Question

When the imidazole ring of Histidine is protonated, the tendency of nitrogen to be protonated (proton migrates from-COOH) is in the order?

A. $\beta > \gamma > \alpha $
B. $\gamma > \beta > \alpha $
C. $\gamma > \alpha > \beta $
D. $\beta > \alpha > \gamma $

Answer 1

Hint: We know that protonation is the process of transfer of transfer of proton $\left( {{{\rm{H}}^{\rm{ + }}}} \right)$ from one atom to another. Here, we have to check the order in which three nitrogen atoms are protonated.

Complete step by step answer:
Here, $\gamma $ nitrogen is nearby to the COOH. And we know that protonation takes place first in the nearby atom from which hydrogen atoms get protonated. So, protonation occurs first in $\gamma $ nitrogen.

In $\beta $ nitrogen, the protonation results in counter resonance as from both sides double bonds try to break due to which resonance is nullified.

In $\alpha $ nitrogen, the lone pair of N is in resonance.

We know that the resonance makes a compound stable.Therefore, protonation occurs before $\alpha $ nitrogen than $\beta $ nitrogen.

So, the order of protonation is $\gamma > \alpha > \beta $.

So, the correct answer is Option C.

Additional Information:
Protonation is a known fundamental chemical reaction. It is a step of many catalytic and stoichiometric processes. If a molecule or ion can undergo more than one protonation it is termed as polybasic acid. Deprotonation is the reverse of protonation reaction. In deprotonation reaction, a proton is removed from a compound. This reaction occurs in most acid-base reactions. Substances that can protonate another substance are termed as Bronsted Lowry acid.

Note: Always remember that resonance stabilizes a compound. Resonance is the phenomenon as a result of which a molecule can be expressed in different forms, none of which can explain all the properties of the molecule. The actual structure of the molecule is termed a resonance hybrid.