Transcript Chemistry 3.5 - CashmereChemistry

Chemistry 3.5
Describe the structure and
reactions of organic compounds
containing selected organic
groups
You will notice the first thing the 3.5
standard says is :
You will be expected to know the
principles of Organic Chemistry you
covered in AS 2.5
How well do you know your stuff
from last year?
The standard then goes on to say….
Organic compounds described are limited
to those containing no more than 8 carbon
atoms
Don’t think you get off that easy
It then goes on to say ………
Larger organic molecules may be used in
questions involving the application of
organic principles (e.g. the identification of
functional groups)
So lets start with Alkanes
1. What’s the general formula for an
alkane?
CnH2n+2
Can you draw the structural formula and
molecular formula for the following?
Methane
Molecular Formula - CH4
Structural Formula
Drawing a 3 dimensional Structure
How about drawing the 3 dimensional
structure for methane?
Use the molymods to make a 3 D model
Any ideas on how to draw it?
These two lines
represent bonds
on the same
plain as the
paper
These lines represent
the bond going behind
the paper
This wedge
represents the
bond coming
out from the
paper
109
o
What is the bond angle
between each of the
atoms?
109 o
Describe the shape of this molecule
Tetrahedral
Is CH4 polar or non polar?
Non polar
Can you give a reason why?
Alkane Nomenclature - Give the names, molecular and
condensed structural formula for the first ten alkanes
Name
Molecular
Condensed Structural Formula
methane
CH4
CH4
ethane
C 2H 6
(CH3)2
propane
C 3H 8
CH3CH2CH3
butane
C4H10
CH3(CH2)2CH3
pentane
C5H12
CH3(CH2)3CH3
hexane
C6H14
CH3(CH2)4CH3
heptane
C7H16
CH3(CH2)5CH3
octane
C8H18
CH3(CH2)6CH3
nonane
C9H20
CH3(CH2)7CH3
decane
C10H22
CH3(CH2)8CH3
Can you remember the alkane
order?
A simple mnemonic isMany
Methane
Ethane
Elderly
Propane
People
Buy
Butane
Pent
Pentane
Hexane
Houses
Heptane
High
Octane
Over
Nonane
North
Dakota
Decane
Cyclic Alkanes
These are Alkanes with at least 1 ring of carbons eg cyclohexane
Draw the structure of cyclohexane in your book?
Cyclohexane
Draw cyclopropane
The general formula for
an alkane with one ring is:
CnH2n
Do you remember how to name a branched alkane?
methyl
Pentane (parent
chain)
methyl
Steps
1. Find the longest chain of continuous carbons (called the parent
chain) and name it:
ie 5 carbons – name pentane
2. Identify any side branches or functional groups
ie methyl if there are more than one of the same type use prefixes
di (2) , tri (3) etc. 2 methyls = dimethyl.
So the name so far is dimethylpentane
3. Number the parent chain from the end that gives the side branch
groups the lowest number
Steps continued
3. Number the parent chain from the end that gives the
side branch groups the lowest number ie from the left
methyl
1
2
3
4
5
methyl
4. Separate numbers by comma’s and separate numbers
from words with a dash – now you can name it
2,3-dimethylpentane
Exercise – Draw the structures for the following
a) 3 – ethylheptane
b) 2,2,4-trimethylpentane
Lets look at making some structure
Turn to Expt 1 in your booklet
In pairs make and draw the structures for
Q 1 to Q 3
Answer the questions
Structure
CH3CH2C
Classification
CCH3
CH3CH2CHCH2CH2CH2CH2CH3
alkyne
O
CH3CH2CH C
CH3
OH
H
H
C C
H3C
CH2CH3
CH3CHCH2CH3
Br
CH3CH2CH2CH2CHCH2CH3
CH3
pent-2-yne
alcohol
(secondary)
OH
Name
carboxylic
acid
octan-3-ol
2-methylbutanoic
acid
alkene
cis-pent-2-ene
haloalkane
2-bromobutane
alkane
3-methylheptane
Structure
CH3CH2
Classification
Name
CH3
C
H3C
C
alkene
2,3-dimethylpent-2-ene
CH3
CH3
CH3CH2CH2C OH
CH3
alcohol
2-methylpentan-2-ol
(tertiary)
O
H C
O CH2CH2CH2CH3
CH3
CH3CH2CHCH2CHCH3
CH3
ester
butyl methanoate
alkane
2,4-dimethylhexane
Structure
CH
C
CH
CH3
Classification
alkyne
Name
3-methylbut-1-yne
CH3
CH3CHCH2CHCH3
CH3
haloalkane
2-chloro-4-methylpentane
Cl
CH3CH2CH2CH2CH2CH2COOH
carboxylic
acid
heptanoic acid
O
CH3CH2 C
O CH2CH2CH3
ester
propyl propanoate
Cholesterol is a major component of gallstones.
