Topic 10 Organic Chemistry
10.1
Organic chemicals exist in groups where there is a common general formula…ex CxH2x+2
.they will behave in a very similar way chemically and such a group is called an homologous series
where the chemicals differ only in the value of x.
The members of a homologous series will show a steadily increasing boiling point as the
value of x rises due to increased VDW’s….the increase becomes less as x rises since the % rise
in the Mr is less. These members will differ in their formula by CH2 as x increase by 1 every time,
since all that is being changed is that carbon chain is being increased by one …and that brings in
another 2 hydrogens.
Structural, empirical and molecular formulae
The empirical formula shows the simplest ratio of atoms in the molecule…Ex CH2 .
The molecular formula is a simple whole number multiple of the empirical and shows the
actual # of each type of atom in the molecule. Example = C4H8 .(Note that the empirical formula of
this is CH2 )
Whereas hundreds of chemicals could have the same empirical formula and several chemicals
could have the same molecular formula, only ONE chemical will have a certain structural formula.
The structural formula could be written out in “condensed form”..Our molecule C4H8 could be
CH3CHCHCH3 or written out in “displayed form” as shown here:
H
H
l
I
H--C – C = C-- C--H
I I
I I
H H
H H
Note that you can spot the various parts of the condensed formula in the displayed form. Try to be
able to handle condensed formulae, but always try to use the displayed form
Crazy though it is, you are told to write complete structural formulae for organics and these
means that you MUST put on all the H’s. In class we will often try to save time by writing various
forms of skeleton formulae:
l
I
-C – C = C- CI I
I I
Or even the more skeletal : C- C = C- C
Know the prefixes for 1(6 ie meth; eth; prop; but; pent and hex .
Structural isomerism
Definition : Same molecular formula, but different structural arrangement of atoms. NOTE : do NOT
substitute the word “chemical” for molecular in this…you will not get the mark.
In order to fully appreciate these things, I am going to have to take you through naming organics.
NAMING ORGANICS : The alkanes
This can get very messy, but you have to be able to do this. For every structural isomer
there is, there is a unique name.
The first group we will look at is the simplest ..the alkanes, an homologous series which has the
general formula: CxH2x+2
So the first 6 alkanes are CH4 ; C2H6 : C3H8 : C4H10 : C5H12 : C6H14 .
The alk- bit is the stem and we use the following prefixes to show the # of CARBONS IN THE MAIN
CHAIN
1 = meth2= eth- 3= prop4=but- 5=pent- 6=hexIf the chain is branched, we still use the prefixes above to describe the main chain length along with
–ane to show the homologous series to which it belongs. However we use numbers and add –yl to
the prefix to denote the length and position of a side chain.
Example :
C–C–C–C–C–C
I
C
This is 3-methyl hexane…a 6 carbon chain alkane with a 1 carbon side chain on the 3rd carbon of
that chain. Always number from the end closest to the side chain. If you have identical side chains
, put di(2) , tri(3) or tetra(4) in between the #’s and the side chain…2,3-dimethylpentane.
NOTE: You cannot have a 1-methyl….you counted the main chain length incorrectly. Please be
aware that these things are TETRAHEDRAL around the C and you are drawing them FLAT and
STRAIGHT and they are not. They are as twisted as my snake.
C–C–C–C
I
C
The above is PENTANE and not 1-methylbutane (or indeed 4-methylbutane)
Structural isomers are possible for alkanes where x =4 or more there is butane and 2-methyl
propane where x=4 and for x=5 we have pentane, 2-methyl butane and 2,2-dimethyl propane as
shown:
C
- C – C – C – C - CC – C – C - CI
I
C - C – CC
I
C
Be able to write out condensed formulae too…2-methyl butane is CH3CH(CH3)CH2CH3
Now try to work out and name all the structural formulae for C7H16 .The straight chain one is heptane
by the way..and there are 7 others you are looking for.
If an H is substituted by a halogen, treat it in the same way you have for carbon side chains
EXCEPT use CHLORO: BROMO et cetera.
Example: 1,2-dichlorobutane
Naming organics: Alkenes
Alkenes have the general formulae Cx H2x and MUST contain a C=C bond. Again, structural
isomerism starts with x=4 ….but-1-ene , but-2-ene and 2-methyl propene as shown below:
C = C – C – C-
-C–C=C–C-
-C=C–CI
C
Note that a number often goes IN THE MIDDLE here…this is where the functional group placing
is shown. So but-1-ene means a 4 carbon alkene chain with the C=C attached to carbon #1. You
do not state where the C=C is if there is no choice…ethene and propene
Draw out 4-methylpent-2-ene…and while you are doing it note that the position of the C=C takes
precedence over the side chain..so it is NOT 2-methylpent-3-ene.
