Topic 2.2: Organic chemistry I (introduction, alkanes, alkenes, halogenoalkanes and alcohols

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2.2a Introduction

2.2 (a)(i) Homologous series
These have

• The same functional group
• Similar chemical properties
• The same general formula     
• Gradually changing physical properties

2.2a (ii) Nomenclature
The root name depends on the number of carbon atoms in the longest chain.  Branch names, and functional group names are also included in the full name of an organic compound.

Nomenclature assignment

2.2 (a) (iii) Structural isomers
All isomers are compounds with the same molecular formula. e.g. C₄H₁₀ or C₂H₆O.  Structural isomers have atoms arranged in different orders.
e.g. CH₃CH₂CH₂CH₃ (butane) and CH₃CHCH₃  (2-methylpropane)  or
                                                                  |
                                                                 CH₃
CH₃CH₂OH (ethanol)   and  CH₃OCH₃ (methoxymethane)
Task2.2(a)(iii).1 Draw and name structural isomers for C₅H₁₂, C₃H₇Cl, C₄H₉OH
Task 2.2(a)(iii).2 Make models, draw and name structural isomers of C₄H₈Cl₂.

2.2 (a) (iv) Geometric (cis-trans) isomerism
Geometric isomers exist for some compounds with C=C double bonds in their molecules.  No rotation is possible around the double bond so different compounds result if atoms or groups are in positions on the same side or opposite sides of a double bond.  For example  1,2-dichlorethene can have its two chlorine atoms on the same side (cis) or opposite sides (trans) of the double bond.

Cl     Cl         cis-2,3-dichlorobut-2-ene   Cl            CH₃   trans-2,3-dichlorobut-2-ene
  \     /                                                          \        /
  C=C                                                           C=C
  /     \                                                          /        \
CH₃  CH₃                                                CH₃        Cl
Task 2.2aiv: Draw both geometric isomers of 1,2-dibromopropene, 1,2-dichloroethene, 1,2 -dibromopent-1-ene 

2.2 (a) (v) Classifying reactions
free radical substitution reactions
A typical example of these is between an alkane and a halogen.  E.g. methane and chlorine.  They involve an attack by an atom or group with an unpaired electron called a free radical.  In the above reaction ultraviolet light causes chlorine atoms (free radicals) to form, Cl^. and methyl radicals, CH₃^. are also involved.  The dot shows the unpaired electron.
electrophilic addition reactions
A typical example of these is between an alkene and a hydrogen halide.  E.g. ethene and hydrogen bromide.  They involve an attack by an ion or group with a positive or partial positive charge called an electrophile.  In the above reaction the electrophile is a hydrogen ion H⁺.  Electrophiles attack centres of negative charge like the electrons in a double bond.  These are addition reactions because the two reactants are added together to give one product.  E.g. 
C₂H₄ + Cl₂ ---> CH₂ClCH₂Cl
nucleophilic substitution reactions
A typical example of these is between an alcohol and hydroxide ions.  E.g. ethanol and aqueous sodium hydroxide.  They involve an attack by an ion or group with a negative or partial negative charge called a nucleophile.  In the above reaction the hydroxide ion OH^- is the nucleophile.  A nucleophile attacks a centre of positive charge such as the partial positive charge on the carbon atom holding a halogen C^D+-X.  It is a substitution reaction because the attacking species replaces the species originally attached to the carbon atom under attack.  E.g.
CH₃CH₂Br + OH^- ---> CH₃CH₂OH + Br^-

elimination reactions
A typical example is the formation of an alkene from an alcohol.  E.g. Ethanol reacting to form ethene.  They involve atoms or groups from either side of a single carbon to carbon bond being removed (eliminated) to form a double bond.  E.g.
CH₃CH₂OH  ---> C₂H₄ + H₂O
hydrolysis reactions
These all involve a reaction with water.  E.g.
CH₃CH₂Br + H₂O ---> CH₃CH₂OH + HBr
reduction reactions
These involve a reaction with hydrogen.  A compound gains hydrogen.  This can be hydrogen H₂ or from a reducing agent like LiALH₄ in ether or aqueous NaBH₄.  E.g.
CH₃CHO + 2H ---> CH₃CH₂OH
oxidation reactions
These reactions involve the loss of hydrogen or the gain of oxygen.  An oxidising agent like acidified potassium dichromate (VI) is used.  E.g. 
CH₃CH₂OH  + 2O ---> CH₃COOH + H₂O
polymerisation reactions
These reactions involve the joining of small molecules called monomers to make large molecules called polymers.  A large number (n) of monomer molecules in involved and a long polymer molecule is formed.  E.g.
nC₂H₄ ---> -(C₂H₄)_n-

