Starting from 1,1-dichloropropane, it reacts with 2 #"HCN"#, and the product is then reacted with #"HCl"# in #"H"_2"O"# and heated to yield ethylmalonic acid. Which of these routes is more likely to proceed from heating?

Route 1: forming an anhydride
Route 2: decarboxylation in acidic conditions

1 Answer
Feb 2, 2016

This is an interesting one.

  1. One might think that heating it will dehydrate the dicarboxylic acid and form an anhydride. In other circumstances where a ring larger than four members would form... sure, I'd go with that. But the ring that would form is exactly four members, so it is more favorable to go the other route.

  2. This other route is decarboxylation. You should notice that since the second carboxyl group is #beta# to the first (two carbons away), this #\mathbf(beta)#-keto acid can experience a decarboxylation in acid (you added acid in the previous step).

Here's what I'm seeing:

The 1,1-dichloropropane gets converted to 1,1-dicyanopropane, and the acid does turn the resultant compound into ethylmalonic acid after going through an amide "intermediate".

At this point is where weighing stabilities would help.

Route 1 essentially works out like this:

So you can see how the 4-membered ring would form when you work it out.

Route 2 gives the indicated product below.

And of course, butyric acid and carbon dioxide are more stable than that weird 4-membered ring. :)