Naming other organics
The other ones you need to know are
a) Alcohols, where an O has been added between the C and an H in an alkane:
C – C – C – OH
This is propan-1-ol, which is the first alcohol with a number in it. Since these can form H bonds,
they are a lot less volatile than alkanes due to larger IMF’s and so they are either liquids or solids at
room temps. The H bonding also means that they are a lot more water soluble as well, but as the C
length increases, water solubility falls rapidly as the nonpolar section begins to dominate.
b) Aldehydes: These are alkanes with a C =O at the end.
I
H
In condensed form we would write it as C2H5CHO. This is propanal and no number is required ,
since the CHO must be on the first carbon. These cannot form H bonds, but are pretty polar, so
they have lower bpts than alcohols and are less water soluble
c) Ketones: These have a C=O somewhere in the middle of an alkane chain. The simplest is
propanone, but you can get things like hexan-2-one:
C–C–C–C–C–C
II
O
These are still polar, but not as polar as aldehydes and thus have lower bpts and lower water
solubility.
d) Carboxylic acid: These have –COOH on the end and that does not contain an O – O bond…it is
C=O
I
O
H
Example = butanoic acid
These have a higher molar mass due to the extra O and H bonding, so they are less volatile than
alcohols and have higher bpt per C in the chain. The extra O increases their polarity and so they are
very water soluble when the C chain length is not that great.
10,1,11 Spotting other functional groups
Those are all you need to name, but you are asked to recognize the following groups in structural
formulae:
C6H5 - C – C – C – O – C – C – NH2 .
II
O
On the left we have the BENZENE ring …horrid little thing, which is a 6 carbon ringed hexagon with
an H at every corner except where it is attached to the rest of the molecule.
In the middle we have the totally obnoxious ESTER group, which is just like a carboxylic acid, but
has a C attached to the O instead of an H.
On the far right we have the distinctly smelly AMINE group, which you could say is an ammonia
molecule where one ( or more) of the H’s has been replaced by a carbon chain.
Do not ask me to name this thing…it would be something like 2-aminoethyl-3-phenylpropanoate.
Funcional isomerism: These chemicals have the same formula, but belong to different homologous
series. Examplee = aldehydes and ketones or carboxylic acids and esters. As a result of this the
behave differently ( they FUNCTION differently)
10,1,12
Alcohols exist as primary, secondary and tertiary structures and can be identified by noting how
many carbons are attached to the C that carries the –OH.
A primary has 0 or 1…ie the OH is on the terminal C of the chain. Example= buan-1-ol
A secondary has 2 and so the OH is on a C in the middle of the chain.
Example = pentan-3-ol
A tertiary has 3, and so the C carrying the OH has NO H’s attached to it.
An example = 2-methylpentan-2-ol
If you now swap the OH for a halogen, you can talk about primary, secondary and tertiary
halogenoalkanes as well.
Reactions of organics
COMBUSTION
Hydrocarbons burn , but how they do so depends on the oxygen supply
a) complete combustion…. Needs plenty of oxygen and produces CO2
and H2O ONLY
b) incomplete combustion….when oxygen supply is limited…..yellow
flame due to soot (solid carbon particles)…so products are H2O and
then not all CO2, but some C. Sometimes for example in a car engine,
there is also CO produced.
Examples:
a) C4H8 + 6O2 ----( 4CO2 + 4H2O
b) C4H8 + 4O2 ----( C + 2CO2 + 2CO + 4H2O
Note here I had a choice of how many C’s or CO’s to put…as long as not all the C produces CO2
then it is incomplete combustion.
In all case, energy is produced, but incomplete combustion produces less energy and the products
are more dangerous.
Note that alcohols burn with a cleaner flame since they have an O already in them, BUT since their
boiling point is higher, as the chain length increases they become harder to light.
Reactions of Alkanes
A compound with one or more C=C is called UNSATURATED, whereas those without any C=C are
called “saturated”. The saturated ones are far less reactive.
Generally alkanes are unreactive…..combustion is one of the few things they do. This is because
they are non-polar molecules with (almost) non-polar bonds that are relatively strong for single
bonds. Of course that is why they have been lying underground for millions of years (oil). However
they will undergo reactions with halogens in sunlight, by something called FREE RADICAL
SUBSTITUTION.
