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Lecture Notes 4 Series: Organic Chemistry
Nomenclature
Due to the wide range of compounds that contain carbon, a systematic method has been introduced by IUPAC
to name them.
Alkanes
Alkanes contain only carbon and hydrogen and are termed hydrocarbons. The important thing to remember
about alkanes is that they contain only σ-bonds. The simplest alkane is methane.
Question 4.1
(a)
Predict the shape of the methane molecule.
(b)
What is the hybridisation at the carbon atom in methane?
Methane contains four single carbon-hydrogen bonds. As it contains no β-bonds, methane is described as
being saturated. In fact, all alkanes are saturated.
The next member of the alkane family is ethane, C2H6.
H
H
H
C
C
H
H
H
Now we have a carbon-carbon bond present. You will notice that we are still obeying the fundamental octet
rule. Each carbon is forming four σ-covalent bonds and thus has eight electrons in its valence shell.
Question 4.2
Draw a Lewis dot-cross diagram for ethane.
The next alkane contains three carbons, the next one four carbons, the next one five carbons, and so on. The
names arise directly from the number of carbons. The table below summarizes the first twelve members of
the alkane family.
The Alkanes
No. of C atoms
1
2
3
4
5
6
7
8
9
10
11
12
Formula
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
C11H24
C12H26
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Name
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
Undecane
Dodecane
A couple of important points arise from this table.
• All the names end in ane - thus the alkanes. There is no other way around this-you must learn the
names. It will make life a lot easier when naming other chemical species.
• There is a definite trend in the formula of the alkanes. In fact they all have the general formula
CnH2n+2. For example octane has eight carbon atoms and thus (2 x 8) +2 = 18 hydrogen atoms. We
can thus recognise whether an organic chemical is an alkane or not. If it has the correct ratio of
carbon:hydrogen of n:2n +2 it must be an alkane.
Alkenes
Alkenes also contain only carbon and hydrogen. However, this time the ratio of carbon:hydrogen is n:2n. The
first member of the family is ethene, C2H4. What does the bonding look like in this molecule? Remember
carbon must form four bonds to gain an octet, however this time there are not enough hydrogens to go
around
H
H
x
x
C
C
x
x
H
H
and thus the carbons are only forming three σ-bonds each. Can we do something to overcome this problem?
Easy- form a π-bond between the carbons. Each carbon still has one electron it can donate to the other carbon
and thus a bond can be formed.
H
H
x
x
C
C
x
x
H
H
or
H
H
C
C
H
H
Thus, there is now a double bond between the carbon atoms- one σ-bond and one π-bond. All alkenes
contain a double bond and are thus termed unsaturated.
The next member of the family is propene, C3H6.
H
H
H
H
C
C
C
H
H
Question 4.3
(a)
The next member of the family contains four carbon atoms. Name it.
(b)
An alkene has the formula C8H16. What is its name?
Alkynes
Another family of compounds which contains only carbon and hydrogen exists. This family is called the
alkynes. They have the general formula CnH2n-2. Carbon still obeys the octet rule and thus still forms four
bonds.
Carbon ALWAYS obeys the octet rule and thus ALWAYS forms 4 bonds.
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Question 4.4
The first member of this family is ethyne, C2H2. Draw this molecule, clearly showing the orbitals involved in
the bonding.
The second member of the family is propyne, C3H4.
The alkynes all contain a triple bond between two carbon atoms, and are therefore, like the alkenes,
unsaturated.
H
H
C
C
C
H
H
Question 4.5
(a)
Classify each of the following molecules as an alkane, alkene or an alkyne
(i)
C2H4
(ii)
C7H12
(iii)
C22H44
(iv)
C10H18
(v)
C18H38
(b)
Consider the following molecule
H
1
2
C
C
H
H
H
C
C
C
3
4
H
H
H
C
C
H
H
5
H
What is the hybridisation at each of the numbered carbons?
(c)
Complete the following table
Molecule
Name
Ethane
Hybridisation at carbon
sp
3
C2H2
C2H4
(d)
Name the following molecules
(i)
C4H6
(ii)
C10H20
(iii)
C7H14
Shorthand
Organic chemists can get lazy at times. Instead of drawing out all the atoms in a molecule they sometimes use
a form of shorthand. For instance when drawing out a propane molecule they might write
C
C
C
where it is taken as granted that there are actually hydrogen atoms at the end of the bonds.
This shorthand can be taken a step further where we don’t even draw the carbon atoms! Propane can be
drawn very simply as
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where it is now recognised that there are CH3 groups at the points where the lines stop.
CH2
=
CH3
CH3
This comes in handy when drawing large molecules such as C10H22. Instead of laboriously drawing the
molecule out we can draw –
Question 5.6
What are the molecular formulae of the following compounds?
(i)
(ii)
(iii)
One last thing on shorthand is the use of “R”. When chemists write a molecule as
R
H
H
C
C
H
the “R” simply signifies an organic group which they are not particularly interested in! That is to say that the
“R” group will remain unchanged during a chemical reaction.
Rings
Alkanes, alkenes and alkynes are all linear molecules, where the carbon atoms are in a chain. There is however
another way of joining carbons together and that is in rings. Consider for example the following molecule.
H
H
H
C
H
C
C
H
H
H
H
C
=
C
H
H
C
H
H
Here we have six carbon atoms joined in a ring. The naming of these compounds is still based upon the
number of carbon atoms so this molecule is called cyclohexane. You will notice that the molecular formula for
cyclohexane is C6H12. For all cycloalkanes the carbon:hydrogen ratio is Cn:H2n. This is exactly the same ratio as
for alkenes. However, cycloalkanes are saturated molecules and they will have a very different chemistry from
alkenes with the same molecular formula, so beware!
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One can also get unsaturated rings such as the following molecule
H
C
H
C
C
H
H
C
C
H
C
H
This pattern of a C-C single bond, C=C double bond, C-C single bond, C=C double bond, etc. is very important in
organic chemistry. If this pattern exists, the molecule is said to be conjugated. If you have a conjugated ring,
as in the above molecule, the molecule is described as being aromatic. We normally draw the above molecule
as
or
The second structure is to show that the π-bonding in the molecule is delocalised, i.e., it is the average of the
two resonance forms of the molecule.
This particular molecule is very important in organic chemistry and is called benzene.
Functional Groups
Carbon has a rich chemistry. It can bond to many other atoms apart from hydrogen. These atoms are
generally known as heteroatoms. Common heteroatoms are nitrogen, sulfur and the halides. The most
important one however is oxygen which is not surprising as we live in an oxygen-rich planet. The chemical
reactions that these compounds which contain heteroatoms undergo depend crucially on which groups of
atoms are present. For instance, a compound which contains a
C
O
group will have a very different chemistry from a compound which contains the group .
C
OH
We can thus classify compounds into families which contain the same grouping of atoms or functional groups.
(a)
Alcohols (or Alkanols)
Alcohols have the functional group
C
OH
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The simplest alcohol is CH3OH, methanol. The suffix –ol is used to describe this functional group. The naming
of the alcohols will come from the number of carbon atoms in the molecule. Thus C2H5OH is ethanol. The next
one is:
H H
H
H
C
C
C
H
H
H
OH
which is propanol and so on.
(b)
Aldehydes
Aldehydes contain the functional group
H
C
O
The first three members of the family are shown below.
H
C
CH3 C
O
methanal
(c)
H
H
H
CH3
O
ethanal
CH2
C
O
propanal
Ketones
Ketones are very similar to aldehydes except they have no hydrogen attached to the carbon which is attached
to the oxygen.
R
C
O
R
Two examples are shown below.