From the following structure of the compound
predict its reaction with ..
CH3
C8H17
CH3
HO
cholesterol
(a) Br2
(b) H 2 with a Pt catalyst
(c) CH3COOH
The reaction with Br2 results in addition of bromine
to the double bonded carbons forming a single
carbon bond. The solution would change colour
from orange to colourless.
CH3
CH3
HO
Br
Br
C8H17
With H2 and a Pt catalyst a hydrogenation reaction
would occur and H atoms would be added across
the double bond forming a single C – C bond.
CH3
CH3
HO
H
H
C8C17
Ethanoic acid reacts with the hydoxy group to form
an ester and water
CH3
CH3
O
H3C
C O
+ H2O
C8H17
We have some new functional groups to
learn this year
Title your page Functional groups
Use the photocopied sheet and copy
the complete the table neatly into your
exercise book
Yes the whole table!
We must learn these!
Complete the task on the handout , glue into
your lab book
Use your chart to help you classify and name
the listed compounds (complete in pencil)
Answers to the left hand column on handout
CH3CH2CH2CH2CHCH2CH3
Cl
Class haloalkane Name 3-chloroheptane
O
CH3CH2CH2 C
Cl
Class acyl chloride Name butanoyl chloride
HO
CH3
C CH2CHCH3
O
Class carboxylic acid Name 3-methylbutanoic acid
Answers to the left hand column continued
CH3CH2CH2CH2NH2
Class amine
Name 1- aminobutane
O
CH3CH2CH2CH2CH2C
NH2
Class amide Name hexanamide
CH3CH2CH2CH2 O
Class
ester
C CH2CH2CH2CH3
O
Name butyl pentanoate
Answers to the Right hand column
CH3CH2CH2COCH2CH3
Class ketone
Name hexan-3-one
CH3CH2CH2CH2CH2OH
Class alcohol
Name pentan-1-ol
O
CH3CH2CH2CH2CH2C
H
Class
aldehyde
Name hexanal
Answers to the right hand column continued
CH3CH2CH2CH2CH2COCl
Class acyl chloride
Name hexanoyl chloride
CH3CH2CHCH3
NH2
Class amine Name 2-aminobutane
CH3CH2CH2CH2CH2CH2CH2CONH2
Class
amide
Name octanamide
Structural Isomers (are also called constitutional
isomers)
These are molecules with the same molecular formula but
different structural formula.
The isomers of a particular molecule will have different
physical properties e.g. melting and boiling points. They may
also have different chemical properties.
Draw and name the structural isomers of C4H10
Name: butane
Boiling point 36.1o C
Name: 2-methylpropane
Boiling point -11.7o C
Task – in pairs use the models to
make hexane
Draw the structural formula for
hexane
Then make as many structural
isomers of hexane as you can
Name and draw each one
There should be 5?
Structural Isomers of Hexane
Hexane
H
H
H
C6H14
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
CH3 H
H
H
C
C
C
C
C
H
H
H
H
H
H
H
2-methylpentane
Structural Isomers of Hexane
H
H CH3 H
H
C
C
H CH3 H
H
C
C
H
2,2-dimethylbutane
CH3
H
H CH2 H
H
C
C
C
C
H
H
H
H
H
3-methylpentane
Structural Isomers of Hexane
H
H CH3 H
H
C
C
C
C
H
H CH3 H
H
2,3-dimethylbutane
Geometric (cis and trans) Isomers
Geometric Isomers will only occur if….
