Hydrocarbons Qs

AQA

AS - Module 3

Foundations of Organic Chemistry

Foundation Organic Chemistry

2006/7

j:\courses\as and a2\module #3 - introduction to organic chemistry - practical\module book\2006 module #3 - introductory organic.doc


1

M3 - Nomenclature & Functional Groups

All organic compounds contain carbon, usually in the form of a number of carbon atoms in a chain, so the basis of the compound's name is the number of carbon atoms in the molecule's chain. The simplest organic compounds are hydrocarbons. Hydrocarbons contain only two elements, hydrogen and carbon. A saturated hydrocarbon or alkane is a hydrocarbon in which all of the carbon-carbon bonds are single bonds. Each carbon atom forms four bonds and each hydrogen atom forms a single bond to a carbon.

H H H | | |

H—C—C—C—H | | | H H H

These compounds for what is known as a homologous series. This is a series of compounds in which the adjacent members differ in their formula by a –CH2- unit. Since each member of a series contains similar bonds they exhibit similar chemical properties. However, they show a gradual change in physical properties.

Using the following table draw out the stick structure for each of the following:

Methane Ethane Propane


2

Butane Pentane Hexane

Carbon Name Molecular Formula

Structural Formula

1 Methane CH4 CH4

2 Ethane C2H6 CH3CH3

3 Propane C3H8 CH3CH2CH3

4 Butane C4H10 CH3CH2CH2CH3

5 Pentane C5H12 CH3CH2CH2CH2CH3

6 Hexane C6H14 CH3(CH2)4CH3

7 Heptane C7H16 CH3(CH2)5CH3

8 Octane C8H18 CH3(CH2)6CH3

9 Nonane C9H20 CH3(CH2)7CH3

10 Decane C10H22 CH3(CH2)8CH3

Above is a table that gives the names of the straight chain alkanes. It's a good idea to commit this table to memory. The general formula for an alkane is:

CnH2n+2

where n is the number of carbon atoms in the molecule. There are several ways of writing a condensed structural formula. For example, butane may be written as:


3

CH3CH2CH2CH3 or CH3(CH2)2CH3 or C4H10

Rules for Naming Alkanes

o The parent name of the molecule is determined by the number of carbons in the longest chain.

o In the case where two chains have the same number of carbons,

the parent is the chain with the most branches. o The carbons in the chain are numbered starting from the end

nearest the branches first. o When more than one of a given branches is present, a prefix is

applied to indicate the number of branches. Use di- for two, tri- for three, tetra- for four, etc. and use the number assigned to the carbon to indicate the position of each branches.

Name each of the following compounds and state which compounds,

if any, are the same. 1)

CH3 CH CH CH3

CH3 CH3

CH3 CH CH CH3

CH3 CH3

CH3 C

CH3

CH3

CH2 CH3

CH3 CH CH CH3

CH3

CH3 A

B C D

2)

CH3 CH2 CH CH CH3

CH3CH3

CH3 CH2 C CH2

CH3CH3

CH3

CH3 CH2 CH CH

CH3 CH3

CH3

CH3 CH CH CH2 CH3

CH3

CH3 E

F G H


4

3)

CH3 CH2 CH2 CH CH3

CH2

CH3

CH3 CH2 CH CH2 CH2 CH3

CH3

CH3 CH

CH3

CH2 CH2 CH2 CH3

C H 3 C H 2 C H2 CH CH2 CH3

CH3

I

J K L

4)

CH3 CH CH

CH3 CH2

CH3

CH2 CH2 CH3

CH3 CH2 CH CH2 CH CH3

CH3

CH3

CH2

CH2

CH3 CH2 CH CH2 CH CH3

CH3

CH2

CH2

CH3

CH3 CH CH

CH2

CH3

CH2 CH2 CH3

CH3

M

N O P

Compounds with Functional Groups Organic compounds, as well as having a carbon chain (with hydrogen atoms attached) can also usually have a functional group. This is the usual point of reaction of a molecule. For example, all alcohols have the -OH group in the molecule. Compounds having the same functional group have similar physical and chemical properties. They like the examples above also form a ‘family’ of compounds called an homologous series, like most families, they have a family name (or actually a name ending). Alcohols all have names ending in -ol. Putting the two parts of the name together, we can see that the alcohol with 2 carbon atoms will be called Ethanol. The Eth tells us that there are 2 carbons in the chain and the -ol ending tells us that the compound is an alcohol.


5

There are many other homologous series, these include:

Alkanes - with a ‘family’ name ending -ane

Alkenes - with a ‘family’ name ending -ene

Aldehydes - with a ‘family’ name ending -anal

Ketones - with a ‘family’ name ending -anone

Carboxylic Acids - with a ‘family’ name ending –anoic Acid

Amines - with a ‘family’ name ending -ylamine

The aldehydes and ketones are very similar compounds and are often grouped together under the general name of Carbonyl compounds. Alkyl Halides (Halo-alkanes) Halo-alkanes gain the prefix of the name of the halo group i.e. Chloro, Fluoro, Bromo etc., this is followed by the corresponding alkane chain name. Notice from the examples that the position of the halo group must be included if it is not on C1 (carbon number 1). Example:

CH3-CH2-CHCl CH3-CHCl-CH2

Chloropropane 2-Chloropropane

Draw the structure of: Iodohexane 1,1,1-Trichloroethane 1,1-Dichloro-2-fluorobutane


6

Alkenes The general structure within all alkenes is:

C CCH3

CH3CH3

CH3

Alkenes are known as being unsaturated since not all of the carbon atoms are completely saturated with hydrogen. If a long alkyl chain had many C=C double bonds it would be known as being polyunsaturated, as in polyunsaturated oils. Alkenes drop the ending –ane of the corresponding alkane and become –ene, notice from the examples that the position of the double bond must be included if it is not between C1=C2. Example:

CH3-CH2-CH2-CH=CH2 CH3-CH2-CH=CH-CH3

Pent-1-ene Pent-2-ene

Draw the structure of:

Propene Hexa-1,3-diene

The general formula for an alkene is:

CnH2n Alcohols Alcohols drop the ending –e of the corresponding alkane and become –ol, notice from the examples that the position of the double bond must be included if it is not on C1 (carbon number 1).