O
CH3
C
O
CH3
CH3CH2 C
propanone
(d)
CH3
butanone
Carboxylic acids
Carboxylic acids contain the functional group
O
C
OH
The first two members of the family are methanoic acid and ethanoic acid.
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O
H
C
O
OH
methanoic acid
CH3
C
OH
ethanoic acid
Question 4.6
Calculate the oxidation number of carbon in the following compounds.
(a)
methane
(b)
methanol
(c)
methanol
Question 4.7
Name the following molecules.
(a)
CH3CH2CH2OH (b)
(e)
CH3CH2CH2COOH
(c)
(d)
methanoic acid
CH3CH2CH2CH2CHO
Esters
Esters are formed when a carboxylic acid reacts with an alcohol. For example
O
CH3COOH
+
CH3CH2OH
CH3
C
O
CH2CH3
+
H2O
ethyl ethanoate
The naming of esters comes from the identity of the R′ and R″. If for instance R″ = CH3 then we have a methyl
substituent. If R″ = C2H5 then we have an ethyl substituent. The fragment is named after the total number of
carbon atoms in this fragment. Thus if we have CH3COO- , the fragment is called “ethanoate”. If instead we
have CH3CH2COO-, we have propanoate and so on. The ester we formed in the above reaction is thus called
ethyl ethanoate.
Question 4.9
(a)
Name the following substituent groups.
(ii)
(i)
-C3H7
(b)
Name the following esters.
(i)
CH3CH2CH2COOCH3
(ii)
(f)
-C4H9
CH3COOCH2CH2CH3
Ethers
Ethers also contain an oxygen heteroatom. The functional group is R′-O-R″. If R′ = R″ = -C2H5 we will have
diethyl ether. (When most people talk of “ether” they are actually talking about diethylether, which used to
be employed as an anaesthetic during surgical procedures).
(g)
Amines and Nitriles
These two functional groups both contain nitrogen but in very different forms as shown below.
R
R
NH2
amine
C
N
nitrile
There are in fact different types of amines and they are differentiated by the number of carbons bonded to the
nitrogen
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H
R
N
o
1 amine
R
R
H
R
N
o
R
H
N
R
o
2 amine
3 amine
o
If for instance we have (CH3)3N we would name this 3 amine, trimethylamine.
The naming of nitriles again comes from the number of carbons. Therefore CH3CN is ethanenitrile (which is
better known to the older generation of chemists as acetonitrile).
(h)
Amides
Not to be confused with amines! Amides contain the functional group
O
C
N
So for instance the compound:
O
H
CH3CH2CH2CH2CH2C
N
H
would be called hexanamide (as there are six carbons in the compound).
(i)
Arenes
It might seem strange to include arenes in this list as they contain only carbon and hydrogen. Arenes are alkyl
-substituted benzene rings. Examples are shown below
CH3
CH2CH2CH2CH3
It is important to include arenes in this list of functional groups as the benzene ring substituent (known as the
phenyl group) undergoes a wealth of chemistry. A table of all these functional groups (and a few more) is
shown overleaf (don’t try and remember the extra ones!).
Naming Compounds
The title of this section might seem a little strange to you. Haven’t we just done nomenclature in the previous
section? Yes we have, but knowing the vocabulary and putting it into action are two different skills. You might
feel very comfortable in naming this compound-
H
H
H
H
H
H
O
C
C
C
C
C
C
H
H
H
H
H
OH
- as hexanoic acid. However, how would you name the following compound?
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Cl
H
H
H
CH3
H
O
C
C
C
C
C
C
H
CH3
H
H
H
C
H
H
C
H
OH
H
Gets a bit complicated doesn’t it? Luckily a set of rules have been put into place. So all we need to do is follow
the rules and we should be able to name any organic compound.
Branching in Alkanes
Lets consider a compound with a molecular formula of C4H10. Its an alkane, so we must have butane. The
compound must be
H H H
H
H
C
C
C
C
H
H
H
H
H
However there is another molecule we can draw with the same molecular formula of C4H10.
H
H
H
H
C
C
C
H
H
H
H
C
H
H
What we have done is introduce a branch in the carbon chain. Now we must be able to distinguish between
these two compounds by giving them different names. In this case it is quite straightforward. We call the
straight-chain molecule n-butane and the branched molecule iso-butane. However, when alkanes have more
than one branch (as many do) we really do need a systematic way of naming them.
Rule 1: Choose the longest carbon chain to base the name upon.
Rule 2: Number the carbons in this longest chain beginning at the end nearest the first branch point.
Rule 3: Identify and number the branches.
Hydrocarbon branches
-CH3
-CH2CH3
-CH2CH2CH3
-CH2CH2CH2CH3
Methyl
ethyl
propyl
butyl
CH3
CH3CHCH3
CH3CHCH2CH3
CH3
sec-butyl
C
CH2
CH3
C
CH3CH2
CH3
C
CH3
neopentyl
tert-pentyl
Example 4.1
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CH2 CH2
isopentyl
CH3
CH2
CH3 C
H
isobutyl
tert-butyl
CH3
CH3
H
CH3
isopropyl
pentyl
CH3
CH3 C
-CH2CH2CH2CH2CH3
Question:
Name the following molecule.
H
H
H
H
H
C
C
C
H
H
H
C
H
H
C
H
H
H
H
C
C
C
C
H
H
H
H
H
H
Answer:
When naming organic compounds use hyphens to separate numbers and letters, and
commas to separate numbers.
Example 4.2
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Question:
Name the following molecule.
CH2
CH3
CH3
H
H
CH2CH3
C
C
C
CH
H
H
CH3
CH2CH2CH2CH3
Answer:
When deciding which substituents to put first, simply put them in alphabetical order. Ethyl thus comes
before methyl (Note: we don’t count the prefix, di- in this case, when deciding this order).
Question 4.10
Name the following molecules.
(a)
C
C
C
C
C
C
C
C
C
C
C
(b)
C
C
C
C
C
C
C
C
C
C
c)
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CH3
CH3
CH3
C
C
C
C
C
CH2
CH3
Question 4.11
(a)
Draw the structure of the following molecules
(i)
3-methylhexane
(ii)
3-ethyl-2-methylhexane
(iii)
3-ethyl-4,7-dimethylnonane
(b)
Give a systematic name to isobutane.
A result of branching in alkanes is that some carbons are now “different” from other carbons in the molecule.
In the following molecule for instance –
CH3
CH3
C
o
CH2
CH3
x
CH
o
CH2
CH3
CH3
some carbons are only bonded to only one other carbon atom (those marked ). Other carbon atoms are
o
bonded to 2 carbon atoms (o). Those bonded to only one carbon atom are termed primary carbon atoms (1 )
o
and those bonded to two are termed secondary carbon atoms (2 ). Both these types of carbon atoms are
encountered in straight chain alkanes. However, with branching we now have two different carbon atoms,
those bonded to three carbon atoms (x) and those bonded to four carbon atoms (•). These are termed
o
o
tertiary carbon atoms (3 ) and quaternary carbon atoms (4 ) respectively.
Naming Compounds with Functional Groups
The names of most organic compounds can be viewed as having three main parts.
Prefix-Parent-Suffix
Prefix: identifies the where and what for the substituents.
Parent: this is the backbone of the molecule, the name of which depends upon the number of carbon atoms.
Suffix: this is particularly important as it describes the functional group present.
Again, we establish the rules and follow them rigorously when naming these compounds.
Rule 1: The parent hydrocarbon chain is the longest continuous carbon chain containing the highest priority
functional group.
Some compounds might contain more than one functional group and thus a priority list is required. For
example, is the following compound named as an alcohol or as a carboxylic acid?