The compound has a double or triple bond
where there can be no rotation around the
C
C bond
Remember alkanes don’t exhibit geometric
isomerism because there is rotation around
the single C C bond
To exist as geometrical isomers the C atoms at
both ends of the double bond must each have
two different groups (or atoms attached).
e.g. The Geometric Isomers of but-2-ene
cis-but-2-ene
trans-but-2-ene
Bpt 3.7o C
Bpt 0.88 o C
* Geometric isomers have similar chemical properties
but different physical properties
Geometric isomerisim does not occur if one of
the carbon atoms in the fixed (ie the double
bond) has two identical atoms or groups of
atoms attached
But-1-ene does not have geometric isomers, because
it has two groups, H atoms, attached to the C atoms
on either side of the double bond
flip it over and it’s the
same as
Therefore but-1- ene does not have geometric
isomers
Geometric isomers are a form of
stereoisomerisim
Stereoisomerism – are where the atoms are bonded in the
same sequence but are arranged differently in space in a
molecule.
e.g. but-2-ene
Same sequence of atoms
Two different geometric
isomers exist where
atoms are arranged
differently in space
cis-but-2-ene
trans-but-2-ene
Bpt 3.7o C
Bpt 0.88 o C
Exercise
Draw the structures of the following alkenes and decide
which of them can exist as cis-trans isomers
a) 2-methylbut-2-ene
forms no cis/trans isomers
b) 3 – methylpent-2-ene
cis-methylpent-2-ene
Occurs as cis trans isomers
trans-methylpent-2-ene
Identify whether cis trans isomers occur
with in the following molecules
H3C
CH3
C
Cl
C
Cl
C
CH3
No
H3C
H
C
Cl
C
CH3
Yes
HO
H
C
CH3
Yes
H3C
H
C
H
H3C
C
CH3
C
CH3
H
C
H
Geometric isomers have the same chemical
properties, but different physical properties.
Why?
Because cis isomers have bulky side groups
and cannot be packed closely together, this
causes weaker intermolecular forces between
molecules and therefore lower melting points
than the trans version
Cl
H
C
H
Cl
C
Cl
C
Cl
H
C
H
However cis forms are sometimes polar
(as above) and therefore have stronger
intermolecular forces between molecules
causing higher melting points.
Testing for Cis - Trans Isomers
Weigh out 2 grams of maleic acid into a 50ml beaker
Add 4mls of water and warm slightly to dissolve the acid
Pour this into a pear shaped flask
Carefully add 5mls of concentrated HCl
Place a condenser on top of the flask and secure it in a retort stand with a water
bath. Then warm the solution until a solid forms.
Cool the solution to room temperature by placing the flask in a cold water bath
ie a 250ml beaker of cold water
Pour the solution into an evaporating dish, pouring off the excess liquid then
carefully rinse with water – Then complete task B on page 159
COOH COOH
C
C
H
H
Maleic acid
(cis isomer)
H
H
O
C
C
C
H
C
H
Give the names and structural formula for the following
substances from their condensed structural formulae.
(c)
H
(d)
CH3CH2CHCHCH2CH2CH3
H
H
H
H
H
H
H
C
C
C
C
C
C
C
H
H
H
H
H
H
hept-3-ene
(CH3)3COH
CH3
CH3
C
CH3
OH
2-methylpropan-2-ol
Give the names and structural formula for the following
substances from their condensed structural formulae.
(a)
CH3CH2CHClCH2CH3
H
(b)
H
H
Cl
H
H
C
C
C
C
C
H
H
H
H
H
H
3-chloropentane
(CH3)2CHCH2CH2CHCH2
CH3
CH3 H
H
H
C
C
C
C
H
5 -methylhex-1-ene
C
H
H
H
H
Another form of Stereoisomerism is
Optical isomerism
E,Z Nomenclature of bond geometry
In cis and trans – nomenclature the ‘like” groups are identified and
used to specify the type of isomer.
