Mr Cole Chemistry

Introduction to Halogenoalkanes Questions

Introduction to Halogenoalkanes

Halogenoalkanes Worksheet

Naming Halogenoalkanes

1. The following compounds are all halogenoalkanes. State their IUPAC name.
a)
Chemical Structure A

2-chloropropane

b)
Chemical Structure B

2-bromo-2-chloropropane

c)
Chemical Structure C

2-bromo-3-methylbutane

d)
Chemical Structure D

2-bromo-2-chloro-3-ethylpentane

Drawing Halogenoalkanes

2. Draw the following halogenoalkanes.
a) Chloromethane
Structure of Chloromethane
b) 2-methyl-2-chloropropane
Structure of 2-methyl-2-chloropropane
c) 2,2-dimethyl-1-iodobutane
Structure of 1-iodo-2,2-dimethylbutane

Bond Enthalpy & Polarity

3. 1-chloropropane and 1-bromopropane are both halogenoalkanes.
a) State which molecule has the stronger C-X (carbon-halogen) bond and explain why.

The C-Cl bond in chloropropane is stronger than the C-Br bond in bromopropane. This is because Chlorine has fewer shells than Bromine, resulting in a shorter bond length and a stronger attraction between the bonded electrons and the nucleus.

b) State which molecule has the more polar C-X bond and explain why.

The C-Cl bond in chloropropane is more polar. Chlorine is smaller and more electronegative than Bromine. This means the bonded pair of electrons is more strongly attracted to the Chlorine atom than to the Bromine atom.

c) State which molecule is more reactive and explain why.

Bromopropane is more reactive because the C-Br bond is weaker than the C-Cl bond, meaning the activation energy to break it is lower.

d) State which molecule has a higher boiling point and explain why.

Bromopropane has a higher boiling point. Bromine is much larger (has more electrons) than Chlorine, resulting in stronger Van der Waals forces (London forces) between molecules, which require more energy to overcome.

Synthesis of Chloromethane

4. Chloromethane can be formed from chlorine and methane.
a) State the conditions required for the reaction.

UV light

b) Write the overall equation for the reaction.
Cl2 + CH4
→
CH3Cl + HCl
c) Explain why the percentage yield of chloromethane is very low from this reaction.

The chloromethane formed can undergo further substitution reactions with chlorine radicals, resulting in a mixture of products including dichloromethane, trichloromethane, and tetrachloromethane.

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