7

Example:

CH3-CH2-CH2-OH CH3-CH(OH)-CH3

Propanol Propan-2-ol

(propan-1-ol) If alcohol functional groups are combined with other functional groups as shown in the example the alcohol becomes known as hydroxy and is prefixed by the number of the carbon to which it is attached. Example:

CH3CH2CH(OH)CH2COOH

3-hydroxypentanoic acid

Draw the structure of: Propan-2-ol Propan-1,2,3-triol Ethane-1,2-diol Aldehydes Adehydes drop the ending –e of the corresponding alkane and become –al, notice from the examples that the position of the aldehyde functional groups is always at the start of a chain (C1) and as such requires no number. Also note that the carbon of the functional groups is counted as part of the chain.

OH

R1


8

Draw the structure of :

Butanal Methanal Ketones Ketones drop the ending –e of the corresponding alkane and become –one, notice from the examples that the position of the Ketone functional groups is never start of a chain (C1) and as such requires a number to identify its position in a chain with the only exception being propanone.

OR2

R1


Buatan-2-one Hexan-2,4-dione Carboxylic Acids Carboxylic Acids drop the ending –e of the corresponding alkane and become –anoic acid, notice from the examples that the position of the carboxylic acid functional group is always at the start of a chain (C1) and as such requires no number. Also note that the carbon of the functional groups is counted as part of the chain.


9

ROH

O

Example: CH3CH2CH2COOH

Propanoic acid


2-Hydroxy propanoic Acid Amines Amines drop the ending –ane of the corresponding alkane and become –yl amine, notice from the examples that more than one alkyl chain can be attached to the amine.

R NH2

Examples: CH3CH2NH2 (CH3CH2)2NH

Ethyl amine Diethyl amine


Diethyl-hexyl amine Triethyl amine


10

Questions 1) Name each of the following organic compounds:

CH3CH2CH2CH3 ……………………………………………………………

CH3CHBrCHBrCH3 ……………………………………………………………

CH3CH2CH2C(CH3)3 ……………………………………………………………

CH2=CHCH2CH2CH3 ……………………………………………………………

CH3CH(OH)CH(CH3)2 ……………………………………………………………

CH3CHClCH(CH3)2 ……………………………………………………………

(CH3)2CHCHO ……………………………………………………………

(CH3)2C(OH)CH2CH3 ……………………………………………………………

CH3CHO ……………………………………………………………

(CH3)2CHCOOH ……………………………………………………………

2) Draw the structures for the following compounds:

2-methylpentane 2,3-dimethylhexane

pent-2-ene 3-methylbut-1-ene

butanoic acid 2-methylpropan-2-ol


11

butanal butanone

1-chloro-2,2-dimethylbutane 2,2-dimethylbutanoic acid

4-hydroxypentanal 3-hydroxybut-1-ene

Self Study Practice Draw the structural formula of each of the following organic compounds.

a) 3-methylbutanone

b) 4-ethylhex-2-ene

c) 2-bromo-2-iodobutane

d) 4-hydroxybut-1-ene

e) 1,2-dichloropent-1-ene

f) 2,2,3-trimethylbutane

g) 3-hydroxybutanoic acid

h) 3-chlorobutanal

i) 2-chloro-3-hydroxypentanoic acid


12

M3 - Homologous Series

It is not only the formulae of a homologous series of compounds that can be worked out but also the physical properties of the compounds. There is a trend in physical properties as the numbers of carbon atoms in the chain increases and the relative molecular mass of the compound increases: Boiling Point : The more carbon atoms present in the chain, the

higher the boiling point: Name Boiling Point (k) Density (g/cm

3)

Methane(g) 109.1 0.466(liquefied) Ethane(g) 184.5 0.572(liquefied) Propane(g) 231.0 0.585(liquefied) Butane(g) 272.6 0.601(liquefied) Pentane(l) 309.2 0.626 Hexane(l) 342.1 0.660 Densities: As the chain length increases. so does the density (see table

above). Solubilities: for compounds which are soluble (or miscible) with water

such as alcohols and carboxylic acids, the compounds become progressively less soluble as the molecules become larger.

Viscosity: The compounds become MORE viscous as the relative

molecular mass increases. (The liquids get ‘thicker’, becomes more ‘syrupy’)

As far as the chemical properties of the compounds are concerned, these depend on the functional group. All compounds within a homologous series have the same functional group, so they will all have similar chemical properties. Therefore, if you know the chemical properties of one


13

compound, you know the chemical properties of all of the compounds in the series. The only thing to remember is, as the length of the carbon chain increases, there is less likelihood of an attacking species hitting the functional group reduces, so the reactions become slower. The table below shows some physical data (density and boiling

points) for some alkanes. 1) Complete the table

C20 = eicosane

Plot a graph (or graphs) to show how the physical properties change

with increasing carbon chain length. The number of carbons should be on the horizontal axis. Label any unusual points on your graph.


14

How does the size (chain length) affect the boiling points and

densities ? …………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

How does the shape (branching) of the molecule affect these

physical properties ? Consider each of the following in turn: Straight chain alkanes

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

Branched chain alkanes (only one branch)

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

Branched chain alkanes (more than one branch)

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

…………………………………………………………………………………………………………………………..

………………………………………………………………………………………………………………………….


15

Naming Members of Homologous Series In most cases, the ‘family name’ of the name of an organic compound is indicated by the last few letters of the name (the suffix). However, sometimes, the family name is at the start of the name (the prefix): Homologous series

‘Family name’ Functional Group

Example

Alkanes

suffix - ane

none

methane CH4

Alkenes

suffix -ene

C = C

Ethene C2H4

Haloalkanes

prefix chloro, -

bromo- or iodo-

-Cl or -Br or

-I

Bromoethane C2H5Br

Alcohols

suffix -ol

prefix hydroxy-

- OH

Methanol CH3OH 2- hydroxyethanoic acid

Ethers

prefix alkoxy-

-O-R

Methoxymethane CH3-O-CH3

Aldehydes

suffix -al

-C = O | H

Methanal HCHO

Ketones

suffix -one

\ C=O

/

Butanone C2H5COCH3

Carboxylic acids

suffix -oic acid

-COOH

methanoic acid CH3COOH

Amines

prefix amino-

suffix -amine

-NH2

Aminoethanoic acid H2NCH3COOH ethylamine C2H5NH2

Amides

suffix - amide

-CONH2

ethanamide CH3CONH2

Nitriles

suffix -nitrile

-CN

ethanenitrile CH3CN


16

M3 - Isomerism

Isomerism (Definition)

Isomerism is the phenomenon where 2 or more compounds with the same molecular formula have different arrangements of atoms. Structural Isomerism This occurs when compounds have the same molecular formula but different structures. i) Chain Isomerism – Is where the compounds have different carbon

skeletons. Example: C4H10

Name the compounds : a) CH3CH(CH3)CH(CH3)CH3 .............................................................................. b) CH3C(CH3)2CH2CH3 .............................................................................. c) CH3C(CH3)2CH2C(CH3)2CH3 ..............................................................................