With E,Z rules the pair of substituent's at each end of the double
bond are given a priority.
Highest priority = atom of highest atomic no attached directly to
the double bond
Eg 2-chloro-3-methylpent-2-ene
CH3
CH2 -CH3
C
C
Cl
1st pair Cl
highest priority
CH3
2nd
pair CH2CH3
highest priority
This isomer has the
highest priority groups
on opposite sides of
double bond therefore
is the E isomer
The Z isomer has the
highest priority groups
on the same side of the
double bond
Optical Isomerism
Optical isomers involve an asymmetric
carbon – a carbon bonded to four different
atoms or groups of atoms such as:
H, OH, CH3, C2H5, C3H7 etc
A molecule with an asymmetric carbon is
known as a chiral molecule. The two forms
of the chiral molecule are known as
enantiomers or optical isomers
Structural formula of
3-D diagram
butan-2-ol
of butan-2-ol
C 2H 5
C 2H 5
CH3
C*
C*
OH
H
C* = asymmetric carbon
H 3C
OH
H
3-D optical isomers of butan-2-ol
H 3C
C 2H 5
C 2H 5
C*
C*
OH
HO
H
H
CH3
Mirror line
Optical isomers are:
mirror images of each other
Are not superimposable on each other
Optical activity
Glucose is an optically
active compound.
On the straight-chain form
of glucose shown here, all
four of the carbons in the
middle of the chain are
chiral centres – they each
have four different groups
attached to them.
Four different groups can be arranged around a central
atom in two different ways. These are optical isomers and
are mirror images of each other and cannot be
superimposed — just as your left and right hands are
mirror images and cannot be superimposed.
The structure for glucose shows four chiral centres,
which means a total of 16 different forms are possible.
Glucose is just one of those forms.
Optical isomers are
Molecules which have one or more chiral centres
rotate plane-polarised light.
Molecules which are mirror images of each other
rotate plane polarised light in opposite directions.
If we add a second grill at right angles to the
first, it will block the red wave.
Two polarising filters placed at right angles will
block all light.
Two sheets of polarising film are placed on an overhead
projector stage. The arrows on the sheets show their
orientation.
On the left we see that when the two sheets are
orientated in the same direction light passes through
them, but when the sheets are at right angles (right) the
light is blocked.
Beakers of water and sucrose solution (which is cheaper
than glucose and also optically active) are placed on a
sheet of polarising film sitting on the overhead stage. A
second sheet of polarising film is on top of the beakers,
at right angles to the first.
Water Sucrose
Although the film above the water beaker is dark, light
shines through the film above the sucrose solution.
The sucrose has rotated the light waves sufficiently so
that they are able to pass through the second film.
Water Sucrose
When we rotate the top sheet, the film above the
sucrose solution is now dark, while light passes through
everywhere else.
Different wavelengths (colours) of light are rotated by
different amounts, so that as the polarising film is
rotated by different angles, we see these different
colours.
Remember:
• Optically-active solutions rotate plane-polarised light
• optical isomers rotate plane-polarised light by equal
angles in opposite directions.
Optical Isomer Properties
Many organic compounds have optical
isomers including hormones and
enzymes involved in biochemical
reactions.
The shape of a molecule is very
important in these reactions and this
means that the mirror image (ie the
optical isomer) of an enzyme will not
work properly in the body.
Optical Isomer Properties
Optical isomers have identical chemical
and physical properties except that they
rotate the plane of polarised light in
opposite directions.
Optical isomers react differently with
other optically active compounds.
an equal mixture of both enantiomers is
called a racemic mixture
The drug thalidomide, prescribed as a treatment for
morning sickness in early pregnancy in the 1960s,
tragically caused the development of serious birth defects
(badly deformed limbs, or none at all).
