HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumerical
Two moles of benzaldehyde react with one mole of propanone in dilute sodium hydroxide, in a double…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Two moles of benzaldehyde react with one mole of propanone in dilute sodium hydroxide, in a double crossed aldol (Claisen-Schmidt) reaction. Two molecules of water are lost. The product is dibenzylideneacetone, , with a molar mass of . A chemist uses of benzaldehyde (), with propanone in excess, and the yield is . Find the number of carbons in the product, the amount of benzaldehyde and the mass of product.
Need a reminder? The key ideaShow
Wolff-Kishner and Clemmensen turn C=O into CH (base and acid). Perkin: ArCHO, (CHCO)O and CHCOONa at 453 K give ArCH=CHCOOH. Haloform: CHCO- with 3 X and 4 NaOH gives CHX and the carboxylate with one carbon fewer. A crossed aldol with an aldehyde that has no alpha hydrogen is a Claisen-Schmidt reaction. For a gas, with kPa L/(mol K).
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. Carbons. .
Step 2. .
Step 3. Product: , so the theoretical mass is . At the mass is .
Answer: 17 carbons, and .
Why it works. Benzaldehyde has no alpha hydrogen, so it cannot be the enolate. Only propanone forms the enolate, which attacks benzaldehyde twice.
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