Question

Which of the following will give yellow precipitate with \(\mathrm{I}_{2} / \mathrm{NaOH} ?\) 1\. \(\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3} \quad 2 . \mathrm{CH}_{3} \mathrm{COOCOCH}_{3}\) 3\. \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (a) 1,3 (b) 3,4 (c) 1,2 (d) 1,4

Short answer

(d) 1,4

Step-by-step solution

Understand the Reaction

The formation of yellow precipitate in the presence of \(\mathrm{I}_{2}/\mathrm{NaOH}\) is indicative of the iodoform reaction. This reaction occurs with certain types of organic molecules, specifically those containing the \(-\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}\) group or the corresponding ketone form, \(-\mathrm{COCH}_3\).

Analyze Each Compound

We will inspect each option to determine if it contains the conditions for the iodoform reaction.1. \(\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) is a ketone with a \(\text{-COCH}_3\) group, so it will give a yellow precipitate.2. \(\mathrm{CH}_{3} \mathrm{COOCOCH}_{3}\) does not have a \(\text{-COCH}_3\) or a \(-\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}\) group, hence it does not produce the precipitate.3. \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) is an amide and lacks the necessary group.4. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) contains the \(-\mathrm{CH}(\mathrm{OH})\mathrm{CH}_{3}\) group, so it will give a yellow precipitate.

Evaluate the Options

Look at the options provided and find which ones have one or more compounds that fit the conditions for the iodoform reaction. From our analysis, only compounds 1 and 4 give yellow precipitate, consistent with option (d) 1,4.

Key concepts

Iodoform Reaction

The iodoform reaction is a fascinating and specific test used in organic chemistry. It's a chemical test that can help us identify certain types of compounds, thanks to its unique yellow precipitate. This reaction involves iodine (\(\mathrm{I}_2\)) and a base such as sodium hydroxide (\(\mathrm{NaOH}\)). The presence of either a methyl ketone group (\(\text{-COCH}_3\)) or an alcohol that can be oxidized to a methyl ketone (\(-\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3\)) triggers this reaction.

• The test is sensitive to compounds like acetone, where the group configuration allows iodine to substitute the hydrogens adjacent to the carbonyl group, ultimately forming iodoform.
• This iodoform is then seen as a distinctive yellow precipitate, confirming the presence of the group.

A classic use of this test is to determine if a molecule possesses such a group, as seen in the exercise provided. This allows chemists to quickly deduce structural components of unknown compounds.

Carbonyl Compounds

Carbonyl compounds are a vast and integral class of compounds in organic chemistry. They are characterized by a carbon atom double-bonded to an oxygen atom (\(\text{C=O}\)). This is known as the carbonyl functional group. These types of compounds include:

• Aldehydes, where the carbonyl group is bonded at the end of a carbon chain.
• Ketones, where the carbonyl group is bonded within the carbon chain.
• Carboxylic acids, esters, and amides, where the carbonyl is part of more complex functional groups.

The polarity of the carbonyl group makes these compounds reactive and susceptible to nucleophilic attacks. In the context of the iodoform reaction, methyl ketones are particularly important, as the reaction relies on transforming the carbonyl compound using iodine into iodoform, a process facilitated by the reactivity of the carbonyl group.

Organic Functional Groups

Understanding organic functional groups is crucial for mastering organic chemistry. Each functional group imparts distinctive chemical properties and reactivity patterns to the molecules that contain them.

• The methyl ketone group (\(\text{-COCH}_3\)) and alcohol group (\(-\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3\)) involved in the iodoform reaction are examples of how functional groups can dictate reactions.
• Functional groups like amides and esters have different reactivities and are affected by the presence of specific atoms or structure.
• Knowing the typical reactions that these groups undergo, such as oxidation of alcohols to ketones in the case of the iodoform reaction, helps in predicting the outcome of chemical tests such as the one demonstrated in the original exercise.

By understanding these groups, one can predict the presence of specific reactive sites within a molecule, aiding in the identification and transformation of organic compounds.