Compound A, C₅ H₁₀ O₅, given a tetraacetate with Ac₂ O and oxidation of A with Br₂-H₂ O gives an acid, C₅ H₁₀ O₆. Reduction of A with HI gives isopentane. The possible structure of A is

Solution

Chemistry ma’am · 3 steps

Each reagent gives one clue: four –OH, a free –CHO, a branched chain. Keep the option with all three.

1. Read the clues. A tetraacetate with Ac₂O means exactly four –OH groups. Bromine water oxidises only an aldehyde, and -CHO → -COOH adds one O, so C₅H₁₀O₅ becomes C₅H₁₀O₆. Bromine water leaves a ketone alone, so A must have a free –CHO.
2. Chain shape. HI removes every oxygen and leaves only the carbon skeleton. Isopentane is CH₃CH(CH₃)CH₂CH₃, a four-carbon chain with one methyl branch. So A must have a branched chain.

   Option	Four –OH (tetraacetate)	Free –CHO (Br₂ water)	Branched chain (isopentane)
   (a)	Four, but two sit on one carbon (unstable gem-diol)	No, a ketone	Yes
   (b)	Yes	Yes	Yes
   (c)	Yes	Yes	No, straight chain gives n-pentane
   (d)	Yes	No, a ketone	Yes

3. So. Only (b), OHC-CH(OH)-C(OH)(CH₂OH)-CH₂OH, has four –OH groups, an aldehyde and a branched skeleton. The answer is (b).

The idea to keep. Read each reagent as one structural fact. Ac₂O counts the –OH groups, Br₂ water confirms an aldehyde, and HI reduction shows the carbon skeleton, straight or branched.

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