2.2 (b) Reactions of organic compounds

2.2 (b) (i) Reactions of alkanes 
CH₄ (g) + Cl₂ (g) --------> CH₃Cl (g) + HCl(g)
C₆H₁₄ (l) + Br₂ (l) --------> C₆H₁₃Br (l) +HBr(g)
Both occur of these free radical substitution reactions occur at room temperatures in inert solvent in the presence of UV light.
Alkanes burn in oxygen (air), their flammability being a hazard.
CH₄ (g) + 2O₂ (g) --------> 2H₂O (l) + CO₂ (g)
Task 2.2(b)(i)
How do you make 2-bromopropane from propane?
How do you make chloromethylbenzene from methylbenzene?
State the conditions for making chloroethane from ethane.
What products are possible from a reaction between Bromine and ethane?

2.2 (b) (ii) Reactions of alkenes
alkenes with hydrogen
Hydrogen adds to an alkene in the presence of a nickel catalyst at 150^oC.
CH₂ = CH₂ + H₂  --------> CH₃CH₃
Task2.2bii.1 draw full displayed structural formulae for the above reaction.
Describe how the following alkanes can be formed from a suitable alkene giving chemical equations; propane, butane, pentane, 2-methylpentane and 2-methylpropane.

alkenes with halogens
H₂C = CH₂ (g) + Cl₂ (g) ---------> ClCH₂CH₂Cl (l)
This reaction occurs in the gas phase.
CH₂ = CH₂ + Br₂  --------> BrCH₂CH₂Br
                      red                colourless
This reaction occurs in an inert solvent like volasil.
Both of these are addition reactions which occur at room temperature.  The decolourisation of bromine seen in the second reaction is a good test for an alkene.
Task2.2bii.2 Draw full displayed structural formulae for the above reactions.
Describe how bromine reacts with propene, but-2-ene, pent-1-ene, 4-methylpent-1-ene, 2-methylpropene.

alkenes with hydrogen halides
Conditions: HX in gas phase or in non-aqueous solvent.
Reaction type: electrophilic addition.
The reactivity of these compounds, changes with bond strengths: HF < HCl < HBr <HI
Symmetrical alkenes undergo a simple reaction.
CH₂ = CH₂ (g) + HBr (g) ----------> CH₃ CH₂Br (g)

When hydrogen halides add to an unsymmetrical alkene, 2 products are possible:
CH₃CH = CH₂ (g) + HBr (g) ----------> CH₃CHBrCH₃ (l)
Note CH₃ CH₂ CH₂Br (l) is not formed
Markownikoff's rule applies which states that the hydrogen attaches to the carbon holding the most hydrogen atoms.
Task2.2bii.3 Draw full displayed structural formulae for the equations above.
Describe how HCl reacts with ethene, propene, but-2-ene, but-1-ene and 3-ethylpent-1-ene.  Name the products in these reactions.

alkenes with manganate(VII)
Conditions: aqueous acidic or alkaline manganate (VII)
Reaction type: oxidation:
Acidified potassium managanate(VII) is decolourised by an alkene an a diol formed.
5CH₂ = CH₂ + 2H₂O + 2MnO₄^- + 6H⁺ -----> 5HOCH₂CH₂OH + 2Mn²⁺
                                                                   ethane -1,2 -diol
Task2.2bii.4 Draw full displayed structural formulae of the above organic compounds.
Describe how to make propane-1,2-diol, butane-2,3-diol and 3-methylpentane-2,3-diol.

2.2 (b) (iii) Reactions of halogenoalkanes
with potassium hydroxide
Conditions: Heat under reflux in aqueous solution. Both NaOH or KOH are suitable. Reaction type: nucleophilic substitution
C₂H₅Br + NaOH (aq) -------------> C₂H₅OH + NaBr
Conditions: Heat under reflux with ethanol as solvent.  Use KOH here. 
Reaction type: elimination
CH₃CH₂Br -------------> CH₂CH₂ + HBr
Task 2.2b(iii).1 Draw fully displayed structural formulae for all of the above organic compounds.  Write a sentence in each case explaining why one reaction is nucleophilic substitution and the other is elimination.

