Describe and explain the solubility of a dissolving polymer based on polyethenol in terms of its molecular structure – insoluble when very many or few internal hydrogen bonds;Soluble when an intermediate number of bonds.

document.docx 1

n

H H

H

C C O H

CH2 CHOH

ethenol monomer

H H

H O

C C

n

polyethenol polymer

Structure of Slime

O C

H C

H C

H

H

H

H

Opolyethenol

H

H

O

B

O

H

O

O

H

C

H C

H C

H

H

HH

O

O

H

H

polyethenolhydrogen bonds

borax

+

-

- +

-

+

H

Dissolving Polymers

Dissolving polymers are made to combat the risk of infection in hospitals. Laundry bags can be made from polyethenol, meaning that soiled laundry does not have to be handled, as the bags can be put straight into the washing machine.

Ethenol (CH2 = CH – OH), the monomer from which polyethanol is made is unstable and does not exist. It can be made from another plastic polyethenyl ethanoate by a process of ester exchange.

The extent of reaction can be controlled by adjusting either the temperature or the reaction time. The plastic’s solubility depends on the percentage of –OH groups present. The plastic is soluble in water as the –OH groups form hydrogen bonds with water. Different solubilities give the plastic different uses.

% OH Groups Solubility in Water Explanation100 – 99 Insoluble

99 -97 Soluble in hot water

96 – 90 Soluble in warm water

Below 90 Soluble in cold water

document.docx 2

Recognise and use the term addition polymerisation and predict the structural formula of the addition polymer formed from given monomers and vice versa.

The double bonds in alkenes can open up and join together to make long chains called polymers. The individual small alkenes are called MONOMERS.

monomer polymer

This is called ADDITION POLYMERISATION. The polyethene is made of many ethene monomers.

To find a MONOMER used to form an addition polymer, take the repeat unit and add a double bond

used to make water pipes, insulation on electric wires.

document.docx 3

H H

H H

CC

CC

CC

CC

n

n

side links show that both sides are attached to other units

Inside the bracket is the repeating unit

H CH3 H CH3 H CH3

H H H H H H

C C C C C C

polymer polypropene

CH3 H

H H

CC

CC

CH3 H

H H

CC

CC

repeat unit monomer propene

CI H

H H

CC

CC

chloroethene

n

CI H

H H

CC

CC

n

H H

H H

polychloroethene

(PVC polyvinylchloride)

document.docx 4

F F

F F

CC

CC

Tetra fluoroethene

n

F F

F F

CC

CC

n

Polytetra fluoroethene(PTFE)

Chemically inert and has non-stick properties

used as a coating for frying pans

Use relevant given date to interpret (and make predictions of) infra red spectra for organic compounds containing a limited range of functional group – hydroxyl, carbonyl and carboxylic acid groups.

To understand infra red (IR) spectroscopy, it is helpful to think of the bond between any two atoms as being like a vibrating spring. Each bond has its own natural frequency of vibration that depends on the types of atoms forming the bond and the type of bond (single, double, triple). When a molecule is exposed to IR radiation, each bond absorbs energy at a particular frequency causing it to vibrate more vigorously. Different bonds absorb different frequencies of IR radiation. Bonds in different places in a molecule absorb different frequencies too – so that the O – H group in an alcohol and the O – H group in a carboxylic acid absorb different frequencies.

This table shows what frequencies different bonds absorb

Functional Group

Where it’s found FrequencyWavenumber (cm-1)

C – H Most organic molecules 2800 – 3100

O – H Alcohols 3200 – 3550

O – H Carboxylic acids 2500 – 3300

C = OAldehydes, ketones,

carboxylic acids1680 - 1750

IR spectra have the following features:

The x – axis shows wavenumber, measured in cm-1, the scale usually starts at around 4000 wavenumbers on the left and descends to about 500 wavenumbers

The y-axis shows percentage transmittance – the baseline is at the top (100% transmittance) and the absorption signals or bands are downward troughs

The part of the IR spectrum below 1500 cm-1 is called the fingerprint region. Every compound has a distinctive fingerprint region which can help in its identification. Absorptions in this region are caused by the bonds in the molecule’s skeleton eg C - C

document.docx 5

An infrared spectrometer produces a graph that shows what frequencies of radiation the molecules are absorbing.

document.docx 6

The absorption of about 3000 cm-1 is caused by the C – H groups

Absorption about 1700 cm-1 shows there is a C = O group

Fingerprint region. Peaks show where radiation is being absorbed

Wavenumber is the measure used for the frequency – (it is 1/wavelength in cm)

The ‘peaks’ on IR spectra are upside-down – they point downwards

document.docx 7

H H

H H

CC CC

H H

H H

CCH C C

H

Recognise and write formulae for alkenes and use systematic nomenclature to name and interpret the names of alkenes.

