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A-Level Substitution and elimination in halogen compounds
What the A-Level syllabus expects for Substitution and elimination in halogen compounds, and how to practise it.
What the syllabus expects
- Recall halogenoalkane chemistry as shown by:
- these nucleophilic substitutions of bromoethane: hydrolysis with NaOH(aq) and heat; conversion to a nitrile with KCN in ethanol and heat; making a primary amine by treating it with ammonia dissolved in ethanol and heating under pressure
- elimination of hydrogen bromide from 2-bromopropane with NaOH in ethanol and heat
- Explain the stereochemistry that results when nucleophilic substitution acts on an optically active substrate:
Scope: see also 11.3(n) - inversion of configuration under the SN2 mechanism
- racemisation under the SN1 mechanism
- Propose diagnostic reactions that separate:
- one halogenoalkane from another
Scope: see also 11.3(f) - a halogenoalkane from a halogenoarene, for instance by hydrolysing and then testing for halide ions
Scope: see also 11.3(g) - Explain how the relative inertness of fluoroalkanes and fluorohalogenoalkanes underpins their uses.
- Recognise how CFCs damage the ozone layer, and that their proposed stand-ins, HFCs and HCFCs, still carry a real environmental cost.
Scope: the step-by-step mechanism of ozone depletion by CFCs and HCFCs is not required
How it's examined
Questions on this topic most often ask you to compare, explain. About 1% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
Phenylmethanol (C6H5CH2OH) may alternatively be made from methylbenzene in a two-step sequence. Propose a route for this synthesis, stating every reagent and condition needed.
Show the worked answer
Step 1 (side-chain substitution): react methylbenzene, C6H5CH3, with chlorine, Cl2, in the presence of ultraviolet light (or heat). This free-radical substitution of a methyl hydrogen gives (chloromethyl)benzene, C6H5CH2Cl. Step 2 (hydrolysis): warm/reflux C6H5CH2Cl with aqueous sodium hydroxide, NaOH(aq). The halogen is replaced by OH, giving phenylmethanol, C6H5CH2OH.
Example 2 (2 marks)
Decide whether thiolates (R-S⁻) are more powerful nucleophiles than alkoxides (R-O⁻), and explain your reasoning.
Show the worked answer
Compare nucleophilicity of R-S^- versus R-O^-. Sulfur sits below oxygen in Group 16, so it is a larger atom with more diffuse, polarisable valence electrons. The negative charge on the thiolate is spread over a larger volume, making S less tightly bound to its electrons than O. These loosely held, polarisable electrons are more readily distorted toward an electrophilic carbon, so thiolates form the new bond more easily. Sulfur is also less electronegative than oxygen, so it holds its lone pair less tightly and donates it more willingly. Therefore thiolates are the stronger (more powerful) nucleophiles.
Example 3 (2 marks)
Treating 2-nitrochlorobenzene with aqueous NaOH, and then adding a suitable acid, produces 2-nitrophenol. Draw the mechanism for this conversion.
Show the worked answer
This is nucleophilic aromatic substitution (addition-elimination). Step 1: the hydroxide ion OH- attacks the ring carbon bearing the Cl atom; the resulting negative charge is delocalised onto the ortho -NO2 group (and the ring), giving a stabilised Meisenheimer-type intermediate. Step 2: the C-Cl bond breaks, expelling Cl- and restoring aromaticity to give the 2-nitrophenoxide ion. Adding acid then protonates the phenoxide to give 2-nitrophenol.
More worked questions on this topic
- Groundwater found to be contaminated with TCE was examined. Analysis revealed three further chl (2 marks)
More A-Level H2 Chemistry topics
The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry