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Questions on planning a synthetic route
The aliphatic and aromatic reactions with the reagents and conditions for every conversion between functional groups; devising a route of two to four steps by comparing carbon skeletons and working backwards from the target; adding a carbon with a nitrile; overall yield, atom economy, hazards, selectivity and the aim of needing no solvent in choosing a route; worked routes from ethene to propylamine and from ethene to ethyl ethanoate; how a route question is marked; and the reactions other boards add: nitrile hydrolysis, acyl chlorides with SOCl₂ or PCl₅, bromination and Friedel–Crafts alkylation of benzene, and Grignard reagents with carbon dioxide, aldehydes and ketones.
23 original questions · 84 marks · the planning a synthetic route notes · Organic synthesis · Flashcards for this unit
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Outline a three-step synthesis of 2-methylpropan-1-amine, (CH₃)₂CHCH₂NH₂, from propene. For each step give the reagents and conditions and name the product, and explain why the route gives a branched chain.
Mark scheme
Indicative content
- step 1: propene reacts with hydrogen bromide at room temperature by electrophilic addition, with no other reagent needed
- the major product of step 1 is 2-bromopropane, CH₃CHBrCH₃, because the secondary carbocation is more stable than the primary one
- step 2: 2-bromopropane is heated under reflux with potassium cyanide dissolved in ethanol and water, in nucleophilic substitution
- the product of step 2 is 2-methylpropanenitrile, (CH₃)₂CHCN, in which the new carbon from cyanide is joined to the middle carbon
- step 3: the nitrile is reduced with hydrogen over a nickel catalyst, or with LiAlH₄ in dry ether; reject NaBH₄, which does not reduce a nitrile
- the product is 2-methylpropan-1-amine, accept 2-methylpropylamine; the branch arises because bromine, and so the cyanide carbon, went onto the middle carbon of propene
Levels
- Level 3 (5–6 marks) all three steps are given with reagents, conditions and the named product of each, the major product of step 1 is justified through carbocation stability, the reducing agent for the nitrile is one that works, and the branch is explained as following from where the bromine went
- Level 2 (3–4 marks) the three steps are given with mostly correct reagents and products; a condition is missing, the nitrile is 'reduced' with a reagent that does not work, or the branch is not explained
- Level 1 (1–2 marks) one or two steps appear, such as HBr or cyanide; 1-bromopropane is typically taken as the major product of step 1, so the route ends at an unbranched amine
Evaluate two routes to butylamine, CH₃CH₂CH₂CH₂NH₂: route A, one step from 1-bromobutane with excess ethanolic ammonia in a sealed tube; route B, two steps from 1-bromopropane through butanenitrile.
Mark scheme
- B1 route A has one step, so it loses less product to repeated steps and takes less time
- B1 route A gives a mixture of primary, secondary and tertiary amines and a quaternary salt, which must be separated, lowering the yield of butylamine
- B1 route B has two steps, so its overall yield is the product of two yields
- B1 route B uses toxic potassium cyanide and flammable hydrogen or violently reactive LiAlH₄, but gives only the primary amine
- B1 a justified judgement: for example, route B is preferred where a pure primary amine is needed, route A where a quick product can be purified afterwards
State the reagents and conditions for each step when 1-bromobutane is converted into pentylamine, CH₃(CH₂)₄NH₂, in two steps through pentanenitrile, and name the type of reaction in each step.
Mark scheme
- B1 step 1: potassium cyanide dissolved in ethanol and water, heated under reflux; reject HCN
- B1 step 1 is nucleophilic substitution, with cyanide as the nucleophile; ignore 'SN2'
- B1 step 2: hydrogen with a nickel catalyst, or LiAlH₄ in dry ether; reject [H] alone
- B1 step 2 is reduction of the C≡N group to CH₂NH₂
Outline a two-step synthesis of phenylamine from benzene, giving the reagents and conditions for each step and naming the intermediate.
Mark scheme
- B1 step 1 uses concentrated nitric acid with concentrated sulfuric acid as the catalyst
- B1 at about 50 °C, kept below 55 °C so that only one nitro group is substituted
- B1 the intermediate is nitrobenzene, C₆H₅NO₂
- B1 step 2 uses tin and concentrated hydrochloric acid heated under reflux, followed by excess sodium hydroxide solution to release phenylamine from its salt
Calculate the overall percentage yield of each of the two routes to the same product whose step yields are given in the table.
Route Step 1 Step 2 Step 3 A 85% 70% 64% B 58% 61% no third step Mark scheme
- M1 route A: 0.85 × 0.70 × 0.64 = 0.3808
- A1 route A overall yield = 38%, to two significant figures to match the data; allow 38.1%
- M1 route B: 0.58 × 0.61 = 0.3538; error carried forward from the method for route A
- A1 route B overall yield = 35%, to two significant figures; allow 35.4%
Calculate the percentage atom economy of making ethyl ethanoate by esterification, CH₃COOH + CH₃CH₂OH → CH₃COOCH₂CH₃ + H₂O, and from ethanoyl chloride, CH₃COCl + CH₃CH₂OH → CH₃COOCH₂CH₃ + HCl. The table gives the relative formula masses.
