Transcript Nomenclature - Clydebank High School

Homologous Series
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A group of Hydrocarbons with the same
General Formula and similar chemical
properties.
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Examples – Alkanes, Alkenes and
Cycloalkanes.
Alkanes
Alkenes
Cycloalkanes
CnH2n+2
CnH2n
CnH2n
C-C single
bonds
At least 1
C=C
doublebond
Unsaturated
Ring structure
Saturated
Saturated
Naming hydrocarbons Nomenclature
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The prefix tells us how many carbons in
molecule – e.g. meth = 1, eth = 2, prop=3.
We number the carbon atoms, number 1 is
always at the end closest to the main functional
group.
A functional group is a group of atoms with
characteristic features e.g. A C=C bond is a
functional group – showing the molecule is an
alkene - unsaturated.
Naming alkenes
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Example
H H H H
I
I
I
I
C = C – C - C– H This is but –1- ene
I
I
I
H
H H
H H H H
I I
I
I
H - C – C = C – C - H This is but –2- ene
I
I
Alkynes
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General Formula C n H2n –2
They contain at least one C to C triple bond.
They are unsaturated.
They start with the usual prefix e.g. ethyne,
propyne etc.
They are named in the same way the alkenes
are – in the main chain the carbons are
numbered – number 1 is closest to the main
functional group i.e. the triple bond.
Example: but – 1 - yne
Naming branched chain Alkanes
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1. Select longest single straight chain – and
name it.
2. Number C atoms in chain – number 1 is
closet to branch.
3. Name branches: CH3 – methyl. C2H5 – ethyl,
C3H7 – propyl etc.
Example
CH3
I
CH3 – CH2 – CH2 – CH3 2 METHYL BUTANE
1
2
3
4
Branched Chain Alkenes
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1. Select longest chain
2. Number Carbons – number 1 closest to C=C
3. Name any branches
Example
CH2=CH.CH3CH2CH2CH3
( CH3 = branch)
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 1
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2
3
4
5
2 methyl pent – 1 - ene
Isomers
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Compounds with the same molecular formula
but different structural formula.
Alkenes and Cycloalkanes are isomers.
Adding branches to chains increases the
number of possible isomers.
Example
CH3CH2CH2CH2CH3 (C5H12)
is an isomer of
CH3CH.CH3CH2CH3 (C5H12)
Alkanols ( alcohol)
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Another homologous series.
They contain the functional group – OH
This is called a Hydroxyl group
They start with the usual prefix and end in ol.
Example
H
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Methanol CH3OH
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Ethanol CH3CH2OH
I
H-C–H
I
OH
Naming Alkanols
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Carbon number 1 is the C closest to the
Hydroxyl functional group. ( - OH )
Example
CH3CH2CH2OH : propan – 1 ol
CH3CH.OHCH3: Propan- 2 – ol.
Primary Alcohols
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The functional group – OH is on a carbon
which also has 2H atoms attached.
Example
Butan – 1- ol
CH3 CH2CH2CH2OH
Primary alcohols undergo oxidation
reactions to form alkanals ( aldehydes)
Secondary Alcohols
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The functional group – OH is on a carbon
with which is also bonded to 1 H atom.
Example
Butan – 2- ol.
CH3 CH2 CH(OH)CH3
Secondary alcohols oxidise to produce
alkanones (Ketones)
Tertiary Alcohols
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The functional group – OH is attached to a
carbon with NO H atoms attached.
Example
2 methyl propan – 2 – ol.
C(CH3)3OH
Tertiary alcohols can not be oxidised.
Alkanals (aldehydes)
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Formed from the oxidation of primary alkanols.
2 H atoms are removed.
Acidified potassium dichromate acts as an
oxidising agent. Mix with alcohol – heat in water
bath. Colour change orange – blue. ( Different
smell)
Alkanals are another example of a homologous
series.
They contain a carbonyl functional group.
C=O ( at the end of a molecule)
Example
Ethanol will oxidise to produce Ethanal.
 CH3CH2OH
—> CH3CHO
H H
H H
I
I
I
I
H - C – C –OH
—> H - C – C=O
I
I
I
H H
H
C=O is the carbonyl functional group
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Alkanones (Ketones)
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These are formed from the oxidation of
secondary alkanols.
The functional group – the carbonyl group
– C=O is in the middle of the chain.
I H atom is removed.
Example
Propan – 2 – ol will be oxidised to
Propan – 2 –one.
Example
CH3CH(OH)CH3 —>CH3COCH3
H OH H
H O H
I
I
I
I II I
H – C - C - C – H —> H – C - C –C - H
I
I
I
I
I
H H H
H
H
- C=O is the Carnbonyl functional group
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Alkanoic Acids
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Alkanoic acids are formed by the oxidation
of alkanals (aldehydes)
They are a subset of the Carboxylic acid
group
They contain a carboxyl functional group
C=O
( COOH)
I
OH
Alkanoic Acids
Examples
 Ethanal will oxidise to give Ethanoic Acid
 CH3CHO —> CH3COOH
H H
H OH
I I
I
I
H - C – C =O —> H – C – C =O
I
I
H
H
 Alkanones can not be oxidised further.
