10.1 Introduction
10.1.1 Describe the features of a homologous series.
The compounds of a homologous series:
1.
Have the same general formula
e.g. alkanes – CnH2n+2, alkenes
– CnH2n
2.
If neighbors, differ by one CH2 group
e.g. methane (CH4) and ethane (C2H6)
3.
Have similar chemical properties
4.
Gradually changing physical properties
e.g. the b.p. of alkanes increases with the
number of carbons and with their length
10.1.2 Predict and
explain the trends in boiling points of members of a homologous series.
The boiling point increases as the chain of a hydrocarbon
gets longer. The larger the boiling
point is, the smaller the effect of each added carbon on the boiling point.
10.1.3 Distinguish
between empirical, molecular and structural formulas.
Structural formula
is one, which shows unambiguously how the atoms are arranged together. A full structural formula shows every atom
and bond:
Structural formula
of ethane
Sometime, we can use condensed
structural formula, which shows only atoms and omits bonds. In such formulas we
can use:
1.
R as the symbol for an alkyl group;
2. 
for representing the benzene ring.
for representing the benzene ring.
Empirical formula
shows the overall number of each atom in the compound.
C2H6
Molecular formula of ethane
10.1.4 Structural
isomers
Structural isomers are compounds with the same molecular
formula but different structural formula (different arrangement of atoms).
The isomers of
butane
10.1.5 Deduce
structural formulas for the isomers of the non-cyclic alkanes up to C6.
10.1.6 Apply IUPAC
rules for naming the isomers of the non-cyclic alkanes up to C6.
Number of carbons
|
IUPAC Prefix
|
1
|
Meth
|
2
|
Eth
|
3
|
Prop
|
4
|
But
|
5
|
Pent
|
6
|
Hex
|
To name alkanes we use an IUPAC prefix and the suffix –ane.
Ex: Empirical formula – C6H14, hexane
10.1.7 Deduce
structural formulas for the isomers of the straight-chain alkenes up to C6.
Each carbon can have only four single bonds. If a carbon is
double-bonded to another carbon, it can have only two more single bonds.
10.1.8 Apply IUPAC
rules for naming the isomers of the straight-chain alkenes up to C6.
To name alkenes we use an IUPAC prefix and the suffix –ene.
Ex: Empirical formula – C3H6, this
can be prop-1-ene or prop-2-ene, i.e. the double bond is the second or the first
bond in the alkene.
We always use the smallest numbers in IUPAC nomenclature.
E.g. it is prop-1-ene and not prop-3-ene.
10.1.9 Deduce
structural formulas for compounds containing up to six carbon atoms with one of
the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and
halide.
Single hydrocarbon can contain more than one functional
group. However, the functional groups are treated as different entities only if
they are separated by a -CH2- group!
The functional groups have certain formulas, which are
unique for all of them. These formulas are used to represent the functional
groups in the organic compounds. The formulas of the functional groups are:
1.
Alcohol – OH;
2.
Aldehyde – CHO;
3.
Ketone – CO;
4.
Carboxylic acid – COOH;
5.
Halide – F/Cl/Br/I.
10.1.10 Apply IUPAC
rules for naming the compounds containing up to six carbon atoms with one of
the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and
halide.
Single hydrocarbon can contain more than one functional
group. However, the functional groups are treated as different entities only if
they are separated by a -CH2- group!
1. Alcohol
We recognize primary, secondary and tertiary alcohols.
Primary alcohols – the OH functional group is attached to
carbon with only one more carbon
attached to it.
Secondary alcohols – the OH functional group is attached to
carbon with two more carbon attached
to it.
Tertiary alcohols – the OH functional group is attached to
carbon with three more carbon
attached to it.
Alcohols are shown in text by the suffix –ol.
2. Aldehyde
Aldehydes have a hydrogen and alkyl group attached to a
carbonyl function (RCHO). Aldehydes are shown in text by the suffix –al. They exist only on the ends of a
compound.
3. Ketone
Ketones have a pair of alkyl groups attached to a carbonyl
function (RCOR). Ketones are shown in text by the suffix –one. They can never be on the ends of a compound, only on 2nd
and (n-1)th carbon, where n is the number of carbons in the
hydrocarbon. Thus, the smallest ketone is propanone.