In fact, only one of the two optical isomers of
thalidomide appear to cause these birth defects, although
it could be that once ingested each isomer readily
changes into the other form. Today thalidomide is being
used successfully as a treatment for leprosy (although
not for pregnant women).
Exercise
Draw the structure of each of the following molecules and then
decide which ones are optically active.
Mark the chiral carbon with an asterisk.
(a) 2-chlorobutane
H3C *CH CH2 CH3
Cl
(b) 2-methylpropanoic acid
O
H3C HC
C
OH
CH3
(c) 3-methylpentanal
O
*
C CH2 CH CH2 CH3
H
H3C
No optical
isomers
(d) 2-aminobutanoic acid
*
O
H3C CH2 CH C
NH2
OH
Homework
Read unit 28 page 111 in pathfinder
Complete Q’s 4 and 5 page 114
Complete Enantiomers on BestChoice
before next Friday please
Answer questions on alkanes and alkenes
page 172 -174
Group work exercise
Each group is to work on problems
giving great detailed answers. But all
people in the group must have the
answers written in their lab books
Random people from each group will
be asked for their groups answer
Turn to page 6 in your booklet
Properties of alkanes and alkenes
Demo Alkene addition
Turn to page 145 in the year 13 lab
book
Comparison: Alkanes vs Alkenes
Complete Structural isomer
starter
Complete Worksheet two
Q’s 1 and 2 in organic booklet
Alkane Reactions
Alkanes are used as fuels and undergo combustion
reactions
In excess air (oxygen) they form the products H2O and CO2
In limited air (oxygen) they form the products H2O and C or
CO2
Because the carbon atoms in alkane molecules
are saturated with hydrogens (ie they don’t have
any double or triple bonds) they are called
saturated hydrocarbons.
Properties of Alkanes
Insoluble in water
Soluble in non polar solvents
Don’t conduct – no free electrons
Float on water because H2O is relatively denser
Boiling/melting point increases with chain length
because as molecular mass increases the
intermolecular forces between molecules
increases
Questions
1. Name the type of intramolecular bonding
and the type of intermolecular bonding in:
a) Methane
b) Liquid pentane
2. Explain in terms of bonding why:
a) Methane gas can be collected over water
b) Petrol floats on water
c) Oil dissolves in petrol
Alkane Reactions
Because alkanes have no double bonds they react slowly
with halogens in the presence of UV light.
This reaction is called a substitution reaction in which an
H atom is replaced by a halogen atom (eg Cl or Br)
UV
light
butane
+
bromine
bromobutane + hydrogen
bromide
What’s missing in this reaction?
The bromine solution decolourises slowly and the HBr
formed is an acidic gas that turns moist blue litmus red
Alkenes (CnH2n) and Alkynes ( Cn H2n - 2 )
These are unsaturated hydrocarbons
This means they have at least one double or triple
covalent bond
These types of bonds are called functional groups
because it’s at these bonds that reactions occur
Alkenes and Alkynes
Like alkanes these unsaturated hydrocarbons
are non polar and insoluble in water.
They undergo combustion the same as
alkanes giving the same products
They have very similar physical and chemical
properties to alkanes
Form addition reactions because of the
reactive double or triple bond
Alkenes exhibit a different form of isomerism
called geometric isomerisim
Starter A
Draw and name the 2 possible products formed when HCl is
added to 2-methylbut-2-ene.Name the products and identify
which is the major product.
H
H
H
H
H
C
C
C
C
H
Cl
CH3 H
H
2-chloro-3-methylbutane
H
H CH3 H
H
C
C
C
C
H
Cl
H
H
H
2-chloro-2-methylbutane – major product
Hydrogenation of an Alkene (Addition reaction)
Hydrogen can be added across the double C bond
The conditions for this reaction are:
• platinum catalyst
• 150 - 200 O C
• pressure of 4 atmospheres
The reaction is
C2H4(g)
+
ethene
H2(g)
CH3CH3(g)
ethane
The Good Oil
This hydrogenation process is
used to harden plant oils to commercially
produce margarine.
