Question

Hybridizations of the atoms indicated with the asterisk (*) in the following compounds sequentially are

• A. \(sp^2\), \(sp^2\), \(sp^3\), \(sp^2\)
• B. \(sp^2\), \(sp^3\), \(sp^3\), \(sp^2\)
• C. \(sp^3\), \(sp^3\), \(sp^3\), \(sp^2\)
• D. \(sp^2\), \(sp^2\), \(sp^3\), \(sp^3\)

Hint:

If an atom with a lone pair is adjacent to a multiple bond or part of an aromatic ring, it typically adopts \(sp^2\) hybridization to enable resonance.

Answer 1

The Correct Option is A

{Step 1: Understanding the Concept:}
Hybridization depends on the number of sigma bonds and lone pairs. A lone pair that is delocalized via resonance is considered to be in a p-orbital, lowering the hybridization index.
{Step 2: Detailed Explanation:}
1. Phenyl acetate: The oxygen lone pair is in resonance with the carbonyl group. $\rightarrow$ \(sp^2\).
2. Vinyl ether: The oxygen lone pair is in resonance with the adjacent double bond. $\rightarrow$ \(sp^2\).
3. Dioxolane: The oxygen has two sigma bonds and two non-delocalized lone pairs in a saturated ring. $\rightarrow$ \(sp^3\).
4. Furan: The oxygen is part of an aromatic ring and must contribute a lone pair to the \(\pi\) system. $\rightarrow$ \(sp^2\).
{Step 3: Final Answer:}
The sequence is \(sp^2, sp^2, sp^3, sp^2\).