Which of the following compounds would undergo aldol condensation,which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
$(i)$ Methanal $(ii)$ $2-$Methylpentanal $(iii)$ Benzaldehyde $(iv)$ Benzophenone $(v)$ Cyclohexanone $(vi)$ $1-$Phenylpropanone $(vii)$ Phenylacetaldehyde $(viii)$ Butan$-1-$ol $(ix)$ $2,2-$Dimethylbutanal

(N/A) Aldehydes and ketones having at least one $\alpha-$hydrogen undergo aldol condensation. The compounds $(ii)$ $2-$methylpentanal,$(v)$ cyclohexanone,$(vi)$ $1-$phenylpropanone,and $(vii)$ phenylacetaldehyde contain one or more $\alpha-$hydrogen atoms. Therefore,these undergo aldol condensation.
Aldehydes having no $\alpha-$hydrogen atoms undergo Cannizzaro reactions. The compounds $(i)$ Methanal,$(iii)$ Benzaldehyde,and $(ix)$ $2,2-$dimethylbutanal do not have any $\alpha-$hydrogen. Therefore,these undergo Cannizzaro reactions.
Compound $(iv)$ Benzophenone is a ketone having no $\alpha-$hydrogen atom and compound $(viii)$ Butan$-1-$ol is an alcohol. Hence,these compounds do not undergo either aldol condensation or Cannizzaro reactions.
Aldol condensation $(ii)$: $2CH_3CH_2CH_2-CH(CH_3)-CHO \xrightarrow{dil. NaOH} CH_3CH_2CH_2-CH(CH_3)-CH(OH)-CH(CHO)-CH_2CH_2CH_3$ (product: $3-$Hydroxy$-2,4-$dimethyl$-2-$propylheptanal).
Cannizzaro reaction $(i)$: $2HCHO + conc. KOH \to CH_3OH + HCOOK$.
Cannizzaro reaction $(ix)$: $2CH_3CH_2-C(CH_3)_2-CHO \xrightarrow{conc. NaOH} CH_3CH_2-C(CH_3)_2-CH_2OH + CH_3CH_2-C(CH_3)_2-COONa$.