactone under basic conditions is reversible but the reformation of the lactone is favoured. A mild basic hydrolysis - 1eq LiOH in 19:1 THF/H2O - should give you your target.

Self condensation is always a possibility with acetaldehyde - this is why few reactions go with 100% yield in real life

  • $\begingroup$ Thanks but any better way to obtain the same product? $\endgroup$
    – Shivansh J
    Commented Sep 23, 2018 at 17:07
  • $\begingroup$ Because I want to get as pure product as possible and therefore avoid self condensation. $\endgroup$
    – Shivansh J
    Commented Sep 23, 2018 at 17:08
  • $\begingroup$ One step condensation of Succinic anhydride with acetaldehyde:- generate the anion at low temp with non- nucleophilic base such as NaH, add acetaldehyde, allow to warm to rt, work up with NHCl soln. $\endgroup$
    – Waylander
    Commented Sep 23, 2018 at 17:16
  • $\begingroup$ If you want clean product from your route, do column chromatography on the ester/lactone before the hydrolysis. $\endgroup$
    – Waylander
    Commented Sep 23, 2018 at 17:22
  • $\begingroup$ Whoa! Wait I don't think I fully understood your method, can you please explain it in detail as an answer (that I will accept) (I guess I'm a chem novice). As I said I'm much more interesting in finding a better method. Thanks in advance! $\endgroup$
    – Shivansh J
    Commented Sep 23, 2018 at 17:24

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