The sunlight is needed to break the weak Hal-Hal bond and when this is split evenly
(HOMOLYTICALLY) it will result in two isolated halogen atoms. Now these are group VII atoms and
so have 7 electrons in their valence shell and of these, one is unpaired. We show this by using a
dot:
Stage 1: initiation..formation of free radical
Cl2 --------( 2Cl .
The Cl atoms are called FREE RADICALS and they attack almost anything.
Stage 2: continuation (propagation)…one free radical reacts to leave behind another..no
change in # of free radicals
It is the free radical that attacks the alkane and removes an H leaving behind an alkyl free radical
and HCl.
Cl
.
+ CH4 --------( CH3 . + HCl
The chlorine atom could also attack a molecule of chloromethane as well and if it did so, we would
get di ,tri substitution et cetera.
Stage 3: termination..two free radicals bond together
In the simplest scenario, the alkyl free radical will bond to the chlorine free radical to produce a
halogenoalkane:
Cl
.
+ CH3 . ------( CH3Cl
However it would also happen that two methyl free radicals would combine to form a longer chain
alkane.
And so overall we get CH4 + Cl2---( CH3Cl + HCl and a Cl in this case has been substituted for an H.
However this is the simplified scenario and we would actually get a range of organic compounds
produced, including longer chain alkanes and multi substituted halogenoalkanes
10.3 : Reactions of alkenes
Alkenes are far more reactive than alkanes and react by ADDITION…..the C=C breaks and the
chemical adds on across the double bond to produce an alkane. This happens for a couple of
reasons:
a) The C=C bond is an electron dense area and so is attacked by +ve species
b) The double bond is NOT made up of two identical C-C bonds, but rather a strong “sigma” bond
and a much weaker “pi” bond.
C=C–C
+ X2 -------(
C-C–C
I I
X X
1,2 di –X- propane ( Ex 1,2dichloropropane)
If X2 is a halogen, the reaction is fast and occurs in the dark (see reactions of alkanes for
comparison) . Bromine solution, which is yellow, is quickly decolorized in this reaction, which is
used as a test for C=C. In this reaction we tend to get an OH group attached to one end of the C=C
bond and a Br to the other, but I do not think you need to know that.
If X = H, then the product is an alkane, but a catalyst (Ni) must be added. This process is often
used to “hydrogenate” plant oils to create a solid margerine
Typically you are asked to add on H – X where X is a halogen or OH (water) , where the product is
a monosubstituted alkane or an alcohol. This is how industrial ethanol is produced.
In this reaction, if the alkene is symmetrical you only get one product:
C-C=C–C
+ HX -------(
C-C-C–C
I I
X H
But-2-ene produces 2-X-butane (2-bromobutane for example)
However unsymmetrical alkenes are more awkward, since there are 2 potential products:
C=C–C
+ H-OH -------(
C-C–C
OR C – C - C
I
I
OH
OH
Thus the product with propene and steam would be a mix of propan-1-ol and propan-2-ol, BUT the
main product would be the -2-ol. (This is not on SL)
Addition reactions are used in several ways: In order to turn a vegetable oil into a solid, it
can be hydrogenated…ie hydrogen added on to the C=C bonds….the more saturated the fat is, the
higher its boiling point is. BUT unsaturated fats are better for you, as saturated are more likely to
clog up your arteries (heart attacks are more likely)
Industrial alcohol can be made by the addition of steam to ethane with a catalyst.
Addition polymerization
Alkenes can be polymerised, here the molecules add on to each other to produce a macromolecule
called a polymer…the original molecule is called a monomer. Polythene, the commonest plastic is
a polymer of ethene, while PVC is a polymer of chloroethene. The whole process is started by
adding an initiator molecule.
I
I
I I
I I I I
I
C = C Polymerisation------. C – C – C – C – C – C – C etc.
I
I
I
I I
I I
I I
X
X
X
X
X
X can be many different things and it will affect the properties of the plastic. Example if X is C6H5
you have Styrofoam and if it is CH3 you have polypropene.
Note that the identity of the monomer can be determined by looking at the repeating unit of the
polymer. In the above we have :
H H
I
-C – CI
I
X H
This process is of MASSIVE importance to the human race…where would we be without plastics
and , as these are derivatives of OIL, their price is going progressively upwards adding to inflation.