Compound Mr CH₃COOH 60.0 CH₃CH₂OH 46.0 CH₃COCl 78.5 CH₃COOCH₂CH₃ 88.0 Mark scheme
- M1 esterification: total mass of reactants = 60.0 + 46.0 = 106.0
- A1 atom economy = 88.0 ÷ 106.0 × 100 = 83.0%
- M1 from ethanoyl chloride: total mass of reactants = 78.5 + 46.0 = 124.5
- A1 atom economy = 88.0 ÷ 124.5 × 100 = 70.7%, to three significant figures
Suggest a two-step synthesis of 1-phenylethanol, C₆H₅CH(OH)CH₃, from benzene, and suggest why the product does not rotate plane-polarised light.
Mark scheme
- B1 step 1: benzene with ethanoyl chloride and an aluminium chloride catalyst, anhydrous, heated under reflux
- B1 the intermediate is phenylethanone, C₆H₅COCH₃
- B1 step 2: phenylethanone with NaBH₄ in water or methanol gives 1-phenylethanol
- B1 the hydride attacks either face of the planar C=O equally, so a racemic mixture forms and the rotations cancel
Suggest a two-step synthesis of 2-hydroxypropanenitrile, CH₃CH(OH)CN, from ethanol, with reagents and conditions, and suggest one hazard of the second step.
Mark scheme
- B1 step 1: ethanol with acidified potassium dichromate(VI), warmed, distilling the ethanal off as it forms
- B1 the intermediate is ethanal, CH₃CHO
- B1 step 2: ethanal with potassium cyanide, followed by dilute acid, gives 2-hydroxypropanenitrile
- B1 potassium cyanide is acutely toxic and releases toxic hydrogen cyanide gas with acid, so it is used in a fume cupboard
Identify the reagents and conditions for making methyl propanoate from propanoic acid, and from propanoyl chloride, and identify one advantage and one disadvantage of the second method.
Mark scheme
- B1 propanoic acid with methanol and a concentrated sulfuric acid catalyst, heated under reflux
- B1 propanoyl chloride with methanol at room temperature, with no catalyst needed
- B1 advantage: the acyl chloride reaction is fast and goes to completion, while esterification reaches an equilibrium
- B1 disadvantage: the acyl chloride is corrosive and releases hydrogen chloride fumes, so it needs a fume cupboard; accept a lower atom economy
Suggest a two-step synthesis of propyl propanoate, CH₃CH₂COOCH₂CH₂CH₃, in which propan-1-ol is the only organic starting material.
Mark scheme
- B1 cutting the ester at its C(=O)–O link shows it is made from propanoic acid and propan-1-ol, each with three carbons
- B1 step 1: part of the propan-1-ol is heated under reflux with excess acidified potassium dichromate(VI) to give propanoic acid
- B1 reflux with excess oxidant is needed, because distilling the product off as it forms would give propanal
- B1 step 2: propanoic acid is heated under reflux with the rest of the propan-1-ol and a concentrated sulfuric acid catalyst to give propyl propanoate
Outline how bromoethane is converted into propanoic acid using a Grignard reagent, and explain why water must be excluded.
Mark scheme
- B1 bromoethane is heated under reflux with magnesium in dry ether
- B1 this gives the Grignard reagent ethylmagnesium bromide, CH₃CH₂MgBr
- B1 the reagent reacts with solid carbon dioxide, followed by dilute acid, to give propanoic acid with one more carbon
- B1 water protonates the Grignard reagent, converting it into ethane, so it is lost before it can react with CO₂
Suggest how a Grignard reagent could be used to make 2-methylpropan-2-ol, (CH₃)₃COH, naming the halogenoalkane and the carbonyl compound, giving the conditions, and writing an equation for the formation of the Grignard reagent.
Mark scheme
- B1 bromomethane is heated under reflux with magnesium in dry ether; accept chloromethane or iodomethane
- B1 CH₃Br + Mg → CH₃MgBr, with the halogen matching the halogenoalkane named
- B1 the carbonyl compound is propanone, added in dry ether, because a ketone gives a tertiary alcohol
- B1 dilute acid is then added to convert the O–MgBr product into the alcohol, 2-methylpropan-2-ol
Justify the order of the steps in the route from ethene to 2-hydroxypropanenitrile, CH₃CH(OH)CN: ethene to ethanol with steam, ethanol to ethanal with acidified potassium dichromate(VI) and distillation, then ethanal to the hydroxynitrile with potassium cyanide followed by dilute acid.
Mark scheme
- B1 the carbon chain is lengthened by nucleophilic addition of CN⁻ to a C=O, so a carbonyl compound must be made before the cyanide step; ethene and ethanol have no δ+ carbonyl carbon for the cyanide ion to attack
- B1 the C=O can only be made by oxidising an alcohol, so the hydration of ethene, with steam and a phosphoric acid catalyst, has to come first to put the OH group on the chain
- B1 the oxidation is run by distilling the ethanal off as it forms, because heating under reflux would oxidise it on to ethanoic acid, which does not undergo nucleophilic addition
- B1 the cyanide step comes last and gives the product directly, as a racemic mixture because CN⁻ attacks the planar C=O from either face; the dilute acid is added only after the cyanide, so that toxic hydrogen cyanide is not released from the potassium cyanide in excess acid
Give the reagents and conditions needed to convert propan-1-ol into propanal, propan-1-ol into propanoic acid, and propan-2-ol into propanone.
