Homologous Series A group of Hydrocarbons with the same General Formula and similar chemical...
-
Upload
annice-thornton -
Category
Documents
-
view
217 -
download
1
Transcript of Homologous Series A group of Hydrocarbons with the same General Formula and similar chemical...
Homologous Series
A group of Hydrocarbons with the same General Formula and similar chemical properties.
Examples – Alkanes, Alkenes and Cycloalkanes.
Alkanes Alkenes Cycloalkanes
CnH2n+2 CnH2n CnH2n
C-C single bonds
At least 1 C=C doublebond
Ring structure
Saturated Unsaturated Saturated
Naming hydrocarbons - Nomenclature
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
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 H H
Alkynes
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
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
1. Select longest chain 2. Number Carbons – number 1 closest to C=C 3. Name any branches Example CH2=CH.CH3CH2CH2CH3 ( CH3 = branch) 1 2 3 4 5
2 methyl pent – 1 - ene
Isomers
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)
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
I
Methanol CH3OH H - C – H I OH Ethanol CH3CH2OH
Naming Alkanols
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
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
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
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)
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
Alkanones (Ketones)
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
Alkanoic Acids
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.
Esters
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
More Examples
Ethanol+ Butanoic Acid —> Ethyl butanoate + Water
Propanoic Acid + Methanol —> Methyl
propanoate + Water
Uses of Esters
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
Ethyl methanoate Methyl ethanoate
H H O H O H
I I II I II I
H – C –C - O -C – H H – C – O – C - C - H
I I I I
H H H H
CH3CH2OCOH 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
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
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 benzenesNote : Although benzene contains delocalised
electrons – they are contained within the ring – benzene does not conduct electricity.
Reactions of Carbon Compounds
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
Alkynes can undergo a 2 stage addition reaction to become saturated.
Example Ethyne + Hydrogen —> Ethene
(Unsaturated) Ethene + Hydrogen —> Ethane
( Saturated)
Addition reactions with Halides
Ethyne + Bromine —> Bromoethene
( unsaturated) Bromoethene + Bromine —>Bromoethane
( saturated) We can use Bromine solution as a test for
unsaturated compounds on alkenes and alkynes.
Ethanol
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 I I H H H H
Dehydration of alcohols
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
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. Balanced equation
CH3OH + CH3COOH <=> CH3COOCH3+ H2O2. Number of moles
I mol Methanol----> 1 mol Methyl ethanoate 3. Put in mass - Theretical
32g ----------------> 74g 4. Actual mass 5g -----------------> 74/32 x 5
11.56g = Theoretical Yield 5. Actual Yield = 9.6g 6. % Yield = Actual/Theoretical x 100 = 9.6/11.56 x 100
= 83%