Organic Chemistry
Organic chemistry is the study of carbon-containing compounds, covering how they are named using IUPAC rules, the different types of isomerism they show, and the mechanisms (free-radical substitution, electrophilic addition, nucleophilic substitution and elimination) by which alkanes, alkenes and halogenoalkanes react. It is one of the largest topics in A Level Chemistry, spanning both years of the course.
Method
- To name an organic compound, identify the longest continuous carbon chain containing the principal functional group, number the chain to give the functional group and substituents the lowest possible locants, then name substituents as prefixes in alphabetical order.
- To identify the type of isomerism between two compounds, first check they share the same molecular formula; if the carbon skeleton differs it is chain isomerism, if the functional group position differs it is positional isomerism, and if the functional group itself differs it is functional group isomerism.
- To draw a curly-arrow mechanism, identify the electrophile (an electron pair acceptor) and the nucleophile (an electron pair donor), then draw arrows starting from a lone pair or bond and ending where the new bond forms.
- For a free-radical substitution mechanism, write the initiation step (homolytic fission of the halogen molecule under UV light) as a stand-alone step, then write the two propagation steps in sequence, checking a radical is regenerated in the second propagation step.
- For electrophilic addition to an unsymmetrical alkene, apply Markovnikov's rule: predict the major product by identifying which carbocation intermediate (secondary/tertiary rather than primary) is more stable, because more alkyl groups donate electron density and disperse the positive charge.
- When comparing reaction conditions, match the reagent and solvent to the mechanism required: aqueous conditions with heat favour nucleophilic substitution of halogenoalkanes, while ethanolic (alcoholic) conditions with heat favour elimination.
Worked example
An alkene, Z, is a gas at room temperature and contains only carbon and hydrogen. Complete combustion of 0.420 g of Z produces 1.32 g of carbon dioxide and 0.540 g of water. Given that Z has a relative molecular mass (Mr) of 42, calculate the molecular formula of Z.
- Calculate moles of CO2 (= moles of C): 1.32/44 = 0.0300 mol.
- Calculate moles of H2O and hence moles of H: 0.540/18 = 0.0300 mol H2O, so moles H = 0.0600 mol.
- Check for oxygen: mass of C + mass of H = (0.0300 x 12) + (0.0600 x 1) = 0.360 + 0.0600 = 0.420 g, equal to the sample mass, confirming Z contains no oxygen.
- Find the simplest mole ratio C : H = 0.0300 : 0.0600 = 1 : 2, giving empirical formula CH2 (empirical mass = 14).
- Divide the given Mr by the empirical formula mass: 42/14 = 3, so the molecular formula is 3 x CH2 = C3H6.
- Final answer: molecular formula of Z = C3H6
Practice questions
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Q1Define the term structural isomers.Show answer
Answer: Compounds with the same molecular formula but a different arrangement of atoms (different structural formula)
Q2State the molecular formula of butan-2-ol.Show answer
Answer: C4H10O
Q3Write an equation for the initiation step when methane reacts with bromine in the presence of UV light.Show answer
Answer: Br2 -> 2Br* (homolytic fission of the Br-Br bond, initiated by UV light)
Q4But-2-ene shows E/Z isomerism, but propene does not. State why propene does not show E/Z isomerism.Show answer
Answer: One of the carbon atoms of the C=C double bond in propene (CH2=CHCH3) is attached to two identical hydrogen atoms, so no two distinguishable geometric arrangements are possible
Q5When 2-methylpropene reacts with hydrogen bromide, the major product is 2-bromo-2-methylpropane rather than 1-bromo-2-methylpropane. Explain why, in terms of carbocation stability.Show answer
Answer: H+ adds to the terminal CH2 carbon, forming a tertiary carbocation rather than a primary one; the tertiary carbocation is more stable because three electron-donating alkyl groups disperse the positive charge, so it forms preferentially, giving 2-bromo-2-methylpropane as the major product
Q61-chloropropane, 1-bromopropane and 1-iodopropane are each warmed with aqueous silver nitrate dissolved in ethanol. State and explain the order in which precipitates form.Show answer
Answer: 1-iodopropane forms a precipitate fastest, then 1-bromopropane, then 1-chloropropane slowest; the C-I bond has the lowest (weakest) bond enthalpy and C-Cl the highest (strongest), so the weaker the C-X bond, the faster the substitution/hydrolysis
Exam-style questions
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Give the IUPAC name of the following compound: CH3-CH(CH3)-CH2-CH2-OH
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Propan-1-ol is heated with excess concentrated sulfuric acid at a high temperature, undergoing acid-catalysed dehydration. Name the type of reaction mechanism, give a suitable catalyst, write the structural formula of the alkene product, and explain why only one alkene product (not a mixture) is formed.
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A student prepared a sample of bromoethane from ethanol using the following method: ethanol is heated under reflux with a mixture of sodium bromide and concentrated sulfuric acid; the crude product is separated by distillation and purified. The student reacted 9.20 g of ethanol (Mr = 46.0) with excess sodium bromide and concentrated sulfuric acid, and obtained 14.5 g of pure bromoethane (Mr = 109.0). Write an equation for the formation of bromoethane from ethanol and hydrogen bromide, and calculate the percentage yield of bromoethane.
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