Natural plant oils contain many double bonds
per molecule, and because several double
bonds exist after hydrogenation, the margarine
is said to be polyunsaturated.
Bromination of an Alkene (Addition reaction)
Alkenes and alkynes undergo addition reactions with halogens to
form a dihaloalkane (or tetrahaloalkane).
The common test for an unsaturated hydrocarbon (ie a
hydrocarbon with a C C or C C bond) is therefore the rapid
decolourisation of an orange bromine solution in the absence of
sunlight
The reaction is
CH2H4(g)
ethene
+
Br2(l)
CH2 Br CH2 Br (l)
1,2-dibromoethane
Alkene molecules can create polymers
(plastics) by addition reactions where many
alkene monomer units are joined together
in the presence of heat and a catalyst
The process involves the breaking of one of the bonds in
the double bond in each alkene molecule.
Can be
drawn as
H
H
H
H
C
C
C. . C
H
H
H
H
Each of the two electrons from the bond go to each end
of the molecule to create a bond with another molecule
that has undergone the same process.
This creates long chains of joined monomers to create
a polymer.
Representation of adddtion polymer reaction
H
H
H
H
H
H
C. . C
C. . C
C. . C
H
H
H
H
Repeating
monomer
unit
H
H H
H
H
H
Heat and a catalyst added
H
H
H
H
H
H
H
C
C. . C
C
C. . C
C
C
C . .C
C
H
H
H
H
H
H
H
3 ethene
monomers
H
H
H
H
Polyethylene
polymer
Changing the side chain of the monomer in the
reaction gives different polymers ie
Cl
H
Cl
H
Cl
H
C. . C
C. . C
C. . C
H
H
H
H
Repeating
monomer
unit
Cl
Cl H
H
H
H
3 vinyl
chloride
monomers
Heat and a catalyst added
Cl
H
H
Cl
H
H
Polyvinylchloride
C
C. . C
C
C. . C
C
C
C . .C
C
Polymer
H
H
H
H
H
Aka PVC
H
H
H
H
H
H
This year we will also look at polymer
reactions involving condensation reactions
•Addition polymers are formed when alkene
monomers undergo addition to form a polymer
eg. polythene from ethene, P.V.C. from vinyl
chloride (chloroethene), polypropene from
propene.
Haloalkanes (alkyl halides) RX
where X = F, Cl, Br, I
Named as a chloroalkane or bromoalkane etc,
with the position of the halogen given by the
appropriate number of the carbon that it is
attached to in the chain.
Exist as primary, secondary, tertiary
The haloalkanes can be classified as:
primary - the C atom to which X is attached
is only attached to one other C atom
eg
H
Carbon
attached to
Br is
attached to
one carbon
H
H
C
C
H
H
Br
Secondary haloalkane - the C atom to which
X is attached is attached to two other C atoms
eg
CH3
H
C
CH3
Br
Carbon
attached to
Br is
attached to
two other
carbons
Tertiary haloalkane - the C atom to which X is
attached is attached to three other C atoms.
eg
CH3
CH3
C
CH3
Br
Carbon
attached to
Br is
attached to
three other
carbons
The Lucas Test
The Lucas test is used to distinguish between
the primary, secondary and tertiary alcohols
The Lucas reagent consists of ZnCl2 in
concentrated HCl
The zinc chloride catalyses a substitution
reaction between the alcohol and the
concentrated HCl
Chloroalkanes form and appear as a cloudy
suspension in the water because they are
insoluble
The Lucas test
*Important
The rates at which the different types of
alcohol react to form chloroalkanes enable
them to be classified as follows:
The Lucas Test
Type of
alcohol
Example
Primary
Butan-1-ol
Secondary
Butan-2-ol
Tertiary
Observation
No
cloudiness,
very slow if
at all
Cloudiness
after 5-15
minutes
2-methylpropan-2-ol Cloudiness
after 1-2
minutes
Product
1-chlorobutane
2-chlorobutane
2-methyl-2chloropaopane
Starter: Write out and fill in the missing
words
The s_____
trength of the C-OH b____
ond
i_______
ncreases from tertiary to secondary to
primary alcohols (which have the
strongest C-OH bond.)
