Question

On warming with \(\mathrm{I}_{2}\) and aqueous \(\mathrm{NaOH}\), iodoform and sodium succinate are formed. The formula of the compound should be (a) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{COC}_{6} \mathrm{H}_{5}\) (c) \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{COCH}_{3}\)

Short answer

The correct answer is (c) \( \mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH} \).

Step-by-step solution

Understand the Iodoform Reaction

The iodoform reaction is a chemical test for the presence of a methyl ketone group \( \text{R}-\text{CO}-\text{CH}_3 \) or secondary alcohols with the formula \( \text{RCH(OH)CH}_3 \). The reaction with iodine and sodium hydroxide produces iodoform (\( \text{CHI}_3 \)) and other products.

Analyze Options for Methyl Ketone or Alcohol

Examine each compound to determine if it contains a methyl ketone group \( \text{R}-\text{CO}-\text{CH}_3 \) or a secondary alcohol with \( \text{RCH(OH)CH}_3 \) that can undergo the iodoform reaction. Only compounds with the \( \text{CH}_3-\text{COX} \) structure will form iodoform.

Evaluate Each Compound

(a) \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \): Has a methyl ketone group (\( \text{CH}_3\text{CO} \)).(b) \( \text{CH}_3\text{COC}_6\text{H}_5 \): Also has a methyl ketone group (\( \text{CH}_3\text{CO} \)).(c) \( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{COOH} \): Does not have methyl ketone.(d) \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{COCH}_3 \): Has a methyl ketone group (\( \text{CH}_3\text{CO} \)).

Consider Formation of Sodium Succinate

Sodium succinate formation suggests the original compound can be broken down into a two-carbon chain with carboxylate ends. Compound (c), \( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{COOH} \), upon breakage can form two acetate groups which can further form succinate, even though it doesn't fit the methyl ketone iodoform criteria perfectly.

Conclude the Correct Answer

Given the succinate production, compound (c) is the only one that upon oxidative cleavage will leave a four-carbon dicarboxylic acid, appearing as succinate in reactions, aligning with given conditions.

Key concepts

Methyl Ketone Group

The methyl ketone group is a specific type of functional group in organic chemistry characterized by the presence of a carbon atom double-bonded to an oxygen atom (this is known as a ketone group), with an adjacent methyl group (\( \text{CH}_3 \)). This setup is written as \(\text{R}-\text{CO}-\text{CH}_3\). The importance of methyl ketones in the iodoform reaction is that they can react with iodine and sodium hydroxide to produce iodoform (\( \text{CHI}_3 \)), a yellow solid with a distinct odor. This reaction is commonly used in organic chemistry to identify the presence of methyl ketones. Not all ketones can give a positive iodoform test. Only those containing the specific methyl group connected to the carbon adjacent to the carbonyl (C=O) group will react. This is why evaluating the structure of a compound is crucial when predicting its behavior in iodoform reactions.

Sodium Succinate Formation

Sodium succinate formation during the iodoform reaction suggests specific transformations within the compound of interest. Succinate is a 4-carbon dicarboxylic acid in its ionized form that stems from oxidative cleavage, which breaks a longer carbon chain into smaller parts.In the context of the given exercise, we need to consider the structure available in option (c), \( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{COOH} \), which upon oxidative cleavage, can lead to the formation of two acetate groups. These acetate groups can link and transform into a succinate chain. This transformation aligns with the original question condition, as sodium succinate is found alongside disintegration products of the original compound post-reaction with iodine and NaOH. It's essential to consider not only the presence of target functional groups but the potential transformation of these groups in a reaction environment.

Oxidative Cleavage Chemistry

Oxidative cleavage is a process in organic chemistry where carbon-carbon bonds are broken, leading to the formation of smaller, typically more oxidized compounds. This often involves the introduction of functional groups such as keto or carboxyl groups at the points of cleavage, fundamentally changing the structure of the original molecule.In the iodoform reaction discussed, oxidative cleavage allows the transformation of a longer carbon chain into smaller, functionally significant fragments. For example, the compound \( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{COOH} \) will, through this mechanism, break into two distinct acetate ions. These can combine and form sodium succinate, the product observed in the reaction environment.Thus, understanding oxidative cleavage is not just about potential changes in structure, but also about predicting the types of new functional groups and molecules that might emerge from complex reactions.

Identifying Functional Groups in Organic Compounds

Identifying functional groups is fundamental to understanding organic compounds. Functional groups determine the chemical reactivity and properties of molecules. Knowing how to locate and interpret these within a complex structure is a key skill for any chemist.In the context of the iodoform reaction, this means pinpointing groups like methyl ketones that react under specific conditions. When analyzing a compound, the first step is visualizing its chemical structure to see which functional groups are present.Key steps include:

• Look for carbonyl groups (\(C=O\)), especially those next to methyl groups (\(CH_3\)). These indicate potential for iodoform reaction.
• Assess other groups (like hydroxyl groups in alcohols or carboxylic acids) that impact potential reactions and transformations.
• Consider how these groups might change or interact under reaction conditions (such as in oxidative cleavage or with iodine/\(NaOH\)).

Mastering this not only aids in theoretical problem-solving but also provides essential knowledge for laboratory practice and chemical analysis.