Topic 18

Question 1

How are Polyamides made from

Answer 1

DiAmine (a compound with two amino groups
A diacarboxylic acid (a compound with two carboxylate groups) or a dioyl dichloride (a compound with two acrylic chloride groups)

Question 2

Making amines

Answer 2

Made from halogenoalkanes or nitriles

Haolgenolalkanes:
1-bromopropane reacts with excess ammonia to form propylamine.
1.
CH3CH2CH2BR + NH3 ——> CH3CH2CH2NH3 + Br-
2.
CH3CH2CH2NH3 + NH3 ——> CH3CH2CH2NH2 + NH4+
3.
CH3CH2CH2Br + 2NH3 ——> CH3CH2CH2NH2 + NH4Br

Amines from nitriles
CH3CH2CN + 4[H] ——> CH3CH2CH2NH2
Reducing agent

Question 3

Making Amides

Answer 3

Can be formed acyl chloride and amine

CH3COCl + NH3 —-> CH3CONH2. + HCL
Ethanoyl chloride. Ethanamide

Question 4

Friedel- Crafts Reaction (Alkylation). Name the electrophile and write the equation

Answer 4

Benzene + CH3Cl (Chloromethane) —> Benzene(CH3) + HCL
Electrophile the alkyl group CH3+
Generated from CH3Cl with an AlCl3 catalyst:
CH3Cl + AlCl3 —-> [AlCl4]- + CH3+

Question 5

Acylation

Answer 5

Benzene + CH3COCl(ethanoyl chloride) —> benzene (O=C-CH3) + HCl
The electrophile for this reaction is the CH3CO + ion. It is generated from CH3COCl with an AlCl3
Catalyst
CH3COCl + AlCl3 —> [AlCl4]- + CH3CO+

Question 6

Grignard Reagents (Reactions with carbonyl compound) and co2

Answer 6

R-MgBr + O=C=O ——> O=C-R-OH. + MgBrCl
Dry ether, dilute acid

Carbon dioxide
R- MgBr. + R’-C=O-R’. ——> R’——-C—R——-R’ —-OH. +MgBrCl

Question 7

(i) Explain whether phenol is likely to be less or more reactive than benzene with
the chloroalkane from (i) 3marks

Answer 7

Phenol is likely to be more reactive becuase
Lone pair on oxygen delocalises
Which increases electron density of the ring
Making phenol more susceptible to an electrophilic atttack

Question 8

Esterification

Answer 8

Carboxhylic acid +alcohol —> Ester and water
Ethanoic acid + ethanol —> ethylethanoate + water

Question 9

Halogenation

Answer 9

Carboxhylic acid + phosphorus (v) chloride —> acyl chloride + phosphorus trichloride oxide + HCL

Question 10

Reduction

Answer 10

Carboxhylic acid + LiAlH4 4[H]—> primary alcohol butane-1-ol + water

Question 11

Neutralisation

Answer 11

Carboxhylic acid+ salt—-> carboxylate salt + water

Question 13

This question is about amines.
Compare and contrast the basicity of phenylamine and butylamine.
(3)

Answer 13

Both are basic because they have a lone pair of electrons on the nitrogen atom which accepts a proton.
In C6H5NH2 the lone pair of electrons of the nitrogen atom becomes incorporated with the delocalised ring of electrons and so less able to accept a proton hence a weaker bass
The alkyl group is C4H9 is electron-releasing

Question 14

How aare grignard reagents made

Answer 14

You take a halogenoalkene react it with magnesium and dry ether and you make your grignard reagent

Question 15

How can carboxylic acids be made from grignard reagnets

Answer 15

reacting reagent with co2 to get carboxylic acid

Question 16

How can ketones and aldehydes be made from grignard reagnets

Answer 16

reacting reagent with carbonyl to get ketones

Question 17

How can primary alcohols be made from grignard reagent

Answer 17

adding methanal

Question 18

How can secondary alcohols be made from grignard reagent

Answer 18

adding aldehyde not methanal

Question 19

How can tertiary alcohols be made from grignard reagent

Answer 19

adding ketone

Question 20

Hydrolysis

Answer 20

tep 3: Hydrolyze the intermediate with water or acid to replace the magnesium halide with a hydroxyl group (-OH), forming an alcohol.
CH₃CH₂OMgBr + 𝐻2𝑂 → CH₃CH₂OH + Mg(OH)Br