ALKENES are a family of hydrocarbons which contain a double bond between carbon atoms – C = C –

The names of the molecules end in ‘ENE’

They have a general formula CnH2n

Number of Name Molecular formula Structural formulacarbon atoms

2 Ethene C2H4

3 Propene C3H8

4 Butene C4H8

document.docx 8

H H

H

CC

H C C

H

CC

H

HH

Naming Alkenes

1. Choose the longest chain of carbon atoms that contains the double bond.

2. Number the carbon atoms in the chain from the end nearest the double bond.

3. Pick the carbon atom with the lowest number to describe the position of the double bond

but – 1 – ene

CH3CH2CH2=CH2

but – 2 - ene

CH3CH CHCH3

CH3CH2CH2CH2CH CH2

document.docx 9

H H

H H

CC

H C C

H

CC

H

HH

4 3 2 1

H H

H

CC

H C C

H

CC

H

H

4 3 2 1

H

6 5 4 3 2 1

hex – 1 ene

CH2 CHCH2Br

document.docx 10

1 2 3

3 bromo prop – 1 - ene

propene CH3CH CH2

CH3CH CHCH2CH2CH2CH3 hept – 2 -ene

CH3CH CHCH3 2methyl but – 2 - ene

CH3

CH2 CHCH2CH CHCH2CH3

hepta – 1,4 - diene

CH3 C C C CH3

2, 4 dimethylpent – 2 - ene

H

CH3 H CH3

cyclopentene

Recognise members of the following homologous series aldehydes, ketones, carboxylic acids.

Aldehydes and ketones are carbonyl compounds.

They have the functional carbonyl group C = O.

Their general formula is CnH2nO

Aldehydes have a hydrogen and one alkyl group attached to the carbonyl carbon atom

eg

Ketones have two alkyl groups attached to the carbonyl carbon

eg

document.docx 11

H

C O

CH3

(hydrogen)

carbonyl carbon

(alkyl group) methyl

Ethanal

R C

O

H

alkylgroups

CH3

CH3

C O

carbonyl carbon

propanone

R

C O

R

R

C O

R

Ketones

document.docx 12

CH3

C O

CH3

propanone

CH3CH2

C O

CH3

butanone

CH3CH2CH2

C O

CH3

Pentan-2-one

Aldehydes

R C O

document.docx 13

H

C O

H

methanal

H

C O

CH3

ethanal

H

C O

CH3CH2

propanal

H

H

C O

CH3CH2CH2

butanal

Carboxylic Acids

R C OH

document.docx 14

O

H C OH

O

methanoic acid HCOOH

CH3 C OH

O

ethanoic acid CH3COOH

CH3CH2C OH

O

propanoic acid CH3CH2COOH

CH3CH2CH2 C OH

O

butanoic acid CH3CH2CH2COOH

Explain and use the terms thermoplastic, thermoset and co-polymer

Thermoplastic – is a polymer without cross-links (covalent bonds) between the chains. The intermolecular forces between the chains are weak. These intermolecular forces between the chains are weak. These intermolecular forces of attraction can be overcome by warming. The polymer chains can slide over each other and the polymer can be deformed ie change shape. On cooling the weak intermolecular forces between the polymer chains can reform and the thermoplastic holds its new shape – e.g. polyethene,

Thermoset – this type of polymer has extensive cross linking between the different polymer chains. The cross-links (covalent bonds) between the chains are much stronger than intermolecular forces of attraction found in thermoplastics. The cross-links (covalent bonds) cannot be broken by warming. The polymer chains cannot move relative to each other and the polymer cannot change shape. If heating continues the polymer just chars and burns e.g. Bakelite.

A co-polymer is made when two different monomers become incorporated into a polymer chain.

e.g. ethene and propene.

document.docx 15

H H

H H

CC

C C

H

CC

H

HH

CC

H

+

+

H

(A) (B)

H H

H H

CC

C C

H

CC

H

CH3H

H

H H

CC

C C

H

CC

H

CH3H

ethene propene

AC

A B BC

H

H

H

H

OC C

H

HC

H

ethanol

(excess)

Explain and use the term elimination reaction

Describe and explain the following properties of alcohols- oxidation of alcohols to carbonyl compounds (aldehydes and ketones) and

carboxylic acids with acidified potassium dichromate (VI) solution, including the importance of the conditions (reflux or distillation) under which it is done

- dehydration of alcohols to form alkenes using heated AI2O3 or refluxing with concentrated H2SO4

Alcohols can be oxidised using potassium dichromate (VI) K2Cr2O7 acidified with H2SO4.

The reaction conditions for the oxidation of alcohols are heating the alcohol under reflux with acidified potassium dichromate.