O
HO
CH2
CH2
CH2
59
CH2
C
OH
The answer is as a carboxylic acid as the –COOH group has the highest priority. The full priority list is shown
below.
Functional Group
-COOH
-COOR
-COX
-CONH2
-CHO
RCOR’
-OH
-NH2
CnH2n-2
CnH2n
CnH2n+2
Name
carboxylic acid
highest priority
ester
acid halide, X = Cl, Br
amide
aldehyde
ketone
alcohol
amine
alkyne
alkene
alkane
An alkyl substituent (CnH2n+1) has the lowest priority. Halides are always named as substituents on the parent
hydrocarbon. Common non-alkyl substituents are –
Substituent
-F
-Cl
-Br
-I
-NO2
Name
Fluoro
Chloro
Bromo
Iodo
Nitro
Rule 2: The chain is numbered in the direction that gives the highest priority functional group the lowest
possible number.
Of course for some functional groups by their very nature have to come at the end of a chain. Examples of
these are
C
C
O
OH
H
aldehydes
O
carboxylic acids
When numbering chains which include these functional groups start at the carbon attached to the
heteroatoms. For example;
CH3 CH2 CH2 COOH
CH3 CH2 CH2 CH2 CH2 CHO
4
6
3
2
butanoic acid
1
5
4
3
hexanal
2
1
Rule 3: If the same number for the functional group is obtained in both directions the chain is numbered in
the direction that gives the substituent the lowest number.
For example the following molecule.
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H
H
H
OH Cl H
C
C
C
C
C
H
H
H
H
H
H
is 2,2-dichloropentan-3-ol (not 4,4-dichloropentan-3-ol).
Example 4.3
Question:
Name the following compound
CH3
CH
CH
Cl
CH3
CH2
C
OH
O
Answer:
The parent hydrocarbon is five carbons long
1
CH3
2
3
CH
CH
Cl
CH3
4
CH2
5
C
OH
O
The suffix will be –oic acid. The substituents are 3-methyl and 4-chloro.
Thus we have 4-chloro-3-methylpentanoic acid.
Example 4.4
Question:
Name the following compound
OH
CH3
C
O
CH2
CH2
CH2
CH3
Answer:
Example 4.5
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CH
Question:
Name the following compound.
OH
CH3
CH
CH2
CH
CH3
Cl
Answer:
Question 4.12
Name the following compounds.
(a)
(b)
CH3CH2 CCH3
CH3CHClCH2COOH
(c)
ClCH2COOH
(d)
ICH3
O
Question 4.13
In the introduction to this section we mentioned it might be a bit complicated to name the following molecule.
Cl
CH2
CH
CH2
C(CH3)2
CH2
COOH
C2H5
Well, now you know the rules, show that is in fact quite simple by giving it a name!
Question 4.14
o
o
o
List the following alcohols as 1 , 2 or 3 alcohols.
CH3
(a)
CH3CH2OH
(b)
CH3CH2CH(OH)CH3
(c)
CH3
C
CH3
Isomers
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OH
Isomers are compounds that have the same numbers and kinds of atoms but differ in the way they are
arranged. We have already seen that the molecular formula C4H10 can give rise to two different molecules-
H3CCH2CH2CH3
H 3C
CH
CH3
CH3
n- butane
isobutane
These two molecules are isomers of each other.
Question 4.15
In how many ways can you arrange the atoms for a molecular formula of C5H12 i.e. how many isomers can you
draw?
There are in fact many different types of isomerism that can occur in organic chemistry and this section will
review them under two umbrella titles- constitutional isomers and stereoisomers.
Constitutional Isomers
This type of isomerism occurs when the atoms are connected differently. Butane and isobutane have a
different way of connecting the carbon atoms and are thus constitutional isomers (as are all the isomers of
C5H12- hope you got them all!).
Another form of constitutional isomerism is when a molecular formula can give rise to different functional
groups. For example if we have a molecular formula of C2H6O, how many ways can we connect the atoms
together? The answer is two, as shown below.
H3COCH3
or
H3CCH2OH
dimethyelther
ethanol
(Try this for yourselves: there is no other way of connecting these atoms while satisfying their valencies.)
For larger molecules the number of different ways of combining the atoms together will increase, and another
type of constitutional isomerism will arise:- that of the position of the functional group. For instance for C3H8O
we have the following possibilities
H3C
O
CH2CH3
H3C
ethylmethyl ether
CH2 CH2OH
H3C
CH
CH3
OH
propan-1-ol
propan-2-ol
Question 4.16
(a)
Draw all the constitutional isomers for the molecular formula C4H10O.
(b)
Name all the compounds in part (a).
[Hint: this question combines the three forms of constitutional isomerism mentioned- different
carbon skeletons, different functional groups, and the different position of the functional groups.]
The presence of unsaturation in a molecule also introduces the possibility of isomerism. This is really just an
extension of the idea of functional groups having different positions. Consider the molecule butene. Given
only this description we can actually draw two possible structures.
CH3CH
CHCH3
or
CH2
CHCH2CH3
We then need to specify the exact position of the double bond. The two molecules would therefore be but-2ene and but-1-ene.
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Question 4.17
Draw all the possible constitutional isomers for the molecular formula C6H12.
[Hint: remember cycloalkanes have the same carbon:hydrogen ratio as alkenes and you can draw more than
one cycloalkane here!].
Stereoisomers
Stereoisomers are compounds where the atoms are connected in the same order but the molecules have
different geometries. When talking of geometries, or shapes, we need to start thinking in three dimensions
(3D). We will discuss three types of stereoisomer in this section: conformational isomers, geometric isomers
and enantiomers. When thinking in 3D we need to remember the two very important concepts
• There is rotation possible around a carbon-carbon single bond.
• There is no rotation possible around a carbon-carbon double-bond.
(a)
Conformational Isomers
Organic chemists have various ways of drawing 3D structures on a 2D page. One way is to show bonds coming
out of a page and one going behind the page. Thus for ethane we can draw the following structure.
H
H
H
H
C
C
H
H
If we now change our point of perspective when looking at the ethane molecule to an end-on view, so we’re
looking directly along the C-C bond axis, then what we see is only one carbon atom. Attached to this carbon
atom we see three hydrogen atoms. We can draw this as:
H
H
H
where the dot in the middle is the carbon atom that we can see.
o
The apparent HĈH angle is 120 C when we look at the molecule like this.
We now draw a circle to represent the other carbon atom (the one we can’t see)H
H
back carbon
front carbon
H
and then we put into the diagram the three hydrogens attached to this carbon.
H
H
H
H
H
H
This type of representation is known as a Newman Projection. Newman projections are very useful when
looking at what happens to the shape of a molecule when we rotate around a carbon-carbon bond. Upon
rotation we can change the relative positions of the hydrogens on the back carbon with respect to the
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hydrogens on the front carbon. There are an infinite number of possibilities here but two extremes stand out.
These are shown below.
H
H
H
H
H
H
H
H
H
H
H
H
eclipsed
staggered
In the eclipsed form one can only see the front carbon and the three hydrogens attached to it. The back
hydrogens are hidden from view behind the front hydrogens. The back-hydrogens are said to be eclipsed. In
the staggered conformation we can see all six hydrogen atoms as the back-hydrogens are staggered exactly ½
way between the front-hydrogens. It is important to note that the two molecules depicted above have
different shapes and thus they are strictly speaking, stereoisomers of each other. However the difference in
energy between the molecules is so small (the staggered conformation is only 12 kJ/mol more stable than the
eclipsed conformation) that they cannot be separated from each other. The eclipsed and staggered forms of
ethane are said to be conformational isomers or conformers.
