Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A colourless liquid A, of formula , decolourises a solution of bromine in tetrachloromethane. A sample of of A is treated with ozone, and the ozonide is then reduced with zinc and water. Two organic compounds are obtained. One is propanone, , with a molar mass of . The other, B, forms a silver mirror with Tollens' reagent. Heating A with hot acidified potassium permanganate gives propanone and butanoic acid instead. The chemist uses the two sets of results together to fix the position of the double bond in A. The products in each case come from the same two halves of the molecule. The carbon atoms of the two products add up to the number in A.
The molar mass of A is . Assume that A reacts completely and that every molecule gives one molecule of each product. The molar mass of B is .
Ozonolysis with zinc and water cuts C=C: gives methanal, an aldehyde and a ketone. Hot acidified permanganate gives carbon dioxide, an acid and a ketone. A terminal alkyne has an acidic hydrogen and gives precipitates with ammoniacal silver or copper(I). Hydration of an alkyne gives a ketone, or ethanal from ethyne.
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. B. carbon atoms. B is an aldehyde, since it gives a silver mirror. Hot permanganate gives butanoic acid from it. So B is butanal, .
Step 2. A. Join the carbon atom of the in propanone to the carbon atom of the in butanal with a double bond: . This is 2-methylhex-2-ene.
Step 3. Moles. of A. It gives of each product.
Step 4. Masses. Propanone: . Butanal: .
Answer: A is 2-methylhex-2-ene and B is butanal. The masses are and .
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