I
The poor biodegradability of plastics is very problematic and our planet is suffering due to plastic.
This includes a large danger to wildlife such as marine turtles which mistake plastic bags for
jellyfish and the plastic floating rubbish patch out in the middle of the pacific.
10.4:
Reactions of alcohols
The only one we need to know is oxidation, BUT due to the presence of the OH grouping, these
things are pretty reactive.
Ethanol can be oxidised to ethanoic acid by powerful oxidising agents such as Cr2O72- The
dichromate ion is orange/yellow and is reduced to green Cr3+ . (the intermediate is ethanal, which is
usually not isolated)……so wine can go off if exposed to air…..it will turn to vinegar!
C2H5OH-----( CH3CHO + 2H+ + 2e --(further oxidation + H2O ---( CH3COOH + 2H+ + 2 e
The ethanal could be easily distilled off, since it has no H bonds and thus its boiling point is much
lower than that of either ethanol or ethanoic acid. Ethanoic acid would be the only product if lots of
oxidising agent was present and the mix was refluxed for a while.
This is the classic two step oxidation of a primary alcohol….
Alcohol ----( aldehyde---( carboxylic acid
However secondary alcohols can be easily oxidized as well. Here the H’s attached to the O and the
C carrying the O are removed to produce a ketone:
C–C–C
---( 2H+ + C – C – C + 2e
I
II
OH
O
The ketone is the end product however, since it does not have the weak C- H bond that the
aldehyde does ( weakened by the C=O right next to it)
A tertiary alcohol is either immune to attack by dichromate, or if you heat the heck out of it, the
molecule will shatter because you cannot create a C=O without breaking a C – C …since you
would have a C with 5 bonds…which tends not to bother some students BUT IT DOES NOT EXIST
…other than for fractions of a second as we will see soon!
10.5 Reactions of halogenoalkanes
The fact that this is in standard level is disgusting as far as I am concerned because SL students
will not have the background to cope!!
Nucleophilic substitution: This is the attack of a -ve species on the +ve C in an organic molecule
that results in the substitution of one group or atom for another group or atom. The positive C is as a
result of an electron withdrawing group or atom attached to that C…Cl or OH for example.
Halogenoalkanes can react with OH1- (and others) to produce alcohols and Hal
substitution in two ways
–
by nucleophilic
SN1 = substitution nucleophilic first order……the Hal leaves as a –ve ion and then
the carbocation reacts with the nucleophile. This happens best with tertiary structures due to the
greater stability of the carbocation:
C4H9Br -----( C4H9+ + Br1- (slow)
then C4H9+ + OH - ---( C4H9OH (fast)
You need to know the attack mechanism and this is shown using curly arrows
a) The loss of the halogen atom as a halide ion:
C – Br
----( C+ + Br1- .
b) The attack by the lone pair on the nucleophile on the +ve C
C+
:OH1- ----( C – OH
An SN1 mechanism is favored by tertiary halogenoalkanes and not favored by primary:
C
I
C-C-C is more stable than C-C-C-C+
+
Due to the +ve charge being more widely distributed over the structure than with the primary, where
it remains on the end C and so tertiary halogenoalkanes go this way better than primary.
SN2 = second order and is more likely in primary structures since the +ve C is
more exposed. The Hal is still attached as the nucleophile moves in:
C4H9X + OH- ----( C4H9XOH- (slow)
C4H9XOH- -----( C4H9OH + X- (fast)
Note in the middle organic compound we have a C with 5 bonds …an activated complex and thus
very unstable. One bond (from the OH to the C) is being formed as the other bond (C-Hal ) is being
broken as the Hal takes on the electrons in that bond.
I will draw this out by hand below to show the stages you are supposed to know and with
the curly arrows you are supposed to use:
Note that the syllabus states the above for the reaction with any halogenoalkanes and SODIUM
hydroxide. However since any strong hydroxide would exist as its component ions in solution, you
could use any of them..say LiOH or KOH, so do not let this throw you
10.6: REACTION PATHWAYS
If you are fully conxersant with the above notes, then you have completed the Standard level organic
section. However you may have to combine some of the above.
Example : How could you convert but-2-ene into butanone?
You should know that you get butanone by the oxidation of butan-2-ol using acidified
dichromate and that you can get butan-2-ol by the reaction of but-2-ene and steam.
Alternatively you could go from 2-bromobutane + NaOH to butan-2-ol and then on to
butanone
Give two different pathways that could be used to create
a) chloroethene
b) ethanal c) 1,2- dichloropropane
In c) one method should produce a range of organics whereas the other should only produce the
deired product. Explain.