Why should the staggered confirmation be more stable? It is thought to be due to the fact the hydrogens on
the front-carbon are at their furthest distance from the hydrogens on the back carbon. Thus the hydrogen
electron clouds are at their furthest distance apart and thus (as electron clouds repel each other) they are at
their lowest possible energy.
The picture becomes slightly more complicated for larger alkanes. For instance if we look at butane, we now
have to specify which carbon-carbon bond we have chosen to draw our Newman projection.
If we choose the C1-C2 bond then our Newman projection will look like this:
H
H
H
H
C2H5
H
If instead we choose the C2-C3 bond then we will have:
H
H
CH3
H
CH3
H
Now when we rotate the C2-C3 Newman projection we will have two different “forms” of eclipsed structures
and two different staggered structures. The two eclipsed forms are shown below.
CH3
H
H
CH3
H
CH3
CH3
H
H
H
65
H
H
The two staggered forms are –
H
CH3
H
H
H
H
H
CH3
H
H
CH3
CH3
Again each of the conformers can be converted into another by rotation around the C2-C3 bond. This is true
for all conformers.
Conformers are stereoisomers which can be made equivalent by rotation around carbon-carbon
single bonds.
(b)
Geometric Isomers
To return to the second point we listed at the start of the section- there is no rotation possible around a
double bond.
Ethene is a planar molecule. In fact the geometry immediately around an alkene functional group is always
2
planar due to the fact that the carbons are sp hybridised. If we now look at but-2-ene we can draw the
molecule as
H3C
CH3
C
C
H
H
(I)
However there is another possible structure. We could just as easily have drawn the molecule as
H
CH3
C
C
H
H 3C
(II)
Note that we cannot form this molecule from the other one by simply rotating around the C=C bond, because
no rotation is possible!
The atoms are still connected in the same order but the shapes of the molecules are different. This means that
we have another form of stereoisomerism. As the geometry around the double bond is different these are
known as geometric isomers. How do we differentiate between them? Molecule (I) has the two methyl
groups on the same side of the double bond. This is called the cis – isomer or the Z – isomer. The other
isomer has the methyl groups on the opposite sides of the double bond. This is known as the trans – isomer or
the E – isomer.
H3C
C
H
H
CH3
CH3
C
C
H 3C
H
66
C
H
cis- isomer (Z- isomer)
trans- isomer (E- isomer)
Question 4.18
Write down and name the geometric isomers possible for the following molecules.
(a)
1,2-dichloroethene
(b)
1-bromopropene
(c)
pent-2-ene
Question 4.19
(a)
Draw the two geometric isomers:- trans-1,2-dimethylcyclopropane and cis-1,2-dimethylcyclopropane.
(b)
Does the molecule 2-methylbut-2-ene exhibit geometric isomerism? If so, draw the isomers.
(c)
Is there a possibility of geometric isomerisation for the molecule 1,2-dichlorocyclohexane?
As well as having the same carbon:hydrogen ratio cycloalkanes and alkenes also share the ability to
exhibit geometric isomerism.
Unlike conformers, geometric isomers can be separated from each other as they have quite distinct physical
properties from each other.
(c)
Enantiomers
3
Another form of stereoisomerism arises from molecules which contain a chiral centre. A chiral centre is an sp
hybridised carbon centre which is attached to four different substituents. An example would be the following
molecule.
CH3
H
Cl
I
Believe it or not there is another way of drawing this molecule! If we imagine a mirror beside the molecule
then we can envisage what the molecule will look like in the mirror (or through the mirror plane).
CH3
I
CH3
H
H
Cl
Cl
I
mirror plane
These two molecules are not the same! If you try and superimpose the two molecules on top of each other
you will find it impossible to do! The two different isomers are called enantiomers.
How do we differentiate between the two compounds? Once again we need to give each isomer a different
name. The naming of enantiomers is quite a complicated business and not within the scope of this course. It
is sufficient to recognise that when a chiral centre is present in a molecule, the isomers are known as the Risomer and the S-isomer.
Lets now try the same thing for a molecule which does not contain a chiral centre. An example would be
CHCl2I.
CH3
CH3
Cl
I
Cl
Cl
67
Cl
I
mirror plane
Here we do not have four different substituents on the C atom. When we reflect through the mirror plane we
will get a molecule that is superimposable upon the original.
Therefore for this type of isomerism to occur we require the presence of a chiral centre. Chiral centres are
also known as asymmetric centres or stereogenic centres.
Enantiomers that occur in a 50:50 mixture with each other are called a racemic mixture or a racemate.
Enantiomers however can be separated by various techniques. Indeed organic chemists spend an awful lot of
time and energy trying to selectively synthesise one enantiomer. Why do they do this? Well, many
biologically active compounds, although containing chiral centres, exist only in one enantiomeric form in
nature. Therefore if one wants to mimic a naturally-occurring compound, say with a drug, only one
enantiomer will be effective.
Question 4.20
Consider the following molecule
(a)
(b)
(c)
(d)
CH3CH(NH2)CH2CHCHCH(CH3)CH3
Draw an extended structure for the molecule, i.e. show the bonds.
Identify the functional groups present in the molecule.
Will this molecule display geometric isomerism? Explain your answer.
Does this molecule have a chiral centre(s)? Indicate them clearly on your drawing of the structure.
An Introduction to Reaction Mechanisms
68
Organic Chemistry is largely about making new carbon-containing compounds. To do this bonds in existing
compounds must be broken and new bonds formed. If we understand the nature of how these two processes
occur we will be well on our way to understanding reaction mechanisms, and perhaps even to predict how
chemicals might react together.
Bond Breaking and Bond Making
Question 4.21
How many electrons are in the following bonds?
(i)
C
(ii)
C
C
(iii)
C
C
C
When a single covalent bond breaks, two different outcomes are possible.
One way is for the two electrons in the bond to divide equally. This is known as homolytic fission and results
in the formation of free radicals.
X + Y → X• + Y•
Each of X and Y now has an odd electron (as indicated by ). For instance if we homolytically split the carboncarbon bond in ethane then two methyl radicals will result.
C2H6
→
2 CH3•
methyl radical
We denote the movement of single electrons with a single-headed arrow
reaction mechanism for the homolytic splitting thus
H3 C
CH3
→
. Thus we can draw the
2 CH3•
Radicals are, in general, very reactive species.
The other way to break a single covalent-bond is heterolytic fission, where the two electrons in the bond go to
one of the atoms. The movement of two electrons is shown by a double-headed arrow
X
Y
→
X
+
+ Y
.
-
Now we have a cation and an anion forming. If a carbon-carbon bond breaks in this fashion then we have a
carbocation and a carbanion resulting. The carbocation species is very important in organic chemistry and will
be returned to later.
Question 4.22
(a)
How many electrons are around carbon in each of the following species
+
(i)
CH3•
(ii)
CH3
(iii)
CH3
(b)
Would you expect these species to be reactive? Explain your answer.
Bond making is as equally important as bond breaking. To form a single covalent bond we need two electrons.
This can also happen in two ways.
X
Y
69
→ X–Y
or
X+
Y-
→ X–Y
i.e. the reverse of the dissociation mechanisms we have just talked about. If we look at the following bond
formation mechanism:
+
Cl + CH3 → CH3Cl
then we could draw the mechanism as
Cl-
+
CH3
→
CH3Cl
where the chloride ion has donated two electrons to carbon.
The chloride ion is termed a nucleophile and the carbocation is an electrophile. The formation of the bond
comes from nucleophilic attack on an electrophile.
A nucleophile is a “positive loving” species and will, itself, be negatively charged or electron rich. An
electrophile is a “negative loving” species and will, itself, be positively charged or electron deficient.
