3‐cyclopentylpropanal from (chloromethyl)cyclopentane

I am not sure on how to start, but I think of these two possible ways:

  1. Reacting the starting material with Grignard reagent and then oxidise the product with PCC to produce the aldehyde;
  2. Hydrolysis the starting material to produce alcohol and then react the alcohol with PCC to form aldehyde.

But these two ways didn't give the desired product.


3 Steps:

Step 1 - react the starting material with the preformed anion of ethyl cyanoacetate (NaOEt/EtOH should do this comfortably)

Step 2 - mild base hydrolysis of the product (NaOH/EtOH) followed by acidification to remove the ethyl ester group.

Step 3 - Dibal-H reduction of the nitrile to aldehyde 1.


You can proceed with the following idea:

React the given starting material with Mg turnings in ether to form Grignard reagent. Now react this with sodium 2-chloro ethanoate. Now add lithium aluminium hydride (LAH) to this and then oxidise it to aldehyde using PCC.

Hope it helps!

  • $\begingroup$ Do you have a reference for the reaction of the Grignard with 2-Chloroethanoic acid? $\endgroup$ – Waylander Dec 9 '20 at 11:30
  • $\begingroup$ @Waylander my organic chem teacher once gave the reaction...will look for a citable reference and share once I get it :) $\endgroup$ – Adiboy Dec 9 '20 at 12:17
  • $\begingroup$ Please correct me if there's any mistake in it $\endgroup$ – Adiboy Dec 9 '20 at 12:19
  • $\begingroup$ Grignard reagents will surely react with carboxylic acids, just not in the way you think. It's just an acid-base reaction:$$\ce{RMgBr + R'COOH -> RH + R'COOMgBr}$$ $\endgroup$ – orthocresol Dec 9 '20 at 12:24
  • $\begingroup$ @orthocresol I agree, that's why I would like to see a reference. I don't think this will work with the acid or the sodium salt $\endgroup$ – Waylander Dec 9 '20 at 12:31

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