Question
The correct order of basicity for the following heterocycles is
The correct answer is
$\text{C} > \text{B} > \text{A}$
To determine the correct order of basicity for the given heterocycles, we need to consider the availability of the lone pair of electrons on the nitrogen atom, which determines its ability to accept protons and thus its basicity.
- The structure of compound A is pyrrole. In pyrrole, the nitrogen's lone pair is delocalized as part of the aromatic system, reducing its availability to bond with protons, thus making it the least basic.
- Compound B is imidazole. Imidazole has one nitrogen atom with a lone pair involved in aromaticity (similar to pyrrole) and another nitrogen with a non-aromatic lone pair, increasing basicity compared to pyrrole but less than a pyridine structure.
- Compound C is pyridine. In pyridine, the nitrogen's lone pair is not involved in the aromatic π system, making it readily available for protonation and consequently, more basic than both pyrrole and imidazole.
Therefore, the correct order of basicity is C > B > A, based on the availability of the nitrogen lone pair for bonding with protons.
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Important Questions from Heterocyclic Compounds
- Compound A on Hofmann exhaustive N-methylation procedure (involving two cycles), followed by ozonolysis (i. $O_3$; ii. $Me_2S$) gives benzaldehyde, formaldehyde and 2,2-dimethylpropanedial. The structure of A is
- Isoquinoline on sequential reaction with benzoyl chloride and KCN followed by heating with aq. NaOH gives
The major products A and B formed in the following reaction sequence are:
- The major product formed in the reaction of indole with NaNH$_2$ and PhSO$_2$Cl is
The major product formed in the following reaction is
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