Home / iGCSE Chemistry 0620 Extended – 11.2 Naming organic compounds- Exam Style Questions Paper 4
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 11.7 — Carboxylic acids
• Topic 3.1 — Formulae
• Topic 11.2 — Naming organic compounds
(a) You should draw a circle strictly around the $-C(=O)-O-$ functional group. This specifically includes 1 carbon atom double-bonded to 1 oxygen atom, and single-bonded to a 2nd oxygen atom that links to the next alkyl group.
(b) Ethyl butanoate.
Explanation: The portion originating from the alcohol is ethanol (which gives the prefix “ethyl”). The portion originating from the carboxylic acid has 4 carbons (butanoic acid), giving the suffix “butanoate”.
(c) $C_{3}H_{6}O$.
Explanation: The complete molecular formula of ethyl butanoate is $C_{6}H_{12}O_{2}$. The empirical formula is the simplest integer ratio of these elements. Dividing all subscripts by their greatest common divisor (which is 2) yields $C_{3}H_{6}O$.
(d)(i) Butanoic acid and ethanol.
(d)(ii) An acid catalyst. (Typically concentrated sulfuric acid is used to both catalyze the reaction and drive the equilibrium forward by absorbing water).
(e) 4.
Explanation: By systematically altering the length of the carbon chains on either side of the ester linkage while keeping a total of 5 carbons, the unbranched isomers are: methyl butanoate, ethyl propanoate, propyl ethanoate, and butyl methanoate.
(f)
$2C_{5}H_{10}O_{2} + 13O_{2} \rightarrow 10CO_{2} + 10H_{2}O$
Explanation: A complete combustion reaction always yields $CO_{2}$ and $H_{2}O$. Starting with 1 mole of ester: $C_{5}H_{10}O_{2} + O_{2} \rightarrow 5CO_{2} + 5H_{2}O$. Next, balance the oxygen. The products have $(5 \times 2) + 5 = 15$ oxygen atoms. The ester provides 2 oxygen atoms, leaving 13 to be provided by the $O_{2}$ gas, so we need $6.5 O_{2}$. To eliminate the fraction, multiply the entire equation by 2.