Question
The major product formed in the reaction of indole with NaNH$_2$ and PhSO$_2$Cl is
The correct answer is
Indole Reaction Mechanism
The reaction between indole, sodium amide ($NaNH$_2$), and benzenesulfonyl chloride ($PhSO$_2$Cl$) proceeds via nucleophilic substitution.
- Step 1: Deprotonation
Sodium amide ($NaNH$_2$) is a strong base that abstracts the acidic proton from the indole nitrogen. This generates the indolate anion, a potent nucleophile.
Reaction: Indole-NH + NaNH$_2$ $\rightarrow$ [Indole-N]$^-$ Na$^+$ + NH$_3$ - Step 2: Nucleophilic Attack
The indolate anion attacks the electrophilic sulfur atom of benzenesulfonyl chloride ($PhSO$_2$Cl$). - Step 3: Substitution
The chloride ion (Cl$^-$) is eliminated as a leaving group, forming the N-sulfonylated indole derivative.
Product: Indole-N-SO$_2$Ph (N-(phenylsulfonyl)indole)
Product Identification
The strong base $NaNH$_2$ directs the reaction to the nitrogen atom of indole. The electrophile $PhSO$_2$Cl$ then reacts at this nitrogen. Therefore, the phenylsulfonyl group ($PhSO$_2$-$) attaches to the N1 position of the indole ring.
Option 4 correctly depicts this N-substitution product, N-(phenylsulfonyl)indole.
| Option | Structure | Analysis |
| 1 |  |
Incorrect (C3 substitution) |
| 2 |  |
Incorrect (C2 substitution) |
| 3 |  |
Incorrect (Benzene ring substitution) |
| 4 | |
Correct (N1 substitution) |
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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 correct order of basicity for the following heterocycles is
The major product formed in the following reaction is
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