Question

Iodoform test is not given by:
A) 2-Pentanone
B) Ethanol
C) Ethanal
D) 3-pentanone

Answer 1

The reaction of ketone, aldehyde, or the alcohol when heated with the aqueous solution of sodium hydroxide or the sodium carbonate in presence of iodine forms yellow precipitate which is known as iodoform $\text{ C}{{\text{H}}_{\text{3}}}\text{I }$ . The carbonyl compounds which contains the $\text{ C}{{\text{H}}_{\text{3}}}\text{CO}-\text{ }$or the $\text{ C}{{\text{H}}_{\text{3}}}\text{C(H)(OH)}-$compounds undergo the iodoform reaction.

Complete answer:
The formation of yellow of iodoform is used as a test for certain aldehydes and ketone which have a methyl group bonded to carbonyl group $\text{ C}{{\text{H}}_{\text{3}}}\text{CO}-\text{ }$ or compounds that can form this group on oxidation. This test is performed by heating the substance with aqueous sodium hydroxide and iodine solution. On warming the reaction mixture, a yellow precipitate of iodoform indicates positive iodoform etc. Give this test
Remember that compounds containing $\text{ C}{{\text{H}}_{\text{3}}}-\text{CO}-\text{ }$or the $\text{ }\begin{matrix} {} & {} & \text{H} & {} \\\ {} & {} & | & {} \\\ \text{C}{{\text{H}}_{\text{3}}} & \- & \text{C} & \- \\\ {} & {} & | & {} \\\ {} & {} & \text{OH} & {} \\\ \end{matrix}\text{ }$ group give the iodoform test.
Let's now see which of the given does not show the iodoform test.
A) 2-Pentanone:
2-Pentanone contains the $\text{ C}{{\text{H}}_{\text{3}}}-\text{CO}-\text{ }$group. This group is at the terminal of the chain. Reaction of 2-Pentanone with the iodine in sodium hydroxide give iodoform .the reaction of the 2-Pentanone is as shown below,
$\text{ C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{ + 3}{{\text{I}}_{\text{2}}}\text{ + 4NaOH }\to \text{ C}{{\text{H}}_{\text{3}}}\text{I(yellow ppt) + C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COONa + 3NaI + 3}{{\text{H}}_{\text{2}}}\text{O }$

B) Ethanol:
Ethanol contains the $\text{ C}{{\text{H}}_{\text{3}}}\text{C(H)(OH)}-$group. This group is at the terminal of the chain. Reaction of with the iodine in sodium hydroxide give iodoform .the reaction of the ethanol is as shown below,
$\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH + 3}{{\text{I}}_{\text{2}}}\text{ + 4NaOH }\to \text{ C}{{\text{H}}_{\text{3}}}\text{I(yellow ppt) + HCOONa + 5NaI + 5}{{\text{H}}_{\text{2}}}\text{O }$

C) Ethanal:
Ethanal contains the $\text{ C}{{\text{H}}_{\text{3}}}\text{CO}-\text{ }$group. This group is at the terminal of the chain. Reaction of with the iodine in sodium hydroxide give iodoform .the reaction of the ethanal is as shown below,
$\text{ C}{{\text{H}}_{\text{3}}}\text{CHO + 3}{{\text{I}}_{\text{2}}}\text{ + 4NaOH }\to \text{ C}{{\text{H}}_{\text{3}}}\text{I(yellow ppt) + HCOONa + 3NaI + 3}{{\text{H}}_{\text{2}}}\text{O }$

D) 3-Pentanone:
The 3-pentanaone structure is as follows,$\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{COC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{ }$ .The ketone does not have a $\text{ C}{{\text{H}}_{\text{3}}}-\text{CO}-\text{ }$or the $\text{ C}{{\text{H}}_{\text{3}}}\text{C(H)(OH)}-$group. Thus the 3-pentanone does not undergo the iodoform reaction. It is represented as follows,
$\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{COC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{ + 3}{{\text{I}}_{\text{2}}}\text{ + NaOH }\to \text{ No reaction }$
Thus, out of the given organic compounds, 3-pentanone does not undergo the iodoform test.

Hence, (D) is the correct option.

Note:
Note that, this reaction is also known as the haloform reaction.it may be noted that the iodoform reaction is the oxidation reaction. But this does not affect the carbon-carbon double bond if present in the molecule. It also needs to remember that certain alcohols like ethyl alcohol and the secondary alcohols which can be converted into acetaldehyde or methyl ketones give the iodoform test.