We haven't learned this in class yet, but after looking online, and through my textbook, I can't seem to find a definite answer. I am hopelessly and utterly lost.

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2 Answers 2


The problem can be solved using retrosythesis.

You look at the product and imagine how that was made.

It's like looking at Lego construction, trying to identify the pieces - or looking at a pair of jeans, trying to identify how it was stitched together.

Then you go back and do the same for the previous building blocks.

The final product apparently is an imine. I'm sure we have seen that before among one of the exercises you worked on - in that case it was an intramolecular reaction.


Imines are usually made from carbonyl compounds and primary amines under acidic conditions. This one is made from propanal and ethylamine.

Looks good here, I see $\ce{H2NCH2CH3,\ H3O+(cat)}$ among your reagents.

What about the aldehyde then? The number of carbon atoms in the aldehyde and the amide (propionamide is the starting material) are the same, just the (formal) oxidation of the carbonyl/carboxyl group are different.

Let's think about a direct reduction of the amide with $\ce{LiAlH4}$:

amide reduction

OK, that isn't it. Just two steps would not have been enough anyway ;-)

But we do remember that

  • esters can be reduced to alkanols with $\ce{LiAlH4}$
  • alkanols can be oxidized to aldehydes with PCC (pyridiniumchlorochromate)

All this is among the list of reagents. Nice, we know to make the propanal from a propionic ester!

How to turn propionamide into a propionic ester then?

Direct reaction with methanol under acidic conditions does not work, but after alkylation of the amide with an alkyl halide, we should be ok.

Lets's summarize our five steps then:

full synthesis

  • $\begingroup$ The reagent for the first step it says is incorrect, it says the amide must be turned into a carboxylic acid first $\endgroup$ Commented Mar 22, 2014 at 23:50
  • $\begingroup$ This is probably not the only way: acidolysis of the amide, followed by esterification is the alternative. $\endgroup$ Commented Mar 23, 2014 at 0:06

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solved! All it took was a little reading and understanding :)


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