3.11 Amines

Question 1

Name compound

Answer 1

Phenylamine

Question 2

Name compound

Answer 2

N-butyl-N-methylamine

Question 3

Name compound

Answer 3

2-aminobutane
Doesn’t have amine at end of name as N is in middle of chain

Question 4

Name compound

Answer 4

Tetramethyl ammonium chloride

Question 5

What can all amines act as

Answer 5

A base as they all have a lone pair on the nitrogen

Question 6

In nucleophilic substitution, how to produce a higher yield of primary amine

Answer 6

Use large excess of ammonia

Question 7

What does further substitution require in nucleophilic substitution

Answer 7

Excess haloalkane

Question 8

If a primary amine is the desired product, what could be done instead of nucleophilic substitution with ammonia

Answer 8

Use KCN, in ethanolic/alcoholic and aqueous conditions
- nitrile formed can then be reduced to an amine using hydrogen and a nickel catalyst

Question 9

How to produce phenylamine from benzene

Answer 9

Question 10

Use for
a) quaternary ammonium salts
b) amines

Answer 10

a) hair conditioner
b) dyes

Question 11

Answer 11

Step 1 :
Conc HNO3 & CONC H2SO4
Step 2 : HCl and Sn

Question 12

Reagent and conditions for step 4

Answer 12

Reagent : Chlorine
Conditions : UV light

Question 13

Amine A formed in step 2 and Amine B formed in step 5. Explain why yield of B in step 5 is less than yield of A in step 2

Answer 13

In step 5 further sub could occur forming other amines, while amine A is only one formed in step 2

Question 14

Explain why amine B is a stronger base then amine A

Answer 14

In B positive inductive effect of alkyl group releases electron density forwards N atom
Lone pair on N more available

Question 15

Explain why an amine connected to a benzene is a weaker base compared to one that has no benzene on it

Answer 15

Lone pair on electrons on N atom delocalises into benzene ring
Lone pair of electrons on N is less available
Weaker base compared to

Question 16

Skip part a

Answer 16

Amide and amine

Question 17

Outline mechanism for reaction of 2-bromopentane with an excess of ammonia

Answer 17

Question 18

Answer 18

Question 19

Answer 19

Electrophilic substitution

Question 20

Answer 20

+ 3H2 … + 2H2O

Question 21

Answer 21

Question 22

Why is ethanol chloride not used in industrial synthesis

Answer 22

It is corrosive as it forms HCl Fumes

Question 23

Phenylamine can be prepared by a process involving the reduction of nitrobenzene using tin and an excess of hydrochloric acid.
Give an equation for the reduction of nitrobenzene to form phenylamine. Use [H] to represent the reducing agent.
Explain why an aqueous solution is obtained in this reduction even though phenylamine is insoluble in water.

Answer 23

C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
C6H5NH2 present as ionic salt

Question 24

CH3CH2CH2Br reacts with NH3 to form CH3CH2CH2NH3
State Mechanism and state conditions
b) (c) Calculate the mass, in grams, of CH3CH2CH2NH2 produced from 25.2 g of CH3CH2CH2Br assuming a 75.0% yield.
Give your answer to the appropriate number of significant figures.

Answer 24

Nucleophilic substitution
excess NH3
b) Amount of CH3CH2CH2Br 25.2/122.9 (=0.205) (mol)
Amount of CH3CH2CH2NH2 M1 × 0.75 (= 0.154) (mol)
Mass CH3CH2CH2NH2 0.154 × 59.0 = 9.07g Must be 3sf

Question 25

CH3CH2CH2Br reacts with excess NH3 to form CH3CH2CH2NH2 in nucleophilic subsitution. However conditions changed and a compound with molecular formula C9H21N is produced. State name of this compound

Answer 25

N,N,N-tripropylamine

Question 26

Name and outline mechanism for reactions 3 Identify reagents and conditions used for this reaction

Answer 26

NaOH and ethanol

Question 27

Answer 27

- Base strength depends on availability of lone pair on Nitrogen atom - In E, Nitrogen is next to benzene ring, electrons are delocalised into the ring, lone pair on Nitrogen in E is less available - Both F & G, Nitrogen is next to an alkyl group so there is positive inductive effect, as electrons pushed towards Nitrogen atom - F & G have lone pair more available to accept protons - F has higher base strength as G has Nitrogen closer to benzene ring compared to F F strongest, E weakest

