L-isomer of a compound 'A' (C₄H₈O₄) gives a positive test with [Ag(NH₃)₂]⁺. Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO₃ respectively. Compound (A) is :

Solution

Chemistry ma’am · 3 steps

Every aldotetrose passes the first three tests; the nitric acid product and the L label pick the answer.

1. What A is. A positive Tollens' test means a free -CHO group. A triacetate means three -OH groups. So C₄H₈O₄ is an aldotetrose, OHC-CHOH-CHOH-CH₂OH, and all four options fit this.
2. The two oxidations. Bromine water oxidises only -CHO to -COOH. The two ends then differ, so B is always optically active. Nitric acid oxidises both ends to -COOH, giving tartaric acid. That acid is optically inactive only when it is the meso form, which needs both -OH groups on the same side of the Fischer projection.

   Option	OH at C2, C3	D or L (OH at C3)	Product with HNO₃
   (a)	left, left	L	meso, inactive
   (b)	right, left	L	chiral tartaric acid, active
   (c)	right, right	D	meso, inactive
   (d)	left, right	D	chiral tartaric acid, active

3. So. Only (a) is an L-sugar that gives inactive meso-tartaric acid. It is L-erythrose, so the answer is (a).

The idea to keep. When both ends of a sugar become the same group, look for a mirror plane: OH groups on the same side give an inactive meso product. For D or L, read only the lowest chiral carbon.

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