Propanal on reaction with dilute $\mathrm{NaOH}$ forms

1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$
2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$
3. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CHO}$
4. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}ight) \mathrm{CHO}$

Solution

Note that it is the $\alpha$ -carbon (and $\operatorname{not} \beta$ -) that is adding on the carbonyl oxygen of the other propanal molecule. ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more ALDEHYDES, KETONES AND CARBOXYLIC ACIDS questions on Aicharya