Question

Hybridisation change involved at C-1 in the above reaction

• A. sp³ to sp
• B. sp³ to sp²
• C. sp² to sp³
• D. sp to sp²

Answer 1

The Correct Option is B

This is a reaction involving the conversion of CH₃-CH=CH-CH₂OH (butenol) to CH₃-CH=CH-CHO (butenal). The reaction is carried out using PCC (Pyridinium chlorochromate), which oxidizes the alcohol (OH) group to an aldehyde (CHO) group.

Regarding the hybridization change at C-1:

In the starting compound (CH₃-CH=CH-CH₂OH), the C-1 is part of a C=C double bond, making it sp² hybridized. When this compound is oxidized to CH₃-CH=CH-CHO, C-1 is still part of the double bond, but now it is adjacent to the carbonyl group, which maintains its sp² hybridization.

Therefore, C-1 does not undergo a change in hybridization, and it remains sp² hybridized throughout the reaction.

Conclusion:

The correct answer is Option 2: sp³ to sp². The carbon in C-1 undergoes a shift from sp³ hybridization (in the alcohol form) to sp² hybridization (in the aldehyde form) due to the formation of the double bond with the oxygen atom.

Answer 2

In the given reaction, we have a compound that is an alcohol (sp³ hybridized) undergoing oxidation with PCC (Pyridinium chlorochromate) to form an aldehyde. The reaction is: \[ \text{CH₃-CH=CH-CH₂OH} \xrightarrow{\text{PCC}} \text{CH₃-CH=CH-CHO} \] In this reaction, at C-1, the alcohol group (\( -OH \)) is converted to a carbonyl group (\( -CHO \)). To undergo this conversion, the hybridization of C-1 must change from sp³ (in the alcohol group) to sp² (in the aldehyde group). This is because the carbonyl group (C=O) requires sp² hybridization.

Therefore, the hybridization change at C-1 in this reaction is from sp³ to sp². Thus, the correct answer is: \[{\text{(B) sp³ to sp²}} \]