HardChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsConceptualNumerical
Six compounds are given: ethanal, propanal, benzaldehyde, pentan-2-one, pentan-3-one and butan-2-ol…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Six compounds are given: ethanal, propanal, benzaldehyde, pentan-2-one, pentan-3-one and butan-2-ol. Decide which of them give a silver mirror with Tollens' reagent, a red precipitate with Fehling's solution, and a yellow precipitate with iodine and sodium hydroxide. How many compounds give each result?
Need a reminder? The key ideaShow
NaBH, LiAlH and H with a metal reduce C=O to an alcohol. Zn-Hg with conc. HCl (Clemmensen) or NHNH then KOH (Wolff-Kishner) reduce it to CH. Tollens' reagent gives 2 Ag for each aldehyde group, and Fehling's solution gives 1 CuO. The iodoform test is given by CHCO- groups and CHCH(OH)- groups.
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. Tollens. The three aldehydes: ethanal, propanal and benzaldehyde. That is 3.
Step 2. Fehling. Aliphatic aldehydes only: ethanal and propanal. That is 2.
Step 3. Iodoform. Ethanal has CHCO-H. Pentan-2-one has CHCO-. Butan-2-ol has CHCH(OH)-, and it is oxidised to butanone in the test. That is 3. Propanal, benzaldehyde and pentan-3-one have no such group.
Answer: 3, 2 and 3.
Check. Ethanal passes all three tests, so it is counted in each.
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