Question

Which of the following does not give iodoform?
(A) Acetic acid
(B) Lactic acid
(C) Acetophenone
(D) Propionic acid

Answer 1

Hint: Iodoform test is used to check the presence of carbonyl compounds with a structure of $$R-CO-C{H}_3$$ or alcohols with structure $$R-CH\left(OH\right)-C{H}_3$$ in a given unknown substance.
- A positive result is the appearance of a very pale yellow precipitate of triiodomethane or $$CH{I}_3$$ which is commonly known as iodoform. Apart from its colour, this can be recognised by its fairly “medical” or antiseptic smell.

Complete Solution :
Option A: acetic acid- acetic acid is a carboxylic acid, that is, it has the $$-COOH$$ bond. Due to delocalisation of electrons, there is presence of $$o$$ adjacent to carbonyl. Therefore, acetic acid does not give iodoform test.

Option B: Lactic acid- lactic acid gives iodoform reaction. The equation is given by:
$$C{H}_3-\stackrel{\stackrel{OH}{\text{I}}}{C}-\stackrel{\stackrel{O}{\text{II}}}{C}-OH\stackrel{}{\to }\underset{\underset{CH{I}_3+Na-\stackrel{\stackrel{O}{\text{II}}}{C}-\stackrel{\stackrel{O}{\text{II}}}{C}-OH}{\downarrow I_2/NaOh/H_{2}O}}{C{H}_3}-\stackrel{\stackrel{O}{\text{II}}}{C}-\stackrel{\stackrel{O}{\text{II}}}{C}-OH$$

Option C: Acetophenone- This has the formula - $$C_6{H}_{5}COC{H}_3$$. Since there is a presence of $$C{H}_{3}CO$$ attached to the carbon, Acetophenone gives the iodoform test.

Option D: Propionic acid- As we can see a $$-COOH$$ group in which electrons are delocalised, hence, it also does not give the iodoform test.
So, the correct answer is “Option A and D”.

Note: 1. A pale yellow precipitate of iodoform is formed when iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or secondary alcohol with a methyl group in the alpha position. This test can be used to identify aldehydes or ketones.
2. Ethanol is the only aldehyde to give the iodoform test.