Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A teaching laboratory tests ten compounds with four reagents: 2,4-dinitrophenylhydrazine, Tollens' reagent, Fehling's solution and iodine with sodium hydroxide. The compounds are methanal, ethanal, propanone, pentan-3-one, benzaldehyde, acetophenone, ethanol, propan-2-ol, propan-1-ol and phenol. Methanal and ethanal give a red precipitate with Fehling's solution, but benzaldehyde does not. Assume that each test is positive wherever the lesson says it is. Count five things. First, the compounds that give the iodoform test. Second, the ketones among them. Third, the compounds that give Tollens' but not Fehling's. Fourth, the compounds with exactly one positive test. Fifth, the compounds with no positive test.
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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. Iodoform. Ethanal, propanone, acetophenone, ethanol and propan-2-ol: 5.
Step 2. Ketones with iodoform. Propanone and acetophenone: 2. Pentan-3-one has no methyl group on the carbonyl carbon.
Step 3. Tollens' but not Fehling's. Methanal and ethanal give both. Only benzaldehyde gives the silver mirror alone: 1.
Step 4. Exactly one positive test. Pentan-3-one (2,4-DNP only), ethanol (iodoform only) and propan-2-ol (iodoform only): 3.
Step 5. No positive test. Propan-1-ol and phenol: 2.
Answer: 5, 2, 1, 3 and 2.
Check. Positive tests per compound: 3, 4, 2, 1, 2, 2, 1, 1, 0, 0. The total is 16, the same as from the columns.
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Bromine adds across one double bond for each molecule of , and a phenol takes three. Tollens' reagent gives for each , and Fehling's solution gives one . The iodoform test is positive for and .