Download Alkanes CH4 + Cl2 → CH3Cl + HCl CH3CH3 + Cl2 → CH3CH2Cl +

Organic I (Reactions)
Alkanes
Free-radical substitution
Halogenation of alkanes is a free radical reaction. It results in –H bonds being replaced for
–Hal bonds. The reagents and conditions are to mix the alkane with bromine or chlorine
and expose to ultra-violet light.
The ultra-violet light provides enough energy to break the halogen bond, which forms
species called free radicals.
When the Hal-Hal bond breaks, it does so with one electron from the bond going with
each atom. This is what leads to production of the free-radicals mentioned.
Definition: Homolytic fission is when a bond breaks and each atom in the bond takes one
electron from the bond.
Definition: A free-radical is a species with an unpaired electron.
Definition: A substitution reaction is one in which an atom or group of atoms in a
molecule is replaced by another atom or group of atoms.
Free radicals are very reactive species, as energy is released when they pair up their
electrons to make new covalent bonds.
CH4 + Cl2
è
CH3Cl + HCl
CH3CH3 + Cl2
è
CH3CH2Cl + HCl
Free radical reactions such as this one are often very random and unpredictable. For
example, there is no way on controlling which –H bonds are replaced, and further
substitution may also occur…
CH3Cl + Cl2
è
CH2Cl2 + HCl
So, a mixture of products is obtained, which makes this an inefficient synthesis of
halogenoalkanes. In addition, mixtures of alkanes and chlorine are often explosive when
radiated with ultra-violet light.
Combustion
Complete combustion of alkanes gives CO2 and H2O.
CH4 + 2O2
è
CO2 + 2H2O
These reactions are very exothermic and burning alkanes and other hydrocarbons are
commonly used as fuels.
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Organic I (Reactions)
Naming of alkanes
Alkanes, and all other organic compounds, are named according to the longest unbroken
chain of carbon atoms they possess. The suffix ‘ane’ indicates that the compound is an
alkane.
Compound
CH4
CH3CH3
CH3CH2CH3
CH3CH2CH2CH3
CH3CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2 CH2CH2CH2CH3
Name
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
NB Octane, as here named systematically, is not the same as octane in the motor industry.
That ‘octane’ is actually ‘iso-octane’, a chain isomer of the octane shown above.
If alkanes have side chains, then these are denoted with prefixes and, if necessary,
numbers to indicate where they are on the chain.
Sidechain
CH3CH3CH2CH3CH2CH2etc
Prefix
methyl
ethyl
propyl
etc
The prefixes only have to be numbered if leaving them out would make the name
ambiguous. It would be very helpful to draw these out as displayed formulae (where you
can see every bond, or at least the C-C bonds).
Compound
CH3CH(CH3)CH3
(CH3)3CH
(CH3)4C
CH3CH2C(CH3)2CH3
CH3CH(CH3)CH(CH3)CH3
Name
methyl propane
also methyl propene
dimethyl propane
2,2-dimethyl butane
1,2-dimethyl butane
The carbon chain should be numbered so as to give smaller numbers to the substituents.
Compound
CH3CH2CH2CH2C(CH3)3
Name
2,2-dimethylhexane
Alkanes can also exist as cyclic compounds, in which the atoms join up in a ring. If so, the
prefix ‘cyclo’ is used to indicate them.
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Organic I (Reactions)
Alkenes
Addition reactions
Alkenes undergo addition reactions with hydrogen, halogens (such as bromine),
hydrogen halides and potassium manganate(VII).
Definition: An addition reaction is one in which two species combine to make single
product. The reactions with bromine and potassium manganate are used as tests for
alkenes.
Tests for alkenes
• Test 1 – Mix the alkene with a solution of bromine.
ü Decolourisation, from brown to colourless.
Bromine can be used either in organic solvents such as tetrachloromethane or hexane,
alternatively it may be used in aqueous solution. In any case, the positive result is
decolorisation of the solution. It would be good to indicate some practical awareness of
the mixing would occur:
§
§
§
gaseous alkenes – bubble alkene through bromine solution.
liquid alkene – shake alkene with bromine solution.
solid alkene, dissolve alkene in inert solvent, then shake with bromine solution.
If bromine water is used, then the product obtained is a bromo-alcohol, as water
participates in the reaction as a nucleophile in the second step of the mechanism.
CH3CH=CH2 + Br2 + H2O
è
CH3CH(OH)CH2Br + HBr
If bromine in an inert solvent is used, then the product obtained is a dibromoalkane.
CH3CH=CH2 + Br2
è
CH3CHBrCH2Br
These addition reactions could also be classified as an oxidation of the alkene.
