Transcript ORGANIC CHEMISTRY

ORGANIC CHEMISTRY
TIERS 5 & 6
TIER 5
•Predict and explain trends in boiling points of members of a homologous series
•Discuss the volatility and solubility in water of compounds containing the functional groups: compounds
up to six carbon atoms with one of the following functional groups: alcohols, ketones, aldehydes,
carboxylic acid, and halides .
•Describe using equations the complete and incomplete combustion of alkanes
•Describe using equations the reactions of methane and ethane with chlorine and bromine
•Describe using equations the reaction of alkenes with hydrogen and halogens
•Describe using equations, the reactions of symmetrical alkenes with hydrogen halides and water
•Distinguish between alkanes and alkenes using bromine water
•Describe using equations the complete combustion of alcohol
•Describe using equations the oxidation of alcohols
•Describe using equations the oxidation of primary and secondary alcohols
•Describe using equations the substitution reactions of halogenoalkanes with sodium hydroxide
TIER 6
•Explain the reactions of methane and ethane with chlorine and bromine in terms of a free radical
mechanism
•Outline the polymerization of alkenes
•Outline the economic importance of the reaction of alkenes
•Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN1 and SN2
mechanisms
•Deduce reaction pathways given the starting materials and the product.
TRENDS IN PHYSICAL PROPERTIES
•Predict and explain trends in boiling points of members of a
homologous series
•Discuss the volatility and solubility in water of compounds
containing the functional groups: compounds up to six carbon
atoms with one of the following functional groups: alcohols,
ketones, aldehydes, carboxylic acid, and halides .
BOILING POINT TRENDS
As the hydrocarbon chain gets larger, the increase in number of electrons increases
the temporary dipoles causing stronger van der Waals’ forces. Two features that
influence the boiling point of alkanes are:
•the number of electrons surrounding the molecule, which increases with the
alkane's molecular weight
•the surface area of the molecule
As a rule of thumb, the boiling point rises 20–30 °C for each carbon added
A straight-chain alkane will have a boiling point higher than a branched-chain alkane
due to the greater surface area in contact with a straight chain, thus the greater van
der Waals forces, between adjacent molecules.
INFLUENCE OF FUNCTIONAL GROUPS ON
BOILING POINT TRENDS
Functional groups which are polar will develop dipole-dipole interactions and
thus will have higher boiling points
Functional groups which enable hydrogen bonding between molecules will
have even higher boiling points
The effect of functional groups on boiling points are as follows:
Most volatile (How easily to change to a gas)
Least volatile
Alkane > Halogenalkane > Aldehyde > Ketone > Alcohol > Carboxylic acid
Van der Waals
dipole-dipole
Increasing strength of intermolecular forces
Increasing boiling point
hydrogen bonding
BOILING POINT
The boiling points of other types of
organic compounds are shown on
the graph to the right
IMPORTANCE OF BOILING POINT
TRENDS
For example, fractional distillation is
used in oil refineries to separate
crude oil into useful substances (or
fractions) having different
hydrocarbons of different boiling
points. The crude oil fractions with
higher boiling points:
have more carbon atoms
have higher molecular weights
are more branched chain alkanes
are darker in color
are more viscous
are more difficult to ignite and to
burn
MELTING POINT TRENDS
The melting point of the alkanes follow a similar trend to boiling point and for the same
reason That is, (all other things being equal) the larger the molecule the higher the melting
point.
However, due to the rigidity of solids, they require more energy to break the intermolecular
forces holding the molecules together The more ordered the molecule the more energy
require to over come the intermolecular forces.
Odd-numbered alkanes have a lower trend in melting points than even-numbered alkanes
because even-numbered alkanes pack well in the solid phase, forming a well-organized
structure, which requires more energy to break apart. The odd-number alkanes pack less
well and so the "looser" organized solid packing structure requires less energy to break
apart.