17

Draw the structure of the following compounds and tell me what you notice about them:

3-methylhexane 2,2-dimethylpentane 2,2,3 -trimethylbutane

Draw all of the possible the chain isomers for C5H12

II) Positional Isomerism –Same carbon skeleton but functional groups

attached to different carbon atoms. Example:

Propan-1-ol Propan-2-ol Example:

2 Chloro methyl benzene 3 Chloro methyl benzene 4 Chloro methyl benzene


18

III) Functional Group Isomerism – Same molecular formula but different functional groups.

Example: C3H6O

Aldehyde Ketone Stereo-isomerism This is shown by isomers with similar structure BUT different spatial orientation (3D shape). Geometric Isomerism

This occurs in compounds containing double bonds.

Why is this (use the diagram above to help you) ? ………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

Compounds with the functional groups on the same side are called cis- and on opposite sides trans-


19

Make and draw using the molymod™ kits both of the geometric isomers of but-2-ene.

Questions 1) Draw the structural formula of each of the following compounds and

identify its functional group(s). a) 2-chlorobutane e) pentan-2-one

b) 3-methylpent-2-ene f) butanoic acid

c) butanal g) ethylamine

d) 2,3-dimethylbutan-2-ol h) 3-chloro-3-ethyl-4-hydroxy-4-

methylpent-1-ene

2) Name the following organic compounds.

a)

CH3

CH3

CH3

CH3

g)

CH3

OCH3

CH3

CH3

b)

CH3CH2C(CH3)2CH2CH3

h)

CH3CH2CH2CH(OH)CH3

c)

(CH3)2C=CH2

i)

CH3CH2CH2CH2CHO

d)

CH3

BrOH

j)

O

HCH3

CH3


20

e)

CH3

CH3 OH

O

k) CH3CH(C2H5)CH2CH2COOH

f)

CH3

CH3

CH3

CH3

OHCH3

l)

CH3CH2COOCH2CH2CH2CH3

3) Draw the structure and name each of the following compounds:

a) a branched chain alkane with five carbon atoms

b) an alkene with four carbon atoms that does not have a geometric isomer

c) a bromoalkane with three carbon atoms

d) a straight chain aldehyde with four carbon atoms

e) a branched chain ketone with five carbon atoms

f) a cycloalkane with six carbon atoms

g) an alkene with five carbon atoms that does have a geometric isomer

h) an amine with two carbon atoms

i) an alcohol with three carbon atoms with a chlorine atom on a carbon

atom that is not adjacent to the alcohol group carbon

4) Draw the structure and name each of the compounds with the formula

C7H16.

5) Draw the structure and name one aldehyde and one ketone with

formula C5H8O.


21

M3 - Petroleum and Alkanes

Deposits of crude oil and natural gas usually occur together and they are formed by the slow decay of marine animals and plants, over millions of years, under heat and pressure in the absence of air. Although the exact composition of crude oils varies around the world, all are a complex mixture consisting mainly of alkanes (including cycloalkanes, some aromatics and other compounds containing some Sulfur and Oxygen). Crude oil has no use in its raw form, so to provide useful products its components must be partly separated (and if necessary modified) - the separation uses the differences in the physical properties of alkanes. Fractional distillation The compounds in crude oil have different boiling points and this is used to separate them by fractional distillation at an oil refinery. C-H bonds are virtually non-polar, so there are only van der Waals' forces between molecules. As the alkane chain gets longer the melting and boiling points increase due to greater van der Waals' forces. Note that a branched chain alkane has a lower boiling point than the straight chain alkane of the same formula, and the more branching the lower the boiling point (less branching allows closer contact between molecules and so stronger van der Waals’ forces). The basic idea is that crude oil is vaporised at high temperature passed into a tower, which is hot at the bottom and cold at the top. As it rises up the tower it condenses at different levels. It is collected at the height it condenses.


22

Diagram:

FUEL GASESCalor gas, LPG

25°C

GASOLINEpetrol for cars

40°C

NAPHTHApetrochemicals

110°C

KEROSENEjet fuel

200°C

DIESEL (GAS OIL)fuel for diesel engines

300°C

RESIDUEdistilled further at lowerpressures to give FUEL OIL,LUBRICATING OIL, WAXESand BITUMEN

350°C

> C20

C15-C19

C11-C15

C7-C14

C4-C12

C1-C4

vaporisedcrude oil

The residue from the primary distillation (first distillation) contains useful substances, such as fuel oil, lubricating oil, waxes and bitumen, that boil above 350°C at atmospheric pressure. However, they would decompose at such high temperatures, so they separated further by distillation at lower pressure (vacuum distillation). The fractions are: NAME OF FRACTION Bpt/ oC MAJOR USE chain length LPG (Refinery Gases) up to 25 Calor gas, camping gaz 1 - 4 Petrol (Gasoline) 40 - 100 petrol 4 - 12 Naphtha 100 - 150 feedstock for the chemical

industry 7 - 14

Kerosene (Paraffin) 150 -250 jet aircraft fuel 11 - 15 Gas Oil (Diesel) 220 - 350 heavy vehicle fuel 15 - 19 Mineral Oil over 350 lubricating oil 20 - 30 Fuel Oils over 400 ships + power stations 30 - 40 Wax And Grease over 400 candles, polish etc. 40 - 50 Bitumen over 400 roofing, roadworks over 50

As the C chain gets longer, the hydrocarbons:

• become more viscous

• harder to ignite

• less volatile

• have higher boiling points


23

Cracking The petroleum fractions with shorter C chains (e.g. petrol and naphtha) are in more demand than larger fractions. To make you of excess larger hydrocarbons and supply demand for shorter ones, longer hydrocarbons are cracked. C-C bonds are broken. In general:

longer alkanes → shorter alkanes + alkenes (+ hydrogen) Example:

C14H30 → C7H16 + C3H6 + 2 C2H4

C14H30 → C12H24 + C2H4 + H2

The mixture of products from cracking are separated by fractional distillation. The shorter alkanes are in great demand as fuels making petrochemicals. The alkenes are used to make polymers, such as ethene used to make poly(ethene). Thermal cracking The alkanes are heated to high temperatures (up to 900°C) and pressures (up to 7000 kPa). Thermal cracking produces a large proportion of alkenes. The higher the temperature, the chain breaks closer to the end, giving low Mr alkenes. The reaction has a free radical mechanism.

