Problem 89 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Compound ' \(\mathrm{X}\), containing chlorine on treatment with strong ammonia gives a solid ' \(Y\) ' which is free from chlorine. 'Y' analysed as \(\mathrm{C}=49.31 \%, \mathrm{H}=\) \(9.59 \%\) and \(\mathrm{N}=19.18 \%\) and reacts with \(\mathrm{Br}_{2}\) and caustic soda to give a basic compound ' \(\mathrm{Z}^{\prime} .{ }^{'} \mathrm{Z}\) ' reacts with \(\mathrm{HNO}_{2}\) to give ethanol. Suggest structures for ' \(\mathrm{X}\) ', ' \(\mathrm{Y}^{\prime}\) and \(^{6} \mathrm{Z}^{\prime}\).

Step-by-Step Solution

Answer

'Y' is propylamine, 'Z' is ethylamine, and 'X' is propionyl chloride.

Step 1: Analyze the empirical formula of compound 'Y'

The percentages by mass of the elements in compound 'Y' are given as: \( C = 49.31\% \), \( H = 9.59\% \), \( N = 19.18\% \). Calculate the empirical formula by converting these percentages into moles per 100 g of the compound. The remaining percentage must be oxygen, since the sum should be 100%.Calculate the number of moles:- For Carbon (C): \[ \frac{49.31 \, \text{g}}{12.01 \, \text{g/mol}} \approx 4.105 \text{ moles} \]- For Hydrogen (H): \[ \frac{9.59 \, \text{g}}{1.008 \, \text{g/mol}} \approx 9.518 \text{ moles} \]- For Nitrogen (N): \[ \frac{19.18 \, \text{g}}{14.01 \, \text{g/mol}} \approx 1.370 \text{ moles} \] - Calculate remaining Oxygen (O) percentage: \( 100\% - (49.31\% + 9.59\% + 19.18\%) = 21.92\% \).- For Oxygen (O): \[ \frac{21.92 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.370 \text{ moles} \]The simplest whole number ratio that fits all elements is found to give the empirical formula, which matches C : H : N : O : 3 : 9 : 1 : 1. This empirical formula is \( C_3H_9NO \), possibly corresponding to an amine.

Step 2: Identify the structure of 'Y' as an amine

Given that compound 'Y' reacts with \( \mathrm{Br_2} \) and caustic soda to release a basic compound 'Z', indicates that 'Y' is likely a primary amine. Considering the empirical formula \( C_3H_9NO \) and properties given, determine the actual identity.Since 'Y' contains only one nitrogen, and reacts further as a primary amine, suggest that 'Y' could be propylamine (\( CH_3CH_2NH_2 \)) with the required functional groups aligned.

Step 3: Propose the structure for compound 'Z'

The fact that 'Z' reacts with \( \mathrm{HNO_2} \) to produce ethanol suggests that 'Z' must be an ethylamine or an amine group directly related to ethyl. Br2/NaOH treatment is known as Hofmann bromamide reaction converting amides into amines.Therefore, think that 'Z' from 'Y', which is a propyl derivative, would be ethylamine (\( C_2H_5NH_2 \)) or any derivative that allows for reduction of carbon, hence suggesting ethylamine as more consistent with the given reactions.

Step 4: Suggest the structure of compound 'X'

Compound 'X' initially contains chlorine and after treatment yields 'Y', which is free from chlorine. Since 'Y' is propylamine, 'X' could be a derivative with chlorine replaced, such as propionyl chloride (\( CH_3CH_2COCl \)), which can react to provide the amine (propylamine) upon substitution or elimination of chlorine.This implies that initial treatment with ammonia replaces the chlorine in compound 'X', finalizing as 'Y' (propylamine).

Key Concepts

Empirical Formula

To find an empirical formula, you first convert the percentage composition of each element in a compound into moles. For the exercise mentioned, elements of compound 'Y' are given in percentages for carbon, hydrogen, nitrogen, and the rest is assumed to be oxygen. To determine the moles:

• Divide the mass percentage of each element by its atomic mass.
• For example, calculate carbon using its mass percentage (49.31%) divided by carbon's atomic mass (12.01 g/mol).
• Do the same for hydrogen and nitrogen, using their respective masses, 9.59% and 19.18%.

Subtract the sum of these percentages from 100% to find oxygen's percentage and convert it to moles. Once all values are obtained, find the simplest ratio to get the empirical formula, here 'C_3H_9NO'. This formula suggests it may be an amine due to the composition.

Chemical Reactions

Chemical reactions often indicate the nature of compounds and their transformations. In the exercise: - Compound 'Y' reacts with bromine and caustic soda. - This behavior is typical for amines, which indicates a primary amine nature. Primary amine properties include forming salts with acids and amides with acid chlorides. Reactions can help deduce structures or formulas, such as amines releasing strong bases when reacted with halogens and alkalies. These specifics of reactions guide us to understand that compound 'Y' is transforming into another compound 'Z.'

Structural Analysis

Understanding molecular structures is crucial in organic chemistry. Structural analysis involves the interpretation of molecule arrangement and components. In the given task, 'Y' was identified as having the empirical formula 'C_3H_9NO' and behaves like an amine in reactions. Through analysis: - Structural formulas show how atoms are bonded. - Primary amines typically have an NH2 group attached to an alkyl chain. In our case, since 'Y' is likely propylamine, its structure is 'CH_3CH_2NH_2'. Recognizing functional groups like NH2 helps deduce likely reactions and transformations.

Primary Amine

Primary amines are a category of functional groups in organic compounds containing an amino group (-NH2) attached to an alkyl group. Learn about these properties: - They undergo condensation reactions and can form amides. - Reacts with nitrous acid, turning amines into alcohols. In the problem, 'Y' interacting with bromine and caustic soda to form 'Z,' indicates 'Y' is a primary amine and transforms under specific reaction conditions. As deduced, propylamine is a primary amine, confirming observed analytical outcomes.

Hofmann Bromamide Reaction

The Hofmann bromamide reaction is a method to transform primary amides into primary amines. This reaction involves: - Bromine and strong bases like sodium hydroxide. - Results in the loss of the carbonyl group from amides, extending carbon chains. In the context of the exercise, understanding this reaction is crucial. The treatment of 'Y' aligns with this reaction, suggesting 'Z' results from an amide being converted to an amine, supporting the transformation to an ethylamine structure. This knowledge emphasizes reaction mechanisms in organic chemistry, aiding compound identification.