Which process is used to make an alkene from a long-chain alkane?
A) combustion
B) condensation
C) cracking
D) polymerisation
▶️ Answer/Explanation
Ans: C
Cracking is the process where long-chain alkanes are broken down into smaller molecules, including alkenes. This is typically done by heating the alkane in the presence of a catalyst. The other options don’t fit:
– Combustion (A) burns hydrocarbons to produce CO₂ and H₂O
– Condensation (B) joins molecules together with the loss of a small molecule like water
– Polymerisation (D) joins small molecules (like alkenes) to form large polymer chains
Only cracking produces alkenes from alkanes by breaking carbon-carbon bonds in the long chain.
Alkanes undergo substitution reactions with chlorine in the presence of ultraviolet light.
Which equation shows a reaction of this type?
A) \( C_3H_6 + Cl_2 \rightarrow C_3H_6Cl_2 \)
B) \( C_3H_8 + Cl_2 \rightarrow C_3H_6Cl_2 + H_2 \)
C) \( C_3H_8 + Cl_2 \rightarrow C_3H_7Cl + HCl \)
D) \( C_3H_6 + Cl_2 \rightarrow C_3H_5Cl + HCl \)
▶️ Answer/Explanation
Ans: C
The question asks about substitution reactions of alkanes with chlorine. Key points:
1. Alkanes have the general formula \( C_nH_{2n+2} \), so \( C_3H_8 \) is an alkane while \( C_3H_6 \) is not (it’s an alkene).
2. In substitution reactions, a hydrogen atom is replaced by a chlorine atom, forming HCl as a byproduct.
Option C shows this correctly: \( C_3H_8 \) (propane) reacts with \( Cl_2 \) to form \( C_3H_7Cl \) (chloropropane) and HCl.
The other options are incorrect because:
– A shows addition to an alkene (not substitution)
– B shows formation of \( H_2 \) (not typical for chlorine substitution)
– D involves an alkene (\( C_3H_6 \)) rather than an alkane
Which statement explains why ethanoic acid is saturated?
A) The molecule dissociates completely in water.
B) There is a carbon-oxygen double bond in the molecule.
C) The carbon-carbon bond in the molecule is a single bond.
D) All the carbon-hydrogen bonds in the molecule are single bonds.
▶️ Answer/Explanation
Ans: C
A saturated compound is one where all carbon-carbon bonds are single bonds, with no double or triple bonds between carbon atoms. Ethanoic acid (CH₃COOH) has a single bond between its two carbon atoms, making it a saturated compound.
Option A is incorrect because dissociation relates to acidity, not saturation. Option B is incorrect because while there is a C=O bond, saturation refers to carbon-carbon bonds. Option D is incorrect because saturation specifically refers to carbon-carbon bonds, not carbon-hydrogen bonds.
Which row shows the properties of methane?
| soluble in water | state at room temperature | gives a positive test with aqueous bromine | |
|---|---|---|---|
| A | no | gas | no |
| B | no | gas | yes |
| C | yes | liquid | no |
| D | yes | liquid | yes |
▶️ Answer/Explanation
Ans: A
Methane (CH₄) has the following properties:
1. Solubility: It is not soluble in water because it’s non-polar while water is polar (eliminates options C and D).
2. State: At room temperature, methane is a gas (eliminates options C and D which say it’s liquid).
3. Bromine test: Methane doesn’t react with aqueous bromine because it’s a saturated hydrocarbon (eliminates option B).
Therefore, option A correctly describes all properties of methane.