with potassium cyanide
Reaction type: nucleophilic substitution
Reflux solution of halogenoalkane and potassium cyanide in ethanol.  This adds a carbon atom to the chain and forms a nitrile.
C₂H₅Br + KCN ------------> C₂H₅CN + KBr
                                         propanenitrile
Task 2.2b(ii).2 Draw fully displayed structural formulae for each organic compound above and explain how this is a nucleophilic substitution.

with ammonia
reaction type: nucleophilic substitution
Amines are formed by heating the halogenoalkane with concentrated ammonia in a sealed tube.
C₂H₅I + NH₃ ---------> C₂H₅ NH₂ + HI
                                   Ethylamine (amine)
In the sealed container other products include: (C₂H₅)₂NH and (C₂H₅)₃N
Task 2.2b(iii).3 Draw full structural formulae of above organic compounds and explain why this is nucleophilic substitution.

test for halogenoalkane
Heat sample of halogenoalkane with aqueous hydroxide ions.
Acidify with dilute aqueous nitric acid.
Add a few drops of aqueous silver nitrate.
A white precipitate soluble in dilute aqueous ammonia, indicates chloride
A buff precipitate insoluble in dilute aqueous ammonia but soluble in concentrated aqueous ammonia, indicates bromide.
A yellow precipitate insoluble in concentrated aqueous ammonia indicates iodide.
        
Task 2.2(b)(iii).4 describe how to change 1-bromopropane into propan-1-ol and propene.
What is formed when 1-bromobutane is heated under reflux with alcoholic potassium cyanide?
Write equations to show all of the products which form when 1-iodopropane reacts with ammonia.
Describe how you could distinguish between a sample of 1-chloropropane and 1-bromopropane.

2.2 (b) (iv) Reactions of alcohols
primary secondary and tertiary alcohols with dichromate(VI)
Reaction type: oxidation
Potassium dichromate (VI) solution is acidified with sulphuric acid. It produces a colour change from orange to blue-green to show that the primary and secondary alcohols have undergone oxidation reactions.
Primary alcohol (mild conditions of dil sulphuric acid and dichromate, heat and distil) ---> aldehyde
C₂H₅OH  + O     ----->    CH₃CHO  + H₂O
ethanol (primary alcohol)  ethanal (aldehyde)
Cr₂O₇^2- ----------> Cr ³⁺
  Dichromate ions Cr₂O₇^2-  are orange but chromium III ions Cr ³⁺   are blue

Primary alcohol (forcing conditions of conc sulphuric acid and dichromate, heat under reflux) ---> carboxylic acid
C₂H₅OH  + 2O      ----->   CH₃COOH  + H₂O
ethanol (primary alcohol)    ethanoic acid (carboxylic acid)

Secondary alcohol (forcing conditions of conc sulphuric acid and dichromate, heat under reflux) --> ketone
CH₃CH(OH)CH₃  + O       ----->        CH₃COCH₃   +  H₂O
propan-2-ol (secondary alcohol )         propanone (ketone)

Tertiary alcohol ---> no reaction
Task 2.2biv.1 Describe conditions, draw structural formulae for the organic products and reactants and give equations for the following reactions:
Making propanal  from an alcohol.
Making butan-2-one from an alcohol
Oxidation of methanol under forcing conditions

alcohols with dehydrating agents
Concentrated sulphuric acid is a dehydrating agent, at 170^oC an alkene is formed in an elimination reaction.
C₂H₅OH + H₂SO₄ -------> C₂H₄ + H₂O + H₂SO₄
If excess alcohol is used and the reaction temperature kept to 140^oC an ether is formed in a nucleophilic substitution reaction.
2C₂H₅OH + H₂SO₄ -------> C₂H₅ -O-C₂H₅ +  H₂SO₄  + H₂O
At 0^oC an alkyhydrogensulfate is formed.
R-OH + H₂SO₄ ---> R-O-SO₃-OH + H₂SO₄ + H₂O

Another dehydrating agent is conc. phosphoric acid.

