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Organic Chemistry: Analysis and Applications

Welcome to this comprehensive course on organic chemistry fundamentals that are frequently tested in university‑level quizzes. The material below expands on each quiz question, turning the multiple‑choice format into a structured learning experience. By the end of the lesson you will be able to:

• Identify members of a homologous series using molecular weight data.
• Calculate the degree of unsaturation (also called the index of hydrogen deficiency) for compounds containing carbon, hydrogen, oxygen, chlorine, and nitrogen.
• Perform combustion analysis to determine empirical and molecular formulas.
• Predict the number of possible substitution products for alkanes undergoing free‑radical halogenation.
• Apply gas‑law concepts to combustion reactions in closed vessels.

1. Homologous Series and Molecular Weight Calculations

Consider a mixture X that contains two consecutive members of a homologous series of hydrocarbons. The total molecular weight of the mixture is 252 g mol⁻¹ and the heavier member has exactly twice the molecular weight of the lighter one. To find the formula of the lighter hydrocarbon, follow these steps:

• Let the molecular weight of the lighter member be M. Then the heavier member weighs 2M.
• The sum of the two molecular weights is M + 2M = 3M = 252 g mol⁻¹, giving M = 84 g mol⁻¹.
• For a hydrocarbon with the general formula CₙH₂ₙ (an alkene) the molecular weight is 14n (12 n for carbon + 2 n for hydrogen). Setting 14n = 84 yields n = 6, which corresponds to C₃H₆ when the series is considered as CₙH₂ₙ with n = 3 (the “n” in the series is the number of carbon atoms, not the factor 2).

The correct answer is C₃H₆. This example illustrates how a simple algebraic relationship can pinpoint the exact member of a homologous series.

2. Degree of Unsaturation (DU) – Counting π Bonds and Rings

The degree of unsaturation (also called the double‑bond equivalent, DBE) tells us how many rings or π bonds a molecule possesses. The formula for a compound containing C, H, N, O, and halogens (X) is:

DU = (2C + 2 + N – H – X) / 2

Oxygen does not affect the calculation because it contributes two valence electrons, just like a carbon‑oxygen double bond.

Applying the formula to C₅H₉O₂Cl:

• C = 5, H = 9, N = 0, X (Cl) = 1.
• DU = (2·5 + 2 + 0 – 9 – 1) / 2 = (10 + 2 – 10) / 2 = 1.

Thus the molecule has exactly one element of unsaturation – either a double bond or a ring.

3. Combustion Analysis – Determining Empirical Formulas

Combustion analysis is a classic technique for finding the empirical formula of an unknown organic compound. The process involves burning a known amount of the sample, capturing the resulting CO₂ and H₂O, and measuring their masses.

**Example 1:** 0.20 mol of an unknown hydrocarbon X yields 20 g of CaCO₃ after reacting the combustion gases with Ca(OH)₂, and a second precipitation step gives an additional 10 g of CaCO₃.

• Each mole of CaCO₃ corresponds to one mole of CO₂.
• Total CO₂ produced = (20 g + 10 g) / 100 g mol⁻¹ = 0.30 mol.
• Since 0.20 mol of X produced 0.30 mol of CO₂, the ratio CO₂/X = 1.5, indicating that each molecule of X contains 1.5 carbon atoms – a non‑integer that points to a simple molecule with a single carbon atom per molecule of CO₂. The only hydrocarbon that fits is CH₄ (methane), which yields exactly one CO₂ per molecule.

**Example 2:** 0.12 mol of an open‑chain organic compound X consumes 10.08 L of O₂ at STP and produces 70.92 g of CaCO₃.

• At STP, 1 L of gas = 0.0446 mol, so 10.08 L O₂ = 0.450 mol O₂.
• 70.92 g CaCO₃ corresponds to 0.71 mol CaCO₃, i.e., 0.71 mol CO₂.
• From the stoichiometry of complete combustion, each mole of X yields one mole of CO₂, so the empirical formula must contain three carbon atoms (0.71 mol CO₂ / 0.12 mol X ≈ 5.9 ≈ 6, adjusted for the presence of nitrogen and oxygen). Detailed balancing leads to the molecular formula C₃H₇O₂N.

4. Closed‑Vessel Combustion and Pressure Changes

When a combustible ester A (general formula CₙH₂ₙO₂) is burned in a sealed container that initially contains twice the stoichiometric amount of O₂, the final pressure after the reaction and cooling back to the original temperature can be used to deduce n.

The balanced combustion equation is:

CₙH₂ₙO₂ + (3n/2) O₂ → n CO₂ + n H₂O

Because the container is closed, the total number of moles before and after the reaction determines the pressure change. If the final pressure is 0.95 atm (a 5 % drop), the number of gas moles must have decreased by 5 %. Solving the mole balance gives n = 3, so the ester formula is C₃H₆O₂.

5. Nitrogen‑Containing Compounds – Amines

When an amine X containing a single nitrogen atom is combusted, the products are CO₂, H₂O, and N₂. By measuring the masses of CO₂ and H₂O and the volume of N₂ at STP, the elemental composition can be derived.

• From the given data: 10.56 g CO₂ → 0.240 mol C, 7.56 g H₂O → 0.420 mol H, and 1.344 L N₂ → 0.060 mol N₂ → 0.120 mol N atoms.
• These ratios correspond to a C:H:N proportion of 2:5:1, giving the molecular formula C₂H₅NH₂ (ethylamine).

6. Halogenated Compounds – Determining Molecular Formulas

For a compound X containing carbon, hydrogen, and chlorine, the products of combustion with AgNO₃ (formation of AgCl) provide the chlorine content, while CO₂ and H₂O give the carbon and hydrogen amounts.

• From the data: 0.22 g CO₂ → 0.005 mol C, 0.09 g H₂O → 0.005 mol H, and 1.435 g AgCl → 0.010 mol Cl.
• The molar mass of X is calculated from the sum of the elemental masses, yielding approximately 84 g mol⁻¹. The relative molecular mass compared to hydrogen (42.5) confirms the formula CH₂Cl₂ (dichloromethane).

7. Free‑Radical Halogenation – Predicting Substitution Products

When an alkane undergoes radical chlorination under light, each distinct set of equivalent hydrogen atoms can give a different monochlorinated product. For the formula C₆H₁₄, the structural isomer that yields exactly two distinct monochloro products is 2,3‑dimethylbutane. The molecule has two nonequivalent carbon positions (C‑2 and C‑3) that each bear a unique set of hydrogens, leading to two possible substitution patterns.

8. Summary of Key Concepts

• Homologous series: Members differ by a CH₂ unit; algebraic relations can pinpoint specific members.
• Degree of unsaturation: DU = (2C + 2 + N – H – X)/2; each unit corresponds to a double bond or a ring.
• Combustion analysis: Convert masses of CO₂ and H₂O to moles of C and H; use O₂ consumption to verify the formula.
• Closed‑system gas calculations: Apply the ideal‑gas law and mole balances to relate pressure changes to molecular composition.
• Amines: One nitrogen atom appears as N₂ after combustion; the C:H ratio is derived from CO₂ and H₂O.
• Halogenated organics: AgCl precipitation quantifies chlorine; combine with carbon and hydrogen data for the molecular formula.
• Radical halogenation: Count nonequivalent hydrogen sets to predict the number of monochloro products.

By mastering these analytical techniques, you will be equipped to solve a wide range of organic‑chemistry problems that appear in exams, laboratory reports, and real‑world applications such as environmental analysis and pharmaceutical synthesis.