We have already seen electrophiles and nucleophiles in action in the formation of dative (coordinate) bonds in
st
the 1 semester (we just didn’t refer to them as nucleophiles and electrophiles!).
Consider again, for example, the reaction between BCl3 and NH3.
Cl3B
NH3 → Cl3B – NH3
Question 4.23
(a)
Identify the nucleophile and the electrophile in the previous reaction.
(b)
Identify each of the following species as an electrophile or a nucleophile.
+
(i)
I
(ii)
HO
(iii)
CH3
(iv)
CH3
(v)
H
+
(iv)
AlCl3
In real life it is very difficult to break carbon-carbon bonds or in fact any bonds in a molecule. Stable molecules
will require an energy input to break bonds or a reactive centre that will encourage either nucleophilic or
electrophilic attack.
Ethane for instance will not spontaneously dissociate into two methyl radicals. Nor will it spontaneously
+
dissociate into CH3 and CH3 .
Free radicals are commonly formed in photochemically induced reactions i.e. the interaction between
molecules and light. The bromine molecule, for instance, can give rise to two bromine radicals thus
Br2 + hυ → 2 Br•
We show the photon as “hυ”. In this case the photon must have enough energy to break the bromine-bromine
single bond in Br2.
Question 4.24
-1
If the bond energy of Br2 is 229 kJ mol , what is the wavelength of the lowest energy photon which can
dissociate Br2?
Functional groups will give rise to reactive centres within molecules. For instance if we have a ketone
containing the functional group
O
C
70
then we have a possible site for reaction. Consider the above functional group.
The carbon is less electronegative than the oxygen and thus the polarity of the bond will be
+
-
δ
δ
C
O
We now have a positive centre in the molecule which will be liable to attack from a nucleophile (Nu−).
Nu
C
O
However if carbon forms a bond to the nucleophile then it will be pentavalent and therefore break the octet
rule. Instead carbon will also give two electrons away, in this case to oxygen. We thus have the reaction
mechanism:
Nu
Nu
C
O
C
O
This is called a concerted reaction mechanism where bonds are forming and breaking at the same time.
Question 4.25
+
Identify areas of positive charge and areas of negative charge by using δ and δ terminology in the following
molecule.
H 2N C
C CH CH2 CH
CH2
Cl
Question 4.26
Will alkenes and alkynes be susceptible to nucleophilic or electrophilic attack? Explain your answer.
Carbocations
Consider the following reaction.
CH3
CH3
H3C
C
H3C
Cl
C
+
Cl-
CH3
CH3
The chloride ion “leaves” the molecule and is therefore simply called a leaving group. It leaves behind a
carbocation. The carbon which used to be bonded to the chlorine has lost two electrons and thus has only six
electrons around it. Six electrons are arranged into three electron pairs and thus the hybridisation at the
3
2
carbon has gone from sp in the parent molecule to sp in the carbocation.
The carbocation formed in the above reaction will look like this.
H3C
+
CH3
C
o
CHand
3
The central carbon has a vacant unhybridised p-orbital
the angles between the C- C bonds are 120 .
71
Carbocations are very important intermediates in many organic reactions. They have a very short lifetime due
to the fact that they are very powerful electrophiles i.e. the positive carbon centre wants to regain another
two electrons to achieve a full valence shell.
Question 4.27
Consider the following molecule.
H
H
C
Br
H
Suggest ways that this molecule might undergo bond breaking.
Some carbocations are more stable than others. This comes from the so-called INDUCTIVE EFFECT. The
carbon centre which is positively charged will try to pull electrons towards it from other parts of the molecule.
Looked at another way the groups attached to the positively charged carbon can donate electron density
through the σ-bonds to the electron-deficient centre. We draw this as follows
C
CH3
where the methyl substituent donates electron density towards the carbocation centre. Alkyl groups such as
the methyl group are said to be electron-donating or to have a positive inductive effect.
The inductive effect is the shifting of electron density within a σ-bond towards a positive centre in a molecule.
If a substituent group is electron-donating it has a positive inductive effect and if it is electron withdrawing it is
said to have a negative inductive effect.
If electron density is “pushed” towards the carbocation centre then the carbocation will be stabilised.
Therefore, the more electron-donating groups attached to the electron-deficient carbon the more stable it will
be. Hydrogen atoms do not display any appreciable inductive effect.
Generally,
(i)
the more alkyl groups attached to the electron-deficient carbon, the greater the inductive effect; and
(ii)
the larger the alkyl group substituent the more electron-donating it will be and therefore the greater
the inductive effect.
Question 4.28
(a)
Arrange the following carbocations in order of increasing stability.
+
+
+
+
CH3
CH3CH2
(CH3)3C
(CH3)2CH
(b)
o
o
o
Label each of the carbocations in the question above as 1 , 2 or 3 .
Question 4.29
Which of the following carbocations is the more stable?
+
ClCH2CH2
or
Explain your answer clearly.
Question 4.30
72
+
CH3CH2
Are carbon radicals more like carbocations or carbanions? Thus, does their stability sequence resemble that of
carbocations or carbanions?
Free Radical Substitution of Alkanes
Alkanes are generally very unreactive (as they have no functional groups) but one particular reaction type is
very important. This involves the substitution of a hydrogen within the alkane by a halide atom. The reaction
between a halogen, such as Cl2, and an alkane, e.g. methane, proceeds via a free-radical mechanism. The
reaction can be broken down into three steps; initiation, propagation and termination.
Step1
Initiation: Generation of free radicals.
Photochemically induced homolytic fission results in two Cl radicals.
Cl2
+ hυ → 2 Cl•
H
H
Cl
H
C
H
Cl
H
C
+
H
H
H
A chlorine radical abstracts a
hydrogen atom from methane
Hydrogen chloride and a
methyl radical result
Step 2
Propagation
Cl
Cl
CH3
Cl
+
Cl
CH3
Chloromethane is formed and
another chlorine free radical forms
which attracts another methane and
the reaction continues
The methyl radical the reacts
with the chlorine molecule
Step 3
Termination
For a free radical substitution to stop, free radicals must react with each other.
Cl
Cl
CH3
Cl
Cl Cl
CH3
Cl CH3
C2H6
CH3
Question 4.31
The previous reaction can also give rise to dichloromethane. Write a suitable reaction to form this product.
When we consider the free radical substitution of larger alkanes then more than one free radical alkyl species
is possible. Consider for example the reaction between a chlorine radical and n-butane.
73
(i)
Cl
H
CH2
CH2
CH2
CH2
CH3
CH2
CH2
CH3 + HCl
CH
CH2
CH3
CH3
(ii)
Cl
H
CH
CH2
CH2
CH3
CH3
o
+ HCl
o
Reaction (i) gives rise to a 1 free radical and reaction (ii) gives rise to a 2 free radical. Which free radical is
more stable? Free radicals, like carbocations are electron-deficient (free radicals require one more electron to
obey the octet rule, carbocations require two more electrons). Therefore free radicals will have the same
o
o
o
order of stability as carbocations – 1 <2 <3 . Thus reaction (ii) will be favoured over reaction (i). The favoured
product of this reaction will be 2-chlorobutane.
Question 4.32
o
o
o
(a)
What order of stability would you predict for 1 , 2 and 3 carbanions? Justify your answer.
(b)
What would be the favoured product of the reaction between isobutane and Cl2? Again, justify your
answer.
(c)
The reaction between Cl2 and butane also gives rise to 3,4-dimethylhexane as one of the minor
products. Suggest a reaction mechanism which will give rise to this product.
Electrophilic Addition Reactions of Alkenes
An addition reaction is one in which the bond order between two carbon atoms decreases. For instance, we
can add a species across a double bond.
C
C
XY
+
X
Y
C
C
In the above reaction the bond order between the two carbon atoms has decreased from two to one.
