Statement I : Compound (X), shown below, dissolves in NaHCO₃ solution and has two chiral carbon atoms
Statement II : Compound (Y), shown below, has two carbons with sp³ hybridization, one carbon with sp² and one carbon with sp hybridization
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Check each claim on its own: the COOH and the two ring stereocentres in X, then every carbon of Y.

1. Acid test. X has a -COOH group. A carboxylic acid is a stronger acid than carbonic acid, so it reacts with NaHCO₃, releases CO₂ and dissolves as its sodium salt. The ketone group does not react, but one -COOH is enough.
2. Chiral carbons. The ring carbon carrying -COOH has four different groups: H, -COOH, the ring CH next to it, and a ring CH₂. The ring carbon carrying the -CO-C₆H₄Br group is chiral for the same reason. The other ring carbons are CH₂ and the benzene carbons are sp², so X has exactly two chiral carbons and Statement I is true.

   Carbon in CH₃-CH₂-CH=C=O	Bonds it makes	Hybridisation
   CH₃	4 single bonds	sp³
   CH₂	4 single bonds	sp³
   =CH⁻	1 double bond, 2 single bonds	sp²
   =C= (ketene carbon)	2 double bonds, to C and to O	sp

3. So. Y has two sp³, one sp² and one sp carbon, so Statement II is also true. Both statements are true, which is option (b).

The idea to keep. A carbon with two double bonds, as in C=C=O or an allene, is sp. For a chiral ring carbon, compare the two ways round the ring: if they differ, they count as two different groups.

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