Mark scheme
- B1 propanal: acidified potassium dichromate(VI), warmed, with the aldehyde distilled off as it forms
- B1 propanoic acid: excess acidified potassium dichromate(VI), heated under reflux
- B1 propanone: acidified potassium dichromate(VI), heated under reflux; allow K₂Cr₂O₇ with H₂SO₄ for the reagent throughout
Explain why sodium tetrahydridoborate rather than LiAlH₄ is used to reduce butanone to butan-2-ol in a school laboratory, giving the conditions for the reduction.
Mark scheme
- B1 NaBH₄ in water or ethanol at room temperature; accept methanol as the solvent
- B1 LiAlH₄ reacts violently with water, so it needs dry ether as the solvent and is a fire hazard
- B1 NaBH₄ is stable enough to use in water and still reduces aldehydes and ketones, so it gives the same product more safely
Explain, by comparing carbon skeletons, why a synthesis of butylamine, CH₃CH₂CH₂CH₂NH₂, from 1-bromopropane needs a nitrile step, while a synthesis of propylamine from 1-bromopropane does not.
Mark scheme
- B1 butylamine has four carbons and 1-bromopropane three, so a new carbon–carbon bond must be made
- B1 substitution with cyanide is the reaction that adds one carbon to the chain, giving butanenitrile, which is reduced to butylamine
- B1 propylamine has the same three-carbon skeleton as 1-bromopropane, so substitution with ammonia is enough
Explain why propylamine made from 1-bromopropane and ammonia is impure, and why the route through propanenitrile avoids the problem.
Mark scheme
- B1 propylamine has a lone pair on nitrogen, so it is itself a nucleophile
- B1 it attacks unreacted 1-bromopropane to give secondary and then tertiary amines and a quaternary ammonium salt, so a mixture forms even with excess ammonia
- B1 reducing propanenitrile gives only the primary amine, as no halogenoalkane is present for the amine to attack
Suggest a two-step synthesis of propanone from propene, giving the reagents and conditions and naming the intermediate.
Mark scheme
- B1 step 1: propene with steam and a phosphoric acid catalyst at about 300 °C and 60 to 70 atm, or concentrated sulfuric acid followed by water
- B1 the intermediate is propan-2-ol, the major product because the more stable secondary carbocation forms
- B1 step 2: propan-2-ol heated under reflux with acidified potassium dichromate(VI) gives propanone
Compare the reactions of 2-bromopropane with potassium hydroxide dissolved in water and with potassium hydroxide dissolved in ethanol.
Mark scheme
- B1 in water, warmed, hydroxide acts as a nucleophile and substitutes the bromine to give propan-2-ol
- B1 in ethanol, heated under reflux, hydroxide acts as a base and eliminates HBr to give propene
- B1 both use the same reagent and both release bromide ions; the solvent and the temperature decide which product dominates
Explain why chemists aim to design processes that do not need a solvent.
Mark scheme
- B1 a solvent does not appear in the equation or the product, yet it has to be bought, heated, separated from the product and recovered or disposed of
- B1 that costs energy and money and produces waste, which is not shown by the atom economy
- B1 most organic solvents are flammable and many are toxic, so removing the solvent also removes a hazard
Suggest a two-step synthesis of butanoic acid from 1-bromopropane that passes through a nitrile, giving the reagents and conditions for each step.
Mark scheme
- B1 step 1: 1-bromopropane heated under reflux with potassium cyanide in ethanol and water
- B1 the intermediate is butanenitrile, CH₃CH₂CH₂CN, which carries the extra carbon
- B1 step 2: butanenitrile heated under reflux with dilute hydrochloric or sulfuric acid hydrolyses the C≡N to COOH, giving butanoic acid and ammonium ions
Explain why the carbon bonded to magnesium in ethylmagnesium bromide acts as a nucleophile, and why with methanal it gives a primary alcohol but with propanal a secondary alcohol.
Mark scheme
- B1 magnesium is much less electronegative than carbon, so the carbon of the C–Mg bond is δ− and attacks a δ+ carbon
- B1 the ethyl group bonds to the carbonyl carbon, which keeps the groups it already had; methanal's carbon has two hydrogens, so the C–OH carbon carries one alkyl group: propan-1-ol
- B1 propanal's carbonyl carbon already carries an ethyl group, so the C–OH carbon ends with two alkyl groups: pentan-3-ol, a secondary alcohol
Explain why acidified potassium dichromate(VI) gives propanal from propan-1-ol if the product is distilled, but propanoic acid if the mixture is heated under reflux.
Mark scheme
- B1 propanal has a lower boiling point than propan-1-ol, so distillation removes it from the flask as it forms, before it can be oxidised further
- B1 under reflux the propanal condenses back into the flask and stays in contact with the oxidising agent, so it is oxidised on to propanoic acid