H H H H | | | | —C—C— —C. . C— | | | | H H H H


24

The process begins with the breakage of a C-C bond (C-C bonds are weaker than C-H), producing two alkyl free radicals (homolytic fission of the C-C bond). Catalytic cracking (cat cracking) Cracking can be carried out at a lower temperature of about 450°C and much lower pressures in the presence of a catalyst. The catalysts are usually zeolites (crystalline aluminosilicates). Catalytic cracking produces mainly branched alkanes (useful for petrol), cycloalkanes and aromatics. The mechanism is different and involves the formation of carbocations. Combustion Alkanes readily burn in the presence of oxygen, this combustion of alkanes being highly exothermic, explaining their use as fuels. The products of complete combustion are CO2 and H2O. Example:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H = -890 kJ/mol

In general:

CxHy + (x + y/4) O2 → x CO2 + y/2 H2O If there is not enough oxygen then incomplete combustion occurs, producing carbon monoxide, which is very toxic and/or C: Example:

CH4(g) + 3/2 O2(g) → CO(g) + 2 H2O(l)

CH4(g) + O2(g) → C(s) + 2 H2O(l)


25

Sulphur containing impurities are found in petroleum fractions which produce SO2 and SO3 when they are burned, which lead to acid rain. The internal combustion engine Petrol (gasoline fraction, which consists of liquid alkanes) is used in the internal combustion engine where the alkanes are vaporised and combusted with air. A number of substances are found in the exhaust gases: carbon dioxide, CO2

water vapour, H2O carbon monoxide, CO, (from incomplete combustion) hydrocarbons (some do not burn) nitrogen oxides, NOx, (from the reaction between N2 and O2 in the high temperatures) Catalytic converters These remove CO, NOx and unburned HC's from the exhaust gases, turning them into CO2, N2 and H2O.

2 CO + 2 NO → 2 CO2 + N2

hydrocarbons + NO → CO2 + N2 + H2O

Converters have a ceramic honeycomb coated with a thin layer of catalyst metals (Pt, Pd, Rh) – to give a large surface area.


26

M3 - Organic Chemistry Overview

Mechanisms Are a way of describing the order of events when chemical react with each other (like a cartoon). They depend on :

1) The nature of the substance itself 2) The nature of the attacking species 3) The condition under which the reaction is taking place

There are only four basic types of reaction in organics chemistry:

1) Substitution – Involves the direct displacement of an atom or group by another atom or group.

CH4 + Cl2 CH3Cl + HCl

2) Addition reaction – Involve the reaction between unsaturated

species and another reactant to give one product.

CH2=CH2 + Br2 BrCH2-CH2 Br 3) Elimination Reactions – The opposite of Addition Reactions.

CH3-CH2-CH2 Cl + KOH CH3CH2=CH2 + KCl + H2O

4) Rearrangement Reactions – Involve a redistribution of the

fragments within a molecule. NB : Condensation reactions are a special example of addition-

elimination reactions.


27

Organic Reagents Can be classified according to one of 2 main types : Nucleophilic reagents – these are species with a ____ or _______ ________ charge that attack carbon atoms with a full or partial ________ charge. Examples include :-

Br-, OH-, CN-, H2O, ROH, NH3, etc

These species are attracted to positive centres and are thus ‘nucleus loving’ Electrophilic reagents – these are species with a ____ or _______ ________ charge that attack carbon atoms with a full or partial ________ charge. Examples include :-

Br+, RN2

+, NO2+

These species are attracted to negative centres and are thus ‘electron loving’ Curly Arrow Notation Curly Arrows represent the flow of electrons. The head of the arrow shows where the electrons are going, the tail shows where they have come from (not from some indiscriminate space). The head of the arrow has significance too.

What does a single headed arrow as shown imply ?

What does the double headed arrow shown imply ?


28

Questions: 1) What is meant by each of the following terms? a) nucleophile e) elimination reaction b) electrophile f) hydrogenation c) addition reaction g) dehydrogenation d) substitution reaction h) hydration (8) 2) What would be produced from each of the following electron

shifts: a)

Cl Cl e)

CH2 CH CH2 Br b)

R

C

R

R+X-

f)

X- R

C

R

YR

c)

Br ClR+

g)

CH2 CH CH2

+

d)

CH3

C C

CH3

H

H

CH3

+

h) O

H

H

NO2+

(8) 3) For each of the following give a compound that contains four

carbon atoms that fits the question. You may use compounds more than once. You should name the compound and draw its structure.

• a branched chain alkene

• a saturated chloroalkane


29

M3 - Organic Chemistry & Alkanes

Alkanes The alkanes form a homologous series of hydrocarbons (compounds which contain carbon and hydrogen atoms only). They have the general formula CnH2n+2. These compounds are obtained from crude oil (see above). The major use of these compounds is as a fuel, as when they react with the oxygen of the air, a lot of energy is released. In a car engine, a combustion reaction takes place between octane and oxygen to form carbon dioxide and water:

C8H18 + 12.5 O2 → 8 CO2 + 9 H2O As air is only about 20% oxygen, this means that one mole of petrol (octane) requires 62.5 moles of air for complete combustion. When a car engine is properly 'tuned', this can happen but usually the air/petrol mixture is not ideal. In such cases, incomplete combustion takes place, with the production of a percentage of carbon monoxide and some unburned hydrocarbons released to the atmosphere. In a recent uprating of the MOT test, cars' exhausts have to be checked for the percentages of these gases and oxides of nitrogen (NOx). These oxides are formed by reaction of oxygen and nitrogen of the air at the high temperatures in car engines. A way of removing these compounds from the exhaust gases is to use a catalytic converter. This contains a mixture of valuable transition metals such as platinum, palladium and rhodium. These convert the nitrogen oxides to nitrogen and oxygen, and this oxygen will then convert the carbon monoxide to carbon dioxide. When the catalyst is up to working temperature, hardly any pollutant gases are released to the atmosphere but before it reaches this temperature it is useless.


