Question

Compound A undergoes Cannizzaro reaction and undergoes positive iodoform test. Therefore:

• A. $A=Acetaldehyde;B=1-pentanal$
• B. $A=C_6{H}_5{CH}_{2}CHO;B=3-pentanone$
• C. $A=Formaldehyde;B=2-pentanone$
• D. $A=Propionaldehyde;B=1-pentanol$

Answer 1

Answer: (C)

$A=Formaldehyde;B=2-pentanone$

Solution:- (C) A = Formaldehyde; B = 2-pentanone

According to Cannizzaro, aldehyde or ketones not having any $\alpha$-hydrogen atoms undergo intermolecular oxidation-reduction to form an alcohol and acid.

In iodoform test, any alcohol, aldehyde or ketone, when treated with iodine $\left(I_2\right)$ and sodium hydroxide $\left(NaOH\right)$ give iodoform. Therefore, the correct combination of A and B must be formaldehyde and 2-pentanone. Their reactions are as follows:

Cannizzaro reaction-

$2\underset{\left(A\right)}{HCHO}+NaOH⟶2C{H}_{3}OH$

Iodoform test-

$\underset{\left(B\right)}{\stackrel{\underset{|}{C{H}_3}}{\underset{\stackrel{|}{C_3{H}_7}}{C}}=O}\stackrel{NaOH+I_2}{\to }\stackrel{\underset{|}{I_{3}C}}{\underset{\stackrel{|}{C_3{H}_7}}{C}}=O\stackrel{NaOH}{\to }{C}_3{H}_{7}COONa+\underset{\left(\text{Iodoform}\right)}{CH{I}_3}$