Identify A in the following reaction.

Solution

Chemistry ma’am · 4 steps

Work backwards from the KMnO4 pieces: oxalic acid means a =CH–CH= unit, so both C=C sit in one ring.

1. Skeleton from H2. A takes up exactly 2 H₂ and gives decalin. So A has the decalin skeleton with two C=C bonds.
2. Read the fragments. Hot KMnO₄ cuts every C=C, and each =CH– end becomes –COOH. Oxalic acid, HOOC–COOH, can only come from a two-carbon unit cut on both sides, that is =CH–CH=. So the two C=C are conjugated and lie in the same ring.
3. Intact ring. The other ring survives as cyclohexane with –COOH on two adjacent carbons. These are the ring-fusion carbons. They must be sp3 CH carbons, each joined by a C=C to the diene ring, so each one ends up carrying a –COOH.

   Option	Where the C=C bonds are	Cleavage gives
   (a)	Conjugated diene in one ring, next to both sp3 fusion carbons	cyclohexane-1,2-dicarboxylic acid + oxalic acid
   (b)	One C=C at the ring fusion, one in the same ring	both rings opened, ketone at fusion carbons
   (c)	One isolated C=C in each ring	both rings opened, no oxalic acid
   (d)	One C=C in each ring	both rings opened, no oxalic acid

4. So. Only (a) gives both products. Its two outer CH carbons become the two –COOH groups on the saturated ring, and its middle =CH–CH= unit leaves as oxalic acid.

The idea to keep. For oxidative cleavage, rebuild the alkene by joining each pair of C=O carbons back with a C=C. A ring that survives cleavage held no C=C, and oxalic acid points to a conjugated =CH–CH= unit.

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