The melting points of branched-chain alkanes can be either higher or lower than those of the
corresponding straight-chain alkanes, again depending on the ability of the alkane to form
more organized structures.
Melting
point [°C]
Density
[g·cm−3]
(at 20 °C)
Alkane
Formula
Boiling
point [°C]
Methane
CH4
-162
-182
gas
Ethane
C2H6
-89
-183
gas
Propane
C3H8
-42
-188
gas
Butane
C4H10
0
-138
gas
Pentane
C5H12
36
-130
0.626
(liquid)
Hexane
C6H14
69
-95
0.659
(liquid)
Heptane
C7H16
98
-91
0.684
(liquid)
Octane
C8H18
126
-57
0.703
(liquid)
Nonane
C9H20
151
-54
0.718
(liquid)
Decane
C10H22
174
-30
0.730
(liquid)
SOLUBILITY IN WATER
There are two factors to consider when determining the
solubilty of an organic compound in water:
•The length of the hydrocarbon chain
•The nature of the functional group
Solubility decreases as the length of the chain increases
Solubility of functional groups depend on their ability to form
hydrogen bonds with water
Lower members of alcohols, aldehydes, ketones, and carboxylic
acids are quite soluble in water
Halogenoalkanes are not soluble in water despite their polarity
b/c they do not form hydrogen bonds with water
CHEMICAL REACTIVITY
Due to the nature of the strong C-C and C-H non-polar bonds, alkanes will only react in
the presence of a strong source of energy.
Thus alkanes have low reactivity. However, they create highly exothermic reactions due
to the fact that a large amount of energy is released forming the double bonds in CO2
and the bonds in H2O.
Alkanes burned in the presence of excess oxygen produce CO2 and H2O by the following
reaction:
CxHy +O2  CO2 + H2O
If oxygen supply is limited then the reaction is :
CxHy +O2  CO + H2O
If oxygen supply is extremely limited then the reaction will be:
CxHy +O2  C + H2O
Describe using equations the reactions of
methane and ethane with chlorine and bromine
Step 1: Initiation
Step 2: Propagation
UV
Known as homolytic fission where the halogen
molecule is broken down by UV light to produce
two“free radicals “
Step 3: Termination
This chain reaction both use and produce “free
radicals” and so allow the reaction to continue.
OVER ALL EQUATION:
OR
CH4 + Cl2  CH3Cl + HCl
These reactions remove the free radicals and cause them
to pair up their electrons
EXAMINER’S HINT: Be sure to understand that a free radical has an
unpaired electron but no net charge and an ion carries a charge
Describe using equations, the reactions of symmetrical alkenes with hydrogen
halides and water
&
Describe using equations the reaction of alkenes with hydrogen and halogens
Distinguish between alkanes and alkenes
BROMINE WATER TEST:
Since alkenes readily undergo addition reactions ,whereas alkanes will not (they
will only undergo substitution reactions in UV light), brome water is used to
distinguish between the two homologous series.
Bromine water is a reddish-brown liquid that will become colorless in an alkene
but remain a reddish-brown liquid in an alkane.
FLAME TEST:
Alkenes burn with a much dirtier, smokier flame that alkanes due to the fact that
they have a higher C-H ratio. This causes more unburned carbon to burn in
alkenes.
Benzene rings burn even dirtier
Describe using equations the complete combustion of
alcohol
Alcohols react with oxygen to produce carbon dioxide and water
EXAMPLE: Combustion of Ethanol
2C2H5OH + 6O2  4CO2 + 6H2O
Describe using equations the oxidation of alcohols
&
Describe using equations the oxidation of primary and
secondary alcohols
The oxidation of alcohols often involves acidified potassium dichromate as the
oxidizing agent. The Cr 6+ is reduced to Cr 3+. In the reaction below, the
oxidizing agent is represented [O]-
Describe using equations the substitution reactions of
halogenoalkanes with sodium hydroxide