Question

Which of the following statements regarding chemical properties of acetophenone are wrong? I. it is reduced to methylphenylcarbinol by sodium and ethanol II. it is oxidized to benzoic acid with acidified \(\mathrm{KMnO}_{4}\) III. it does not undergo electrophillic substitution like nitration at meta position. IV. it does not undergo iodoform reaction with iodine and alkali. (a) I and II (b) II and IV (c) I and III (d) III and IV

Short answer

Statement IV is incorrect, making option (b) correct.

Step-by-step solution

Analyze Statement I

To verify if acetophenone is reduced to methylphenylcarbinol by sodium and ethanol, we should recognize that sodium and ethanol typically perform a reduction process such as a Clemmensen or Wolff-Kishner reduction. This reduction can yield the corresponding alcohol, but a more common result is formation of an alkane, indicating statement I could be plausible but needs more specific reaction conditions to confirm its truth.

Analyze Statement II

To determine if acetophenone is oxidized to benzoic acid with acidified \(\mathrm{KMnO}_{4}\), we need to note that \(\mathrm{KMnO}_{4}\) is a strong oxidizing agent capable of cleaving side chains to form carboxylic acids. Acetophenone's methyl side group can be oxidized to benzoic acid, making this statement correct.

Analyze Statement III

Regarding acetophenone undergoing electrophilic substitution like nitration at the meta position, we should remember that the carbonyl group is a meta-directing deactivator in electrophilic aromatic substitution reactions. This makes statement III correct, as acetophenone typically results in meta substitution.

Analyze Statement IV

To check if acetophenone undergoes the iodoform reaction, we need to recognize that the iodoform test is positive for compounds with a methyl ketone group, like acetophenone. Therefore, acetophenone does undergo the iodoform reaction with iodine and alkali, making statement IV incorrect.

Identify Wrong Statements

From the previous analysis, statements I, II, and III are correct, whereas statement IV is incorrect. Thus, any provided options that claim statement IV is wrong are worth considering. Thus, option (b) II and IV is correct.

Key concepts

Reductive Reactions with Sodium and Ethanol

In organic chemistry, reduction processes are a way to break down molecules and form simpler compounds. Acetophenone, a common ketone, can undergo a reduction process. A typical scenario involves using sodium and ethanol to reduce acetophenone to methylphenylcarbinol. This reaction often results in the formation of an alcohol group but can sometimes directly form an alkane, depending on the conditions. Here is a quick overview of the process:

• Sodium acts as the reducing agent, providing the electrons necessary for the reaction.
• Ethanol serves as a solvent and sometimes a mild reducing agent.
• The end product could be a reduction of the carbonyl group to form an alcohol or an alkane.

This reduction is more common in specific conditions and is not always a straightforward transformation to methylphenylcarbinol. However, in many cases, sodium and ethanol provide a strong enough reducing environment to achieve the desired chemistry.

Oxidation with Potassium Permanganate

Oxidation reactions occur when a substance loses electrons, often facilitated by an oxidizing agent. Potassium permanganate (\(\mathrm{KMnO}_{4}\) ) is a potent oxidizing agent. It has the ability to convert side chains into carboxylic acids. In the case of acetophenone, it can be oxidized to benzoic acid by cleaving the methyl side chain. Here’s a breakdown of how it works:

• Potassium permanganate provides the necessary oxygen to convert the methyl group into a carboxylic group.
• The reaction typically occurs in acidic or neutral conditions to facilitate the oxidation.
• The end product, benzoic acid, is a carboxylic acid, exhibiting properties like solubility in water and an acidic character.

Remember, not every oxidation will yield benzoic acid, as the conditions must favor the cleavage of side chains.

Electrophilic Aromatic Substitution

Aromatic compounds undergo characteristic reactions due to their stable ring structure. Electrophilic aromatic substitution is one such reaction where an electrophile replaces a hydrogen atom on an aromatic ring. Acetophenone contains a carbonyl group, which influences the substitution pattern. Here’s how it affects reactions:

• The carbonyl group is an electron-withdrawing group, making the ring less reactive compared to other alkylbenzenes.
• This leads to a directing effect, favoring substitution at the meta position.
• Common substitutions include nitration, sulfonation, and halogenation.

Despite the deactivating nature, acetophenone can still participate in these reactions, with major products formed at the meta position due to the carbonyl group's influence.

Iodoform Reaction

The iodoform reaction is a notable test in organic chemistry, used for detecting methyl ketones. Acetophenone can indeed partake in this reaction due to the presence of its ketone group. Let's understand how this reaction works:

• The reaction requires iodine and an alkali, such as sodium hydroxide.
• For a successful iodoform reaction, the compound must contain a \(\text{-C(=O)CH}_3\) , which acetophenone has because of its methyl ketone structure.
• This transforms into triiodomethane (iodoform), which appears as a yellow precipitate.

If such yellow precipitate forms, it confirms that the compound has a methyl ketone group. Acetophenone typically gives a positive iodoform test, underlining its chemical properties.