Question

Two organic compounds \(\mathrm{X}\) and \(\mathrm{Y}\) on analysis gave the same percentage composition, namely \(\mathrm{C}=\left(\frac{12}{13}\right) \times\) \(100 \%\) and \(\mathrm{H}=\left(\frac{1}{13}\right) \times 100 \% .\) However, compound "X" decolourises bromine water whilc compound "Y" docs not. Two compounds " \(\mathrm{X}\) " and "Y" may be respectively (1) ethylene and benzene (2) acetylene and ethylene (3) toluene and benzene (4) benzene and styrene

Short answer

Ethylene (X) and benzene (Y).

Step-by-step solution

Calculate the molecular formula based on composition

First, determine the molecular formula of the given compounds. With C as \(\frac{12}{13} \) times 100% and H as \(\frac{1}{13} \) times 100%, the empirical formula can be calculated. Since the percentage composition is given, the empirical formula is CH (one carbon atom for every hydrogen atom).

Identify experimental differences

Next, consider the behavior of compounds 'X' and 'Y' with bromine water. Compound 'X' decolourises bromine water, indicating it has a carbon-carbon double bond (C=C). Compound 'Y' does not, suggesting it does not have a double bond.

Compare possible compound pairs

Now, evaluate the given options based on their molecular structures and reactivity with bromine water. Ethylene (C2H4) decolourises bromine water due to its C=C bond, while benzene (C6H6) does not. Ethylene and acetylene (C2H2) both decolourise bromine water which disqualifies option 2. Toluene (C7H8) and benzene do not match the empirical formula CH. Benzene and styrene (C8H8) also do not both decolourise bromine water.

Conclude the correct compounds

From the remaining options, ethylene (C2H4) and benzene (C6H6) are the correct compounds matching the given conditions. Ethylene decolourises bromine water due to the presence of a double bond while benzene does not.

Key concepts

percentage composition

Percentage composition in chemistry indicates the proportion of each element in a compound relative to the total mass of the compound. For example, a compound with a percentage composition of \(\text{C} = \frac{12}{13} \times 100\text{\textpercent}\) and \(\text{H} = \frac{1}{13} \times 100\text{\textpercent}\) shows the mass percentage of carbon and hydrogen in the compound. This means that out of the total mass, \(\frac{12}{13}\) is carbon and \(\frac{1}{13}\) is hydrogen. Understanding these percentages can help in determining the empirical formula and thereby identifying the compound.

• Percentage tells how much of each element is present.
• Useful in calculating the empirical formula.

By knowing the percentage composition, we can move on to finding the empirical formula which is a key step in the identification of the compound.

empirical formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. From the previous step, we calculated the mass percentages of carbon and hydrogen. By dividing the percentages by the atomic masses of carbon (12) and hydrogen (1), we determine the mole ratio of these elements. Since carbon and hydrogen showed a \( \frac{12}{13} \) and \( \frac{1}{13} \) ratio, it simplifies to a 1:1 ratio, giving us the empirical formula CH.

• Simplest form of the molecular makeup.
• Gives a blueprint for the molecule’s composition.

Understanding the empirical formula is crucial in distinguishing between different compounds with similar compositions.

reactivity with bromine water

Bromine water is a standard test for the presence of carbon-carbon double bonds (C=C) in organic compounds. When a compound containing a C=C bond is mixed with bromine water, the bromine adds across the double bond, resulting in a colorless solution.

• Shows the presence of double bonds.
• Indicator reaction changes the color of bromine.

In this exercise, compound X decolourises bromine water, indicating it has a C=C bond. Compound Y does not affect bromine water, suggesting the absence of such a bond. This clue helps narrow down the possible identities of the compounds.

carbon-carbon double bond

A carbon-carbon double bond (C=C) is a covalent bond where two carbon atoms share two pairs of electrons. This bond can undergo addition reactions such as the decolourisation of bromine water, confirming its presence.

• Signifies reactivity and unsaturation in organic compounds.
• Displayed as two lines connecting carbon atoms in structural formulas.

In our problem, Compound X has a C=C bond, hence it reacts with bromine water and decolourises it. Identifying such bonds enables us to understand and predict the reactivity of organic molecules more effectively.
Recognizing these structural features ties back to their chemical properties and behaviors, essential in organic chemistry problem-solving.