(I) is more acidic than
The negative charge of the conjugate base of `(II)` cannot be delocalised through the benzene ring.

Options

A. If both `(A)` are `( R)` are true and `( R)` is the correct explanation of `(A)`.
B. If both `(A)` and `( R)` are true but `( R)` is not the correct explanation of `(A)`.
C. If `(A)` is true but `( R)` is false.
D. If `(A)` is false but `( R)` is true.

Answer

D

Explanation

`PhCOOH (II)` is more acidic than `(I)` due to `(-I)` and `(+R)` effects of `(Ph)` group. `(I)` is less acidic due to `(+I)` effect of its `R^, s`of the sides of the ring, so `(A)` is false.

can delocalise to `(C = O)` group but not with the benzene ring because `O^(Ө)` is not in conjugation with `(C = C)` bond of benzene ring.