All Free Chemistry MCQs with Answers

201. The IUPAC name of CH3CH2CHClCH3 is

• A. 1-chlorobutane
• B. 2-chlorobutane
• C. 3-chlorobutane
• D. 2-chloropropane

Explanation: The longest chain has four carbons, giving butane, and numbering from the end that puts the chlorine on the lower numbered carbon makes it position 2 rather than position 3. Naming it 3-chlorobutane means numbering from the wrong end, which is the standard error. The rule is always to give substituents the lowest possible locants.

202. Aryl halides such as chlorobenzene are far less reactive towards nucleophilic substitution than alkyl halides because

• A. the benzene ring repels all nucleophiles equally
• B. chlorine is not electronegative in aromatic compounds
• C. a lone pair on the halogen is delocalised into the ring, giving the carbon to halogen bond partial double bond character
• D. aryl halides are ionic

Explanation: Overlap between the halogen lone pair and the ring pi system shortens and strengthens the bond, so the halogen is a much poorer leaving group and substitution requires extreme conditions of temperature and pressure. The carbon involved is also sp2 hybridised and holds its electrons more tightly. This is why chlorobenzene does not give a precipitate with silver nitrate on warming while chloroethane does.

203. Warming an alkyl halide with aqueous silver nitrate produces a precipitate of silver halide. This reaction is used to

• A. measure the boiling point of the halide
• B. convert the halide into an alkene
• C. prepare a Grignard reagent
• D. compare the relative rates at which different carbon to halogen bonds break

Explanation: The halide ion must be released before it can precipitate with silver, so the speed at which the precipitate appears reflects how easily the carbon to halogen bond breaks, and the iodide gives a yellow precipitate fastest while the chloride is slowest. Ethanol is added as a co solvent so that the halide and the aqueous reagent mix. The colour of the precipitate also identifies which halogen is present.

204. Chlorofluorocarbons are damaging to the ozone layer because in the stratosphere they

• A. dissolve ozone directly
• B. release chlorine radicals under ultraviolet light, and each radical destroys many ozone molecules in a chain reaction
• C. react with oxygen to form carbon dioxide
• D. absorb all incoming ultraviolet radiation harmlessly

Explanation: Ultraviolet light breaks the carbon to chlorine bond homolytically, and the chlorine radical produced converts ozone to oxygen and is then regenerated, so a single radical can destroy thousands of ozone molecules before it is removed. Their very stability in the lower atmosphere is what allows them to reach the stratosphere intact. This is why the Montreal Protocol phased them out in favour of hydrofluorocarbons.

205. Alkanes are relatively unreactive towards most reagents because

• A. their carbon to hydrogen bonds are strong and almost non polar, offering nothing for a nucleophile or electrophile to attack
• B. they contain a double bond
• C. they are ionic compounds
• D. they have very high molar masses

Explanation: Carbon and hydrogen have similar electronegativities, so the bonds carry almost no dipole and there is no region of high or low electron density to attract a reagent. This is why alkanes are called paraffins, meaning little affinity, and why their main reactions are combustion and free radical substitution, both of which need harsh conditions. Alkenes react far more readily because the pi electrons are exposed.

206. The hybridisation of each carbon atom in an alkane such as ethane is

• A. sp
• B. sp2
• C. sp3
• D. unhybridised p

Explanation: Four sigma bonds require four equivalent orbitals, so one s and three p orbitals mix to give four sp3 hybrids pointing to the corners of a tetrahedron with bond angles of 109.5 degrees. Alkenes use sp2 with 120 degree angles and alkynes sp with a linear 180 degree arrangement. Counting the sigma bonds and lone pairs on an atom is the quickest way to assign hybridisation.

207. The chlorination of methane in the presence of ultraviolet light proceeds by

• A. nucleophilic substitution
• B. electrophilic addition
• C. free radical substitution
• D. elimination

Explanation: Ultraviolet light splits the chlorine molecule homolytically into two radicals, which then abstract hydrogen from methane and set up a chain reaction of propagation steps until two radicals combine and terminate it. Because the chain continues, the reaction does not stop at chloromethane but produces a mixture of all four chlorinated products. Substitution is characteristic of saturated compounds, while addition belongs to unsaturated ones.

208. In the free radical substitution of methane, the initiation step is

• A. the reaction of a chlorine radical with methane
• B. the combination of two methyl radicals
• C. the reaction of a methyl radical with chlorine
• D. the homolytic splitting of a chlorine molecule into two chlorine radicals by ultraviolet light

Explanation: Initiation is the step that creates radicals where there were none, and only the chlorine to chlorine bond is weak enough to be broken by ultraviolet light. Propagation steps then consume one radical and produce another, so the chain continues, while termination removes radicals by combining them. Identifying which step is which by counting radicals on each side is the reliable method.

209. Which reaction is characteristic of alkenes?

• A. Free radical substitution
• B. Electrophilic addition
• C. Nucleophilic substitution
• D. Electrophilic substitution

Explanation: The exposed pi electrons of the double bond attract electrophiles, which add across the bond and convert an unsaturated compound into a saturated one, as happens with bromine, hydrogen halides and water. Alkanes undergo free radical substitution and benzene undergoes electrophilic substitution, which preserves its delocalised ring. Addition is favoured because the pi bond is weaker than a sigma bond.

210. The reagent used to distinguish an alkene from an alkane in the laboratory is

• A. sodium metal
• B. silver nitrate solution
• C. bromine water, which is decolourised by the alkene
• D. limewater

Explanation: The orange colour of bromine water disappears as bromine adds across the double bond to form a colourless dibromo compound, and an alkane leaves the colour unchanged in the dark. Acidified potassium permanganate, which loses its purple colour, is the other standard test. Silver nitrate is used to test for halide ions and limewater for carbon dioxide.