A chemist mixes two carbonyl compounds in dilute sodium hydroxide…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A chemist mixes two carbonyl compounds in dilute sodium hydroxide. Case one: ethanal with propanal. Case two: ethanal with benzaldehyde. Case three: benzaldehyde with methanal in dilute base. How many aldol products can form in each case? Name the aldol condensation product that is most useful in case two.
Need a reminder? The key ideaShow
An alpha hydrogen is needed for the aldol reaction (dilute base). With no alpha hydrogen, concentrated base gives the Cannizzaro reaction. A crossed aldol of two enolisable compounds gives four products. The carbonyl group is meta-directing. For the masses, use moles = mass / molar mass and the equation ratio.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Case one. Both can form enolates, and each enolate can attack either compound. products.
Step 2. Case two. Only ethanal has an alpha hydrogen, so only one enolate forms. It can attack ethanal or benzaldehyde. products.
Step 3. Case three. Neither has an alpha hydrogen, so there are no enolates and 0 aldol products. Methanal and benzaldehyde would try a Cannizzaro reaction in concentrated base.
Step 4. In case two, the useful product is the one from attack on benzaldehyde, which gives 3-phenylprop-2-enal, cinnamaldehyde, after dehydration.
Answer: 4, 2 and 0.
Discussion (0)
Log in to join the discussion.
Loading…