Given below are two statements:
Consider the following reaction
R-C(∥)(O)-R + H₂O ⇌(K) R-C(╲ ╱)(HO OH)-R
Statement (I): In the case of formaldehyde
is about 2280, due to small substituents, hydration is faster.
Statement (II) : In the case of trichloro acetaldehyde
is about 2000 due to -I effect of -Cl .
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Ask what makes the hydrate stable: less crowding helps formaldehyde, and electron pull helps chloral.

1. What K measures. Hydration turns a flat sp² carbonyl carbon into a crowded sp³ carbon with two OH groups. A large K means the hydrate is strongly favoured. Two things raise K: small groups that avoid crowding, and electron-withdrawing groups that make the carbonyl carbon more positive.

   Carbonyl compound	Groups on C=O	Approximate K	Why
   Acetone	two CH₃	about 10⁻³	+I groups, more crowding
   Acetaldehyde	H and CH₃	about 1	one small H
   Formaldehyde	two H	about 2280	smallest groups, no +I donation
   Chloral, CCl₃CHO	H and CCl₃	about 2000	strong -I effect of three Cl

2. Statement I. In HCHO both groups are H, so the hydrate CH₂(OH)₂ has almost no steric strain and there is no alkyl group feeding electrons to the carbon. Hydration is strongly favoured, K ≈ 2280. True.
3. Statement II. In CCl₃CHO the three Cl atoms pull electron density away through the -I effect. The carbonyl carbon becomes very electrophilic and the C=O is destabilised, so water adds readily and K ≈ 2000. This is why chloral hydrate is a stable, isolable solid. True.
4. So. Both statements are correct, with correct reasons, so the answer is (b).

The idea to keep. Hydrate formation is favoured by small groups and by electron-withdrawing groups on the carbonyl carbon. Alkyl groups, with their bulk and +I effect, push the equilibrium back to the carbonyl.

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