question_answer

Compound A undergoes Cannizaro reaction and B undergoes positive iodoform test. Therefore,

A)  A = acetaldehyde B = 1-pentanal

B)  \[A={{C}_{6}}{{H}_{5}}C{{H}_{2}}CHO\]  B = 3-pentanone

C)  A = formaldehyde  B = 2-pentanone

D)  A = propionaldehyde B = 1-pentanol

Correct Answer: C

Solution :

According to Cannizaro, aldehyde or ketones having no any a-hydrogen atoms undergo intermolecular oxidation-reduction to form an alcohol and acid. Again, in iodoform test, any alcohol, aldehyde or ketone, when treated with iodine \[({{I}_{2}})\] and sodium hydroxide (NaOH) give iodoform. Therefore, the correct combination of A and B must be formaldehyde and 2-pentanone. Their reactions are as follows : (i) \[\underset{\begin{smallmatrix}  formaldehyde \\  (B) \end{smallmatrix}}{\mathop{2HCHO}}\,+NaOH\xrightarrow{{}}C{{H}_{3}}OH\] (ii) \[NaOH+{{I}_{2}}\xrightarrow{{}}NaOI+NaI+{{H}_{2}}O\] \[\xrightarrow{NaOH}{{C}_{3}}{{H}_{7}}COONa+\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,\]