The number of aldol reaction(s) that occurs in the given transformation is `CH_(3) CHO + 4 HCHO overset (conc. aq. NaOH) to`

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

`CH_(3)CHO` contains three `alpha`-H atoms but HCHO does not, therefore in the presence of conc. Alkali, three aldol condensation reaction take place to yield (I)
`CH_(3)-CHOoverset(HCH"O"//OH^(-))underset("First alcohol")rarrHOCH_(2)CH_(2)CHO`
`{:(" CH"_(2)OH),(" |"),(overset("HCHO"//OH^(-))underset("second alcohol")rarrHO-CH_(2)-CH-CHO):}`
`{:(" CH"_(2)OH),(" |"),(overset("HCHO"//OH^(-))underset("third alcohol")rarrHO-CH_(2)-C-CHO),(" |"),(" CH"_(2)OH):}`
`overset(bar(O)H)undersetunderset("reaction")("Cross Cannizzaro's")rarr underset("Formate ion")(H-overset(O)overset(||)C-bar(O)+`

Futher since HCHO both do not contain `alpha-H`, therefore, in the presence of conc. `OH^(-)`, both these aldehydes undergo Cross Cannizzaro reaction to give pentaerythritol(II) and formate ion