Lesson 18B.2
18B.2 Preparing amines and amides Quiz: Pearson Edexcel Chemistry, Unit 18
20 questions
In partnership with Revision Ninja
Lesson 18B.2, Preparing amines and amides: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 18: Organic Chemistry III, written with Revision Ninja.
Host it live on the board and students join with a game code on their own devices, or revise alone with Free Play. The answers are revealed in the game.
The 20 questions
-
Which reagent reduces an aromatic nitro compound to an amine in the laboratory?
- Acidified potassium manganate(VII) under reflux
- Tin and concentrated hydrochloric acid, followed by alkali
- Sodium borohydride in water at 0 C
- Dilute sulfuric acid at room temperature
-
What is the product of reducing nitrobenzene with tin and concentrated hydrochloric acid, followed by alkali?
- Phenol, C6H5OH
- Phenylamine, C6H5NH2
- Benzoic acid, C6H5COOH
- Benzene-1,4-diamine
-
Which reagent converts a halogenoalkane into a primary amine?
- Lithium tetrahydridoaluminate in dry ether, which reduces the halogen atom
- Excess concentrated ammonia in ethanol, heated in a sealed tube
- Concentrated sulfuric acid at 170 C, which eliminates HX to form an alkene
- Sodium hydroxide in ethanol, heated under reflux with the halogenoalkane
-
Which reagents make an amide from an acyl chloride?
- Concentrated ammonia or a primary amine, at room temperature
- Hydrogen gas with a nickel catalyst at 150 C, which reduces the acyl group
- Acidified dichromate(VI) ions, heated gently with the acyl chloride
- Aqueous sodium hydroxide, heated under reflux with the acyl chloride
-
Which compound is a primary amine?
- Ethanamide, CH3CONH2
- Trimethylamine, (CH3)3N
- Propan-1-amine, CH3CH2CH2NH2
- N-methylethanamine, CH3NHCH2CH3
-
Calculate the relative molecular mass of phenylamine, C6H7N, using Ar: C 12, H 1, N 14.
- 94
- 77
- 107
- 93
-
What is the by-product when ethanoyl chloride reacts with ammonia to form an amide?
- Sodium chloride
- Water
- Ammonium chloride
- Ethanoic acid
-
When nitrobenzene is reduced with tin and excess acid, what forms before alkali is added?
- Benzoic acid, formed by oxidation of the ring under the acidic reaction conditions
- Phenol, which forms directly on adding alkali to the acidic reaction mixture before any separation
- A phenylammonium salt, which is converted to phenylamine when alkali is added
- Benzene, formed by loss of the nitro group from the ring as nitrogen dioxide gas
-
Calculate the relative molecular mass of N-butylethanamide, C6H13NO, using Ar: C 12, H 1, N 14, O 16.
- 113
- 115
- 129
- 101
-
Which amide forms from ethanoyl chloride and methylamine?
- N-methylethanoate, CH3COOCH3
- Ethanenitrile, CH3CN
- Methylethanamine, CH3CH2NHCH3
- N-methylethanamide, CH3CONHCH3
-
What is the role of concentrated hydrochloric acid in the tin reduction of nitrobenzene?
- It precipitates tin as a white solid, which then catalyses the reduction of the nitro group in the mixture
- It provides the acidic conditions for the reduction and forms a salt with the amine product
- It converts nitrobenzene into phenol before the reduction starts, so the amine forms from the phenol
- It acts as an oxidising agent that removes hydrogen from the amine, which then forms a neutral product
-
Which functional group is reduced by tin and hydrochloric acid?
- The nitro group, -NO2, which is reduced to an amine group, -NH2
- The hydroxyl group, which is reduced to an alkane
- The benzene ring, which is reduced to a cyclohexane ring
- The carbonyl group, which is reduced to a primary alcohol
-
Why is excess ammonia used when preparing a primary amine from a halogenoalkane?
- Excess ammonia reduces the amine product to a primary alcohol, so a large excess is needed for conversion
- Excess ammonia makes it more likely that a halogenoalkane meets ammonia rather than the amine product
- Excess ammonia removes the halogenoalkane as a salt, so no amine can form until the excess is used up
- Excess ammonia makes the reaction exothermic, so the mixture heats up and the reaction proceeds more quickly
-
Why is alkali added at the end of the tin and hydrochloric acid reduction?
- To oxidise the tin, which removes it from the product
- To neutralise the acid and liberate the free amine from its salt, so it can be separated
- To precipitate the product as an insoluble salt that cannot be extracted
- To convert phenylamine back into nitrobenzene
-
Why are acyl chlorides preferred to carboxylic acids for making amides?
- Carboxylic acids cannot react with any nitrogen compound at all, so an acyl chloride is the only route available
- Acyl chlorides react with ammonia at room temperature, giving a good yield of amide without heat or a catalyst
- Carboxylic acids react with amines to give an ester instead of an amide, which is why acyl chlorides are used
- Acyl chlorides are less toxic and so avoid the formation of any by-product, which makes purification simpler
-
Calculate the mass of phenylamine from 0.0200 mol of nitrobenzene at 100% yield, using Mr = 93 for phenylamine.
- 3.72 g
- 0.93 g
- 2.46 g
- 1.86 g
-
Which conditions best prepare butylamine from butanenitrile?
- Heat with concentrated sulfuric acid to form butene, then add water to form the amine product
- Hydrolyse with dilute sodium hydroxide at room temperature, followed by acidification with dilute acid
- Reduce with lithium tetrahydridoaluminate in dry ether, then add water to work up
- Oxidise with acidified dichromate(VI) under reflux, then distil to isolate the amine product
-
A student makes an amide using aqueous ammonia but gets a low yield. Which reason is most likely?
- The dilute aqueous ammonia lets water hydrolyse the acyl chloride, competing with the amine reaction
- The ammonia is too basic, so it decomposes the amide product immediately after it is formed in the flask
- The amide is insoluble in water, so it escapes as a gas from the flask during the reaction
- The ammonia reduces the acyl chloride to an aldehyde, which then fails to react with the remaining reagent
-
For ethanoyl chloride with excess ammonia, how many moles of ammonia are needed per mole of amide formed?
- 0.5
- 2
- 1
- 3
-
What is the relative molecular mass of 1-bromobutane, C4H9Br, using Ar: C 12, H 1, Br 79.9?
- 152.9
- 120.9
- 135.0
- 136.9
Related quizzes
- Bonding and stability of benzene Quiz · 18A.1 · 20 questions
- Electrophilic substitution of benzene Quiz · 18A.2 · 20 questions
- Phenol and its bromination Quiz · 18A.3 · 20 questions
- Amines: identification, basicity and reactions Quiz · 18B.1 · 20 questions
- Polyamides and amino acids Quiz · 18B.3 · 20 questions
- Proteins and the peptide bond Quiz · 18B.4 · 20 questions
- Deducing formulae from data Quiz · 18C.1 · 20 questions
- Planning multi-step reaction schemes Quiz · 18C.2 · 20 questions
- Practical preparation and purification Quiz · 18C.3 · 20 questions
- Sub-atomic particles, isotopes and relative masses Quiz · 1.1.1 · 20 questions