The test for the C-OH strength is called
the l____
ucas test which uses concentrated
____
HCl and Z_________
inc chloride as the catalyst.
The speed of this s_________
ubstitution reaction
of the OH for a Cl indicates the type of
Alcohol
Properties of haloalkanes
Haloalkanes do not form hydrogen bonds, so they
have lower boiling points than alcohols and are not
miscible in water.
However, they are polar compounds, so have
higher boiling points than their parent alkanes.
The lowest mass haloalkanes are gases at room
temperature, but the rest are volatile liquids.
To make a haloalkane we can substitute the OH
on an alcohol using eg. PCl3, PCl5,SOCl2 or conc
HCl/ZnCl2
C2H5OH
ethanol
C2H5OH
ethanol
C2H5OH
ethanol
PCl3 or PCl5
SOCl2
C2H5Cl
chloroethane
C2H5Cl + HCl +SO2
chloroethane
Conc HCl/ZnCl2
C2H5Cl
chloroethane
• Haloalkanes are relatively nonpolar overall
(despite the polarity of the C-X bond) and are
insoluble in water.
A monohaloalkane eg. 2-bromopropane can be
formed by:
a) addition of HBr to propene (forming only one
product)
CH3
H
C
H
+ HBr
C
H
H
H
H
H
C
C
C
H
H
H
Br
b) substitution of propane using Br2.
(forming two products, the bromoalkane
and HBr)
H
H
H
H C
C
C H + Br2
H
H
H
H
H
H
H C
C
C H + HBr
H
H
Br
Lab book Page 201 Preparation of a
Haloalkane
Substitution of haloalkanes to form alcohols
Like tertiary alcohols, tertiary haloalkanes are
easily substituted.
A tertiary haloalkane will react with cold water to
form an alcohol:
R—X + H2O → R—OH + HX
We can tell whether this reaction has taken
place by the presence of the X– ions, which will
form precipitates with silver nitrate solution.
summary!
•
Tertiary haloalkanes form alcohols in cold water
•
Secondary haloalkanes react when the water is warm
•
Primary haloalkanes do not react with water, but react
to form alcohols with aqueous sodium hydroxide.
Nucleophiles
A nucleophile is any species which loves nuclei,
that is anything attracted to a positive charge.
Nucleophiles are therefore species that carry a
negative charge or a lone pair of electrons,
eg OH- , H2O and NH3 (in alcohol)
The C -X bond is polar and the slighty positive C
atom is prone to attack from a negative
nucleophile.
eg
R X(l) + OH- (aq)
R
OH + X- (aq)
Elimination to form an alkene
Haloalkanes can also undergo an elimination
with hydroxide to form an alkene:
R—X + alcOH– → R’=C + HX
Use the above reaction as a template to
write your own formation of an alkene from
a haloalkane
When deciding where to put the double bond in an
elimination reaction, apply the rule ‘the poor get poorer’.
One carbon of the double bond will be the carbon that lost
the halogen.
To decide whether the bond goes to the left or right of
that carbon, look at the number of hydrogen atoms on each
of those carbons.
The carbon to lose the hydrogen atom (and thus become
the other half of the double bond) is that carbon which
has the fewer hydrogen atoms already bonded to it.
H
H
Cl H
H
C
C
C
C
H
H
H
H
H
(Alc)KOH
H
H
H
C
C
C
C
H
H
H
H
H + HCl
Draw the structural formula for the reaction of the
tertiary haloalkane 2-chloro, 2-methylpropane
with water
H
CH3 H
H C
C
C H + H2O
H
Cl
H
H
H C
H
CH3 H
C
C H + HCl
OH H
2-methylpropan-2-ol
Tertiary alcohol
Amines (aminoalkanes)
Amines are named as substituents eg aminomethane, CH3NH2.
These may be classed as primary, secondary or tertiary, but
unlike the haloalkanes the classification depends on the
number of C atoms attached to the N atom.
