Questions on amines: preparation, basicity and nucleophilic reactions

Question 1

Explain the order of base strength of phenylamine, ammonia and ethylamine, starting with the weakest.

Accepted Answer

Indicative content

the order is phenylamine weakest, then ammonia, then ethylamine strongest, all compared at the same concentration
base strength depends on how available the lone pair on the nitrogen is to accept a proton from water or an acid
in ethylamine the ethyl group releases electron density onto the nitrogen, the positive inductive effect; reject 'larger group'
so the lone pair on ethylamine's nitrogen is more available than ammonia's, and ethylamine is the stronger base
in phenylamine the lone pair on nitrogen overlaps with the delocalised π system and is partly delocalised into the benzene ring
so less electron density remains on the nitrogen, its lone pair is less available to accept a proton, and phenylamine is weaker than ammonia

Levels

Level 3 (5–6 marks) the order is correct and both comparisons are explained through the availability of the nitrogen lone pair: the positive inductive effect of the ethyl group and the delocalisation of phenylamine's lone pair into the ring; the answer does not rely on size or on 'more alkyl groups'
Level 2 (3–4 marks) the order is right and one comparison is fully explained through the lone pair; the other is asserted, or put down to the size of the molecule
Level 1 (1–2 marks) the order is given or the lone pair mentioned; phenylamine is often placed above ammonia because the ring is taken to push electrons towards the nitrogen

Question 2

Outline the mechanism for the reaction of bromoethane with an excess of ammonia to form ethylamine.

Accepted Answer

B1 a curly arrow from the lone pair on the nitrogen of NH₃ to the carbon bonded to bromine B1 a curly arrow from the C–Br bond to the bromine atom B1 the structure of the intermediate ion, CH₃CH₂NH₃⁺, with the positive charge on nitrogen B1 a curly arrow from the lone pair on a second NH₃ molecule to an H atom bonded to the N⁺ B1 a curly arrow from that N–H bond to the nitrogen, giving ethylamine and NH₄⁺

Question 3

Name the compounds CH₃CH₂CH₂NH₂ and (CH₃)₄N⁺Cl⁻, and explain why the ion in the second compound can act neither as a base nor as a nucleophile.

Accepted Answer

B1 CH₃CH₂CH₂NH₂ is propylamine; allow propan-1-amine B1 (CH₃)₄N⁺Cl⁻ is tetramethylammonium chloride B1 the nitrogen has used its lone pair to form a fourth bond to carbon B1 with no lone pair it cannot accept a proton or donate a pair of electrons to a carbon atom

Question 4

Compare making ethylamine from bromoethane and ammonia with making it from bromomethane by way of a nitrile.

Accepted Answer

B1 the ammonia route is one step, while the nitrile route has two: KCN in ethanol and water, then reduction with LiAlH₄ or H₂ and nickel B1 the ammonia route gives a mixture of primary, secondary and tertiary amines and a quaternary salt, lowering the yield B1 the nitrile route gives only the primary amine, because no halogenoalkane remains when the amine forms B1 the nitrile route uses potassium cyanide, which is toxic, and adds a carbon, so it starts from a halogenoalkane with one carbon fewer

Question 5

Describe the reaction of ethanoyl chloride with methylamine, naming the type of reaction and the organic product.

Accepted Answer

B1 the reaction is vigorous at room temperature, with white fumes or a white solid as the HCl meets excess amine B1 it is a nucleophilic addition–elimination; reject 'nucleophilic substitution' B1 the organic product is N-methylethanamide, CH₃CONHCH₃ B1 the HCl formed reacts with a second methylamine molecule: CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻

Question 6

Identify each of these as a primary, secondary or tertiary amine: (CH₃)₂CHNH₂, (CH₃CH₂)₂NH, (CH₃)₃N.

Accepted Answer

B1 (CH₃)₂CHNH₂ is a primary amine, with one carbon group on the nitrogen B1 (CH₃CH₂)₂NH is a secondary amine, with two carbon groups on the nitrogen B1 (CH₃)₃N is a tertiary amine, with three carbon groups on the nitrogen

Question 7

Explain why a solution of methylamine is alkaline, giving an equation with state symbols.

Accepted Answer

B1 the nitrogen atom in methylamine has a lone pair of electrons that accepts a proton, so it is a Brønsted–Lowry base B1 CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq); the equilibrium sign is required B1 the proton comes from water, so hydroxide ions are left in solution and the solution is alkaline

Question 8

Explain why the reaction of bromoethane with ammonia gives a mixture of products.

Accepted Answer

B1 the ethylamine formed has a lone pair on its nitrogen, so it is also a nucleophile, and a better one than ammonia B1 it attacks further bromoethane to give diethylamine, which in turn gives triethylamine B1 triethylamine reacts to give the quaternary salt, tetraethylammonium bromide, so all four products are present

Question 9

State the reagents used to convert nitrobenzene into phenylamine, and the purpose of the final reagent.

Accepted Answer

B1 tin and concentrated hydrochloric acid, heated under reflux, as the reducing agent; allow iron for tin B1 sodium hydroxide solution is then added B1 the sodium hydroxide removes a proton from the phenylammonium ion, releasing free phenylamine

Question 10

Give the equation, using [H] for the reducing agent, for the reduction of propanenitrile, and name the reducing agent used in the laboratory and the one used in industry.

