Problem 66 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An organic compound \(X\), on analysis gives \(24.24\) per cent carbon and \(4.04\) per cent hydrogen. Further, sodium extract of \(1.0 \mathrm{~g}\) of \(X\) gives \(2.90 \mathrm{~g}\) of silver chloride with acidified silver nitrate solution. The compound \(X\) may be represented by two isomeric structures, \(Y\) and \(Z\). \(Y\) on treatment with aqueous potassium hydroxide solution gives a dihydroxy compound while \(Z\) on similar treatment gives ethanal. Find out the molecular formula of \(X\) and give the structures of \(Y\) and \(Z\).

Step-by-Step Solution

Answer

Molecular formula of \(X\) is \(\text{C}_2\text{H}_4\text{Cl}_2\); \(Y\) is 1,2-dichloroethane, \(Z\) is 1-chloro-2-chloropropane.

Step 1: Calculate Molar Mass of Chlorine

Given that 1.0 g of compound \( X \) gave 2.90 g of silver chloride (AgCl), first we calculate the molar mass of AgCl. The atomic masses are \( \text{Ag} = 107.87 \text{ g/mol} \) and \( \text{Cl} = 35.45 \text{ g/mol} \). Therefore, the molar mass of AgCl is \( 143.32 \text{ g/mol} \).

Step 2: Determine Mass Percentage of Chlorine in X

From the result, 2.90 g of AgCl is equivalent to \( \frac{2.90}{143.32}\text{ mol} \approx 0.0202 \text{ mol} \) of AgCl. Since AgCl has a \(1:1\) ratio, chlorine is also \(0.0202 \text{ mol}\). The mass of chlorine is \(0.0202 \text{ mol} \times 35.45 \text{ g/mol} \approx 0.715 g \). So, the mass of chlorine in \( X \) is \( 71.5 \text{ per cent chlorine}. \)

Step 3: Determine Empirical Formula of X

The weight percentages in \( X \) are: C: 24.24%, H: 4.04%, Cl: 71.5%. To determine the empirical formula, convert these percentages to moles per 100 g of \( X \): \( \frac{24.24}{12.01} = 2.02 \text{ mol C} \), \( \frac{4.04}{1.008} = 4.01 \text{ mol H} \), \( \frac{71.5}{35.45} = 2.02 \text{ mol Cl} \). The simplest ratio is \( C:H:Cl = 1:2:1 \), so the empirical formula is \( \text{CH}_2\text{Cl} \).

Step 4: Identify Possible Structures for Y and Z

Given that Y forms a dihydroxy compound with KOH and Z forms ethanal, the most likely molecular formula for \( X \) is \( \text{C}_2\text{H}_4\text{Cl}_2 \) as both hydration and hydrolysis reactions can align with the empirical formula \( \text{CH}_2\text{Cl} \).The dihydroxy results in 1,2-dichloroethane, which reacts with KOH to give ethylene glycol. Another structure for \( Z \) could be 1-chloro-1,2-dichloroethane, which upon reaction gives ethanal.

Key Concepts

Isomerism

Isomerism is an important concept in organic chemistry. It refers to compounds that have the same chemical formula but different arrangements of atoms in space. These different structures are called isomers. In our exercise, compound \( X \) can be represented by two isomeric structures: \( Y \) and \( Z \). This means that \( X \) has more than one way that its atoms can be arranged. Although \( Y \) and \( Z \) share the same molecular formula, they behave differently in chemical reactions.

• \( Y \) forms a dihydroxy compound when treated with potassium hydroxide.
• \( Z \) yields ethanal on treatment with the same reagent.

Understanding isomerism is crucial because it explains why different compounds can have the same formula but exhibit varied properties. These different properties make isomers useful in different applications and essential for creating various products in chemistry.

Empirical Formula

The empirical formula represents the simplest whole number ratio of elements in a compound. To find this for compound \( X \), we analyze the given percentages of carbon, hydrogen, and chlorine. From the exercise, these are found to be \( 24.24 ext{ ext{ q{}}% C, 4.04 ext{ ext{ q{}}% H},\) and \( 71.5 ext{ ext{ q{}}% Cl} \). Converting these percentages to moles allows us to establish the ratio of the atoms in \( X \):

• Carbon: \( 2.02\text{ mol} \)
• Hydrogen: \( 4.01\text{ mol} \)
• Chlorine: \( 2.02\text{ mol} \)

The simplest whole-number ratio is \( 1:2:1 \), giving an empirical formula of \( \text{CH}_2\text{Cl} \). This formula provides insight into the compound's basic composition but does not indicate how atoms are bonded. It forms the foundation for identifying the molecular formula and possible isomeric structures.

Molecular Structure

Molecular structures refer to the fixed arrangements of atoms within a molecule, specifying not just which atoms are present, but how they are bonded. In our scenario, the empirical formula \( \text{CH}_2\text{Cl} \) indicates a certain arrangement. However, two differing molecular structures are possible: \( Y \) and \( Z \).

• \( Y \) could be structured as 1,2-dichloroethane \( \text{C}_2\text{H}_4\text{Cl}_2 \), which forms ethylene glycol upon reacting with potassium hydroxide.
• \( Z \) might be structured as 1-chloro-1,2-dichloroethane, which results in ethanal after the reaction.

These structures determine how the compound interacts during chemical reactions. They highlight the significance of isomerism in yielding distinctive products, although the empirical formula remains unchanged.

Hydrolysis Reaction

A hydrolysis reaction involves the splitting of a molecule by the addition of water. This type of reaction plays a vital role in understanding the behavior of compounds \( Y \) and \( Z \) in our exercise. The reaction with potassium hydroxide (KOH) involves a mechanism where the compound is cleaved, and new products are formed.

• Compound \( Y \) undergoes hydrolysis to form a dihydroxy compound (ethylene glycol). This reaction typically progresses via a geometry that allows the water molecule to break specific bonds in \( Y \).
• For compound \( Z \), hydrolysis with KOH yields ethanal. The difference in structure makes \( Z \) susceptible to forming this specific aldehyde as a product.

Recognizing how hydrolysis reactions transform different structures helps predict the outcomes of chemical reactions and provides insight into potential applications of each isomer in chemical processes.