Primary RNH2, secondary R2NH, tertiary R3N.
H
H
H C
N
H
H
H
Aminomethane
Primary amine
H H
H C
N
H
H
CH3
N-methylaminomethane
Secondary amine
H C C
H H
N
CH3
CH3
N,N-dimethyaminoethane
Tertiary amine
Amines (aminoalkanes)
Amines have an unpleasant “fishy” smell.
The smaller amines, up to C5, are soluble in water but larger
amines are insoluble, as the size of the non-polar hydrocarbon
chain cancels out the effect of the polar amino functional
group.
Like ammonia itself, water soluble amines form
alkaline solutions. They react with water by
proton transfer to form OH- ions. This means
aqueous solutions of amines turn litmus blue.
RNH2 + H2O

RNH3+ + OH
Amines also react with acids to form salts.
CH3NH2 + HCl
aminomethane

CH3NH3+ Cl
methyl ammonium chloride
The formation of an ionic salt increases the
solubility of the amine in acidic solutions
(compared to their solubility in water).
This is why we put lemon juice on
our fish to get rid of the amine
smell
Formation of amines
Another nucleophilic substitution reaction occurs
between haloalkanes and alcoholic ammonia:
R—X + NH3(alc) → R—NH2 + HX
amine
Why do you think the ammonia has to be
alcoholic?
It must be alcoholic ammonia: if water is
present the alcohol could be formed
instead.
Write these out give the compound class and name
CH3CH2CH2CH2CH2CH2COCl
Class acyl chloride
Name heptanoyl chloride
CH3CH2CHCH2CH3
NH2
Class amine
Name 2-aminopentane
CH3CH2CONH2
Class
amide
Name propanamide
Write these out give the compound class and name
CH3CH2CH2CH2COCH2CH3
Class ketone
Name heptan-3-one
CH3CH2CH2CHOHCH2
Class secondary alcohol
Name pentan-2-ol
O
CH3CH2CH2CH2CH2C
H
Class
aldehyde
Name hexanal
Draw and name the structural isomers of C4H10O
CH3CH2CH2CH2OH
H3C
CH
CH2
CH3
OH
butan-1-ol
butan-2-ol
H
H3C
C
OH
CH2 OH
H3C
C
CH3
CH3
2-methylpropan-1-ol
CH3
2-methylpropan-2-ol
Alkene Reactions
Alkenes react readily by adding small molecules
across the double C C bond.
These reactions are known as addition
reactions because molecules add “across” the
double bond
Write these out give the compound class and name
CH3CH2CH2CH2CH2NH2
Class amine
Name 1- aminopentane or pentylamine
O
CH3CH2CH2C
NH2
Class amide Name butanamide
CH3CH2 O
Class
ester
C CH2CH2CH2CH3
O
Name ethyl pentanoate
When deciding where to put the double bond in an
elimination reaction, apply the rule ‘the poor get poorer’.
H
H
Cl H
H
C
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H + HCl
One carbon of the double bond will be the carbon that lost the
halogen.
To decide whether the bond goes to the left or right of that
carbon, look at the number of hydrogen atoms on each of those
carbons.
The carbon to lose the hydrogen atom (and thus become the
other half of the double bond) is that carbon which has the
fewer hydrogen atoms already bonded to it.
Haloalkanes are molecular substances, so they do not
contain free X– ions.
When silver nitrate solution is added to 1-bromo-butane
no cream precipitate of silver bromide forms.
When silver nitrate solution is added to 2-chloro, 2-methyl
propane, the water in the solution reacts with the
haloalkane, forming an alcohol and releasing chloride ions
which then react with the silver nitrate to form a white
precipitate.
Primary haloalkanes can also be converted into alcohols,
but a stronger nucleophile is needed: OH–.
Dilute sodium hydroxide solution is added to 1-bromo
butane and shaken.
The excess NaOH is
neutralised with dilute
nitric acid.
When silver nitrate is added
the solution turns cloudy.