If x and y are the pK_b values of p-methylaniline and N,N-dimethylaniline, what is the relation between x and y?

x > y

The correct answer is option 4. x > y.

The \( pK_b \) value of a compound is a measure of its basicity, with lower \( pK_b \) values indicating stronger bases. To determine the relationship between the \( pK_b \) values of p-methylaniline (x) and N,N-dimethylaniline (y), we need to understand the effects of the substituents on the basicity of the amine nitrogen.

p-Methylaniline: p-Methylaniline (p-toluidine) has a methyl group in the para position of the benzene ring. The methyl group is an electron-donating group through both inductive and hyperconjugation effects, which increases the electron density on the nitrogen, making it more basic than aniline. Thus, p-methylaniline is more basic than aniline, which means it has a lower \( pK_b \) value compared to aniline.

N,N-Dimethylaniline:

N,N-Dimethylaniline has two methyl groups directly attached to the nitrogen. The methyl groups are electron-donating through inductive effects, increasing the electron density on the nitrogen significantly more than in p-methylaniline. Therefore, N,N-dimethylaniline is more basic than p-methylaniline, resulting in an even lower \( pK_b \) value.

Since N,N-dimethylaniline is more basic than p-methylaniline, the \( pK_b \) value of N,N-dimethylaniline (y) will be lower than the \( pK_b \) value of p-methylaniline (x). This means: \( x > y \)

Therefore, the correct relationship between the \( pK_b \) values is x > y.