Home / iGCSE Chemistry (0620) Core:11.5 Alkenes: Exam Style Questions Paper 1

iGCSE Chemistry (0620) Core:11.5 Alkenes: Exam Style Questions Paper 1- New Syllabus

Question

Three hydrocarbons are separately mixed with aqueous bromine.
Which row gives the correct observation for each hydrocarbon?
▶️ Answer/Explanation
Correct Option: C

Detailed solution:

The bromine water test distinguishes between alkanes and alkenes. The first two hydrocarbons shown are alkanes—methane and ethane—which are saturated and do not react with bromine water in the dark, so the orange colour stays unchanged. The third hydrocarbon is ethene, an alkene with a carbon-carbon double bond. It is unsaturated and undergoes an addition reaction with bromine, instantly turning the orange bromine water colourless. So the correct sequence is: stays unchanged, stays unchanged, changes from orange to colourless.

Question

Some words used to describe organic compounds are listed.
  1. hydrocarbon
  2. monomer
  3. saturated
  4. unreactive
Which words describe ethene?
A. 1 and 2
B. 1 and 3
C. 2 and 4
D. 3 and 4
▶️ Answer/Explanation
Correct Option: A

Detailed solution:

Ethene (C₂H₄) is a hydrocarbon because it is composed only of carbon and hydrogen atoms. It is a monomer because it can be polymerised to form poly(ethene). However, ethene contains a carbon-carbon double bond (C=C), which makes it an unsaturated compound, not saturated. The double bond also makes ethene fairly reactive, as it readily undergoes addition reactions. So ethene is described as a hydrocarbon and a monomer.

Question

Which products can be formed by the cracking of one molecule of hexane, $\text{C}_6\text{H}_{14}$?
A. $\text{C}_4\text{H}_{10}$ and $\text{C}_2\text{H}_4$ only
B. $\text{C}_{12}\text{H}_{26}$ and $\text{H}_2$ only
C. $\text{C}_3\text{H}_7$ only
D. $\text{C}_2\text{H}_6$ and $\text{C}_4\text{H}_{10}$ only
▶️ Answer/Explanation
Correct Option: A

Detailed solution:

Cracking breaks a large alkane into a smaller alkane and an alkene. Hexane ($\text{C}_6\text{H}_{14}$) has 6 carbons. Option A gives butane ($\text{C}_4\text{H}_{10}$, 4C alkane) and ethene ($\text{C}_2\text{H}_4$, 2C alkene), total 6C — perfect. Option B has more carbons than the original molecule, which is impossible. Option C shows a fragment, not stable molecules. Option D gives two alkanes, but cracking must produce at least one alkene. So only A works.

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