•
Test 2 – Mix the alkene a cold dilute aqueous solution of alkaline potassium manganate
(VII).
ü Decolourisation, from purple to green to colourless. A brown precipitate of MnO2
forms.
In this test, the alkene is turned into a diol.
CH3CH=CH2 + H2O + [O]
è
CH3CH(OH)CH2OH
KMnO4 reacts with many substances, primary and secondary alcohols, aldehydes and
even some ketones. What makes this test specific to alkenes is the conditions. When the
KMnO4 is alkaline, and cold, and dilute, it is such a poor oxidising agent that only alkenes
are sufficiently reducing to react with it.
This addition reaction can also be classified as an oxidation of the alkene.
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Organic I (Reactions)
Other reactions of alkenes:
•
Hydrogen adds to alkenes to form alkanes. Reagents and conditions are pass the gases
over a heated nickel catalyst.
CH3CH=CH2 + H2
è
CH3CH2CH3
This can also be classified as a reduction of the alkene.
•
Hydrogen halides add to alkenes to form haloalkanes. Reagents and conditions are mix
the gases at room temperature.
CH3CH=CH2 + HBr
è
CH3CHBrCH3
When the substance attacking the double bond is unsymmetrical (like HBr), the hydrogen
atom goes on the atom with more H atoms to begin with. This will be studied more in a
later unit, just remember it for now. It is Markovnikov’s rule.
Substances that add to C=C bonds, such as H2, Br2, and HBr are termed electrophiles.
Definition: An electrophile is an electron deficient species that can accept a pair of
electrons from an electron rich region of another molecule.
Naming of alkenes
The suffix ‘ene’ is used to indicate that a compound contains a C=C bond, sometimes with
a number to indicate where it is in the carbon chain. The direction of numbering along the
chain is chosen so as to give smaller numbers to the carbons containing the double bonds.
Compound
CH2=CH2
CH3CH=CH2
CH3CH2CH=CH2
CH3CH=CHCH3
Name
ethene
propene
but-1-ene
but-2-ene
Draw ethene, showing all the bonds.
If there are alkyl groups or halogen atoms to denote in the name, then the position of the
double bond takes precedence.
Compound
CH2=CHCH2CH2CH2C(CH3)3
Name
6,6-dimethyl hept-1-ene
Multiple double bonds are named as follows:
Compound
CH2=CHCH=CH2
CH2=C=CHCH2
Name
buta-1,3-diene
buta-1,2-diene
This a in buta is just to make the name easier to say, and you can probably forget about it.
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Organic I (Reactions)
Halogeno alkanes
Substitution reactions
Definition: A substitution reaction is one in which an atom or group of atoms in a
molecule is replaced by another atom or group of atoms.
Definition: A nucleophile is an electron rich species that is attracted to an electron
deficient region of a molecule and can donate a pair of electrons to form a new covalent
bond. The important point is the ability to donate a pair of electrons.
•
Substitution with hydroxide (hydrolysis)
The reagents and conditions for hydrolysis of halogenoalkanes are boil under reflux with
dilute aqueous sodium hydroxide.
CH3CH2Br + NaOH
•
è
CH3CH2OH + NaBr
Substitution with cyanide (cyanolysis)
This just like hydrolysis but the nucleophile is CN-. The reagents and conditions are to boil
the halogenoalkane under reflux with sodium or potassium cyanide in a solvent such as
ethanol and water, or propanone.
CH3CH2Br + NaCN
•
è
CH3CH2CN + NaBr
Substitution with ammonia (aminolysis)
This is just like hydrolysis but with ammonia as the nucleophile. The reagents and
conditions are heat the halogenoalkane with excess concentrated ammonia in ethanol in a
sealed tube.
CH3CH2Br + NH3
è
CH3CH2NH2 + HBr, or
CH3CH2Br + 2NH3
è
CH3CH2NH2 + NH4Br
In the above reactions, the OH-, CN- and NH3 molecules are behaving as nucleophiles.
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Organic I (Reactions)
Halogeno alkanes
Elimination reactions
Definition: An elimination reaction is one where the elements of a small molecule are
removed from a larger one.
§
§
§
§
§
Halogenoalkanes undergo elimination by removal of the halogen atom and a hydrogen
atom.
The hydrogen atom must come from a carbon atom that is adjacent to the carbon atom
that loses the halogen atom.
The C=C is formed between the two atoms that lose the hydrogen and halogen atoms.
Often there is more than one hydrogen atom that can be removed, in which case the
product will be a mixture of molecules.
