Given: Two statements about compounds (X) and (Y) are to be checked.
Find: Which statement is true and which is false.
For Statement I, solubility in NaHCO3 indicates the presence of an acidic group such as −COOH. The given compound contains a carboxylic acid group, so it reacts with sodium bicarbonate and dissolves.
The structure also has two chiral carbon atoms. As stated in the solution, the fragment −C∗H(OH)C∗H(CH3)COOH has two starred carbons, and each is attached to four different groups. Therefore, Statement I is true.
For Statement II, analyze the hybridization in CH3−CH2−C≡C−C(=O)H.
- The CH3 and CH2 carbons are sp3 hybridized. Count =2.
- The carbonyl carbon C=O is sp2 hybridized. Count =1.
- The two carbons in the triple bond C≡C are both sp hybridized. Count =2.
So the statement claiming only one carbon is sp hybridized is incorrect. Therefore, Statement II is false.
Hence, Statement I is true but Statement II is false. The correct option is D.