Primary alcohols are oxidised to aldehydes and then to carboxylic acids

Secondary alcohols are oxidised to ketones only

Tertiary alcohols are not oxidised

Oxidation of primary alcohols

Primary alcohols are oxidised to aldehydes if the alcohol is in excess

If an excess of oxidising agent is used primary alcohols are oxidised to carboxylic acids.

document.docx 16

H

H

H

C C

O

H

ethanal

(aldehyde)

Cr2O72-

H+

During the reactions the acidified orange dichromate (VI) ions are reduced to green chromium ions

Oxidation of primary alcohol by oxidising agent K2Cr2O7 in dilute H2SO4

Cr2O72−(aq ) + 8H(aq )

+ +3CH 3CH 2OH (C )→Cr (aq )3+ +7H2O(C )+3CH 3CHO(C )

document.docx 17

H

H

H

C C

O

H

ethanal

Cr2O72-

H+

H

H

H

OC C

H

HC

H

ethanol

H

H

H

C C

O

ethanoic acid

(carboxylic acid)

Cr2O72-

H+

O H

Cr2O72-

orange

Cr3+

green

dichromoate

(ORANGE)

ETHANOL

alcohol

chromium ions

(GREEN)

aldehyde

ETHANAL

Oxidation of a primary alcohol, ethanol to an aldehyde, ethanal.

If the alcohol is in excess of the acidified potassium dichromate (VI). The mixture is gently heated in the apparatus below, but with the receiver cooled in ice to reduce evaporation of the ethanal.

Ethanal (boiling temperature 21oC) vaporises as soon as it is formed and distils off. This stops it from being oxidised further to ethanoic acid. Unreacted ethanol remains in the flask.

document.docx 18

DISTILLATION APPARATUS

Oxidation of a primary alcohol, ethanol to a carboxylic acid, ethanoic acid.

For a complete reaction, the acidified potassium dichromate is in excess. The mixture is heated in the apparatus shown below.

document.docx 19

Reflux – means that vapour condenses and drips back into the reaction flask.

While the reaction mixture is refluxing, any ethanol or ethanal vapour will condense and drip back into the flask, eventually, it is all oxidised to ethanoic acid.

After refluxing, the ethanoic acid (boiling temperature 118oC) can be distilled off by rearranging the apparatus to a distillation apparatus.

Oxidation of secondary alcohols

Elimination Reactions

Elimination reactions are ones in which a small molecule leaves the parent molecule. In the case of alcohols, this molecule is water so elimination reactions of alcohols are always dehydrations.

Alcohols can be dehydrated by refluxing with excess concentrated sulphuric acid or by passing alcohol vapour over heated aluminium oxide AI2O3. An alkene is formed.

eg

document.docx 20

H

H

H

C C

propanone

(ketone)

Cr2O72-

H+

H

H

H

CC C

H

HC

OH

propan – 2 - ol

H

H

H

H

C H

O

H

H

H

C C

propene

(alkene)

AI2O3

600K

H

H

H

CC C

H

HC

H

propan – 1 - ol

H

OH

H

H

C

H2O

H

+ H2O

Recognise where E/Z isomerism occurs explaining it in terms of lack of free rotation about C = C bonds when there are two different groups on each carbon.

Draw and interpret diagrams to represent E/Z isomers for alkenes which have the same groups on both sides of the double bond (E diagonally opposite sides of C = C bond; Z- same side of C = C bond).

In such molecules describe E as ‘TRANS’ and Z as ‘CIS’ and extend this CIS-trans nomenclature to other, more complicated, alkenes.

E/Z isomerism is a form of geometric isomerism OR stereoisomerism. Geometric isomers have the same structural formula but a different arrangement of atoms of groups in space. This is because there is a lack of rotation of atoms or groups attached to the double bonded carbon atoms. When double bonded carbon atoms each have two different atoms or groups attached to them, - an E-isomer and a Z-isomer is produced.

For example the double bonded carbon atoms in each but-2-ene molecule, each have an H and a CH3 group attached.

document.docx 21

H

H3C

C C

H

CH3

(trans-but-2-ene)

E-isomerE-but-2-ene

When the same groups are ACROSS the double bond – it is an E-isomer

H3C

H

C C

H

CH3

(CIS-but-2-ene)

Z-isomerZ-but-2-ene

When the same groups are BOTH ABOVE or

BOTH BELOW the double bond –

it is an Z-isomer

E/Z isomerism is sometimes called CIS-TRANS ISOMERISM. Where CIS means the Z isomer, and ‘TRANS’ means the E-isomer. So E but-2-ene can be called trans-but-2-ene, and Z-but-2-ene can be called CIS-but-2-ene.

E/Z isomerism is one type of STEREOISOMERISM. In stereoisomerism, the atoms are bonded in the same order but arranged differently in space in each isomer.

