An aromatic compound has the formula C₈H₈O…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
An aromatic compound has the formula . It gives an orange solid with 2,4-dinitrophenylhydrazine, no silver mirror with Tollens' reagent and a yellow precipitate with iodine in sodium hydroxide. Find its degree of unsaturation. Count the aromatic carbonyl compounds of this formula that have a carbonyl group, a benzene ring and no other ring: the three methylbenzaldehydes, phenylethanal and acetophenone. How many of those five give the iodoform test? Name the compound.
Need a reminder? The key ideaShow
Work backwards from the target and count the carbons first. Cyanide and RMgX with CO add one carbon, an aldol joins two molecules, and soda lime or the haloform reaction removes one. The overall yield of a route is the product of the step yields. The degree of unsaturation is . The tests are 2,4-DNP (any aldehyde or ketone), Tollens (aldehydes), Fehling (aliphatic aldehydes) and iodoform (CHCO-).
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. . A benzene ring takes 4, and the C=O takes 1.
Step 2. The five isomers. The three methylbenzaldehydes (ortho, meta and para), phenylethanal and acetophenone. That is 5.
Step 3. The tests. The four aldehydes give a silver mirror, so they are ruled out. Only acetophenone is a ketone. It has the CHCO group, so it gives the iodoform test. That is 1 isomer.
Answer: 5, 5 and 1. The compound is acetophenone, 1-phenylethanone.
Check. Hot permanganate would take this compound to benzoic acid, as the benzene ring has no other side chain.
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