To determine the product of the electrophilic substitution reaction of pyridine in the presence of KNO3 and concentrated H2SO4 at 300°C, we need to understand the mechanism of the nitration of pyridine.
Pyridine is aromatic and contains a nitrogen atom which is electron-withdrawing due to its higher electronegativity compared to hydrogen. As a result, the electron density of the pyridine ring is lower than that of benzene, making electrophilic substitution on pyridine more challenging.
In nitration, the nitronium ion (NO2+) serves as the electrophile. Pyridine undergoes substitution reactions predominantly at the meta position relative to the nitrogen atom because:
The ortho and para positions to the nitrogen are deactivated due to its electron-withdrawing effect, which destabilizes the cation formed during electrophilic substitution.
The meta position is less hindered by the resonance and thus more stable for electrophilic attack.
Therefore, when KNO3 and concentrated H2SO4 are used to generate the nitronium ion in the reaction with pyridine, the substitution occurs predominantly at the 3-position, leading to the formation of 3-Nitropyridine.
Thus, the correct answer to the given question is 3-Nitropyridine.
Now, let us evaluate the other options:
4-Nitropyridine: Substitution at the 4-position (para-position) is less favorable compared to the meta-position due to resonance instability.
2-Nitropyridine: Substitution at the 2-position (ortho-position) is also disfavored for the same reason as the para-position.
N-nitro pyridinium salt: This product is not typically formed because the reaction conditions favor electrophilic substitution on the carbon atoms of the pyridine ring.
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