The correct order of decreasing basic strength of following compounds is :

Q: The correct order of decreasing basic strength of following compounds is :

The correct order of decreasing basic strength of following compounds is :

Question

Accepted Answer

The correct order of decreasing basic strength of the given compounds is: Pyrrole > Pyridine > Piperidine > Aniline. This is supported by Option 4: $II>III>IV>I$.

1. **Explanation for Option 4:**

- **Pyrrole (II):** Pyrrole is a five-membered aromatic heterocycle with a nitrogen atom. The lone pair of electrons on the nitrogen is involved in the aromaticity of the ring, making it less available for protonation. However, in comparison to the other compounds listed, pyrrole's basicity is relatively higher due to its ability to stabilize the positive charge through resonance when it does accept a proton.

- **Pyridine (III):** Pyridine is a six-membered aromatic heterocycle with a nitrogen atom. The lone pair of electrons on the nitrogen is not involved in the aromaticity of the ring, making it more available for protonation compared to pyrrole. This makes pyridine more basic than pyrrole.

- **Piperidine (IV):** Piperidine is a saturated six-membered ring containing a nitrogen atom. The lone pair on the nitrogen is not involved in any resonance or aromaticity, making it highly available for protonation. This makes piperidine more basic than both pyrrole and pyridine.

- **Aniline (I):** Aniline is an aromatic amine where the lone pair on the nitrogen is delocalized into the benzene ring, reducing its availability for protonation. This makes aniline the least basic among the given compounds.

2. **Explanation for why other options are incorrect:**

- **Option 1 ($I>II>III>IV$):** This option suggests that aniline is the most basic, which is incorrect. Aniline's lone pair is delocalized into the benzene ring, making it less basic than the other compounds.

- **Option 2 ($II>I>III>IV$):** This option suggests that pyrrole is more basic than aniline, which is correct, but it incorrectly places aniline above pyridine and piperidine. Aniline is less basic than both pyridine and piperidine due to the delocalization of its lone pair.

- **Option 3 ($IV>I>III>II$):** This option suggests that piperidine is the most basic, which is correct, but it incorrectly places aniline above pyridine and pyrrole. Aniline is the least basic due to the delocalization of its lone pair into the benzene ring.

In conclusion, Option 4 correctly represents the order of decreasing basic strength as Pyrrole > Pyridine > Piperidine > Aniline, based on the availability of the nitrogen lone pair for protonation and the influence of resonance and aromaticity.