L-isomer of a compound A (C_{4}H_{8}O_{4}) gi | vedclass

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(A) $1$. Compound $A$ has the molecular formula $C_{4}H_{8}O_{4}$.$2$. It gives a positive test with $[Ag(NH_{3})_{2}]^{+}$ (Tollens' reagent),indicat

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Vedclass · Answer

(A) $1$. Compound $A$ has the molecular formula $C_{4}H_{8}O_{4}$.$2$. It gives a positive test with $[Ag(NH_{3})_{2}]^{+}$ (Tollens' reagent),indicating the presence of an aldehyde group.$3$. Treatment with acetic anhydride yields a triacetate,suggesting the presence of three hydroxyl $(-OH)$ groups.$4$. The $L$-isomer configuration means the $-OH$ group on the chiral carbon furthest from the aldehyde group is on the left in the Fischer projection.$5$. Bromine water oxidizes the aldehyde group to a carboxylic acid,forming an aldonic acid. For the $L$-isomer of erythrose (where both $-OH$ groups are on the left),this product is optically active.$6$. Concentrated $HNO_{3}$ oxidizes both the aldehyde and the primary alcohol group to carboxylic acids,forming an aldaric acid. For the $L$-isomer of erythrose,the resulting tartaric acid is meso (optically inactive due to an internal plane of symmetry).$7$. Comparing these properties,compound $A$ is $L$-erythrose,which corresponds to the structure in option $A$.

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