Which among the following aldehydes is most reactive towards nucleophilic addition reactions?
The Correct Option is A
Solution and Explanation
To determine which aldehyde is most reactive towards nucleophilic addition reactions, we need to consider the electronic and steric factors that influence the reactivity of aldehydes.
Conceptual Explanation
Aldehydes are more reactive towards nucleophilic addition reactions compared to ketones due to the less sterically hindered environment around the carbonyl carbon and the greater partial positive charge on the carbonyl carbon.
- Electronic Factors: Aldehydes with electron-withdrawing groups increase the electrophilicity of the carbonyl carbon, making the carbon more susceptible to nucleophilic attack.
- Steric Factors: The reactivity of aldehydes decreases with increased steric hindrance around the carbonyl carbon. Smaller and less hindered aldehydes are generally more reactive.
Analysis of Options
This option represents formaldehyde (HCHO). Formaldehyde is the simplest aldehyde with no alkyl groups attached, leading to minimal steric hindrance. It is also highly electrophilic as there are no alkyl groups to donate electron density to the carbonyl group.
This option represents acetaldehyde (CH3CHO). The presence of one methyl group slightly decreases its reactivity compared to formaldehyde due to the electron-donating effect and steric hindrance it introduces.
This option represents propionaldehyde (C2H5CHO). The ethyl group adds more steric hindrance and electron donation, reducing its reactivity compared to acetaldehyde.
This option represents butyraldehyde (C3H7CHO). With an even larger alkyl group, it has greater steric hindrance and electron donation, leading to the least reactivity in this set.
Conclusion
Based on electronic and steric considerations, formaldehyde (HCHO) is the most reactive towards nucleophilic addition reactions. It has no alkyl groups to cause steric hindrance, and no electron-donating groups to reduce the electrophilicity of the carbonyl carbon.
Thus, the correct answer is:
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