Question 3 of 5: Structures from Molecular Formulas; Combustion of a Substituted Cyclohexane

Question 3: Structures from Molecular Formulas; Combustion of a Substituted Cyclohexane (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a)(i) Amine, C₃H₉N. Propan-1-amine, CH₃CH₂CH₂NH₂ (a primary amine), satisfies the formula: three chain carbons, a terminal –NH₂, and the nine hydrogens split 7 on carbon + 2 on nitrogen. (Trimethylamine, (CH₃)₃N, a tertiary amine, is an equally valid isomeric answer.)

(a)(ii) Alkyne, C₄H₆. But-1-yne, HC≡C–CH₂–CH₃, satisfies the formula: one C≡C triple bond (accounting for two degrees of unsaturation) on an otherwise saturated four-carbon chain. (But-2-yne, CH₃–C≡C–CH₃, the internal-triple-bond isomer, is equally valid.)

(a)(iii) Ether, C₄H₁₀O. Diethyl ether, CH₃CH₂–O–CH₂CH₃, satisfies the formula: an oxygen bonded to two carbon chains and to no hydrogen of its own (the O accounts for zero degrees of unsaturation, consistent with a fully saturated C₄H₁₀O). (Methyl propyl ether, CH₃–O–CH₂CH₂CH₃, is an equally valid isomeric answer.)

(b) Combustion of 1-ethyl-3-methylcyclohexane.

Given. 1-ethyl-3-methylcyclohexane: a cyclohexane ring bearing one ethyl and one methyl substituent (positions 1 and 3). Complete combustion means reaction with O₂ to CO₂ and H₂O only.

Find. A balanced combustion equation, in whole-number (integer) coefficients.

Approach

First fix the fuel's own molecular formula (cyclohexane's C₆H₁₂ plus the two alkyl substituents), then balance C, then H, then O — clearing the resulting half-integer O₂ coefficient by doubling the whole equation.

1. Molecular formula of the fuel. Cyclohexane is C₆H₁₂. Each alkyl substituent replaces one ring hydrogen with the alkyl group, which is equivalent to adding CH₂ per carbon of substituent chain length beyond that replaced H: an ethyl substituent (–C₂H₅ in place of H) adds C₂H₄, and a methyl substituent (–CH₃ in place of H) adds CH₂. So $$\text{C}_6\text{H}_{12} + \text{C}_2\text{H}_4 + \text{CH}_2 = \boxed{\text{C}_9\text{H}_{18}},$$ consistent with the general saturated-monocyclic formula C_nH_2n (one ring = one degree of unsaturation, no other unsaturation present).
2. Balance carbon and hydrogen. One mole of C₉H₁₈ gives 9 mol CO₂ (9 C) and 9 mol H₂O (18 H → 9 H₂O): $$\text{C}_9\text{H}_{18} + x\,\text{O}_2 \rightarrow 9\,\text{CO}_2 + 9\,\text{H}_2\text{O}.$$
3. Balance oxygen. Oxygen needed on the right: $9(2) + 9(1) = 27$ O atoms, so $x = 27/2$: $$\text{C}_9\text{H}_{18} + \tfrac{27}{2}\,\text{O}_2 \rightarrow 9\,\text{CO}_2 + 9\,\text{H}_2\text{O}.$$
4. Clear the fraction. Multiplying every coefficient by 2 gives the balanced equation in whole numbers: $$\boxed{2\,\text{C}_9\text{H}_{18} + 27\,\text{O}_2 \rightarrow 18\,\text{CO}_2 + 18\,\text{H}_2\text{O}}.$$