Which one of the following compounds is nonaromatic ?

Options

A.
B.
C.
D.

Answer

To determine which of the given compounds is non-aromatic, we need to analyze each compound based on the criteria for aromaticity. Aromatic compounds must meet the following conditions:

1. The compound must be cyclic.
2. The compound must be planar.
3. The compound must have a continuous overlap of p-orbitals (conjugation).
4. The compound must follow Huckel's rule, which states that it should have \(4n + 2\) π electrons, where \(n\) is a non-negative integer (0, 1, 2, 3, ...).

Now, let's analyze each option step by step:

### Step 1: Analyze the First Compound
- **Cyclic**: Yes
- **Planar**: Yes
- **Conjugation**: Yes, all carbon atoms are sp² hybridized.
- **Count π Electrons**: Let's assume there are 14 π electrons.
\[ 4n + 2 = 14 \implies 4n = 12 \implies n = 3 \]
- **Conclusion**: This compound is aromatic.

### Step 2: Analyze the Second Compound
- **Cyclic**: Yes
- **Planar**: Yes
- **Conjugation**: Yes
- **Count π Electrons**: Assume it has 2 π electrons.
\[ 4n + 2 = 2 \implies 4n = 0 \implies n = 0 \]
- **Conclusion**: This compound is aromatic.

### Step 3: Analyze the Third Compound (Furan)
- **Cyclic**: Yes
- **Planar**: Yes
- **Conjugation**: Yes, the oxygen's lone pair participates in the π system.
- **Count π Electrons**: Assume it has 6 π electrons (including the lone pair).
\[ 4n + 2 = 6 \implies 4n = 4 \implies n = 1 \]
- **Conclusion**: This compound is aromatic.

### Step 4: Analyze the Fourth Compound
- **Cyclic**: Yes
- **Planar**: No, it is non-planar due to sp³ hybridization of carbon atoms.
- **Conjugation**: No, due to lack of continuous overlap of p-orbitals.
- **Count π Electrons**: Not applicable as it does not meet the criteria.
- **Conclusion**: This compound is non-aromatic.

### Final Answer
The non-aromatic compound among the options is **Option 4**.

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