${\mathrm{CH}}_3\mathrm{CHO}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}\underset{\mathrm{\Delta }}{\overset{\mathrm{NaOH}}{\to }}\mathrm{Product}$. Products are

A) 2-butenal

B) 1-butenal

C) 3-hexenal

D) 2-pentenal

E) 2-methyl-2-pentenal

F) 2-methyl-2-butenal

It is an example of cross aldol condensation Both aldehydes contain α-hydrogen atoms.

The possible products.

$\begin{array}{l}\left(1\right){\mathrm{CH}}_3-\mathrm{CH}=\mathrm{O}+{\mathrm{H}}_2\mathrm{CH}-\mathrm{CHO}\underset{\mathrm{\Delta }}{\overset{{\mathrm{OH}}^{-}}{\to }}{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}+{\mathrm{H}}_2\mathrm{O}\\ \left(2\right){\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{O}+{\mathrm{H}}_2\underset{\stackrel{|}{{\mathrm{CH}}_3}}{\mathrm{C}}-\mathrm{CHO}\underset{\mathrm{\Delta }}{\overset{{\mathrm{OH}}^{-}}{\to }}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\underset{\stackrel{|}{{\mathrm{CH}}_3}}{\mathrm{C}}-\mathrm{CHO}+{\mathrm{H}}_2\mathrm{O}\end{array}$

(1) and (2) are self aldol products.

$\begin{array}{l}\left(3\right){\mathrm{CH}}_3-\mathrm{CH}=\mathrm{O}+{\mathrm{H}}_2\underset{\stackrel{|}{{\mathrm{CH}}_3}}{\mathrm{C}}-\mathrm{CHO}\underset{\mathrm{\Delta }}{\overset{{\mathrm{OH}}^{-}}{\to }}{\mathrm{CH}}_3-\mathrm{CH}=\underset{\stackrel{|}{{\mathrm{CH}}_3}}{\mathrm{C}}-\mathrm{CHO}+{\mathrm{H}}_2\mathrm{O}\\ \left(4\right){\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{O}+{\mathrm{H}}_2\mathrm{CH}-\mathrm{CHO}\underset{\mathrm{\Delta }}{\overset{{\mathrm{OH}}^{-}}{\to }}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}+{\mathrm{H}}_2\mathrm{O}\end{array}$

(3) and (4) are crossed aldol products.