Addition reactions of alkenes proceed via a carbocation intermediate.
Alkenes are nucleophilic, that is, they have a centre of electron density in the double bond. We therefore
require a positive centre to react with it. If we take a polar molecule such as HBr then we will have a positive
center on the H atom.
δ+
δ-
H – Br
Thus the first step of the reaction will be a nucleophilic attack by the alkene at the hydrogen end of the HBr
molecule.
Taking ethane as an example:
C
C
H
Br
As the carbon-hydrogen bond forms the hydrogen-bromine bond will simultaneously break to give –
H
H
C
C
H
H
H
74
+ Br
The bromide ion will now attack the carbocation thus
H
H
C
C
H
H
Br-
H
H3C
CH2Br
The sum total of the reaction is that we have converted an alkene into an alkyl halide.
We need to be careful when dealing with larger alkenes. Consider for example the reaction between propene
and HBr. Two carbocation intermediates are now possible depending on which carbon atom of the double
bond uses the pi electrons to form a bond to the hydrogen atom.
H3C
Br-
CH
CH2
+
CH2
CH2
CH2
H
Br
CH3
CH
CH3
+
Br-
Question 4.33
(a)
Which is the more stable carbocation in the previous example, A or B?
(b)
Which will be the preferred product of this reaction, 1-bromopropane or 2-bromopropane? [Note we
will get both products but one will be the major product and the other the minor product]
(c)
1-bromopropane and 2-bromopropane are isomers. Which type of isomers are they?
For even larger alkenes we’ve got another variable to consider, that of the possibility of the formation of a
different kind of isomers during the reaction. Consider the reaction between cis-but-2-ene and HBr
H3C
CH3
C
H3C
C
H
+
H
Br
H C
H
H
CH3
C
H
+ Br
Only one carbocation intermediate is possible in this case. However, the subsequent nucleophilic attack from
2
the bromide ion can now come from two possible directions. Remember a carbocation is sp -hybridised and
trigonal planar. Thus the bromide ion can attack either from above or below.
Above
H3 C
H C
H
C
CH3
H
H3 C
+ Br
H
C*
C
H
H
Below
H 3C
H C
H
C
CH3
H
H 3C
+ Br-
75
H
C
H
C*
CH3
Br
H
Br
CH3
At first sight these products might not look different from each other. However, if we take a closer look we
will see that the carbon marked * is in fact a chiral centre, and the two products formed are in fact
enantiomers of each other, i.e. non-superimposable mirror images.
As the carbocation intermediate is planar, the attack from the nucleophile is equally likely to attack from
above or below the plane. Therefore we will end up with a 50:50 mixture of both enantiomers or a racemate
(racemic mixture). You will note that a racemic mixture of enantiomers is only possible when a chiral centre
results at the reaction centre.
Question 4.34
How many products will be observed in the following reactions? Justify your answer by drawing suitable
reaction mechanisms for the formation of each product. If more than one constitutional isomer is predicted
indicate which will be the major product.
CH3CH2
CH3
(a)
2-methylbut-2-ene + HBr (b)
trans - but-2-ene + HBr (c)
C C
+ HBr
H3C
CH3
Ozonolysis
One particularly important oxidation reaction of alkenes is their reaction with ozone, O3. This reaction cleaves
the double bond in the alkene to form a carbonyl species i.e. a species containing the C=O group. The reaction
can be summarised as shown below.
R
R
C
O C
+
C O
R
R
R
R
R
R
O3
C
The exact identity of the products of this reaction will depend upon the nature of the R groups. If for instance
we start with the following alkene
H
H3C
x
C*
C
CH3
H3C
then the products will be
H3C
*
C
O
+
x
O
H
C
CH3
H3C
Why is this reaction so important? Well it allows us to identify the alkene starting material if its structure is
unknown.
If for instance we identify the products of ozonlysis as ethanal and butanone then we would know the identity
of the starting alkene by reverse logic.
H
CH3
C
O
O
C
CH2CH3
H 3C
76 O3
C
C
CH2CH3
H
CH3
H
or
C
C
Note we cannot say which of the two geometric isomers is the starting material.
Question 4.35
Ozonolysis of alkene A gave the following aldehydes. Specify the reaction conditions and deduce the structure
of alkene A.
A → CH3CH2C(CH3)2CHO and (CH3)3CCHO
Organic Redox Chemistry
Carbon is in group 14 of the periodic table, with four valence electrons. The rules of assigning formal oxidation
numbers dictates that when carbon combines with oxygen it will exhibit positive oxidation numbers. However
when carbon combines with hydrogen it will exhibit negative oxidation numbers. The two extremes of this
arrangement are CO2, with C in the +4 oxidation state, and CH4, with C in the -4 oxidation state. The oxidation
states in between +4 and –4 will be encountered for carbon compounds which contain both oxygen and
hydrogen.
Question 4.36
Assign an oxidation number to carbon in the following compounds
(i)
CO
(ii)
CH3OH
(iii)
CH2O
(iv)
HCOOH
Therefore if we want to convert an alcohol into an aldehyde then we would consider using an oxidising agent.
If instead we want to convert a carboxylic acid to an aldehyde we would use a reducing agent and so on.
Another example of thinking of organic reactions in terms of oxidation numbers is if we go from an alkene to
an alkane. An example would be –
C2H4 → C2H6
ethane
ethane
If we calculate the change in formal oxidation number of carbon in this reaction we should see that carbon is
going from –2 to –3 i.e. it is being reduced. Thus we would add a reducing agent to ethene to obtain ethane.
Of course we also need to add hydrogen atoms, but that isn’t a problem because if we add H2(g) to ethene we
are adding a source of hydrogen and a reducing agent in one.
Thus organic chemists talk about the addition of hydrogen to a compound as a reduction. If hydrogen is
removed from a compound then they will have effected an oxidation. Similarly if they add oxygen to a
compound they term it an oxidation (as the formal oxidation number of carbon will increase). If oxygen is
removed then we will have reduced the compound.
Inter-Conversion of Alcohols/Aldehydes/Ketones/Carboxylic Acids
o
If we oxidise a 1 alcohol then we will get an aldehyde
77
H
R
C
OH
H
oxidizing agent
C
H
O
R
Question: What would we use as an oxidising agent?
Answer: A species which has a high oxidation state.
In this particular case we use CrO3 in acidic media.
Question 4.37
+
Name the product when propan-1-ol is treated with CrO3/H .
o
If we oxidise a 2 alcohol then we will get a ketone
H
R
C
OH
CrO3 / H+
R
C
R'
O
R'
+
Again we can use CrO3/H as our oxidising agent.
Question 4.38
+
If we react a tertiary alcohol with CrO3/H , what product would you predict (use tert butylalcohol as an
example)? Hint: be careful!
Question 4.39
o
+
In fact, if we treat a 1 alcohol with excess CrO3/H then it will be oxidised all the way through to the
o
+
corresponding carboxylic acid. However, it we treat a 2 alcohol with excess CrO3/H the reaction stops at the
formation of the ketone. Explain.
If, instead, we wish to go from an aldehyde or a ketone to an alcohol we should recognise that this reaction is
a reduction and we should therefore treat the aldehyde/ketone with a reducing agent. A very common
reducing agent in organic chemistry is LiAlH4, (lithium aluminium hydride). LiAlH4 is a source of hydride ions,
H in solution.
