${\mathrm{CH}}_3-\mathrm{CHO}\stackrel{\mathrm{dilute} \mathrm{NaOH}}{\to } \mathrm{A} \stackrel{∆}{\to } \mathrm{B}.$

Regarding the compounds ‘A’ and ‘B’,  incorrect statement is

The reaction is aldol condensation; 

‘A’ is optically active, it is aldol ($\mathrm{\beta }$-hydroxy butyraldehyde)

${\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_2-\mathrm{CHO} \underset{-{\mathrm{H}}_2\mathrm{O}}{\overset{∆}{\to }} \underset{\left(\mathrm{B}\right)}{{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} }$

‘B’ shows geometrical isomerism.

‘B’ is 2-butenal (or) crotonaldehyde ($\mathrm{\alpha },\mathrm{\beta }$-unsaturated carbonyl compound)