Problem 118 - A Complete Resource Book in Chemistry for JEE Main

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) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the context

The compound in question forms a yellow precipitate when warmed with an aqueous solution of NaOH. This points towards a compound that reacts with NaOH to possibly form iodoform (CHI3), which is yellow.

Step 2: Identifying the reaction compatibility

The iodoform test is typically positive for compounds with methyl ketones (RCOCH3) or secondary alcohols with the structure CH3CH(OH)R. Among the options, (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) and (c) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}\) fit these criteria.

Step 3: Calculate the molecular weight

Calculating the molecular weight of each option to compare with the vapor density value. Vapor density = molecular weight / 2. Thus, molecular weight should equal 29 * 2 = 58. - (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\): 58 g/mol- (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\): 58 g/mol- (c) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}\): 60 g/mol- (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\): 60 g/mol.

Step 4: Finalize based on molecular weight and reaction

\(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (option b) has the correct molecular weight (58 g/mol) to match the vapor density and can react to give a yellow precipitate (iodoform) when warmed with \(\mathrm{NaOH}\). Thus, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) is the compound.

Key Concepts

Vapour Density

Vapour density is an important concept in chemistry that helps us understand the properties of a gaseous compound. It is defined as the mass of a certain volume of a gas compared to the mass of the same volume of hydrogen under the same conditions of temperature and pressure. This comparison to hydrogen is useful because hydrogen is the lightest gas and provides a baseline for measurements. The formula for calculating vapour density is given by:\[ \text{Vapour Density} = \frac{\text{Molecular Weight}}{2} \]This formula indicates that vapour density is essentially half the molecular weight of the gas. Therefore, if the vapour density of a compound is known, the molecular weight can be easily calculated by simply multiplying the vapour density by 2. In the context of our exercise, the compound's vapour density is 29. By using the formula, we calculate that the molecular weight should be \( 29 \times 2 = 58 \) g/mol. This calculation helps in identifying the correct compound among several choices.

Methyl Ketones

Methyl ketones are a specific type of organic compound characterized by the presence of a carbonyl group (C=O) bonded to two hydrocarbon groups, one of which is a methyl group (CH3). They are represented as RCOCH3, where R stands for any alkyl or aryl group. Methyl ketones are significant because they undergo the iodoform test, a qualitative analysis used to detect methyl ketones or alcohols with the specific structure CH3CH(OH)R. When a methyl ketone is treated with a solution of sodium hydroxide (\( \text{NaOH} \) ) and iodine (\( \text{I}_2 \) ), it results in the formation of a yellow precipitate of iodoform (\( \text{CHI}_3 \) ), indicating a positive test result. In the given exercise, \( \text{CH}_3\text{COCH}_3 \) is a methyl ketone, as it fits the characteristic group : \( \text{RCOCH}_3 \), making it a suitable candidate for producing a yellow precipitate in the iodoform test.

Sodium Hydroxide Reaction

The reaction of a compound with sodium hydroxide (NaOH) is an interesting and informative process, especially in the context of determining the presence of certain functional groups in organic chemistry. Solutions of NaOH can act as both a nucleophile and a strong base; thus, it can participate in various organic reactions. In the case of the iodoform test, NaOH is involved in facilitating the nucleophilic attack on the carbonyl carbon of a methyl ketone. This initiates the conversion of the ketone into an enolate intermediate, which is further halogenated by iodine to eventually produce iodoform (CHI3), a yellow precipitate. This sequence is specific for methyl ketones due to the arrangement of \( \text{-COCH}_3 \), a methyl group adjacent to a carbonyl group. Besides methyl ketones, secondary alcohols that can be oxidized to methyl ketones also give a positive iodoform test when reacted with NaOH and iodine. This test provides a simple and convenient method for identifying certain organic compounds based on their reactivity with NaOH.

Molecular Weight Calculation

Calculating molecular weight is a fundamental skill in chemistry, as it helps to determine the molecular formula of a compound and infer its chemical properties. The molecular weight is simply the sum of the atomic masses of all the atoms present in a molecule. It is expressed in grams per mole (g/mol). In exercises like ours, knowing the molecular weight is critical for identifying the correct compound from a list. The calculation using vapour density involves understanding that the vapour density (VD) of a compound is equal to half of its molecular weight (MW). Therefore, if we have a known vapour density, the molecular weight can be calculated using the relationship:\[ \text{MW} = 2 \times \text{VD} \]For the case in the exercise, a vapour density of 29 leads to a calculated molecular weight of 58 g/mol.This value ensures that we are looking at the right characteristic along with other tests, such as the iodoform test, to pinpoint a compound such as \( \text{CH}_3\text{COCH}_3 \).Understanding molecular weight assists in predicting how a compound will behave in chemical reactions, especially those involving stoichiometric calculations and synthesis.