Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Propanone () is converted to its oxime () with hydroxylamine, and to its 2,4-dinitrophenylhydrazone () with 2,4-dinitrophenylhydrazine. In each case one mole of propanone gives one mole of product. Find the amount of propanone in and the mass of the 2,4-DNP derivative it can give. Then find the mass of oxime from of propanone. Name the small molecule that is lost in both reactions.
A nucleophile attacks the carbonyl carbon, the carbon changes from to , and the oxygen then takes a proton. Alkyl groups lower the positive charge and crowd the carbon, so methanal > other aldehydes > ketones. HCN gives a cyanohydrin, NaHSO a bisulphite compound, alcohols with dry HCl an acetal, and compounds a C=N compound and water.
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. .
Step 2. 2,4-DNP derivative. .
Step 3. Oxime. , so the mass is .
Step 4. The small molecule. Water is lost in both reactions. The carbonyl oxygen and two hydrogens of the group leave together.
Answer: , , , and water.
Check. The oxime is heavier than propanone. It has one more nitrogen atom and one more hydrogen atom.
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