Home / iGCSE Chemistry (0620) Theory (Core):11.2 Naming organic compounds: Exam Style Questions Paper 3

iGCSE Chemistry (0620) Theory (Core):11.2 Naming organic compounds: Exam Style Questions Paper 3- New Syllabus

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Question

Compound B is a molecule found in a plant called basil.
Fig. 4.1 shows the displayed formula of a molecule of compound B.
(a) Deduce the molecular formula of compound B to show the number of atoms of carbon, hydrogen and oxygen.
(b) On Fig. 4.1, draw a circle around one part of the molecule which shows that the molecule is unsaturated.
(c) A different molecule found in plants has the molecular formula \( \mathrm{C}_9\mathrm{H}_{10}\mathrm{O}_3 \).
Complete Table 4.1 to calculate the relative molecular mass of \( \mathrm{C}_9\mathrm{H}_{10}\mathrm{O}_3 \).
(d) Fig. 4.2 shows the names of some of the fractions obtained from petroleum using a fractionating column.
Using only the fractions shown in Fig. 4.2, name the fraction which contains compounds that:
(i) have the highest boiling point
(ii) have the shortest chain length
(iii) are used to make waxes and polishes.
(e) Table 4.2 shows some properties of alcohols.
(i) The compounds in Table 4.2 are members of the alcohol homologous series.
State what is meant by the term homologous series.
(ii) A molecule of octanol has eight carbon atoms.
Use the information in Table 4.2 to deduce the formula of octanol.
(iii) Use the information in Table 4.2 to predict the melting point of nonanol.
(f)(i) Hexanol is a fuel. In excess oxygen, hexanol undergoes complete combustion to produce water and one other product.
Identify this other product.
(f)(ii) State the colour change observed when water is added to anhydrous copper(II) sulfate.
(g) Ethanol can be manufactured by fermentation.
Name the reactant, and state two conditions needed for the fermentation of ethanol.
(h) Ethanol can be converted into ethanoic acid.
Draw the displayed formula of ethanoic acid. Show all the atoms and all the bonds.

Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):

• Topic 11.1 — Formulae, functional groups and terminology (Parts (a), (b), (e)(i), (e)(ii), (e)(iii))
• Topic 3.2 — Relative masses of atoms and molecules (Part (c))
• Topic 11.3 — Fuels (Part (d))
• Topic 11.6 — Alcohols (Parts (f)(i), (f)(ii), (g), (h))

▶️ Answer/Explanation

(a)
\( \mathrm{C}_{10}\mathrm{H}_{12}\mathrm{O}_1 \) (or \( \mathrm{C}_{10}\mathrm{H}_{12}\mathrm{O} \))

By counting the atoms in the displayed formula shown in Fig. 4.1, there are 10 carbon atoms, 12 hydrogen atoms, and 1 oxygen atom.

(b)
Any \( \mathrm{C=C} \) circled

An unsaturated compound contains one or more carbon-carbon double bonds (\( \mathrm{C=C} \)). Circling any of these double bonds in the diagram correctly identifies the unsaturated part of the molecule.

(c)
Relative molecular mass = \( 166 \)

Using relative atomic masses (C = 12, H = 1, O = 16):
Carbon: \( 9 \times 12 = 108 \)
Hydrogen: \( 10 \times 1 = 10 \)
Oxygen: \( 3 \times 16 = 48 \)
Sum: \( 108 + 10 + 48 = 166 \).

(d)(i)
Lubricating oil

At the bottom of the fractionating column, fractions have high boiling points. Bitumen is highest, but based on the fractions provided in the mark scheme, lubricating oil is the highest boiling fraction listed.

(d)(ii)
Gasoline / petrol

Short chain lengths correspond to low boiling points, which are found in fractions at the top of the column like gasoline/petrol.

(d)(iii)
Lubricating oil

The lubricating oil fraction is specifically used for making lubricants, waxes, and polishes.

(e)(i)
(Family of similar) compounds with similar chemical properties / same functional group.

A homologous series is a group of organic compounds that have the same functional group, similar chemical properties, and a trend in physical properties.

(e)(ii)
\( \mathrm{C}_8\mathrm{H}_{17}\mathrm{OH} \)

The general formula for alcohols is \( \mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH} \). For octanol, \( n = 8 \), so the formula is \( \mathrm{C}_8\mathrm{H}_{17}\mathrm{OH} \).

(e)(iii)
Any value from -15 to 6 inclusive.

The melting points of the alcohols show an increasing trend as the chain length increases. A value within the range specified by the mark scheme (between -15°C and 6°C) is a valid prediction based on the table’s trend.

(f)(i)
Carbon dioxide

The complete combustion of any alcohol (or hydrocarbon) in excess oxygen produces carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2\mathrm{O} \)).

(f)(ii)
From white to blue.

Anhydrous copper(II) sulfate is white. When water is added, it forms hydrated copper(II) sulfate, which is blue. This is the standard chemical test for water.

(g)
Reactant: (aqueous) glucose
Condition 1: 25–35 °C
Condition 2: yeast (or absence of oxygen)

Fermentation uses glucose as a reactant, broken down by yeast enzymes into ethanol and carbon dioxide at a warm temperature (optimum for yeast activity) in the absence of oxygen (anaerobic conditions).

(h)
The displayed formula for ethanoic acid is:

The structure must show the carboxyl functional group (-COOH) correctly bonded to the methyl group (-CH₃).

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