30

It takes up to 10 minutes for a catalyst to get up to working temperature. Until that time more pollutants are being pumped into the air than if the car did not have a catalyst. This is because the catalyst slows down the flow of exhaust gases, making the engine slightly less efficient and using more fuel. If the journey time is significantly longer than 10 minutes, then the advantages will outweigh the disadvantages. Leaded petrol must never be used in cars with catalysts as the lead will 'poison' the catalyst - forming a coating over the surface of the precious metal alloy thereby preventing the exhaust gases from coming into contact with it. If you check the electronegativaties of carbon and hydrogen, you will find that they are very similar (within 0.4 of each other). This means that the dipole in the C-H bond is small, and the fractional positive charge (δ+) on the carbon atom is small. It does not provide a suitable 'target' for negatively charged species such as the OH

- ions from alkalis. As the

carbon atom is slightly positively charged, it will repel all positively charged ions such as the H

+ ions from acids. As you can see, alkanes will

not react with acids and alkalis - in fact they resist reaction with virtually all reagents in solution. The old name for alkanes was ‘paraffins’. A name that means 'having few reactions - having little affinity' in Latin. Chlorination of Alkanes Alkanes are however open to reaction with halogens under the right conditions. Consider Chlorine, the molecule consists of two chlorine atoms joined by a single covalent bond Cl : Cl (the : represents the pair of shared electrons which make up the bond). If energy (such as sunlight or ultra violet light) falls on chlorine molecules, they can split into atoms (homolytic fission). u/v light

Cl : Cl Cl. + Cl.


31

The products are two chlorine free radicals (in this case they are atoms of chlorine with 7 electrons on the outer shell of the atom). They have no charge and but they are very reactive. They try to achieve stability by ripping an electron from anything they happen to blunder into. (think of them as the Tasmanian Devil character from Bugs Bunny cartoons - like a whirlwind with teeth). Alkane molecules may be resistant to chemical reactions but when they meet up with a free radical - they have no chance ! This is what happens when methane reacts with chlorine. STEP 1 - Formation of the free radicals (Initiation) u/v light

Cl : Cl Cl. + Cl. STEP 2 - Reaction of the free radicals with methane (Propagation)

Cl. + CH4 CH3. + HCl

CH3. + Cl2 CH3Cl + Cl.

As you can see, one of the products from the second propagation step is a chlorine free radical. This can then react with another methane molecule in a chain reaction. STEP 3 - Two free radicals joining up together, (Chain Termination)

CH3. + Cl CH3Cl

CH3. + CH3

. C2H6


32

In order for the termination reaction to occur, there should be a third particle in the vicinity to absorb some of the energy released by the bond formation. The whole process :

CH4 + Cl2 CH3Cl + HCl is a free radical substituition reaction

Mechanism (for understanding only !):


33

M3 - Alkenes

Alkenes are hydrocarbons, they contain only the elements carbon and hydrogen. They have the general formula CnH2n

They are all UNSATURATED compounds containing at least one double bond so molecules can react by ADDING across the double bond. Their names all end in the letters ‘-ene’ As there must be a minium of TWO carbon atoms to have a carbon-carbon double bond, the first member of the alkene homologous series is Ethene. The compound methene does not exist !!! Theory The double bond in the alkene molecule is made up of two electron pairs, one pair forming the sigma bond between the carbon atoms and the second pair forms a Pi bond about the sigma bond: Diagram:


34

The concentration of negative electrical charge due to the four electrons in the double bond makes this, and other alkene molecules open to attack by positively charged particles or electrophiles. Alkenes tend to take part in electrophilic addition reactions. Example: The addition of Hydrogen Bromide to ethane: In the example above only one possible product is possible. However, if the same reaction was perfomed with propene more than one product could be formed.

Predict the two products from the reaction of propene with HBr: Markownikoff’s addition There are four electrons making up the double bond of an alkene, all of which have a negative charge. This area of the molecule is likely to be attacked by positively charged particles in an electrophilic reaction. As there is a double bond present, an addition reaction is also likely. Hence, electrophilic addition !. If the molecule hydrogen bromide (HBr) reacts with a simple alkene such as ethene, the positive hydrogen atom is attracted to the electron rich centre of the double bond.


35

A covalent bond forms between the carbon and the hydrogen atoms. The positive hydrogen ion has no electrons to donate to form the bond, so both electrons must come from the alkene. The pi bond electrons are used:

C CH

H H

H

H+

C+

C HHH H

H

Br This leaves a positive charge on the carbon atom on the left, which is then attacked by the negative bromide ion. A lone pair on the bromide ion is used to form the second bond. Suppose the reaction is between propene (an asymetrical alkene) and hydrogen bromide. A quick look back in these notes shows that there are two possible products :

C CH

H H

CH3

HBr

C+

C HHH CH3

H

Br

C C HHH CH3

HBr

C C+

HHCH3

H

HC C HH

CH3

H

H

Br

Br

In this case, the product is 1-Bromopropane. The bromine atom in this case is attached to the first carbon atom in the chain.


36

Which of the two isomers is formed when reaction takes place? Markownikoff found that over 90% of the product that formed was the 2-Bromopropane. This led him to formulate his rule: When a compound of the type H - X is added across a carbon/carbon double bond, the hydrogen atom adds onto the carbon which already has more hydrogen atoms around it. (them as, as gets !!) This is called an empirical Law. Markownikoff did not understand why the products were formed in different amounts. We are now in a position to explain why The addition of HBr or Br2 to an alkene leads us into the area of reaction mechanisms: Formation of Carbocations (Carbonium ions) The first stage in this reaction is the attack of the positive centre of the HBr (the hydrogen) on the negative centre of the alkene. This is the far right carbon atom. When the hydrogen reacts with the pi bond electrons, a positively charged particle is formed. Remember we are dealing with organic, covalent compounds and any sort of charge on a covalent compound is unstable. It’s like you trying to hold onto a hot potato. Unless you can do something to make things more comfortable, you are going to have to drop the potato. In the same way, the molecule will have to ‘drop’ the charged particle unless it can stabilise it in some way (make it more stable). The problem is the positive charge. If there is some way of feeding in negative charge, this will help to stabilise the ion. (A bit like you blowing on the hot potato to cool it down). Where can this negative charge come from ?


37

Remember the tiny C - H dipoles ? They can feed small amounts of charge to help to stabilise the ion, but only those that are close enough to the ion. Let’s take a look at the possible carbocations formed by attacking propene with the positively charged hydrogen :

CC

+C

H

HH

HH

H

H

C+ CC

HH

H

H

H

HH

(A) (B) In case (A) from the last sheet there are a total of only 4 C - H bond dipoles close enough donate electron density to the positive charged carbon, but in case (B) ALL the C - H bond dipoles (7 of them!) can donate to the positively charged central carbon atom. This is the more stable ion and it is more likely to be the one which is formed !!. When the Br

- attacks in the second stage of the reaction, the product is

the 2-Bromopropane. The carbonium ion formed in case (A) with the positive charge at the end is called a Primary carbocation The carbonium ion formed in case (B) with the positive centre on the middle carbon atom is called the Secondary carbocation. You will come across primary and secondary (and tertiary) compounds again.


