Alkanes — practice questions

(a) First, balance the carbons, then hydrogens, then oxygens. $C_4H_{10}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)$ Balance C: $C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O$ Balance H: $C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$ Balance O: There are (4 × 2) + 5 = 13 oxygen atoms on the right. So we need 13/2 $O_2$. $C_4H_{10}(g) + 6.5O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)$ To use whole numbers: $2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)$

(b) According to Avogadro's law, the ratio of volumes of gases is equal to the ratio of their moles in the balanced equation. From the equation: 1 mole of $C_4H_{10}$ produces 4 moles of $CO_2$. Therefore, the volume ratio is $C_4H_{10} : CO_2 = 1 : 4$. Volume of $CO_2$ = 4 × Volume of $C_4H_{10}$ Volume of $CO_2$ = 4 × 120 cm³ = 480 cm³. (Note: At RTP, 1 mole of any gas occupies 24 dm³ or 24000 cm³, but we can use volume ratios directly as both substances are gases).

Initiation: The Br-Br bond is broken by UV light to form two bromine radicals. $Br_2 \xrightarrow{UV} 2Br\cdot$

Propagation: Step 1: A bromine radical abstracts a hydrogen atom from a propane molecule to form an alkyl radical and hydrogen bromide. To form 1-bromopropane, the hydrogen must be removed from a terminal carbon. $Br\cdot + CH_3CH_2CH_3 \rightarrow \cdot CH_2CH_2CH_3 + HBr$ Step 2: The propyl radical reacts with a bromine molecule to form the product and a new bromine radical, which continues the chain. $\cdot CH_2CH_2CH_3 + Br_2 \rightarrow CH_3CH_2CH_2Br + Br\cdot$

Termination: Any two radicals can combine. For example, two propyl radicals could combine. $\cdot CH_2CH_2CH_3 + \cdot CH_2CH_2CH_3 \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3$ (hexane)