L-isomer of tetrose X (C$$_4$$H$$_8$$O$$_4$$) gives positive Schiff's test and has two chiral carbons. On acetylation 'X' yields triacetate. 'X' also undergoes following reactions

There are two open-chain aldotetroses: Erythrose and Threose.

• When reduced with $$\text{NaBH}_4$$, the $$\text{-CHO}$$ group is converted into a primary alcohol ($$\text{-CH}_2\text{OH}$$):
  • Erythrose reduces to erythritol, which has a plane of symmetry and is an optically inactive meso (achiral) compound.
  • Threose reduces to threitol, which has no plane of symmetry and remains an optically active chiral compound.

Since the reaction of 'X' with $$\text{NaBH}_4$$ yields a Chiral compound ('B'), X must be Threose.

for the L-isomer of X.

• In carbohydrate chemistry (Fischer projections), "L" means the $$\text{-OH}$$ group on the bottom-most chiral carbon (Carbon-3 in this case) points to the Left.
• Since it's a Threose, the two middle $$\text{-OH}$$ groups must point to opposite sides. If Carbon-3 points to the left, Carbon-2 must point to the right.

Correct Option B