Question

The increasing order of the acidity of the $\mathrm{\alpha }$-hydrogen of the following compounds

• A. $\left(\mathrm{D}\right)<\left(\mathrm{C}\right)<\left(\mathrm{A}\right)<\left(\mathrm{B}\right)$
• B. $\left(\mathrm{A}\right)<\left(\mathrm{C}\right)<\left(\mathrm{D}\right)<\left(\mathrm{B}\right)$
• C. $\left(\mathrm{C}\right)<\left(\mathrm{A}\right)<\left(\mathrm{B}\right)<\left(\mathrm{D}\right)$
• D. $\left(\mathrm{B}\right)<\left(\mathrm{C}\right)<\left(\mathrm{A}\right)<\left(\mathrm{D}\right)$

Answer 1

The correct option is A

$\left(\mathrm{D}\right)<\left(\mathrm{C}\right)<\left(\mathrm{A}\right)<\left(\mathrm{B}\right)$

1. The acidity character increases when the electron-withdrawing group is present near the atom.
2. Because the electron-withdrawing group attracts the nearer bond towards itself and it leads to the ease of removal of ${\mathrm{H}}^{+}$ ion from the compound nearby the electron-withdrawing group.

The Explanation for the correct option (A):

Compound (B)

• It has the highest acidic character in alpha-hydrogen of the compound.
• This is because alpha–hydrogen is surrounded by two keto groups which have an electron-withdrawing group and therefore it increases the acidity of alpha hydrogen.

Compound (A)

• It has the second highest acidic alpha-hydrogen in the compound and (D) has the least acidic character in the alpha-hydrogen compound.
• This is because (D) and (C) have an electron-releasing group nearer to alpha-hydrogen and it decreases its acidity.
• Thus, in comparing (A), (C), and (D), (A) has high acidic alpha-hydrogen since it won’t contain an electron-releasing group in its structure.

Compounds (C) and (D)

• Among (C) and (D), (D) has the least acidic alpha-hydrogen, since $-{\mathrm{NMe}}_2$ is more highly activating than $\mathrm{OMe}$ and therefore it decreases the acidity.

The correct option is (a).