Question

Pyridine undergoes electrophilic aromatic substitution preferentially at the 3 position as illustrated by the synthesis of 3 -nitropyridine. Under these acidic conditions, the species undergoing nitration is not pyridine, but its conjugate acid. Write resonance contributing structures for the intermediate formed by attack of \(\mathrm{NO}_{2}^{+}\)at the 2,3 , and 4 positions of the conjugate acid of pyridine. From examination of these intermediates, offer an explanation for preferential nitration at the 3 position.

Short answer

Answer: In the resonance contributing structures for the intermediate formed after the attack, the 3-nitropyridinium ion intermediate is more stable than the other two because it has less positive charge on the nitrogen. Since nitrogen is more electronegative than carbon, it is less stable when bearing a positive charge. Therefore, the reaction prefers to proceed via the formation of the 3-nitropyridinium ion intermediate, leading to the preferential synthesis of 3-nitropyridine.

Step-by-step solution

Draw the conjugate acid of pyridine

To understand the reaction and the intermediates formed, we first need to identify the conjugate acid of pyridine. Pyridine is a nitrogen-containing aromatic compound with the formula \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\). Its conjugate acid is formed by protonation of the nitrogen atom, resulting in a positive charge on the nitrogen. The structure of the conjugate acid of pyridine is as follows: 

Draw resonance structures for the attack of \(\mathrm{NO}_{2}^{+}\) at positions 2, 3, and 4

Now, we need to draw the resonance contributing structures for the intermediate formed by the attack of \(\mathrm{NO}_{2}^{+}\) at the 2, 3, and 4 positions of the conjugate acid of pyridine. Here are the resonance structures for each position: 

Examine the intermediates and offer an explanation for preferential nitration at the 3 position

We can now examine the resonance structures of the intermediates to understand why the 3 position is preferred for nitration. In the 2-nitropyridinium ion, we can see that the positive charge is delocalized across the nitrogen and the ortho carbons (positions 2 and 6). In the 4-nitropyridinium ion, the positive charge is delocalized across the nitrogen and the para carbons (positions 4 and 6). However, in the 3-nitropyridinium ion, the positive charge is only delocalized across the meta carbons (positions 3 and 5), which means that the nitrogen atom does not bear a significant positive charge. Since the nitrogen is more electronegative than carbon, it is less stable when bearing a positive charge. The 3-nitropyridinium ion intermediate is more stable than the other two because it has less positive charge on the nitrogen. Consequently, the nitration reaction prefers to proceed via the formation of the 3-nitropyridinium ion intermediate, leading to the preferential synthesis of 3-nitropyridine.