Question

A compound gives a yellow ppt. on warming with the aqueous solution of \(\mathrm{NaOH}\). Its vapour density is 29 . The compound is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Short answer

The compound is (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\).

Step-by-step solution

Understanding Precipitate Formation

A yellow precipitate formed with NaOH is typically indicative of the presence of the iodoform test, which gives a characteristic yellow precipitate. This test is positive for compounds like methyl ketones (e.g., acetone).

Analyze the Options for Compounds

Review the provided compounds: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\), (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), (c) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}\), (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\). From these, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) has a methyl ketone group and would give a positive iodoform test.

Calculate Molar Mass and Compare to Vapor Density

Calculate the molar mass of the compound \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\): it is \((3\times 12) + (6\times 1) + (1\times 16) = 58\). The vapor density is half the molar mass, hence \(\frac{58}{2} = 29\), which matches the given vapor density in the problem.

Conclusion Based on Evidence

Given that \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) both matches the vapor density calculation and will yield a yellow precipitate with NaOH, it fits the conditions of the problem.

Key concepts

Organic Chemistry Reactions

Organic chemistry reactions are vital processes that involve the transformation of organic molecules. Understanding these reactions allows chemists to predict the behavior and properties of different compounds. One common reaction in organic chemistry is the Iodoform Test. This test is a classic example that helps in identifying certain types of organic compounds. When a compound like a methyl ketone reacts with iodine in the presence of a base such as sodium hydroxide (NaOH), it forms a yellow precipitate known as iodoform. This yellow precipitate is a tell-tale sign of the presence of a methyl ketone in the compound.

Many organic reactions follow specific patterns, like addition, substitution, elimination, or rearrangement. Each of these types has its mechanisms that govern the nature and outcome of the reaction. For example, the Iodoform Test involves a substitution reaction where a hydrogen atom in the methyl group is substituted by iodine atoms, facilitating the formation of iodoform.

In summary, organic chemistry reactions are central to identifying, creating, and understanding organic compounds through various established methods and tests.

Methyl Ketones

Methyl ketones are a specific group of organic compounds characterized by the presence of a carbonyl group (C=O) adjacent to a methyl group (-CH₃). They are part of the larger family of ketones.

Methyl ketones often serve as critical indicators in various chemical tests due to their unique reactivity, the most notable being the Iodoform Test. This test is positive for compounds having at least one methyl group attached to the carbonyl carbon, resulting in the formation of a yellow iodoform precipitate. Acetone, with the formula CH₃COCH₃, is a classic example of a methyl ketone.

The importance of identifying methyl ketones lies in their ubiquity in biological systems and use in industrial processes. They are key intermediates in synthesizing numerous pharmaceuticals and are prevalent in metabolism processes. Understanding their reactivity and properties allows chemists to manipulate them for desired reactions efficiently.

Vapour Density

Vapour density is a concept that helps chemists relate the mass of a gaseous compound to its molar mass. It provides insight into the molecular characteristics of the compound under study.

Vapour density is defined as half the molar mass of a given compound. Mathematically, it is represented as \( \text{Vapour Density} = \frac{\text{Molar Mass}}{2} \). This relationship is incredibly useful when analyzing unknown substances. For example, if a compound has a vapour density of 29, its molar mass is calculated as \( 29 \times 2 = 58 \).

This concept enables chemists to verify the identity of compounds through indirect methods, by comparing calculated vapour densities with observed values in experiments. It is a vital tool in both academia and industry for the study and identification of gaseous substances.