Free Alkyl Halides MCQs with Answers

11. Reaction of bromoethane with alcoholic ammonia under pressure gives

• A. ethanol
• B. ethylamine
• C. ethanenitrile
• D. ethane

Explanation: Ammonia has a lone pair on nitrogen and acts as a nucleophile, displacing bromide to give ethylamine, though the product is itself nucleophilic so further substitution to secondary and tertiary amines readily follows. Using a large excess of ammonia limits this. Ethanenitrile would be formed with potassium cyanide instead.

12. The reaction of an alkyl halide with potassium cyanide is synthetically useful because it

• A. shortens the carbon chain by one atom
• B. produces an alkene
• C. lengthens the carbon chain by one carbon atom
• D. removes the halogen without adding anything

Explanation: The cyanide ion substitutes for the halogen, so the nitrile formed has one more carbon than the starting halide, and that nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine. Building the chain one carbon at a time is a standard strategy in organic synthesis. Alcoholic conditions are used to favour substitution over elimination.

13. A Grignard reagent is formed when an alkyl halide reacts with

• A. sodium metal in dry ether
• B. magnesium in dry ether
• C. zinc in aqueous solution
• D. copper in ethanol

Explanation: Magnesium inserts itself into the carbon to halogen bond to give an alkylmagnesium halide, and the ether must be perfectly dry because even a trace of water destroys the reagent, converting it into an alkane. The carbon attached to magnesium is strongly nucleophilic, which makes Grignard reagents extremely versatile for building carbon to carbon bonds. Sodium with an alkyl halide gives the Wurtz reaction instead.

14. Alkyl halides have higher boiling points than the corresponding alkanes because

• A. they contain hydrogen bonds
• B. they are ionic
• C. their molecules are lighter
• D. they are polar and have greater molar mass, so intermolecular forces are stronger

Explanation: The polar carbon to halogen bond adds dipole to dipole attraction on top of the dispersion forces, and the heavy halogen atom increases those dispersion forces as well, so more energy is needed to separate the molecules. Boiling point rises from the fluoride to the iodide for the same alkyl group. Hydrogen bonding is not possible because hydrogen is not bonded to fluorine, oxygen or nitrogen here.

15. Which alkyl halide would react fastest by the SN1 mechanism?

• A. CH3Br
• B. CH3CH2Br
• C. (CH3)2CHBr
• D. (CH3)3CBr

Explanation: SN1 rates follow the stability of the carbocation formed, and that increases from primary through secondary to tertiary because each alkyl group releases electron density and spreads the charge. The tertiary bromide therefore ionises most readily. The order is exactly reversed for SN2, where the methyl halide reacts fastest because it is the least hindered.

16. In an SN2 reaction the configuration at the reacting carbon atom is

• A. inverted, because the nucleophile attacks from the side opposite the leaving group
• B. retained completely
• C. randomised, giving a racemic mixture
• D. unchanged because no bond is broken

Explanation: Back side attack turns the three remaining groups inside out like an umbrella in the wind, so an optically active substrate gives a product of opposite configuration, an effect called Walden inversion. An SN1 reaction gives a racemic mixture instead, because the planar carbocation can be attacked equally from either face. Stereochemistry is therefore the clearest experimental way to distinguish the two mechanisms.

17. 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.

18. 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.

19. 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.

20. 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.