4. Carboxylic acid
Carboxylic acids have an alkyl group attached to a
hydroxyl-carbonyl function (RCOOH). Carboxylic acids are shown in text by the
suffix –anoic.
5. Halide
Hydrogencarbons, which have one or more hydrogens replaced
by a halogen (RX, where X=F, Cl. Br or I). Like with the alcohols halides can
be primary, secondary or tertiary. For explanation refer to the point 1.
10.1.11 Identify the
following functional groups when present in structural formulas: amino (NH2),
benzene ring (
) and esters (RCOOR).
) and esters (RCOOR).
Benzene
With benzene resonance occurs.
Benzene permanently vacillates between two states, so that
neither double bonds, nor single bonds occur.
It is important to remember that the bond in benzene is
neither single nor double. Thus, we use the following symbol to show benzene:
Amines
R—NH2
IUPAC nomenclature: Alkamine
Esters
IUPAC nomenclature: Alk(1)yl
Alk(2)anoate
The alkyl (R') group is named first.
10.1.12 Identify
primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.
In primary alcohol
or halogenalkane the functional group is bonded to a carbon, which has one carbon attached to it.
In secondary
alcohol or halogenalkane the functional group is bonded to a carbon, which has two carbons attached to it.
In tertiary alcohol
or halogenalkane the functional group is bonded to a carbon, which has three carbons attached to it.
10.1.13 Discuss the
volatility and solubility in water of compounds containing the functional
groups listed in 10.1.9.
Volatility and solubility of compounds depends on the
intermolecular forces. We know three kinds of intermolecular forces, whose
strength goes as following:
Van der Waals >
dipole-dipole > hydrogen bonding
Volatility: If
there are strong intermolecular forces (dipoles or hydrogen bonding), a
compound is not very volatile. If there are only weak Van der Waal forces, the
compound probably is volatile.
Solubility: Depends on the difference in
electronegativity of two bonded atoms. If the difference is high, then the
bonding is polar – if the difference is low, then it is non-polar. If we have a
regular structure the polarity may cancel out. For instance, in
tetrafluoromethane, the bond polarity cancels out, so the specie is non-polar.
The fundamental rule: likes
dissolve likes
10.2 Alkanes
10.2.1 Explain the
low reactivity of alkanes in terms of bond enthalpies and bond polarity.
Bond enthalpy: The
bond enthalpy of C-C and H-H bond is very high – this means it takes a lot of
energy to break the bonds. Hence, the C-C and H-H bonds are very strong.
Alkanes are stable in air until sufficient activation energy is introduced.
Polarity: Alkanes
have a very low polarity of the C-H bond (0.4). Thus, they are very stable and
alkanes do not dissolve in polar solvents like water.
10.2.2 Describe,
using equations, the complete and incomplete combustion of alkanes.
All alkanes may undergo combustion with oxygen to form
carbon dioxide and water:
CH4(g) +
2O2(g) → CO2(g) + 2H2O(l)
If there is insufficient oxygen for complete combustion,
alkanes will still burn, but undergo incomplete combustion (this is what occurs
in internal combustion engines such as car engines).
Incomplete combustion may produce elemental carbon, carbon
monoxide or both:
C2H4 + 2O2 → 2CO + 2H2O
C2H4 + O2 → 2C +2H2O
The combustion of unsaturated hydrocarbons is more likely to
be incomplete.
10.2.3 Describe,
using equations, the reactions of methane and ethane with chlorine and bromine.
The
significance of such reactions is that alkanes react readily only with
halogens. The process is called free
radical substitution.
Methane: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
Ethane: CH3CH3(g)
+ Cl2(g) → CH3CH2Cl(g) + HCl(g)
The process requires
UV light, usually from the sunlight.
10.2.4 Explain the
reactions of methane and ethane with chlorine and bromine in terms of a
free-radical mechanism.
Alkanes
react readily only with halogens. The process is called free radical substitution and has three steps:
1.