You can use the CIS-trans system if there are more than two different groups (other than hydrogen atoms) attached around a double bond.

This could be trans-I-bromo-I-fluoropropene because the Br and CH3 are on opposite sides or it could be CIS-I-bromo-I-fluoropropene because the F and CH3 are on the same side.

The E/Z system keeps on working.

Each of the groups, linked to the double bonded carbons is given a PRIORITY. If the two carbon atoms have their ‘higher priority group’ on opposite sides, then it is an E-isomer.

If the two carbon atoms have their ‘higher priority group’ on the same side, then it is a Z isomer (you do not need to know the rules for deciding the order of these priorities).

document.docx 22

F CH3

Br H

C C

F CH3

Br H

C C

E-1 bromo-1-fluoropropene

Br CH3

F H

C C

Z-1- bromo-1-fluoropropene

In the E/Z system, Br has a higher priority than F, so the names depend on where the Br atom is in relation to the CH3 group (which has a higher priority than the H atoms).

document.docx 23

Recall the addition reactions of alkenes with the following

bromine to give a dibromocompound and the use of this as a test for an alkene

hydrogen bromide to give a bromocompound hydrogen in the presence of a catalyst to give an alkane

- nickel catalyst with heat and pressure or platinum catalyst at room temperature and pressure

water in the presence of a catalyst to give an alcohol (concentrated H2SO4 and then add water or steam/H3PO4 under heat and pressure)

Use bromine water to test for C = C bonds.

When you shake an alkene with orange/brown bromine water, the solution quickly decolourises. Bromine is added across the double bond to form a colourless dibromoalkane.

eg

The mechanism is electrophilic addition.

document.docx 24

H

H

C C

ethene

H

+ Br2

Br

H

H

C C

1,2 dibromoethane

colourless

H

Br

H H

orange

brown

H

H

C C

H

H

Br+

Br-

H

Br

C C+

Brx-

H

H H

H

Br

C C

H

H H

Br

carbocation intermediate

x

Hydrogen halides, HCI, HBr and HI, add across a double bond to form a halogen alkane.

Example hydrogen bromide and ethane.

This happens by electrophilic addition.

In a symmetric alkene like ethene or but-2-ene only one product is formed.

If HBr adds to an asymmetric alkene like propene these are 2 possible products.

document.docx 25

H H

H H

C C

H

H

C C

H

H

H+

Br-

H

H

C +C

Br-

H

H

H

H

H

C C

H

H Br

H

+ HBr

hydrogenbromide

ethene

H

H

C C

H

Br

H

H

bromoethane

x

x

H

H

C C

H

CH3

H+

Br-

H

C C

Br

H

H

Br

C C

CH3

H H

H

x

H CH3

H

Adding hydrogen to C = C bonds produces alkanes.

Ethane will react with hydrogen gas to produce ethane. The reaction needs a nickel catalyst and a temperature of 150oC.

Margarine is made by ‘hydrogenating’ unsaturated vegetable oils.

Adding water across a C = C bond produces alcohols.

The reaction is used industrially to make alcohols and is carried out with steam at a suitable temperature 300oC and a pressure of 60-70 atm, using an acid catalyst such as phosphoric acid, H3PO4.

document.docx 26

H

C C

H

H

H H

H

C C

H

H H

H

+H2

Ni150 ∘C

ethene ethane

H

C C

H(g)

H H

+ H2O(g)

ethene

H3PO4

70 atm300oC

CH3CH2OH(g)

ethanol

Recall the difference between primary, secondary and tertiary alcohols from their structures, and identify examples of them.

1. Monohydric Alcohols – alcohols with one OH group

document.docx 27

The carbon C bearing the OH group is in turn bonded to one alkyl group in this case CH3CH2-, ethyl group

The carbon C bearing the OH group is bonded to two alkyl groups e.g. two CH3, methyl groups

Primary Alcohol (1o)

H

C C

H

H

H H

O-HC

H

CH3CH2CH2OH

propan-1-ol

Secondary Alcohol (2o)

H

C C

O

H

H

H H

HC

H

H

H

CH3CH(OH)CH3

propan-2-ol

document.docx 28

The carbon C bearing the OH group is bonded to 3 alkyl groups e.g. three methyl CH3 groups

Tertiary Alcohol (3o)

H

C C

O

H

H

H

H

HC

H

H

2 methyl propan-2-ol

C HH

trihydric alcohol

Polyhydric Alcohols – alcohols with more than one OH group

H C

OH

H

CC

OH

H

CH2OHCHOHCH2OH

propane-1,2,3-triol

dihydric alcoholH C

OH

H

HC

OH

H

CH2OHCH2OH

Ethane-1,2,-diol

H

H

OH

document.docx 29