Homologous Series • A group of Hydrocarbons with the same General Formula and similar chemical properties. • Examples – Alkanes, Alkenes and Cycloalkanes.
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 – CH32METHYLBUTANE 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 —> Ethylmethanoate H H OH + Water I I I H – C – C – OH + O=C I I I H H H
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 • 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 benzenes Note : 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+ H2O 2. 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%