Sometimes the products can exist as cis and trans isomers.
The reagents and conditions for elimination are heat strongly under reflux with
concentrated potassium or sodium hydroxide in ethanol. Under these conditions, the OHion behaves as a base by removing a hydrogen atom from the molecule.
CH3CH2Br + NaOH
è
CH2=CH2 + H2O + NaBr
In the above equation, the “elements of a small molecule” that have been removed are
those of “HBr”, although an actual molecule of HBr is not created.
When more than one product is possible the one with the greater number of alkyl groups
present on the C=C is formed in greater yield. When geometrical isomers are possible, cis
and trans are formed in equal yield.
CH3CH2CHBrCH3
è
è
CH3CH2CH=CH2 (smaller yield)
CH3CH = CHCH3 (greater yield, cis and trans 1:1)
Halogenoalkanes are more prone to elimination if they are more branched, i.e. tertiary
eliminates more easily than secondary than primary. Branching provides more hydrogen
atoms for the reaction to start with, and also sterically hinders the carbon atom thus
preventing the OH- from acting as a nucleophile and causing a substitution reaction
instead.
Very important point: In substitution the OH- behaves as a nucleophile (electron pair
donor), whereas in elimination it behaves as a base (proton acceptor).
Draw 2-bromobutane showing all bonds. Label the H atoms that can be lost to form but1-ene and those that can be lost to form but-2-ene.
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Organic I (Reactions)
Test for halogen atoms in halogeno alkanes, Cl, Br and I
Step one - Free the halide ion from the organic molecule. How this is done depends on the
type of molecule, but for simple halogeno alkanes it is done by heating the halogenoalkane
under reflux with dilute, aqueous sodium hydroxide solution.
Step two – add excess dilute nitric acid, to neutralise excess NaOH or other alkaline
impurities. Any alkaline impurities will give a grey precipitate when silver nitrate is
added, which could be interpreted as a false-positive result.
Step three – add silver(I)nitrate solution.
ü Chloride ions will give a white precipitate
ü Bromide ions will give a cream precipitate
ü Iodide ions will give a yellow precipitate
Step four – filter off the precipitate and test its solubility in ammonia solution:
1. AgCl will dissolve in dilute ammonia solution
2. AgBr will dissolve in concentrated ammonia solution
3. AgI will not dissolve in any concentration of ammonia solution.
See group VII notes for more details of this test for halide ions.
Naming of halogenoalkanes
Halogen atoms are specified with a prefix and a sometimes a number to specify where the
halogen is on the carbon chain. The chain is numbered from the direction that gives the
smaller number of set of numbers for atoms which have halogen atoms attached.
Halogen
F
Cl
Br
I
Prefix
Fluoro
Chloro
Bromo
Iodo
Compound
CH3Cl
CH3CHBrCH3
CH3CHBrCH2CH2CH3
Name
chloromethane
2-bromopropane
2-bromopentane
The prefixes are modified if more than one halogen is present. They are given
alphabetically according the name of the halogen. Alkyl groups can also be included.
Compound
CH3CBr2CBrCl2
CH3CH(CH3)CHF2
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Name
1,2,2-tribromo-1,1-dichloro propane
1,1-difluoro-2-methyl propane
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Organic I (Reactions)
Alcohols
Oxidation of alcohols
Reagents and conditions are to heat with potassium dichromate solution acidified with
dilute aqueous sulphuric acid. You must specify the acid, writing H+ will not get you full
marks.
•
Primary alcohols are oxidised to aldehydes then carboxylic acids.
CH3CH2OH
è
CH3CHO è
CH3CO2H
•
In order to obtain the aldehyde, the reaction is carried out with excess alcohol and the
aldehyde is distilled off as it is formed. This is possible because the aldehyde has the
lowest boiling point of the three compounds, because it does not have hydrogen bonds
between its molecules. Reagents and conditions are add potassium dichromate(VI) in
dilute sulphuric acid to the hot alcohol and allow the aldehyde to distil off.
•
In order to obtain the carboxylic acid, the reaction is carried out with excess oxidising
agent and the heating is carried out under reflux. This is in order to maximise yield.
Reagents and conditions are boil the alcohol with excess potassium dichromate(VI) in
dilute sulphuric acid under reflux.
In organic oxidation reactions, equations can be balanced using the symbol [O].
CH3CH2OH + [O]
è
CH3CHO + H2O
CH3CHO + [O]
è
CH3CO2H
•
Secondary alcohols are oxidised to ketones.
CH3CH(OH)CH3 + [O]
•
è
CH3COCH3 + H2O
Tertiary alcohols are not oxidised under these conditions.
