Aldehyde and Ketone

$ \text { The complete reaction occurs as follows } $

Aldol condensation requires the presence of at least one $\alpha$-hydrogen in the aldehyde. Among the options, acetaldehyde $\left(\mathrm{CH}_3 \mathrm{CHO}\right)$ has $\alpha$-hydrogens and undergoes the reaction.

$ 2 \mathrm{CH}_3 \mathrm{CHO}+\text { dilute } \mathrm{NaOH}\longrightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \text { (3-hydroxy butanal) } $

This $\beta$-hydroxy aldehyde may further dehydrate to form an $\alpha, \beta$-unsaturated aldehyde:

$ \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \longrightarrow \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCHO}+\mathrm{H}_2 \mathrm{O} $

The other compounds either lack $\alpha$-hydrogens or are deactivated (e.g., benzaldehyde, phenylacetylene derivative).

$ \text { The complete reaction sequence is as follow } $

Aldehydes which contain $ \alpha $-hydrogen on a saturated carbon, i.e. $ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} $ will undergo aldol condensation.

Although $ \mathrm{CH}_{2}=\stackrel{\alpha}{\mathrm{C}} \mathrm{H}-\mathrm{CHO} $ also contain a $ \alpha $-hydrogen atom but cannot be easily abstracted by a base to form a carbanion and hence does not undergo aldol condensation.

Acetone does not undergo substitution reaction because they do not contain suitable leaving group.

Right answer is $d$ because after heating $(\Delta)$, dcarboxylation occurs.