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

Question

Compound A undergoes Cannizaro reaction and B undergoes positive iodoform test. Therefore, (A) A = acetaldehyde B = 1-pentanal (B)  A=C6H5CH2CHO B = 3-pentanone (C) A = formaldehyde B = 2-pentanone (D) A = propionaldehyde B = 1-pentanol

Tardigrade Admin · Accepted Answer

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

(N a O H)

give iodoform.
Therefore, the correct combination of

A

and

B

must be formaldehyde and

2

-pentanone.
Their reactions are as follows :
(i)

formaldehyde 2 H C H O + N a O H \to C H_{3} O H

(ii)

N a O H + I_{2} \to N a O I + N a I + H_{2} O

N a O H C_{3} H_{7} COON a + iodoform C H I_{3}

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

A = acetaldehyde B = 1-pentanal

$A = C_{6} H_{5} C H_{2} C H O B = 3 -$ pentanone

A = formaldehyde B = 2-pentanone

A = propionaldehyde B = 1-pentanol

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

A = acetaldehyde B = 1-pentanal

A=C6​H5​CH2​CHOB=3−pentanone

A = formaldehyde B = 2-pentanone

A = propionaldehyde B = 1-pentanol

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

$A = C_{6} H_{5} C H_{2} C H O B = 3 -$ pentanone