Accepted Answer

B1 CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂ B1 in the laboratory, lithium tetrahydridoaluminate(III), LiAlH₄, in dry ether B1 in industry, hydrogen with a nickel catalyst, because LiAlH₄ is too expensive at scale

Question 11

Calculate the mass of phenylamine obtained when nitrobenzene is reduced, using the data given in the table.

Accepted Answer

M1 amount of nitrobenzene = 36.9 ÷ 123.0 = 0.300 mol M1 maximum mass of phenylamine = 0.300 × 93.0 = 27.9 g A1 mass obtained = 27.9 × 65.0 ÷ 100 = 18.1 g, to three significant figures

Question 12

Calculate the percentage atom economy for making ethylamine by CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. Mr: CH₃CH₂Br 108.9, NH₃ 17.0, CH₃CH₂NH₂ 45.0.

Accepted Answer

M1 total Mr of reactants = 108.9 + 2 × 17.0 = 142.9 M1 atom economy = 45.0 ÷ 142.9 × 100 A1 atom economy = 31.5%, to three significant figures

Question 13

Suggest two advantages of selling a medicine that contains an amine group as its hydrochloride salt rather than as the free amine, and what happens when the salt meets a strong base.

Accepted Answer

B1 the salt is ionic, so it dissolves more readily in water and in the body B1 the salt is a stable solid without the smell of the amine, which may be an oily liquid B1 hydroxide ions take the proton from the RNH₃⁺ ion, releasing the free amine

Question 14

Suggest whether phenylmethylamine, C₆H₅CH₂NH₂, is a stronger or a weaker base than phenylamine, C₆H₅NH₂, and explain your answer.

Accepted Answer

B1 phenylmethylamine is the stronger base B1 its nitrogen is separated from the ring by a CH₂ group, so its lone pair cannot overlap with the delocalised ring B1 the CH₂ group releases electron density onto the nitrogen, so its lone pair is more available to accept a proton

Question 15

Calculate the mass of phenylamine obtained when it is made from benzene in two steps, using the data given in the table.

Accepted Answer

M1 amount of benzene = 23.4 ÷ 78.0 = 0.300 mol M1 amount of phenylamine = 0.300 × 0.750 × 0.600 = 0.135 mol A1 mass of phenylamine = 0.135 × 93.0 = 12.6 g, to three significant figures

Question 16

Suggest the reagents and conditions for converting phenylamine into benzenediazonium chloride, and why the temperature must be controlled.

Accepted Answer

B1 sodium nitrite and hydrochloric acid, which together make nitrous acid, HNO₂ B1 the mixture is kept below 10 °C, in ice B1 above about 10 °C the diazonium ion decomposes, giving off nitrogen and forming phenol

Question 17

Describe how an azo dye is made from benzenediazonium chloride and phenol, and why the product is coloured.

Accepted Answer

B1 cold benzenediazonium chloride solution is added to phenol dissolved in sodium hydroxide solution B1 the diazonium ion substitutes at carbon 4 of the phenol ring, giving a yellow-orange precipitate in which an –N=N– group joins the two rings B1 the azo group links the delocalised systems of both rings, and the extended delocalised system absorbs visible light

Question 18

Justify the order of the three steps used to obtain phenylamine from nitrobenzene: heating under reflux with tin and concentrated hydrochloric acid, then adding an excess of sodium hydroxide solution, then separating the organic layer.

Accepted Answer

B1 the reduction needs the acid, and the excess acid protonates the phenylamine as it forms, so the product of the first step is the water-soluble salt C₆H₅NH₃⁺Cl⁻ rather than the free amine B1 sodium hydroxide must come after the reduction, because it would neutralise the acid the reduction needs; added afterwards it removes the proton, C₆H₅NH₃⁺ + OH⁻ → C₆H₅NH₂ + H₂O, releasing the free base B1 phenylamine is only slightly soluble in water, so once freed it forms a separate oily layer that can be removed, which the ionic salt, dissolved in the water, could not be

Question 19

Suggest why a quaternary ammonium salt with a long hydrocarbon chain is used in fabric conditioner.

Accepted Answer

B1 the positively charged head is attracted to the negatively charged surface of wet fabric fibres B1 the hydrocarbon tails point outwards, making the fabric feel smooth and preventing static charge building up

Question 20

Explain why ethanamide, CH₃CONH₂, is a much weaker base than ethylamine, CH₃CH₂NH₂.

Accepted Answer

B1 in ethanamide the lone pair on nitrogen is delocalised over the N–C=O group, drawn towards the electronegative oxygen B1 so it is far less available to accept a proton than the lone pair in ethylamine, which the ethyl group makes more available

Question 21

Give the observations when butylamine solution is added drop by drop to copper(II) sulfate solution until it is in excess.

Accepted Answer

B1 a pale blue precipitate of copper(II) hydroxide forms at first, as butylamine acts as a base B1 in excess the precipitate dissolves to give a deep blue solution of a copper–butylamine complex ion

Quantity	Value
mass of nitrobenzene, C₆H₅NO₂	36.9 g
Mr of nitrobenzene	123.0
Mr of phenylamine, C₆H₅NH₂	93.0
percentage yield	65.0%

Quantity	Value
mass of benzene, C₆H₆	23.4 g
Mr of benzene	78.0
Mr of phenylamine, C₆H₅NH₂	93.0
yield of nitration	75.0%
yield of reduction	60.0%