Dehydration is also possible by passing alcohol vapour over an aluminium oxide catalyst at 300^oC.
C₂H₅OH(g)  -----> C₂H₄(g) + H₂O(g)

Secondary and tertiary alcohols also undergo dehydration.
Task 2.2biv.2 Describe conditions, draw structural formulae for the organic products and reactants and give equations for the following reactions:
The formation of propene from a suitable alcohol

alcohols with halogenating agents
HBr is made in situ from KBr and conc. H₂SO₄ .  The alcohol forms a bromoalkanes during heating under reflux.
C₂H₅OH + HBr ----> C₂H₅Br + H₂O
Alkyl chlorides can be made by refluxing the alcohol with HCl (conc. HCl) in the presence of ZnCl₂
C₂H₅OH (l) + HCl (g) ---------> C₂H₅Cl (l) + H₂O (l)
Phosphorus pentachloride produces white steamy fumes of HCl in forming an alkyl chloride.  This is a test for the hydroxyl group.
C₂H₅OH  + PCl₅  ----->  C₂H₅Cl  + POCl₃ + HCl
Iodoalkanes can be made in a reaction phosporus triiodide made in situ from iodine and red phosporus.
3C₂H₅OH + PI₃ ---> 3C₂H₅I +H₃PO₃
The relative reactivities of alcohols in halogenation are tertiary > secondary > primary alcohol.
Task 2.2biv.3 Describe conditions, draw structural formulae for the organic products and reactants and give equations for the following reactions:
The formation of 1-bromobutane.
The formation of iodomethane.
The treatment of of propan-2-ol with phosphorus pentachloride.

2.2 (c) Bonding and reactivity
Alkenes are much more reactive than alkanes, this is rather surprising at first as a double bond would be expected to be stronger than a single one, and therefore more stable.
The bond energy of a double bond is greater than that of a single bond, though not twice as great, compare 346kJmol^-1 with 598kJmol^-1. This would suggest that the two bonds in C=C may not be identical. In fact, two kinds of covalent bonds are involved:
bonding in ethene - (powerpoint needed)

(i) A bond situated symmetrically between the two carbon atoms, formed by the overlap of two p orbitals; a s bond.

(ii) A bond formed by the “sideways” overlap of two 2p orbitals. Because each p orbital has two lobes, this bond has two regions, one above and one below the plane of the molecule; a p bond.

The electrons of the pi bond are not situated axially between the carbon atoms- the cloud of electrons which forms the p bond lies above and below the plane of the three sp² hybrid orbitals (s bond) formed by each of the unsaturated carbon atoms. This means that they are not “on average” as close to the nuclei of the atoms as the s bonds. Therefore they do not attract the nuclei as strongly, so the p bond is not as strong as the s bond.

conclusion:
The energy used in breaking the one pi bond is more than repaid by the energy released when new bonds are formed during a reaction- thus alkenes are thermodynamically unstable relative to their products in an addition reaction. They are also kinetically unstable, because the high electron density in the double bond tends to attract electron deficient groups- thus the pi electrons are more susceptible than the sigma electrons to attack by an electrophilic reagent.

Halogenoalkanes undergo nucleophilic substitutions reactions in which the halogen atom is replaced by another functional group. An example of such a reaction is that with the hydroxyl (-OH) group:
the halogenoalkanes have a slightly polarised C-Hal bond. Water acts as a nucleophile towards the carbon atom in this bond. As a result , the -OH group substitutes for the halogen, giving an alcohol and a hydrogen halide- a reaction that is sometimes called HYDROLYSIS (splitting with water).
Results of investigations show that the rate of hydrolysis of the halogenoalkanes occurs in the order:
 1-iodobutane > 1-bromobutane > 1-chlorobutane

What determines the reactivity?
The ease of reaction depends on the ease of breaking the C-Hal bond:

Bond : C-I C-Br C-Cl C-F   
Bond enthalpy terms (kJ mol^-1 ): +238 +276 +338 +484   
Thus the ease of bond breaking is, C-I > C-Br > C-Cl > C-F.
Alcohols can also undergo nucleophilic substitution if the concentration of the nucleophile is high as the carbon to oxygen bond is polarised.  This gives the carbon atoms holding the oxygen in an alcohol a partial positive charge making susceptible to nucleophilic attack.

2.2 (d) Quantitative organic chemistry
2.2 (d) (i) empirical and molecular formulae
Empirical formula is the simplest ratio of the number of atoms of each element in a molecule of a substance.  The molecular formula shows the actual number of atoms of each element in a molecule of a substance. If the empirical formula of a substance is C_xH_y then the molecular formula is nC_xH_y if 
n=relative molecular mass of the substance/relative molecular mass of C_xH_y. 
 
