Draw the products formed from the acid–base reaction of KOH with each compound.

a.

b.

c.

d.

Acids and bases undergo neutralization to give salt and water.

The anionic part $\text{(}{}^{-}\text{OH)}$ in the base reacts with the cationic part ${\text{(H}}^{+}\text{)}$ in the acid to form water. This step is called the proton abstraction.

The remaining counter ions of the acid and the base then react together to form a salt.

The $\text{(}{}^{-}\text{OH)}$in the base (KOH) gives its lone pair of electrons to the –H atom in the given compound, which donates its bonded electron density to the compound. Hence, the proton from the acid is transferred to the base in the proton abstraction step. The products formed are salt and water of the respective compounds.

a.

The products are mentioned below:

Reaction of a. with KOH

b.

The products are mentioned below:

Reaction of b. with KOH

c.

The products are mentioned below:

Reaction of c. with KOH

d.

The products are mentioned below:

Reaction of d. with KOH