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Esters
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Esters are formed from a condensation reaction
between an alkanol and an alkanoic acid.
Esterification.
Esters have very distinctive smells
Esters are insoluble in water.
The first part of an ester name comes from the
alkanol – the second part comes from the
alkanoic acid.
Example
Methanol + Ethanoic Acid —> Methyl ethanoate
+ Water
Ester examples
Ethanol + Methanoic acid —> Ethyl
methanoate
H H
OH
+ Water
I
I
I
H – C – C – OH + O=C
I
I
I
H H
H
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Ester formed
H H
O
I
I
II
H - C – C – O – C – H + H2O
I
I
H
H
O
II
This is the ester link
-O–C-
More Examples
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Ethanol+ Butanoic Acid —> Ethyl
butanoate + Water
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Propanoic Acid + Methanol —> Methyl
propanoate +
Water
Uses of Esters
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Esters can be used as non polar solvents.
Example – Ethyl ethanoate is nail polish
remover!
They are used to add flavour and taste to many
substances.
When an ester is made there are 2 very
noticeable changes:
1. The smell
2. It is immiscible with water – we can see the
separate layers.
Esters Shortened Structural
Formula
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Ethyl methanoate
H H
O
I I
II
H – C –C - O -C – H
I I
H H
CH3CH2OCOH
Methyl ethanoate
H
O H
I
II I
H–C–O–C-C-H
I
I
H
H
CH3OCOCH3
More!
H O
H H H
I
II
I
I I
H – C – C –O – C – C – C – H
I
I
I I
H
H H H
CH3COOCH2CH2CH3
Propyl ethanoate
Hydrolysis of Esters
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This is the opposite of a condensation reaction.
We are splitting the ester - back into the
alkanol and alkanoic acid.
We must add back the water which is removed
in the condensation reaction.
This is not very successful with water alone so
we add a dilute acid to catalyse it e.g. HCl or
H2SO4. (Or an alkali.)
They provide H+ ions to catalyse the reaction.
It is a reversible reaction ( PPA – 2)
Aromatic Hydrocarbons
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They are a subset of hydrocarbons.
Benzene is the simplest aromatic compound – C6 H6
Each carbon has 3 ½ filled electron clouds which bond
with the nearest atom – delocalised electrons.
When we replace one of the H atoms with another group
we have a phenyl group
C6 H5 Examples
C6H5 – CH3 = methyl benzene (Toluene)
C6H5 – OH = Phenol
C6H5 – COOH = Benzoic Acid
C6H5 – NH2 = Phenyl amine.
Uses of Benzene
It is an important feedstock.
 It used to produce:
Cylco hexane
Ethyl benzene
Phenol
Alkyl benzenes
Note : Although benzene contains delocalised
electrons – they are contained within the ring
– benzene does not conduct electricity.
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Reactions of Carbon Compounds
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Revision from SG
Cracking – using heat/catalyst to break
heavier fractions into smaller more useful
ones.
Addition reactions – adding atoms to
unsaturated compounds e.g. alkenes.
Example – decolourisation of Br2( aq)
instantly.
Addition reactions - Alkynes
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Alkynes can undergo a 2 stage addition
reaction to become saturated.
Example
Ethyne + Hydrogen —> Ethene
(Unsaturated)
Ethene + Hydrogen —> Ethane
( Saturated)
Addition reactions with Halides
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Ethyne + Bromine —> Bromoethene
( unsaturated)
Bromoethene + Bromine —>Bromoethane
( saturated)
We can use Bromine solution as a test for
unsaturated compounds on alkenes and
alkynes.
Ethanol
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Ethanol can be produced in 2 ways:
1. Fermentation of glucose
2. Addition of water to alkenes using a catalyst –
catalytic hydration.
Example
H
H
H H
I
I
I
I
C = C + H2O —> H – C – C - OH
I
I
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H
H
H H
Dehydration of alcohols
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We can convert Ethanol to Ethene by
dehydration.
We soak mineral wool in the alcohol and heat in
presence of a catalyst.
Aluminium oxide can act as a catalyst in the lab.
Examples
Butan – 1 – ol will become But – 1 - ene.
Butan – 2 – ol can produce both But – 1 - ene
and But – 2 – ene.
% Yield
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The yield is the quantity of the product
obtained.
% yield is when we calculate the actual
yield as a % using the theoretical yield.
Example
5g of Methanol reacts with excess
Ethanoic acid to produce 9.6g of methyl
ethanoate.
Steps
1.
2.
3.
4.
5.
6.
Balanced equation
CH3OH + CH3COOH <=> CH3COOCH3+ H2O
Number of moles
I mol Methanol----> 1 mol Methyl ethanoate
Put in mass - Theretical
32g
---------------->
74g
Actual mass
5g
-----------------> 74/32 x 5
11.56g = Theoretical Yield
Actual Yield = 9.6g
% Yield = Actual/Theoretical x 100 = 9.6/11.56 x 100
= 83%