Question 28

Amine F can be prepared in a 3 step synthesis starting from methylbenzene Suggest structure of the two intermediate compounds For each step, give reagents and conditions only. Equations and mechanisms not required

Answer 28

Product of Step 1 : C6H5CH2Cl Product of Step 2 : C6H5CH2CN reagents and conditions : - step 1 : Cl2 & UV light - step 2 : KCN Alcoholic and aqueous - step 3 : H2 & Nickel catalyst

Question 29

Outline a mechanism for Step 1. Use RNH2 to represent 2,6-dimethylphenylamine

Answer 29

Question 30

Draw structure of compound S For each of steps 3 and 4, give a reagents and conditions, other than heat

Answer 30

S is CH3CH(CN)CH2CH3 Step 3 : KCN, aqueous and alcoholic Step 4 : H2 and Nickel catalyst

Question 31

Answer 31

a) electrophilic substitution b) Sn/HCl (Sn is tin) bii) CH3C6H4NO2 + 6[H] -> CH3C6H4NH2 + 2H2O biii) making dyes or making hair conditioner

Question 32

Answer 32

Quaternary ammonium salt/bromide Reagent : Bromoethane Condition : excess bromoethane Mechanism : nucleophilic substitution

Question 33

Answer 33

N-propylethanamide

Question 34

State one disadvantage of route A and route B

Answer 34

Route A : KCN involved as this is poisonous Route B : further substitution likely, - impure product may be formed

Question 35

How to distinguish between ammonia and methylamine

Answer 35

Measure pH with a meter Methylamine would have a higher pH (higher pH as it’s a stronger base due to positive inductive effect etc.)

Question 36

Name compound

Answer 36

N-methylpropanamide

Question 37

Answer 37

a) N,N-diethylamine b) F = CH3CH2Cl g = CH3CH2NH2 c) Step 1 : electrophilic addition, HCl Step 2 : NH3, nucleophilic substitution Step 3 : CH3CH2Cl, nucleophilic substitution

Question 38

Identify one organic impurity in product of step 3 and give a reason for its formation

Answer 38

Quaternary ammonium salt Further substitution will take place

Question 39

Draw structure of secondary amine formed as by-product in reaction 3

Answer 39

Question 40

Answer 40

H is ethanenitrile S is ethylamine Step 1 : KCN, aqueous and alcoholic Step 2 : H2 & Nickel catalyst

Question 41

Answer 41

Question 42

Explain why butylamine is a stronger base than ammonia

Answer 42

R group in butylamine increased well chronic density so has positive inductive effect Lone pair more available in butylamine

Question 43

Why is forming phenylamine from bromobenzene not same as forming ethylamine from bromoethane

Answer 43

Benzene ring repels nucleophiles such as NH3

Question 44

Equation for reaction of methylamine with water to produce an alkaline solution

Answer 44

Question 45

Suggest a substance that could be added to aqueous methylamine to produce a basic buffer

Answer 45

HCl

Question 46

Draw cation formed when methylamine reacts with large excess of bromoethane

Answer 46

Question 47

If you formed a primary amine and you want to prevent further substitution what do you need

Answer 47

Excess ammonia

Question 48

If you have currently have CH2CN and want to form CH2CH2NH2, what reagents and conditions do you need for this step

Answer 48

H2 & Nickel

Question 49

Answer 49

Use term higher atom economy instead of yield

Question 50

Answer 50

Tip : never give molecular formula Equation has NH4Br not HBr because in nucleophilic sub, for excess amine, another ammonia is introduced only to eliminate a hydrogen so it must be involved in equation * Double check equation as it is missing NH4Br

Question 51

Answer 51

For Cylclic secondary amine, be able to spot that NH2 On right has lone pair and bonds to bromine

Question 52

Define electronegativity

Answer 52

Relative tendency of an atom to attract a pair of electrons in a covalent bond

Question 53

Answer 53

Don't draw Br^- because it only asks for main organic species

Question 54

Answer 54

Question 55

Answer 55

Question 56

Answer 56

You got fooled by the question implying there is only one condition. Forget the grammar and unless it says one, input all other conditions

Question 57

Answer 57

Question 58

Answer 58

Lone pair on Nitrgon in P more available to accept a proton Due to greater positive inductive effect of alkyl groups

Question 59

Answer 59

Question 60

Draw structure of a primary amine with four carbon atoms that cannot be formed from a nitrile

Answer 60