Question

Which carboxylic acid give positive iodoform test? (A) Lactic acid (B) Pyruvic acid (C) Carbolic acid (D) Both (A) and (B)

Short answer

(B) Pyruvic acid

Step-by-step solution

1. Understand the structures of the given carboxylic acids

First, let's examine the structures of the given carboxylic acids to understand whether they possess a methyl ketone group or an active methyl group. (A) Lactic acid: CH3CH(OH)COOH (B) Pyruvic acid: CH3COCOOH (C) Carbolic acid (Phenol): C6H5OH

2. Perform the iodoform test on Lactic acid

Lactic acid (CH3CH(OH)COOH) neither has a methyl ketone group, nor an active methyl group. Thus, it will not give a positive iodoform test.

3. Perform the iodoform test on Pyruvic acid

Pyruvic acid (CH3COCOOH) contains a methyl ketone group (CH3CO-). On reaction with iodine and a base, it forms iodoform (CHI3), which indicates a positive iodoform test.

4. Perform the iodoform test on Carbolic acid

Carbolic acid (C6H5OH) or phenol does not have a methyl group. It doesn't have any chance for the formation of a methyl ketone nor an active methyl group. So, it will not give a positive iodoform test.

5. Determine the correct answer choice

Given the above discussion, only Pyruvic acid (B) gives a positive iodoform test. Therefore, the correct answer is (B) Pyruvic acid.

Key concepts

Lactic Acid and the Iodoform Test

When it comes to the iodoform test, lactic acid typically leaves students scratching their heads. Let's get to the bottom of this. Lactic acid, with its chemical formula CH3CH(OH)COOH, might seem like a candidate for the iodoform test because of its methyl group (CH3). However, it's not just any methyl group that's required for a positive result. The iodoform test specifically looks for a methyl ketone group – that's a methyl group attached to a carbonyl (C=O) function. Lactic acid does not have this structure; the methyl group is attached to a carbon with a hydroxyl group (OH), not a carbonyl. So, despite its methyl group, lactic acid doesn't produce the tell-tale yellow precipitate of the iodoform test. Simply put, no methyl ketone group, no positive iodoform test.

Pyruvic Acid's Reaction in the Iodoform Test

Pyruvic acid grabs the spotlight in the iodoform test. Structurally, pyruvic acid distinguished as CH3COCOOH, is equipped with the elusive methyl ketone group (CH3CO-) that the test screens for. This group is essential because it can be oxidized to form iodoform (CHI3), a yellow precipitate, when treated with iodine in the presence of a base. It's literally a textbook example of a positive reaction. This tells us that pyruvic acid is the substance we're looking for when we want to see a positive iodoform test result. The presence of that specific structure, a methyl group connected to a carbonyl, becomes the star of this chemical show.

Carbolic Acid (Phenol) and Its Relationship to the Iodoform Test

Carbolic acid might sound like it's related to carboxylic acids, but it's much better known by its common name, phenol. Its structure, a benzene ring with an OH group attached (C6H5OH), lacks both a methyl group and a carbonyl group. This structure means that phenol is out of the running for a positive iodoform test from the start. There's no carbon chain ending in a -CH3 group that could be oxidized to form iodoform. Therefore, carbolic acid (phenol) does not fulfill the criteria needed for a positive iodoform test. This differentiation is critical as it aids in understanding the specific requirements for the chemical reactions tested.

Understanding the Methyl Ketone Group in Carboxylic Acids

Within the realm of organic chemistry, a methyl ketone group holds a special place, particularly in the context of the iodoform test. This group has the structure -CH3CO-, and it's a clear signpost for the presence of a compound that will react positively in the iodoform test. When the methyl ketone group reacts with iodine in an alkaline environment, it undergoes a haloform reaction, resulting in the formation of iodoform, which is characterized by a yellow precipitate. It's this methyl ketone group that is present in pyruvic acid, which passes the iodoform test, unlike lactic acid and carbolic acid (phenol) which lack this functional group. Understanding the specific structure of the methyl ketone group is fundamental in predicting the outcomes of the iodoform test and in differentiating various compounds in organic chemistry.