Question

Iodoform test is not given by:

• A. 2-Pentanone
• B. Ethanol
• C. Ethanal
• D. 3-pentanone

Answer 1

The correct option is D

3-pentanone

Explanation for correct answer:

Option (D) 3-pentanone:

1. The iodoform test is used to determine whether a given unknown chemical contains carbonyl compounds with the structural formula $\mathrm{R}-\mathrm{CO}-{\mathrm{CH}}_3$ or alcohols with the formula $\mathrm{R}-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3$.
2. Iodine, a base, and a methyl ketone react to produce a yellow precipitate and an "antiseptic" odour.
3. It additionally exhibits positive results for a small number of certain secondary alcohols with at least one methyl group in the alpha position.
4. 3-pentanone ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CO}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$. It does not have a $-\mathrm{CO}-{\mathrm{CH}}_3$ group. Hence it does not show the iodoform test.
5. Hence correct option.

Explanation for incorrect answer:

Option (A) 2-Pentanone:

1. Its chemical formula is ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CO}-{\mathrm{CH}}_3$, and an iodoform test will reveal this.

Option (B) Ethanol:

1. Iodoform testing demonstrates that ethanol has a chemical formula ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ .

Option (C) Ethanal:

1. Ethanal has a ${\mathrm{CH}}_3-\mathrm{CHO}$ structure, which means that it has a $-\mathrm{CO}-{\mathrm{CH}}_3$ group.

Therefore, 3-pentanone, option D, does not give the iodoform test.