Problem 31 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A certain hydrocarbon \(A\) was found to contain \(85.7\) percent carbon and \(14.3\) per cent hydrogen. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon \(B .1 .00 \mathrm{~g}\) of hydrocarbon \(A\) just decolourized \(38.05 \mathrm{~g}\) of a 5 per cent solution (by weight) of \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\). Compound \(A\), on oxidation with concentrated \(\mathrm{KMnO}_{4}\), gave compound \(C\) (molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) ) and acetic acid. Compound \(C\) could easily be prepared by the action of acidic aqueous mercuric sulphate on 2 - butyne. Determine the molecular formula of \(A\) and deduce the structure of \(A, B\) and \(C\).

Step-by-Step Solution

Answer

A is 2-butene; B is butane; C is 3-butanone.

Step 1: Determine Empirical Formula

First, calculate the moles of carbon and hydrogen in 100 grams of hydrocarbon A. Given that it has 85.7% carbon, the mass of carbon is 85.7 g and for hydrogen, it's 14.3 g. The moles of carbon are \(\frac{85.7}{12}=7.14\) and the moles of hydrogen are \(\frac{14.3}{1}=14.3\). The simplest ratio of moles is approximately 1:2, leading to the empirical formula \(\text{CH}_2\).

Step 2: Analyze Bromine Decoloration Reaction

1.00 g of A decolorizes 38.05 g of a 5% bromine solution in CCl4, which means the total bromine content is \(38.05 \times 0.05 = 1.9025\) g of Br2. The moles of Br2 used are \(\frac{1.9025}{159.808}=0.0119\). Given A decolorizes the equivalent in moles of Br2, it suggests one double bond is being saturated per molecule of A.

Step 3: Determine Molecular Formula

Compound A adds one mole of hydrogen indicating \(\text{C}_n\text{H}_{2n}\) hydrocarbon. As A converts into B by adding 1 H2 molecule, it supports alkene structure. Comparing experimentally moles of reacted substances, C4H8 aligns with empirical formula findings and satisfies reaction conditions.

Step 4: Structural Analysis with KMnO4 Oxidation

Oxidation of A with KMnO4 gives C4H8O and acetic acid, indicating A must have C4 alkene structure and resultants possess a ketone and a carboxylic acid group. Using oxidation cleavage logic, grouping possibilities assert terminal alkene for A: valid candidate "butene" structures like 2-butene.

Step 5: Identify Compound B from Hydrogenation

Hydrogenation saturation of A gives B, consistent with alkene's C-H bond saturation into alkane. Hence, Butane forms as the stable saturated product from 2-butene post hydrogenation.

Step 6: Confirm Compound C Formation

Compound A oxidation identifies 3-butanone as C from KMnO4 processing of terminal alkene butane-2-one, which aligns with known reactions by aqueous mercuric sulphate conversions.

Key Concepts

Empirical Formula

Understanding the empirical formula is essential in analyzing the composition of a compound. For hydrocarbon A, knowing it consists of 85.7% carbon and 14.3% hydrogen is the starting point. To find the empirical formula, we convert these percentages into grams, assuming a 100 gram sample. Thus, we have 85.7 grams of carbon and 14.3 grams of hydrogen.

Next, we calculate the moles of each element. Carbon has an atomic mass of approximately 12 g/mol, while hydrogen is about 1 g/mol. Therefore, the moles of carbon are \(\frac{85.7}{12} \approx 7.14\) and for hydrogen \(\frac{14.3}{1} = 14.3\). The simplest mole ratio here simplifies to approximately 1:2, giving us the empirical formula \(\text{CH}_2\). This ratio indicates that for every carbon atom, there are two hydrogen atoms in the compound's simplest structure.

Bromine Decoloration

The bromine decoloration test is an essential tool for identifying unsaturation in hydrocarbons. When hydrocarbon A reacts with bromine, it causes the characteristic brown bromine color to vanish, suggesting the presence of a double bond.

In our case, 1.00 g of hydrocarbon A decolorizes 38.05 g of a 5% solution of bromine, which equates to 1.9025 g of bromine. The moles of bromine engaged in this reaction are given by \(\frac{1.9025}{159.808} \approx 0.0119\) moles. This supports the hypothesis that hydrocarbon A comprises a single site of unsaturation, as an alkene, because one mole of bromine reacts with each mole of double bond present.

KMnO4 Oxidation

Oxidation reactions with \(\text{KMnO}_4\) are instrumental in hydrocarbon analysis as they provide insight into the structure by breaking the molecule at double bonds and identifying the resulting products. For compound A, oxidation yields compound C (\(\text{C}_4 ext{H}_8 ext{O}\)) and acetic acid, suggesting a specific structural arrangement in A.

This outcome implies that hydrocarbon A bears a terminal alkene structure leading to the formation of a ketone (compound C) and a carboxylic acid. Given A's transformation during the oxidation, it is reasonable to deduce that the original structure of A must include a terminal unsaturation, resembling compounds like 2-butene, which allows for such products upon oxidation mechanism.

Molecular Formula Determination

Determining the molecular formula connects empirical formula insights and reaction data to pin down the exact composition of compound A. We've learned that A transforms into B by adding a mole of hydrogen, suggesting it goes from an alkene structure to an alkane. This consistent with the formula \(\text{C}_n ext{H}_{2n}\) typical of alkenes.

Given the empirical formula is \(\text{CH}_2\), aligning with the alkene nature, compound A can extrapolate to \(\text{C}_4 ext{H}_8\). This formula supports the observed reactions: being decolorized by bromine, oxidized by \(\text{KMnO}_4\) and forming a simple alkane (butane) when hydrogenated. Consequently, understanding these reaction pathways allows for a coherent deduction of A's molecular formula and structure.