38

Industrial uses of Alkenes There are considerable amounts of alkenes produced as by-products from the cracking of alkanes in the petrochemicals industry, so they are readily available and relatively cheap - ideal properties for a ‘feedstock’ for the chemical industry! These uses are based on addition reactions : Formation of Alcohols This can be achieved in the laboratory by using concentrated sulphuric acid followed by water. In industry, water in the form of steam is added across the C = C bond at high temperatures and in the presence of a catalyst of phosphoric acid in celite (a type of clay):

CH2=CH2(g) + H2O(g) CH3-CH2OH Ethanol is formed in the reaction of addition of water to ethene (a process called hydration) Polymerisation Other industrial addition reactions involve the addition of alkene molecules to other alkene molecules to form a long chain compound called a polymer. The term polymer means 'many units' - ethene molecules are the single units (called monomers). The polymer formed is called poly(ethene) or, more commonly, polythene. Use the atomic models to build the first few stages in the formation of the polymer chain of polythene. The equation for the reaction is :

n CH2=CH2 → -(-CH2-CH2-)-n

The 'n' is a large number.


39

Other monomers that react in a similar manner include chloroethane, CH2=CHCl – this forms poly(chloroethane), a more common name for the monomer is vinyl chloride hence the name polyvinylchloridre (PVC).

n CH2=CHCl → -(-CH2-CHCl-)n- If the monomer is phenylethene (see earlier notes about friedel crafts alkylation later) CH2=CHC6H5 the resulting polymer is poly(phenylethene) a more common name for the monomer is styrene hence the name polystyrene. If the monomer is tetrafluoroethane CF2=CF2 the resulting polymer is poly(tetrafluoroethane) a more common name for polymeric material is PTFE more commonly known as Teflon with a big use in non stick materials, apparently Teflon on Teflon has less friction than ice sliding over ice !. All of the above addition polymerisation products are chemically inert and as such are non-biodegradable.

Draw a section of the structure of the polymer chain of polystyrene in the space below:

In order to make these alkene molecules to add together, a certain amount of activation energy is needed. 60 years or so ago, when polythene was first being produced, a temperature of 300oC was used, along with high pressure and an oxygen catalyst. The product from the process was a solid which softened at 100oC, a major problem when it was used to make washing up bowls into which boiling water was poured !


40

Today, special catalysts called Ziegler/Natta catalysts allow polymerisation at 60oC and only double atmospheric pressure. The product also has superior properties.

Questions: Propene reacts with hydrogen bromide by an electrophilic addition reaction forming 2-brompropane as the major product. The equation for the reaction is shown below:

CH3CH=CH2 + HBr → CH3 CH2CH2Br (i) Outline the mechanism for this reaction, showing the structure of

the intermediate carbocation formed. (ii) Give the structure of the alternative carbocation that could be formed in the reaction between propene and hydrogen bromide.

[5] (iii) Explain why the carbocation in (ii) is formed in much lower amounts

than the carbocation in (i) ………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………………………[3]


41

M3 - Epoxyethane

Epoxyethane (ethylene oxide) is a compound formed by reacting oxygen with ethene:

CH2 CH2

CH2 CH2OO+

This reaction is exothermic, producing quite a lot of heat. The oxygen/ethene/epoxyethane mixture is explosive, so the rate of reaction must be controlled carefully. For this reason, silver is used as a catalyst. It is a poor catalyst, but this prevents the reaction proceeding too fast and getting out of hand. The three membered ring has bond angles of about 60

o. The usual bond

angles in carbon compounds are 109.5o. This means that the ring is

‘strained’ and easily breaks open. The compound is highly reactive towards nucleophiles. Reactions of Epoxyethane 1) Reaction with water In working out the products of any reaction with epoxyethane, simply imagine the molecule to be broken :

CH2-CH2-O The attacking species will have an -OH group in it. The hydrogen attaches itself to the oxygen atom in the epoxyethane. The oxygen attaches to the other end of the broken molecule:

-CH2-CH2-O- + H-OH → HO-CH2-CH2-OH The product is ethane-1,2-diol or ethylene glycol which is used to make antifreeze and synthetic fibres.


42

As we will see later, a large excess of water needs to be used to reduce the possibility of reactions forming by products. 2) Reaction with methanol

-CH2-CH2-O- + CH3O-H → CH3-O-CH2-CH2-OH The product has an ether functional group in the middle of the molecule and an alcohol functional group at the end. It is an alkoxyalcohol. In general, the reaction is written :

-CH2-CH2-O- + RO-H → RO-CH2-CH2-O-H The main use for such compounds are as solvents for making paints and printing inks. You will notice that the product has an -OH group and so can react with another molecule of epoxyethane:

-CH2-CH2-O- + RO-CH2-CH2-O-H → RO-CH2-CH2O-CH2-CH2-O-H The product may be written :

RO(CH2-CH2-O)2H This product also contains the -OH functional group as will react further with epoxyethane to form:

RO(CH2-CH2-O)3H

These longer chain complex molecules are used as brake fluids, plasticisers and, if long chain alcohols are reacted with a large excess of epoxyethane, non - ionic detergents are formed.


43

If we go back to the reaction with water:

-CH2-CH2-O- + H-OH → HO-CH2-CH2-OH The product here also has an -OH functional group, and can react with epoxyethane

-CH2-CH2-O- + HO-CH2-CH2-OH → HO-CH2-CH2-O-CH2-CH2-O-H These ‘polyethylene glycols’ are used as solvents, as lubricants or in the manufacture of polyester resins. However, if the concentration of the product is kept low by having a large excess of water, such further reactions are kept to a minimum.


44

M3 - Haloalkanes

All but the lightest haloalkanes are liquids. They have boiling points higher than alkanes and alkenes of the same relative molecular mass. The reason for this is that the haloalkanes molecules are POLAR. The halogen atom (even if it is iodine) has a higher electronegativity than the carbon atom to which it is attached. This means that the electrons in the covalent bond between the two atoms are drawn towards the halogen atom, creating a PERMANENT DIPOLE in the bond: H +→ H - C - Cl H The δ+ carbon of one molecule will be attracted to the δ- of the chlorine in a neighbouring molecule. This dipole-dipole attraction is stronger than the Van Der Waal's forces between alkane and alkene molecules. This stronger attraction means that more energy has to be put in to separate the molecules when the liquid boils, which means a higher boiling point. The haloalkanes are, however, NOT miscible with water. When the two liquids are added together, they separate out into two layers. Water has hydrogen bonding between its molecules and the only organic molecules that will dissolve well in water are those that also have hydrogen bonding between their molecules. The δ+ charge on the carbon (it is an electron deficient centre) makes it a target for nucleophiles.

