Question

Two organic compounds \((\mathrm{X})\) and \((\mathrm{Y})\) both containing only \(\mathrm{C}\) and \(\mathrm{H}\) yields on analysis, the same percentage composition by mass \(\mathrm{C}=92.3 \%\) and \(\mathrm{H}=7.7 \% .(\mathrm{X})\) decolourizes bromine water and \((\mathrm{Y})\) does not. Identify (X) and (Y). (a) \(\mathrm{X}=\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{Y}=\mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(\mathrm{X}=\mathrm{C}_{2}^{2} \mathrm{H}_{4}, \mathrm{Y}=\mathrm{C}_{2} \mathrm{H}_{6}^{\circ}\) (c) \(\mathrm{X}=\mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{Y}=\mathrm{C}_{2} \mathrm{H}_{2}\) (d) \(\mathrm{X}=\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{Y}=\mathrm{C}_{6} \mathrm{H}_{6}\)

Short answer

(X) is \(\mathrm{C}_2\mathrm{H}_2\) and (Y) is \(\mathrm{C}_6\mathrm{H}_6\).

Step-by-step solution

Calculate Empirical Formula

First, determine the empirical formula from the given mass percentages. For each compound, the percentage composition by mass is \(C = 92.3\%\) and \(H = 7.7\%\). These can be treated as 92.3 grams of Carbon and 7.7 grams of Hydrogen per 100 grams of the compound.Calculate the moles of each element:\[\text{Moles of C} = \frac{92.3}{12} \approx 7.69 \\text{Moles of H} = \frac{7.7}{1} = 7.7\]The simplest whole number ratio is 1:1 for C:H, which suggests the empirical formula of \(CH\).

Determine Molecular Formula

Both compounds \((X)\) and \((Y)\) have the same percentage composition, leading to an empirical formula of \(CH\). However, they may have different molecular formulas based on the characteristic reactions or characteristics provided.Consider the possible molecular formulas with the empirical formula \(CH\), such as \(C_2H_2\) or \(C_6H_6\), which are consistent with \(CH\).

Analyze Bromine Decolourization Test

Compound \((X)\) decolourizes bromine water, indicating the presence of a carbon-carbon double or triple bond (a site of unsaturation).Compound \((Y)\) does not decolourize bromine water, suggesting it contains no double or triple bonds, indicating saturation.

Match with Given Choices

From the choices provided, only option (d) where \( \mathrm{X} = \mathrm{C}_2 \mathrm{H}_2 \) and \( \mathrm{Y} = \mathrm{C}_6 \mathrm{H}_6 \), fits the analysis:- \(\mathrm{C}_2\mathrm{H}_2\) (acetylene) decolourizes bromine water due to its triple bond.- \(\mathrm{C}_6\mathrm{H}_6\) (benzene) does not decolourize bromine water because its bonds are delocalized and stable.

Key concepts

Empirical Formula

The empirical formula is the simplest way to express a compound using the smallest whole number ratio of atoms. When you have the percentage composition of a compound, you can use it to deduce its empirical formula. Consider a compound with carbon and hydrogen percentages. By treating these percentages as grams in a 100-gram sample, you can find the moles of each element:

• For carbon (C): Moles of C = \(\frac{92.3}{12}\)
• For hydrogen (H): Moles of H = \(\frac{7.7}{1}\)

From these values, the simplest ratio of C:H is found to be 1:1, suggesting that the empirical formula for this composition is \(CH\). This formula represents how the elements combine at their most basic level, but it doesn't reveal the real number of atoms in a molecule, which the molecular formula does.

Molecular Formula

While the empirical formula gives you the simplest atom ratio, the molecular formula tells you the exact number of each type of atom in a molecule. The molecular formula can be a multiple of the empirical formula. For example, if the empirical formula is \(CH\), the molecular formula could be \(C_2H_2\), \(C_3H_3\), or any other multiple that satisfies both the molecular weight and the characteristics of the compound. It depends heavily on additional information like molecular weight and chemical reactivity to decipher which formula matches the real-world molecule. In our context, both X and Y originate from the empirical formula \(CH\), and thus must be analyzed further to determine their true molecular nature.

Bromine Water Decolourization

Bromine water is a useful test for identifying unsaturation in organic compounds. If a compound decolourizes bromine water, it's highly indicative of an unsaturated bonding structure, either a double or triple bond. This reaction occurs because the bromine molecules add across the double or triple bonds, resulting in the disappearance of bromine's characteristic brown color. For instance, compound X decolourizes bromine water, showing it has sites of unsaturation, like a double or triple bond. This helps identify its molecular structure, such as \(C_2H_2\) which contains a triple bond, making it react with bromine water.

Saturation and Unsaturation in Hydrocarbons

Saturation in hydrocarbons refers to the presence of only single bonds between carbon atoms. Such compounds won't react with bromine water because there are no double or triple bonds for bromine to react with. Examples include alkanes like ethane \(C_2H_6\). In contrast, unsaturated hydrocarbons such as alkenes and alkynes contain double and triple bonds, respectively, allowing them to react with bromine water. This provides a straightforward way to distinguish between the two types of hydrocarbons. In the exercise, Y does not decolourize bromine water, indicating saturation—no reactable double or triple bonds. This detailed distinction is crucial for understanding compound identification through reactions.