Addition Reactions of Alkenes
An addition reaction of an alkene involves the following transformation
C
X
XY
C
Y
C
C
We have already seen that hydrogen halide species, HX, can add across the double bond via a carbocation
intermediate. However other species can add across the double bond to give a variety of products. In some
cases the mechanism might be different but the overall reaction involves transforming an unsaturated species
into a saturated species. Thus the overall effect is a reduction. Below are some examples of addition reactions
(i)
H3C
CH3
C
H3C
C
H
H2O / H+
78
H3C
OH
H
C
C
CH3
CH3
H
Effectively we have added the elements of water across the double bond. This is called HYDRATION and is a
method for preparing alcohols
(ii)
H H
H
H
C
H2
C
H
H
catalyst
H
C
C
H
H
H
Here we have added hydrogen across the double bond to form an alkane. The presence of the ‘catalyst’
enables the HYDROGENATION reaction to proceed at a faster rate.
(iii)
Br Br
H
H
C
Br2
C
H
H
H
C
C
H
H
H
Bromination of the double bond again results in a saturated species.
Elimination Reactions of Alkyl Halides
If addition reactions of alkenes are considered to be reductions then there must be an opposite reaction that
will be considered an oxidation. The opposite of addition is elimination, where a small molecule is eliminated
leaving behind an unsaturated alkene. For example:
H
H
Br
C
C
H
CH3
CH3
H
HBr
CH3
C
H
C
CH3
In the above reaction HBr is eliminated to convert an alkyl halide into an alkene. This type of reaction is
catalysed by a very strong base, namely tert-butoxide:
CH3
H3C
O-
C
CH3
If we look at the alkyl halide we will notice that the bond most likely to be broken is the C-Br bond as it is the
most polar, and Br will be the best leaving group.
H
H
C
H
x Br
C CH
3
CH3
If this happens of course then the carbon marked with an X will be electron deficient. Therefore we require to
move an electron pair towards it.
H
H
C
Br
C
H' CH3
79
CH3
All we’ve got to do therefore is “encourage” the carbon-hydrogen, C-H’ bond to break! How do we do this? By
+
recognizing that if the H’ atom gives up its two electrons then it will become H , or an acid. What better
therefore than to add a base such as our tert-butoxide ion! The full reaction mechanism can then be drawn as
H
H
Br
C
C
CH3
CH3
H
H
CH3
C
-
O
H3C
C
C
+
Br- + (CH3)3C
OH
CH3
H
CH3
CH3
Therefore we end up eliminating H and Br from the alkyl halide and form an alkene. You’ll note that we
require a very strong base to do this as the hydrogen bonded to the carbon is very weakly acidic.
Nucleophilic Substitution Reactions of Alkyl Halides
A substitution reaction occurs when a substituent at a carbon centre is replaced with another, different,
substituent. If we have nucleophilic substitution then we can outline the general reaction as
Nu-
C
+
-
X
C
Nu
+
X-
-
where Nu is the nucleophile and X is the halide leaving group. Two different modes of reaction mechanism
can take place and we will deal with each in turn.
SN1 Reactions
The diagram below shows what SN1 stands for.
SN1
only one reactant is involved in the
formation of the reaction intermediate
Substitution
Nucleophilic
How do SN1 reactions proceed? The answer is via a carbocation intermediate. The first step is therefore the
breaking of the carbon-halogen bond.
R'
R''
R'
C
R''
X
R'''
C
+
X-
R'''
The carbocation thus formed then undergoes nucleophilic attack.
R'
R'
R''
C
R'''
+
Nu
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R''
C
R'''
Nu
An example of SN1 would be the following reaction.
H3C
C
CH3
CH3
CH3
H3C
Cl
C
+
OH-
H3C
C
CH3
CH3
CH3
OH
SN1 reactions go via a carbocation intermediate.
Once again we need to consider the consequences of going through a carbocation intermediate. Consider the
following reaction in a basic medium.
CH3
C2H5
sp
CH3
3
C
Br
C2H5
sp
2
C
+
Br-
C3H7
C3H7
2
Once again we are forming an sp hybridised carbon centre which is trigonal planar. Again (as was the case in
the addition reactions of alkenes) we must recognise that the nucleophile, OH , now can either attack from
below or above.
H5C2
OH
C
CH3
C3H7
CH3
C2H5
-
C
OH
C3H7
H3C
OH
C
C2H5
C3H7
Two enantiomers will thus result and we will have a racemic mixture of products.
If the carbocation centre has three different substituents attached to it then a racemic mixture of the
two enantiomers will result.
Question:
Answer:
What type of alkyl halides will undergo SN1 reactions?
Those that can form stable carbocation intermediates.
Question:
Answer:
Which alkyl halides will form the most stable carbocation intermediates?
o
Tertiary alkyl halides as they will form 3 carbocations.
Tertiary alkyl halides undergo substitution via an SN1 reaction mechanism.
SN2 Reactions
Would we expect CH3Br to undergo an SN1 type reaction? If it did then the first step would be
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H
H
H
C
Br
H
C
+
Br-
H
H
is the least stable carbocation that can form as it has no stabilising inductive effect from alkyl
substituents.
+
CH3
Question 4.40
o
o
o
Designate what type of carbocation (1 , 2 or 3 ) the following alkyl halides will give rise to if they undergo an
SN1 type of reaction.
CH3CH2I
CH3C(CH3)2CH2I
CH3CH2C(CH3)2I
CH3CH2CH(CH3)I
o
o
Generally primary alkyl halides will give rise to 1 carbocations which are not stable. Therefore 1 alkyl halides
do not undergo the SN1 reaction. Rather they undergo an SN2 route.
Question 4.41
How many reactants are involved in the formation of the reaction intermediate in an SN2 type reaction?
o
Instead of forming carbocations 1 alkyl halides form a five-coordinate intermediate. If, for example, we look
at the reaction of iodoethane with bromide ions:
Br-
H3C
C
+
H
I
H
Firstly bromide approaches the central carbon atom from the opposite side to the iodide substituent.
What then happens is the carbon simultaneously starts to form a bond with the bromine and starts to break
the bond with the iodine. We thus have a very short-lived pentavalent-carbon intermediate.
CH3
Br
C
I
H
H
signifies bond forming or breaking
The iodide ion then leaves the intermediate to give:
CH3
+
C
Br
H
I-
H
At no stage has a carbocation intermediate formed as in SN1.
How many products form in SN2 reactions? Well, again it will depend upon the starting alkyl halide. If we start
with an alkyl halide that has three different substituents on the carbon attached to the halide (i.e a chiral
centre) then we need to specify whether the starting material is a pure enantiomer or a racemic mixture. This
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is important, as during an SN2 reaction the attack by the nucleophile is unidirectional i.e. it attacks from one
direction, the direction from opposite the leaving group. This will result in inversion at the chiral C-atom.
To help explain this, consider the following alkyl halide.
H
C
H5C2
I
CH3
-
When this is attacked by a nucleophile, say OH , then the following intermediate will form:
H5C2
H
C
OH
I
CH3
and finally we will get:
H
HO
C
C2H5
CH3
If we therefore started with, say, the R-isomer of the alkyl halide we will get the S-isomer of the final alcohol,
and only the S-isomer. This is called inversion. If instead we started with a racemate of the starting material
then because all the starting material will be inverted to the other enantiomer a racemic mixture of the
product will result.
o
1 alkyl halides undergo SN2 substitution pathways
o
3 alkyl halides undergo SN1 substitution pathways
o
2 alkyl halides undergo either SN1 or SN2 substitution pathways
Acids & Bases in Organic Chemistry
Some organic acids are very easy to spot – they’re not called the carboxylic acids for nothing! Others,
however, are less obvious. Terminal alkynes for instance are considered to be acids, albeit very weak ones.
Carboxylic Acids
Carboxylic acids are weak acids i.e. they do not fully ionise in solution. Rather an equilibrium is established.