What is a nucleophile ? …………………………………………………………………………………………..

………………………………………………………………………………………………………………………………………


45

Dilute Solution of Hydroxide Ions [NaOH]

CH3-Br → CH3-OH + Br-

:OH-

Cyanide ions [KCN]

CH3-Br → CH3-CN + Br :CN

- Ethanenitrile

Nitriles are useful compounds because they may easily be converted to carboxylic acids by warming under reflux with either a strong alkali or a strong acid. The reaction occurs in two stages. In the first stage, an amide is formed :

CH3-CN + H2O → CH3-CONH2 Ethanamide

CH3-CONH2 + H2O → CH3-COOH + NH3

In acidic solution, the ammonia would form the ammonium ion so, overall the reaction would be:

CH3-CN + 2 H2O + H+ → CH3-COOH + NH4

+

Ammonia Although the ammonia molecule does not have a full negative ionic charge, the nitrogen atom has a lone pair. This makes it nucleophillic.


46

When haloalkanes are warmed in a sealed container with ammonia, primary amines are formed: H | CH3-Br → CH3-N

+-H + Br

- → CH3 - NH2 + NH4+

| :NH3 H :NH3 All the above reactions are nucleophilic substitution reactions. There is, however, another type of reaction that can occur between the haloalkanes and sodium hydroxide solution. In this case, the hydroxide ions act as bases, accepting a proton to form water:

CH3CHBrCH3 + OH- → CH3 CH=CH2 + H2O + Br

-

The mechanism is fairly complicated, involving three ‘curly arrows’: The reaction of haloalkanes with sodium hydroxide solution produces a mix of alcohol and alkene. A greater proportion of alkene is formed when the solution is concentrated and hot.


47

Questions: Q1) Name the following compounds: a ) CH3CH2I .............................................................................. b) CH3CHClCH2CH3 .............................................................................. c) CH2ClCHClCH2CH .............................................................................. d) CH3CH(CH3)CHClCH .............................................................................. e) CH3C(CH3)ClCH3 .............................................................................. f) Which of the above compound(s) are primary haloalkanes? .......................................................................................................................................... .......................................................................................................................................... g) Which of the above compound(s) are tertiary haloalkanes? .......................................................................................................................................... h) Which of the above compound(s) have optical isomers? .......................................................................................................................................... Q2) Explain why haloalkanes will not dissolve (are not miscible) in water. ..........................................................................................................................................

..........................................................................................................................................


48

3a) Underline which of the following bonds do you think will be

strongest:

Carbon-chlorine Carbon-bromine Carbon-iodine 3b) Underline the compound you think will react most rapidly with

aqueous NaOH

1- Chlorobutane 1- Bromobutane 1-Iodobutane Elimination Reactions Reactions such as the formation of ethene from the haloalkane with hot, concentrated sodium hydroxide solution are called elimination reactions:

C2H5Br + NaOH → CH2=CH2 + NaBr + H2O The molecule HBr is eliminated from the haloalkane, reacting with the NaOH to form NaBr and water. Primary haloalkanes generally undergo substitution reactions Tertiary haloalkanes generally undergo elimination reactions Secondary haloalkanes generally undergo a mixture of reactions Higher temperatures and stronger basic strength of the nucleophile also favour elimination reactions. Rates of Reaction of Haloalkanes

Although the compound with the largest dipole [one containing the C-F bond] will attract nucleophiles most strongly, it is NOT the substance that reacts most rapidly.


49

A second criteria for the rate of reaction is the strength of the covalent bond. The stronger the bond, the more difficult it is to break and therefore the slower is its reaction.

As both carbon and fluorine atoms are small, the C-F bond is a short bond and is, therefore, very strong. This means that fluoroalkanes do not react very rapidly.

Covalent bond Bond strength kJmol-1

C-F 484 C-Cl 338 C-Br 276 C-I 238

As you can see from the above table, the C-I bond is the weakest of the four. This means that iodo- compounds will react most rapidly. However, if a steady rate of reaction is required, the haloalkane of choice is the bromo- compound. As the compounds are not miscilble with water, reactions with aquoeus solutions, such as sodium hydroxide solution, are quite slow at room temperature. The rate of this reaction can be increased by heating but, to prevent the reactants boiling away, a vertical condenser is fitted to the reaction flask. Any vapours given off are cooled and condensed to liquids that then drop back into the flask. This is known as heating under reflux. If alcohol is used as a solvent (as in the case of the ammonia reaction) this helps the miscibility as the alcohol molecule has a non polar part, like the carbon chain of the haloalkane.


50

M3 - Alcohols

Nomenclature All alcohols have names ending in the letters -ol. Example:

COHH

H

H

CC

OH

H

HH

H

H

Methanol Ethanol

CH3OH CH3CH2OH

Propanol has two isomers - owing to the position of the functional group:

CC

C

H

HH

HH

H

H

OH

CC

CH

H

HOH

HH

H

H

CH3CH(OH)CH3 Propan-2-ol

This is a secondary

alcohol

CH3CH2CH2OH Propan-1-ol

This is a primary

alcohol

Tertiary alcohols have the -OH group attached to a carbon atom in the chain which also carries a branch, such as :

CC

CH

OHH

H

HHCH3

H

CH3C(OH)(CH3)CH3

2-methylpropan-2-ol

This is a tertiary alcohol


51

Alcohols, like haloalkanes, may also have optical isomers** such as butan-2-ol. This will be covered in the second year’s work. There are significant differences in the chemical properties of primary, secondary and tertiary alcohols so it very important that you understand the differences between them and are able to recognise each type. Physical Properties of Alcohols The -O-H bond in alcohols is very polar and, as a result, the intermolecular force between alcohol molecules is hydrogen bonding. This means that the boiling points of alcohols are much higher than expected and appreciably higher than the corresponding alkanes or haloalkanes. The larger the alcohol molecule (the longer the chain), the higher the boiling point. The branched isomers of alcohols (the secondary and tertiary alcohols) have slightly lower boiling points than the corresponding straight chain isomers. Alcohols are also miscible with water, although the heavier molecules, with longer carbon chains, are less miscible. Preparation of Alcohols 1) By reduction of the appropriate carbonyl compound: a) Primary alcohols are obtained by reducing the carboxylic acid with

LiAlH4 or the aldehyde with H2 and Ni catalyst or NaBH4

Example:

COCH3

HC

OHCH3

H

H2 [H]


52

b) Secondary alcohols are formed by reduction of ketones by NaBH4

or H2 and Ni catalyst. Example:

COCH3

CH3

COHCH3

CH3

H2 [H]

2) By hydrolysis of haloalkanes with aqueous strong alkali.

see previous notes on reactions of haloalkanes with aqueous alkali’s the mechanism is by (Nucleophilic substitution).

