Question:medium

Which among the following compounds is a weakest base?

Show Hint

When ranking the basicity of aniline derivatives, remember this straightforward trend: basic strength increases with the number of alkyl groups on the nitrogen atom because of their electron-donating inductive effect. Therefore, the ordering goes: $\text{Aniline (weakest)} < \text{N-Methylaniline} < \text{N,N-Dimethylaniline}$.
Updated On: Jun 12, 2026
  • Phenylmethanamine
  • N-Methylaniline
  • Benzenamine
  • N,N-Dimethylaniline
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: State what controls amine basicity.
Basicity depends on how freely the nitrogen lone pair can grab a proton. If that lone pair is tied up by resonance into a ring, the amine becomes weaker.
Step 2: Set aside the strongest base.
Phenylmethanamine (benzylamine) has its $-\text{NH}_2$ on an $sp^3$ carbon, so the lone pair is not in conjugation with the ring. It stays fully available, making it the strongest base, not the weakest.
Step 3: Look at the three aromatic amines.
Benzenamine (aniline), N-methylaniline and N,N-dimethylaniline all have nitrogen attached directly to the ring, so all three lose lone-pair availability to resonance.
Step 4: Add the alkyl effect.
Methyl groups on nitrogen push electron density in by the $+I$ effect, partly restoring basicity.
Step 5: Rank the substituted anilines.
N,N-dimethylaniline has two methyls (strongest $+I$), N-methylaniline has one methyl (some $+I$), and plain aniline has none.
Step 6: Identify the weakest.
Aniline suffers full resonance loss with no compensating alkyl groups, so its nitrogen is the least electron-rich and it is the weakest base of the three aromatic amines, and of all four options.
\[ \boxed{\text{Benzenamine (aniline), option (3)}} \]
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