Question

Which of the following will not give iodoform test? (a) ethanal (b) benzyl alcohol (c) ethanol (d) isopropyl alcohol

Short answer

Benzyl alcohol will not give the iodoform test.

Step-by-step solution

Understand the Iodoform Test

The iodoform test is a chemical reaction used to determine the presence of a methyl ketone group (CH3CO-) or a secondary alcohol with a methyl group in the alpha position. Compounds that contain either of these structures will undergo halogenation and form iodoform (CHI3), which is a yellow precipitate.

Analyze Ethanal (Option a)

Ethanal (CH3CHO) is an aldehyde where the carbonyl group is attached to a methyl group, fitting the methyl ketone structure. Therefore, ethanal will give a positive iodoform test.

Analyze Benzyl Alcohol (Option b)

Benzyl alcohol has the structure C6H5CH2OH, which is a primary alcohol. It lacks a methyl group next to the alcohol group, so it does not have the structure required for the iodoform test. This means benzyl alcohol will not give a positive iodoform test.

Analyze Ethanol (Option c)

Ethanol (CH3CH2OH) is a primary alcohol. However, during oxidation, it forms acetaldehyde (CH3CHO), which has a methyl group adjacent to a carbonyl group. Thus, ethanol will eventually give a positive iodoform test.

Analyze Isopropyl Alcohol (Option d)

Isopropyl alcohol ((CH3)2CHOH) is a secondary alcohol with a methyl group adjacent to the hydroxyl group. It meets the requirement for the iodoform test to produce a yellow precipitate. Thus, it will give a positive iodoform test.

Determine the Compound Not Giving the Iodoform Test

All options except benzyl alcohol (which is a primary alcohol without a methyl group in the alpha position) can give the iodoform test. Therefore, benzyl alcohol is the compound that will not give a positive iodoform test.

Key concepts

Methyl Ketone

Methyl ketones are an essential group of organic compounds that include a specific structure of interest, namely the

• methyl group (CH₃)
• and a carbonyl group (C=O).

This arrangement can also be represented as

• CH₃CO-.

The presence of a methyl group directly attached to a carbonyl group is critical because it influences certain chemical reactions, namely the iodoform test.

The iodoform test indicates the presence of a methyl ketone by the formation of a characteristic yellow precipitate of iodoform (CHI₃). For example, ethanal (an aldehyde) can be considered as a functional methyl ketone because it has an alpha-methyl group capable of reacting in this test.

In general chemistry, these compounds are known not only for their reactivity but also for being common intermediates or products in chemical syntheses. Paying attention to their unique structural features is essential for identifying them in various reactions including the iodoform test.

Secondary Alcohol

Secondary alcohols are identified by the presence of

• two carbon groups (alkyl groups) attached to the carbon bearing the hydroxyl group (OH).

A prominent characteristic of certain secondary alcohols is their ability to undergo the iodoform test.

This test relies on the presence of a methyl group on the carbon adjacent to the hydroxyl group. Isopropyl alcohol, for example, has the necessary structure. It forms a yellow precipitate when it reacts, confirming its identity as a secondary alcohol capable of undergoing this reaction.

Not all secondary alcohols will give a positive iodoform test. Only those with a methyl group in the alpha position qualify. Knowing the structure of secondary alcohols and their ability to react is crucial. Such structural knowledge provides a deeper understanding of potential reactions and properties.

Chemical Reaction

Chemical reactions involve the transformation of substances through breaking and forming chemical bonds. The iodoform test is a specific type of chemical reaction that serves as a confirmation test for certain groups, such as methyl ketones and eligible secondary alcohols.

This test uses iodine in an alkaline environment to induce a reaction that forms iodoform (CHI₃), clearly marked by its yellow precipitate. The reaction mechanism is detailed:

• The halogenation first involves the substitution of hydrogen atoms adjacent to the carbonyl group with iodine.
• Further reaction with an alkali leads to the cleavage of the carbon-carbon bond, forming iodoform and other byproducts.

Understanding the iodoform test as a chemical reaction provides insight into why specific groups react. This knowledge helps in predicting the outcome of tests and reactions, valuable for laboratory analyses and educational pursuits. By grasping this mechanism, students can better appreciate the interplay of molecular structures and reaction conditions.