Mechanism:

3) From alkenes by reacting with concentrated sulphuric acid followed

by water. The sulphuric acid adds on H-HSO4 according to Markownikoff's rule. This means that, for propene, the product will be propan-2-ol and not propan-1-ol.

Mechanism:


53

4) In industry, there are two major ways of producing ethanol, one

method uses ethene as the raw material and the other, fermentation, uses carbohydrates such as grain and sugars:

H3PO4

CH2=CH2 + H2O CH3-CH2OH

yeast C6H12O6 2 CH3-CH2OH + 2 CO2

There are advantages and disadvantages of each process:

(i) Raw materials Ethene is obtained by cracking crude oil fractions. At present, there is no shortage of crude oil and ethene is freely available and cheap but oil is a non-renewable resource. It will eventually run out. The fermentation method uses sugars or starches as raw materials which ARE re-newable - every year, there is a fresh crop of grain and sugar. ii) Continuous and batch processes The production of ethanol from ethene is continuous, the vapours are passed over the catalyst and are converted to the product (any unreacted reagents are recirculated). This is an efficient way of producing a product as the plant only closes down to replace the catalyst, which does not happen very often. The fermentation method is a batch process. The sugar solution reacts using the biological enzyme catalysts in yeast, in large vats but as soon as the alcohol has reached a concentration at which the yeast starts to die, the process stops. The product is removed for processing and a new batch of sugars and yeast added. iii) Energy considerations The ethene method operates at about 250

oC, the fermentation method at

35 oC. Both processes are exothermic, but more energy has to be used to

maintain the higher temperature of the ethene production.


54

(iv) Purity of product Fermentation produces a product that contains about 10 % ethanol, the ethene method produces a much purer product. If pure alcohol is required, the ethene method is superior. Reactions of Alcohols 1) Dehydration (Formation of alkenes) This is usually done by concentrated sulphuric acid. The strong acid first protonates the alcohol.

CH3CH2OH + H+ → CH3CH2OH2

+

The product is unstable and breaks down to form water and the carbonium ion. The ion then loses a hydrogen ion to form the alkene. An excess of acid is required in the reaction. Mechanism:

CC

O

H

HH

H

HH

CC

O+

H

HH

H

HH

H

CC

+H

HH

H

HC

CH

HH

H

H+

OH2H+ + +


55

2) Esterification Alcohols react with carboxylic acids in the presence of a strong acid catalyst in an addition-elimination reaction to form esters:

OHC

CH3

H

HCH3C

OH

O

CH3C

O

OC

CH3

H

H+ OH2+

CH3COOH + CH3CH2OH → CH3COOCH2CH3 + H2O ethanoic acid + ethanol → ethyl ethanoate + water 3) Reation with PCl5 (Test for OH group) If any OH group is present, HCl fumes are formed with PCl5 (this is a vigorous reaction):

CH3CH2OH + PCl5 → CH3CH2Cl + POCl3 + HCl 4) Oxidation Acidified potassium dichromate(VI) will oxidise primary alcohols to aldehydes then on to carboxylic acids. Secondary alcohols form ketones but tertiary alcohols are not oxidised. a) Oxidation of Primary alcohols If the aldehyde is required as the product, care must be taken that it is not oxidised further to the carboxylic acid. In the preparation of ethanal, ethanol is dripped slowly into hot, acidified potassium dichromate solution and the product distilled off as soon as it is formed.

C2H5OH + [O] → CH3CHO + H2O


56

If the acid is the required product, the mixture is refluxed together, (heated in a flask with a condenser mounted vertically). This ensures that full oxidation takes place:

C2H5OH + 2 [O] → CH2COOH + H2O b) Oxidation of Secondary Alcohols These compounds are only oxidised to the ketone by acidified potassium dichromate(VI). Any further oxidation would involve the breaking of a strong carbon-carbon bond and the oxidising agent is not powerful enough to achieve this:

CH3CH(OH)CH3 + [O] → CH3COCH3 + H2O In each case, the orange colour of the potassium dichromate(VI) changes to green, indication that oxidation has taken place. Can you think of a use of this colour change ? NB: that tertiary alcohols ore not oxidised by acidified potassium

dichromate(VI), as this would involve the breaking of a carbon-carbon bond.

Distinguishing Between Different Alcohols If an alcohol is not oxidised by acidified potassium dichromate solution (there being to colour change from orange to green) the alcohol must be a tertiary alcohol. If the alcohol is oxidised, then tests must be performed on the oxidation product. We need a test to distinguish between aldehydes and ketones. There are, in fact, two tests: a) Tollen’s Reagent Tollen’s reagent is made by adding ammonia solution to silver nitrate solution, producing what is, in effect, a solution of silver oxide in ammonia. It is a weak oxidising agent and will have no effect on a ketone. aldehydes, however, will be oxidised:

CH3CHO + Ag2O → CH3COOH + 2 Ag


57

When the aldehyde is warmed with Tollen’s reagent, the silver forms a mirror like surface on the inside of the test tube. b) Fehling’s Solution When warmed with an aldehyde, the deep blue copper II complex colour changes to a dark red copper I precipitate (Cu2O). There is no reaction with a ketone.

Questions: Why does ethanol dissolve in water ? ………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

Write equations to show the possible oxidation products for:

i) Butan-1-ol ii) Butan-2-ol ………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………

Draw the structures of the two possible organic products from the reaction of butan-2-ol with concentrated sulphuric acid at 170 oC ?


58

Outline a mechanism for the reaction that gives rise to either of these two products. Mechanism:


59

Appendices


60

Date post:	17-May-2017
Category:	Documents
Upload:	sophiestevens
View:	309 times
Download:	2 times

Hydrocarbons Qs

Documents