Q.50. - Organic Chemistry

Question

Question: a. Explain why protonation of pyrrole occurs at C2 to form A, rather than on the N atom toform B.

b. Explain why A is more acidic than C, the conjugate acid of pyridine.

Step-by-Step Solution

Answer

Answer

 

a. An electron-rich center is formed at C2 during resonance, making the protonation of pyrrole take place at C2 to form A rather than the nitrogen atom to form B. 

b. The conjugate base of an aromatic compound is less stable than that of a non-aromatic compound. Hence, A is more acidic than C.

Step 1: Protonation

The generation of a conjugate base by incorporating a proton into acid is termed protonation. Numerous chemical reactions involve protonation.

Step 2: Acidic strength

The charge present in the atom can influence the acidic strength of a particular compound. A compound comprising a greater positive charge is regarded to be acidic.

Step 3: Answers related to pyridine and pyrrole

a. The protonation of pyrrole can be given as:

 

Protonation of pyrrole

 

Protonation of C2 forms conjugate acid A and the positive charge can be delocalized by resonance. The resonance stabilization of the positive charge does not occur in compound B.

 

 

b. The loss of a proton from A gives two electrons to nitrogen and forms a pyrrole that contains  $6\mathrm{\pi }$ electrons. These electrons can delocalize in the five-membered ring and become aromatic.

 

This leads the deprotonation process to be favorable, and compound A becomes acidic.

Compound A

 

Both the compounds C and its conjugate base pyridine are aromatic. The compound C contains $6\mathrm{\pi }$ electrons and is aromatic. Hence, there is less to be gained by deprotonation, and C is less acidic than A.

Compound C