-
RCOOH + H2O ⇌ RCOO + H3O
+
Although carboxylic acids are technically weak acids they’re still, in the realm of the organic chemist at least,
considered to be a good source of protons or hydronium ions. This is due to the stability of the carboxylate
ion.
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O
R
C
O-
Is this the only way we can draw the structure or is there the possibility of another resonance form? How
about this?
O-
O
R
C
R
O-
C
O
The presence of another resonance structure stabilizes the carboxylate ion as now the negative charge is
delocalised and the average structure can be drawn as
O
C
R
O
Alcohols
Alcohols are strange species. They can act either as an acid or as a base. Alcohols are less acidic than water
but they can act as proton donors.
+
R – O – H + base → R – O + base – H
An example would be the reaction of phenol with sodium amide, NaNH2, a strong base.
H
O-
O
+ NH2-
NH3 +
Compounds resulting from substitution of a benzene ring are usually known by their nonsystematic names e.g. phenol. Some other examples are
CH3
COOH
toluene
benzoic acid
NH2
aniline
However alcohols can also accept protons and therefore exhibit basic properties. This arises because of the
lone pairs on the oxygen atom of the alcohol.
H
R
H+
O
R
O
H
H
oxonium ion
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We shouldn’t be too surprised by this behaviour as it is exactly analogous to the behaviour of water, another
amphiprotic species.
H
O
+
OH-
OH-
H2O
+
H
H
O
H+
+
H3O+
H
Would we expect alcohols to react with carboxylic acids? The answer is yes – acids react with bases! An
example would be the following reaction.
CH3CH2COOH
acid
+
CH3CH2OH
base
→
CH3CH2COOCH2CH3
+ H2O
ethyl propanoate
The exact mechanism of the ester formation is quite complicated and we shouldn’t push the acid-base story
too far. Organic chemists in fact call this type of reaction a CONDENSATION as a small molecule, in this case
water, is produced in the reaction.
Terminal Alkynes
Terminal alkynes contain the following group R – C ≡ C – H
The terminal hydrogen is very weakly acidic and therefore we would expect terminal alkynes to react with
strong bases such as sodium amide.
R
C
C
H
+
-
NH2
R
C
C
+ NH3
This is a very important reaction in organic chemistry as it gives a route to form carbon-carbon bonds.
Question 4.42
(a)
Is the anion R – C ≡ C an electrophile or a nucleophile?
(b)
Is the anion R – C ≡ C an acid or a base?
-
The anion, R – C ≡ C is looking to attack a positive centre on another molecule. What type of molecules
contain positive centres? Easy – compounds which contain carbon attached to a more electronegative atom,
such as an halogen (X)
R
R
C
C
C
H
X
R
C
C
CH2
R
+
X-
H
If we think about this reaction type very carefully we should be able to recognise it as a nucleophilic
substitution.
Question 4.43
Would you predict the above reaction to go via an SN1 or an SN2 mechanism? Explain your answer.
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We could also react this powerful nucleophile with CO2 in acidic media.
O
R
C
C
O
C
O
H
+
R
C
C
C
OH
So we have another route possible to prepare carboxylic acids.
Reagents and Reaction Conditions
Now we’ve got a general picture (however small!) of the reactions that can occur, an important issue to
consider is how to actually go about doing these reactions in the lab or in the workplace. This is important
because if one chooses the wrong conditions then
the reaction might not occur at all, or even worse
the reaction will proceed but to the wrong products!
Most organic reactions need to have the following specified.
(a) solvent
(b) temperature
(c) catalyst
(d) reagents to be used
The presence of a catalyst is of particular importance. Catalysts are added to speed things up. However
catalysts can speed some reactions up faster than others and therefore one catalyst might give a different
product from another catalyst with the same reactants!
An example of this would be the reactions of alkynes with hydrogen gas. What would we expect to get?
R – C ≡ C – R’ + H2 → ?
Here we have got an unsaturated molecule reacting with a reducing agent. We would therefore expect to get
ADDITION across the triple bond. Possible products therefore are:
H
H
C
C
R
R'
H
C
R'
R
C
R
H
H
H
C
C
H
H
R'
The actual product formed depends upon the experimental conditions.
(i)
H
H
R
C
C
Lindlar catalyst
R'
H2
C
C
R
R'
cis- alkene
(ii)
R
C
C
R'
Pd/C
2 H2
R
H
H
C
C
H
H
R'
alkane
The so-called Lindlar catalyst stops the reaction at the alkene (note it is also specifically the cis-alkene) while
the use of a palladium on carbon catalyst takes the reaction further. Therefore when we react H2 with an
alkyne we must specify the catalyst used.
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o
Another example of the importance of specifying the reaction conditions is when oxidising 1 alcohols.
R
CH2
CrO3
H+
OH
R
CrO3
H+
CHO
R
COOH
+
The use of CrO3/H will oxidise the alcohol all the way through to the carboxylic acid. However the use of
pyridinium chlorochromate (PCC) in dichloromethane allows the isolation of the aldehyde.
R
CH2
OH
PCC
CH2Cl2
R
CHO
Sometimes we can use our knowledge of the reaction conditions to yield structural information as in the
following example.
Example 4.6
Question:
A hydrocarbon of unknown structure has the molecular formula C8H10. On catalytic hydrogenation over the
Lindlar catalyst, 1 equivalent of H2 is absorbed. On hydrogenation over a Pd/C catalyst 3 equivalents of H2
react. Draw a structure for the hydrocarbon that fits the data.
Answer:
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Question 4.44
Predict the products from reaction of hex-1-yne with H2 over
(a) Lindlar catalyst
(b) Pd/C catalyst
Question 4.45
Predict the products from reaction of dec-5-yne with
(b) H2, Pd/C
(a) H2, Lindlar catalyst
Question 4.46
Predict the products from reaction of pentan-1-ol with
(a) CrO3
(b) PCC in CH2Cl2
Putting it All Together!
The process of learning at the tertiary level can be viewed as having three stages.
• Knowing the facts
• Understanding the facts
• Applying the knowledge gained
It is the last point above that is the most important in chemistry. What can you do with what you have at your
fingertips? We will therefore end this tutorial with some problem-solving exercises in organic chemistry. The
best way to learn how to solve problems is to attempt to solve problems and therefore we will take you
through some worked examples and then leave the rest up to you!
Example 4.7
Question:
An unknown hydrocarbon, A, has a formula of C6H10. On catalytic hydrogenation over Pd/C it reacts with only
1 equivalent of H2. A also undergoes reaction with ozone to give a single product a dialdehyde, B. Suggest a
structure for A and B
Answer:
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Example 4.8
Question:
How could you prepare 2-methylhexane from an alkyl halide and an alkyne?
Answer:
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Question 4.47
In Example 4.8 we broke the C4-C5 bond to choose our starting materials.
(a)
Still by breaking the C4-C5 bond, choose another alkyl halide and another alkyne which will give the
same product.
(b)
Do a retrosynthetic analysis to make 2-methylhexane by breaking the C5-C6 bond.
Example 4.9
Question:
An aldehyde A (C5H10O) can be reduced by LiAlH4 to B. Treatment of B with concentrated H2SO4 gave C.
Ozonolysis of C affords methanal and D (C4H8O). D gave a positive iodoform test which identifies
methylketones. Identify A, B, C and D.
Answer:
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Example 4.10
Question:
A hydrocarbon A adds one equivalent of hydrogen in the presence of a Pd/C catalyst to form n-hexane. When
A is reacted vigorously with KMnO4 a single carboxylic acid containing three carbon atoms is isolated. Give the
structure and name of A.
Answer:
Question 4.48
If we start with hex-2-ene, what would be the products upon vigorous treatment with KMnO4?
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