What is the empircal formula and the molecular formula of a compound which contains 1.2g of carbon and 0.2g of hydrogen if the relative molecular mass of the compound is 42?
                                            carbon                     hydrogen
mass  =                                 1.2g                           0.2g
amount=mass/molar mass=    1.2/12=0.1mol        0.2/1=0.2mol   
ratio(divide by smallest) =                     1         :        2
empirical formula=                                        CH₂
molecular formula = nCH₂    
n = relative molecular mass of substance/relative molecular mass of CH₂
n = 42/14 = 3
Molecular formula =   nCH₂  = 3 * CH₂  = C₃H₆  
Task 2.2(d)(i)  

2.2(d) (ii) theoretical yield and percentage yield
Many organic reactions produce side products and during purification some of the main product is normally lost.  The result is that the amount of product obtained, the yield, is less than the theoretical maximum yield.
The theoretical yield is calculated using the balanced chemical equation.  E.g.
C₂H₅OH + HBr ----> C₂H₅Br + H₂O
1 mol of ethanol forms 1 mol of bromoethane so
46.0g of ethanol forms 109g of bromoethane so if we start with 2.30g of ethanol
1g of ethanol forms 109/46.0g of bromoethane
2.30g of ethanol forms 2.30*109/46.0g = 5.45g of bromoethane so
the theoretical yield of bromoethane is 5.45g if we start with 2.30 g of ethanol.
If only 4.00g are actually formed (actual yield = 4.00g) in the preparation then
percentage yield = 100*actual yield/theoretical yield 
percentage yield = 100*4.00/5.45 = 73.4%
Task 2.2(d)(ii)

2.2e Applied organic chemistry

2.2e (i) Advantages and disadvantages of liquid and gaseous fuels
Liquid fuels have the advantage of having higher densities than gaseous fuels.  More energy may be stored in a liquid fuel than in the same volume of gaseous fuel.  Petrol is a typical liquid fuel which can be thought of as mostly octane.  It has two important properties namely volatility and octane number.  Octane number is a measure of how smoothly the fuel burns.  If a fuel ignites too soon in an engine the engine does not run smoothly or efficiently.  The composition of petrol must be changed during the year to give the best values for these two properties.  Fuels can be compared by looking at their relative energy values, handling characteristics and their environmental impact.

Task 2.2e(i) Hydrogen buses are running on a few select routes in the City of London.  Explain why this is happening and why few cars have so far been manufactured to use hydrogen.
During the second world war (st ba time a severe fuel shortages) a few vehicles could be seen driving around with huge canvas bags of methane on top of them.  Why is methane gas not used like this today and how could it be used?
Butane is the main fuel used by residents in Ameican trailer parks.  What advantages does it have?
What is the main use of octane and why is it valuable for this purpose?
Is ethanol a suitable replacement for petrol?  Explain your answer.

2.2e (ii) Structures, properties and uses of addition polymers
Typically polymers are made by free radical addition reactions at high pressure (1500 atm and 200^oC).

Task 2.2e(iii) Draw sections of repeating structures for all of the four polymers above.
Name suitable polymers, with reasons for their choice, for the following uses: To make plastic toy ducks for children to play with in the bath, a polymer to coat replacement hip joints, a new reuseable supermarket shopping bag that can repeatedly carry heavy shopping, a pipe to carry rainwater from a roof to a drain.

2.2e (iii) The use of halogens in herbicides and pesticides
PVC used as electrical insulator  -[-CH₂-CHCl-]_n-  The C-Cl bond is strong so PVC insulation lasts a long time but when discarded it does not rot (it is not biodegradable)

Freon 12 CF₂Cl₂ is a refrigerant and an example of a chlorofluorcarbon (CFC).  The C-F and C-Cl bonds are very strong.  The result is that it does not decompose easily so lasts for the lifetime of a refrigerator.  It does not decompose quickly when discarded but does so in the upper atmosphere.  The radicals it forms react with ozone.  The loss of ozone leads to an increase in UV radiation reaching the Earth's surface and a corresponding increase in skin cancers in humans.

DDT is an pesticide used to kill mosquitos.
              CCl₃
               |
Cl-C₆H₅-C-C₆H₅-Cl
               |
              H
The strong C-Cl bonds give DDT a long life in the field killing pests.  It is however so long lived that it persists in the environment and builds up in the food chain threatening creatures at the top of the chain.
Task 2.2eiii Describe and explain any advantages and environmental problems associated with using the herbicide called 2,2,3-trichloropropanoic acid. 
Pentachlorophenol is an insecticide approved for use in countries such as the USA and Japan.  Explain in what way it might be similar to DDT.
Explain the chemistry behind why Dieldrin was so useful but is now banned.