Initiation – UV light breaks the halogen gas to create
a radical (homolitic fission)
Cl2→2Cl∙
2.
Propagation – radicals used and re-created
Cl∙ + CH3CH3 → CH3CH2∙ + HCl
Cl2 + CH3CH2∙ → CH3CH2Cl + Cl∙
3.
Termination – radicals removed and reaction proceeds to
the end
Cl∙ + CH3CH2∙ → CH3CH2Cl
OR
Cl∙ + Cl∙ →Cl2
OR
CH3CH2∙+ CH3CH2∙ → C4H10
The process requires
UV light, usually from the sunlight.
Homolitic fission – a covalent bond is
broken so that each atom retains one electron from the broken bond; these atoms
are very reactive free radicals.
10.3 Alkenes
10.3.1 Describe,
using equations, the reactions of alkenes with hydrogen and halogens.
Alkenes have a carbon-carbon double bond of which one bond,
the π bond, can be easily broken. As this bond is broken, two new bonding
positions within the compound are created. This enables alkanes to undergo addition reaction with:
1.
Hydrogen -
hydrogenation
·
Hydrogen gas reacts with alkenes to form alkanes
·
Ex: CH3CHCH2 + H2
--> CH3CH2CH3
2.
Halogen -
halogenation
·
Halogens react with alkenes to produce dihalogeno compounds
·
Ex: CH3CHCH2 + Br2 --> CH3CHBrCH2Br
10.3.2 Describe,
using equations, the reactions of symmetrical alkenes with hydrogen halides and
water.
Alkenes have
a carbon-carbon double bond of which one bond, the π bond, can be easily
broken. As this bond is broken, two new bonding positions within the compound
are created. This enables alkanes to undergo addition reaction with:
1. Hydrogen
halides (Ex: HCl, HBr)
·
Hydrogen halides react with alkenes to produce
halogenoalkanes
·
These reactions are very rapid
·
Reactivity: HI>HBr>HCl because the
hydrogen-halide bond weakens down Group 7
2. Water
·
The
reaction between water and turns the alkene into alcohol
·
Addition
reaction of water and halogen requires a catalyst to take place – such catalyst
is concentrated sulfuric acid
10.3.3 Distinguish between alkanes and
alkenes using bromine water.
Because halogens react with alkenes we can
use bromine water to distinguish alkanes from alkenes. Bromine only reacts with
alkenes and when it does, it loses its orange/yellow color.
This test is useful because the addition
reaction with halogens proceeds very quickly at room temperature. Alkanes react
with Br2 only if UV light is present.
10.3.4
Outline the polymerization of alkenes.
Because the C=C bond in alkenes can be easily broken,
alkenes like to undergo additional reactions. Two alkenes, whose double bond is
broken, can react with each other in an addition
polymerization reaction.
The addition
polymerization forms long hydrocarbons, usually containing 40,000 to 80,000
carbons, called polymers. The
repeating building unit of a polymer is called monomer.
The
polymerization reactions, which we should know are:
1. Additional
polymerization of ethane to form poly(ethane).
2. Additional
polymerization of chloroethene to poly(chloroethane), aka, PVC.
3. Additional
polymerization of propene to poly(propene).
Propene poly(propene)
Do
not forget:
We identify the
polymer as a repeating unit.
E.g. for polyethene -(-CH2-CH2-)n-
10.3.5
Outline the economic importance of the reactions of alkenes.
- The hydrogenation of vegetable oils in the manufacture of margarine
- The process of hydrogenation is used in the margarine industry to convert oils containing many unsaturated hydrocarbon chains into more saturated compounds which have higher melting points
- This is done so that the margarine will be a solid at room temperature.
- Margarine
- The hydration of ethene in the manufacture of ethanol
- The hydration of ethene is of industrial significance because ethanol is a very important solvent and is therefore manufactured on a large scale.
- Solvent of substances intended for human contact or consumption, fuel, alcoholic beverages, antiseptic
- The polymerization in the manufacture of plastics
- Polyethene
- household containers, plastic bags, water tanks and piping
- Polypropene
- clothing and thermal wear for outdoor activities
- Polychloroethene
- construction materials, packaging, electrical cable sheeting
- it is one of the world's most important plastics
10.4 Alcohols
10.4.1 Describe, using equations, the
complete combustion of alcohols.
