Question

In nucleophilic addition reaction of HCN to ethanal, the hybridisation of the carbonyl carbon in the final product is

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

Hint:

In nucleophilic addition reactions of aldehydes and ketones:
- Starting carbonyl carbon: \(sp^2\) (planar, \(120^\circ\)).
- Final product carbon: \(sp^3\) (tetrahedral, \(109.5^\circ\)).

Answer 1

The Correct Option is A

Concept:
Nucleophilic addition is the characteristic reaction of aldehydes and ketones.
During this reaction, the planar \(sp^2\)-hybridized carbonyl carbon is attacked by an incoming nucleophile, converting it into a tetrahedral \(sp^3\)-hybridized carbon in the resulting addition product.

Step 1: Carbonyl Carbon in Starting Ethanal:
The chemical structure of ethanal is \(\text{CH}_3-\text{CH}=\text{O}\).
The carbonyl carbon is bonded to:
- One hydrogen atom by a \(\sigma\)-bond,
- One methyl group by a \(\sigma\)-bond,
- One carbonyl oxygen atom by a double bond (one \(\sigma\)-bond and one \(\pi\)-bond).
With three \(\sigma\)-bonds and no lone pairs, the steric number is \(3\), corresponding to \(sp^2\) hybridization with trigonal planar geometry.

Step 2: Addition of Hydrogen Cyanide:
The addition of \(\text{HCN}\) to ethanal proceeds via base catalysis:
1. The cyanide nucleophile (\(:\!\text{CN}^-\)) attacks the electrophilic carbonyl carbon.
2. The \(\text{C}=\text{O}\) \(\pi\)-bond cleaves, transferring the electron pair to oxygen to form an alkoxide intermediate.
3. Subsequent proton transfer yields ethanal cyanohydrin (2-hydroxypropanenitrile):
\[ \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3-\text{CH}(\text{OH})-\text{CN} \]

Step 3: Determining Hybridization in the Final Product:
In ethanal cyanohydrin, the former carbonyl carbon atom is bonded to four distinct groups:
- A single \(\sigma\)-bond to \(-\text{H}\),
- A single \(\sigma\)-bond to \(-\text{OH}\),
- A single \(\sigma\)-bond to \(-\text{CH}_3\),
- A single \(\sigma\)-bond to \(-\text{CN}\).
With four single \(\sigma\)-bonds and zero lone pairs, the steric number of this carbon atom is \(4\).
Therefore, the hybridization of this carbon atom changes from \(sp^2\) in ethanal to \(sp^3\) (tetrahedral geometry) in the final product.
Final Answer:
The hybridization of the carbonyl carbon in the final product is \(sp^3\), corresponding to option (A).