Reaction of alcohols with dehydrating agents
Alcohols can be dehydrated to form alkenes. The reactions can be classified as
dehydration or elimination.
Reagents and conditions are heat under reflux at 170oC with excess conc. H2SO4, or
heat at 300oC over Al2O3.
You should be able to predict the alkenes that would form from a typical alcohol, such as
butan-2-ol. The OH group is removed, along with any hydrogen atom attached to a carbon
next to the carbon with the OH group. Note that geometrical isomers may form.
CH3CH2CH(OH)CH3 è
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CH3CH=CHCH3, major product, cis and trans.
CH3CH2CH=CH2, minor product.
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Organic I (Reactions)
Reaction of alcohols with halogenating agents
Halogenating agents replace the –OH group with a halogen. You need to be familiar with
three different ones for this unit.
§
Method one – PCl5(s)
Reagents and conditions are carefully add solid PCl5 at room temperature, under dry
conditions.
CH3CH2OH + PCl5
§
è
CH3CH2Cl + POCl3 + HCl
Method two – NaBr(s) and conc. H2SO4
NaBr(s) and conc. H2SO4 react to form HBr (but see group VII notes). Reagents and
conditions add a mixture of the alcohol and conc. H2SO4 to NaBr at room temperature.
CH3CH2OH + HBr
§
è
CH3CH2Br + H2O
Method three – phosphorous and iodine
Phosphorous and iodine react together to form PI3. Reagents and conditions are add the
alcohol to moist red phosphorous and iodine at room temperature.
3CH3CH2OH + PI3
è
3CH3CH2I + H3PO3
Test for –OH groups in alcohols (and carboxylic acids).
The PCl5 reaction above is used to test for an OH group. Note that it works for any
compound that contains an OH group, so both alcohols and carboxylic acids, so it is not a
test for an alcohol specifically. Water also contains an –OH group, so the conditions must
be anhydrous to avoid a false-positive result.
•
Carefully add a small amount of solid PCl5 to a small amount of the sample in a clean
dry test tube, in a fume cupboard at room temperature. The sample must be dry.
ü Effervescence of steamy fumes (HCl) which turn moist blue litmus paper red and give
a dense white smoke (NH4Cl) with an ammonia stopper.
PCl5 + ROH
è
POCl3 + RCl + HCl(g)
HCl(g) + NH3(g)
è
NH4Cl(s)
Using dichromate as a test for alcohols.
If heating a test substance with excess potassium dichromate(VI) in dilute sulphuric acid
turns it green, then the test substance would be a primary or secondary alcohol, but
remember, it could also be an aldehyde.
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Organic I (Reactions)
Naming of alcohols and related compounds
Alcohols are denoted with the suffix ‘ol’, with a number, if necessary, to indicate where
the –OH group is in the molecule.
Compound
CH3OH
CH3CH2OH
CH3CH2CH2OH
CH3CH(OH)CH3
Name
methanol
ethanol
propan-1-ol
propan-2-ol
Draw methanol, showing all the bonds.
Substituents can be included by using prefixes with numbers, if necessary, to indicate their
positions.
Compound
CH3CH(CH3)CH2OH
(CH3)3COH
C6H5CH2CH2OH
CCl3CF2OH
Name
2-methyl propan-1-ol
2-methyl propan-2-ol
2-phenyl ethanol
2,2,2-trichloro-1,1-difluoro ethanol
When primary alcohols are oxidised they initially form aldehydes, which contain the
–CHO group. Aldehydes are denoted with the ‘al’ suffix.
Compound
HCHO
CH3CHO
CH3CH2CHO
CH3CH2CH2CHO
Name
methanal
ethanal
propanal
butanal
Draw ethanal, showing all the bonds.
Secondary alcohols are oxidised to ketones, which contain the >C=O group.
Compound
CH3COCH3
CH3COCH2CH3
CH3CH2COCH2CH3
CH3COCH2CH2CH3
Name
propanone
butanone
pentan-3-one
pentan-2-one
Draw propanone, showing all the bonds.
Aldehydes are further oxidised to carboxylic acids, which contain the –CO2H group. The
–CO2H group is denoted with the suffix ‘oic acid’.
Compound
CH3CO2H
CH3CH2CO2H
Name
ethanoic acid
propanoic acid
Draw ethanoic acid, showing all the bonds.
Substituents can be indicated with prefixes and numbers, numbering from and including
the C in the –CO2H group.
Compound
CH3CH(CH3)CH2CH2CO2H
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Name
4-methyl pentanoic acid
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