Question

The increasing order of extent of $H$-bonding of the alkyl ammonium ions, $RN{H}_3^{+},R_{2}N{H}_2^{+},R_{3}N{H}^{+}$in water is

• A. $R_{3}N{H}^{+}<R_{2}N{H}_2^{+}<RN{H}_3^{+}$
• B. $R_{3}N{H}^{+}<RN{H}_3^{+}<R_{2}N{H}_2^{+}$
• C. $R_{2}N{H}_2^{+}<RN{H}_3^{+}<R_{3}N{H}^{+}$
• D. $RN{H}_3^{+}<R_{2}N{H}_2^{+}<R_{3}N{H}^{+}$
• E. $RN{H}_3^{+}<R_{3}N{H}^{+}<R_{2}N{H}_2^{+}$

Answer 1

Answer: (A)

The ammonium compounds exhibit the following order of solubility in water.
Primary ammonium $\left(RN{H}_3^{+}\right)>$secondary ammonium $\left(R_{2}N{H}_2^{+}\right)>$ tertiary ammonium $\left(R_{3}N{H}^{+}\right)$
Quaternary ammonium salts exhibit the lowest extent of H-bonding of alkyl ammonium ion in water. The reason behind this is more the number of hydrogen bonding, more is the hydration energy of molecule and hence, more is the stability of the ammonium compound.