Identify product [A], [B] and [C] in the following reaction sequence.
CH₃-C ≡ CH →(Pd / C)[A] →((i) O₃)[ B]+[C]

Solution

Chemistry ma’am · 4 steps

Stop at the alkene with one H₂, then cut the C=C in two and put an oxygen on each half.

1. Partial hydrogenation. Propyne takes one mole of H₂ over the Pd catalyst, and the triple bond becomes a double bond. So [A] is propene, CH₃-CH=CH₂. It has to be an alkene, because the next step, ozonolysis, needs a C=C to attack.
2. Reductive ozonolysis. Ozone adds across the C=C and forms an ozonide. Zn and H₂O break the ozonide and remove the extra oxygen, so each carbon of the old double bond ends up as a C=O group. Aldehydes stay as aldehydes. They are not oxidised to acids or reduced to alcohols.

   CH₃-CH=CH₂ →((i) O₃, (ii) Zn, H₂O) CH₃CHO + HCHO

3. Check carbons. Propene has 3 carbons. Cutting it at the double bond gives a 2-carbon piece, ethanal (CH₃CHO), and a 1-carbon piece, methanal (HCHO). That gives 2 + 1 = 3, so no carbon is lost.
4. So. [A] is CH₃-CH=CH₂, [B] is CH₃CHO and [C] is HCHO, which is option (b).

The idea to keep. For ozonolysis, find each C=C, cut it, and put =O on both ends. With Zn/H₂O you get aldehydes and ketones only, and the carbons of the pieces always add up to the alkene's carbons.

Answer from the NTA final answer key. Solution by PrepWiser.