All alcohols burn in presence of oxygen to
make carbon dioxide and water.
Example: Combustion of ethanol
C2H5OH(l) + 3O2 → 2CO2(g) + 3H2O(l)
10.4.2
Describe, using equations, the oxidation reactions of alcohols.
We distinguish partial oxidation and complete
oxidation. Partial oxidation converts
alcohol to aldehyde, while complete
oxidation coverts alcohol to carboxylic acid.
Conditions for oxidation:
1. Excess potassium dichromate
catalyst;
2. The catalyst must be
acidified, for instance, by sulfuric acid (since it is inert);
3. The mixture must be heated;
4. Individual conditions:
§ Partial oxidation to
aldehyde – distill as the reaction goes;
§ Complete oxidation to
carboxylic acid – heat under reflux and then distill.
During the reaction the color of the solution
changes from orange to dark green because orange chromium(VI) is reduced into
green chromium(III).
10.4.3 Determine the products formed by the
oxidation of primary and secondary alcohols.
Primary alcohols are oxidized to form aldehydes and carboxylic acids.
Secondary alcohols are oxidized to form ketones.
10.5 Halogenalkanes
10.5.1 Describe, using equations, the
substitution reactions of halogenoalkanes with sodium hydroxide.
Halogenalkanes react with aqueous hydroxide
ions from sodium hydroxide – the name of this process is nucleophilic substitution. It is called substitution because the hydroxide ion takes place of halogen in a
halogenalkane. It is nucleophlic
because the hydroxide ion is attracted to positive dipole of carbon atom.
The carbon atom has a positive dipole because
it is bonded to a halogen, which has much higher electronegativity than carbon.
The hydroxide ion is a nucleophile because it has a negative charge and thus it
is attracted to positively charged particles.
Substitution does not occur with
fluoroalkanes because the C-F bond is very strong for fluorine’s high
electronegativity.
Example of nucleophilic substitution reaction:
Do
not forget to use the curly arrow notation!
10.5.2
Explain the substitution reactions of halogenoalkanes with sodium hydroxide in
terms of SN1 and SN2 mechanisms.
In nucleophilic substitution, a nucleophile
gets attracted to the localized positive charged of a molecule, repulses
halogen and takes its bonding position.
SN1
The SN1 mechanism is unimolecular
nucleophilic substitution – this means that it starts with one molecule only. It
occurs with tertiary halogenalkanes (tertiary – carbon to which halogen is
attached has three more carbons attached to it). SN1 occurs with
tertiary halogenalkanes because alkyl groups prohibit nucleophiles from
accessing the positive dipole (just imagine a tertiary halogen).
SN1 occurs in two steps:
1. Random collision causes
heterolytic fission of the C-(halogen) bond. Halide ion and a carbocation are
formed. This is the rate determining step because it is very slow.
2. The carbocation reacts
rapidly with the OH- nucleophile.
SN2
The SN2 mechanism is bimolecular
nucleophilic substitution. It occurs with primary halogenalkanes. It occurs in
one step, but for a simplicity we can divide it in multiple phases (remember
that these phases cannot be distinguished):
1. Nucleophile becomes
attracted to the localized positive charge created by uneven sharing of
electrons between a carbon (α+) and halogen (α-).
2. Nucleophile comes to the
localized positive charge and, at the same time, repulses the halogen. Because
the halogen has higher electronegativity than the carbon, as it gets repulsed a
heterolytic fission occurs – the halogen takes both electrons from the
C-(halogen) bond.
3. The halogen leaves the halogenalkane
with both electrons from the C-(halogen) bond making a halide ion. The OH-
nucleophile takes its bonding position.
The position, in which both nucleophile and
halogen are equally strongly attracted to the carbon is called transition state.
Do not forger to use curly arrows to indicate
the movement of electrons!
10.6.1
Deduce reaction pathways given the starting materials and the product.
No comments:
Post a Comment