Create your own flow chart with but-2-ene at its center.
11.3.1
Other functional groups
R = alkyl chain..methyl, ethyl et cetera
a) alcohol
R –O- H (alkanol)
O
II
b)aldehyde R – C – H or RCHO (alkanal)
O
II
c) ketone R – C – R
(alkanone)
OH
d) carboxylic acid
I
R – C = O or R COOH
e) ester R – C – O – R’ or
II
O
RCOOR’
f) amine R – N – H or RNH2
I
H
g) amide R – C = O or RCONH2
I
NH2 .
USE DATA BOOKLET if you forget (Usually chart 11 or 12)
11.3.2
Functional isomerism exists between different homologous series. Examples from above: groups
b) and c) as well as d) and e) and the BIG one ethers and alkanols
Ethanol C – C – O – H
and ethoxyethane
- C – O – C – are both C2H6 O and
because they have different chemical groups, they behave very differently…also alcohols can
produce H bonds and so have much higher boiling points than ethers.
11.3.3
Optical isomerism exists where you find a C with 4 different groups around it. Two
structures can be drawn that are mirror images of each other and differ only in the way they rotate
the plane of plane polarised light. Plane polarized light, is light in one plane (vertical for example)
and now one isomer will cause a clockwise rotation and the other an anticlockwise rotation
Butan-2-ol is a classic example:
H
I
C–C–C–C–
I
OH
So the third C has a CH3 ,a C2H5 , an OH and an H around it An optical isomer is also called an
ENANTIOMER and the carbon with the 4 different groups attached is called ASSYMETRIC or a
CHIRAL CENTER. In living systems, one optical isomer is produced and the other is almost never
seen……D- glucose but never L- glucose (D and L refer to each type of optical isomerism…d =
dextro = right and L = laevo = left)
Beware of any 2-substituted butane, because they are mostly optical…as well as most amino
acids.
Optical isomerism is a form of STEREO ISOMERISM where chemicals possess the same bonds in
the same order, but have different arrangements in space…geometric isomerism is another form of
stereo isomerism.
11.3.4
The solubility of the various groups depend upon whether they form H bonds, whether they are polar
and the length of the non-polar carbon chain.
Solubility in water: The H bonded chemicals , alcohols, carboxylic acids, amides and
amines dissolve well when the C chain is short, the highly polar chemicals, aldehydes, and
ketones are not as soluble, but do dissolve quite well when the C chain is very short. Ethers and
esters are only slightly polar and do not dissolve well at all.
Solubility in non-polar solvents: This is the exact opposite of the above..ie ethers dissolve
well.
Volatility : the ease with which the chemical forms a gas…..ie if its boiling point is low, then its
volatility is high…..so the most volatile ones are ethers and esters…with carboxylic acids and
amides being least volatile, as their van der Waal forces are larger ( for the same length C chain
AND they form H bonds).
Acid / base nature : carboxylic acids are weak acids and amines are weak bases. All others are
essentially neutral. The O – H bond in the carboxylic acid is weak and the H+ can split off. In
amines, N pulls electrons from 2 H’s and a carbon chain….so it is pretty negative……so it attracts
H+ to its lone pair.
11.3.5
Alkenes are far more reactive than alkanes and react by ADDITION…..the C=C breaks and
the chemical adds on across the double bond to produce an alkane. This happens for a couple of
reasons:
a) The C=C bond is an electron dense area and so is attacked by +ve species
b) The double bond is NOT made up of two identical C-C bonds, but rather a strong “sigma” bond
and a much weaker “pi” bond.
C=C–C
+ X2 -------(
C-C–C
I I
X X
1,2 di –X- propane
If X2 is a halogen, the reaction is fast and occurs in the dark (see reactions of alkanes for
comparison) . Bromine solution, which is yellow, is quickly decolorized in this reaction, which is
used as a test for C=C.
If X = H, then the product is an alkane, but a catalyst ( Ni) must be added. This process is often
used to “hydrogenate plant oils to create a solid margerine
Typically you are asked to add on H – X where X is a halogen or OH (water) , where the product
is an alcohol. This is how industrial ethanol is produced
11.3.6
Addition reactions are used in several ways: In order to turn a vegetable oil into a solid, it
can be hydrogenated…ie hydrogen added on to the C=C bonds….the more saturated the fat is, the
higher its boiling point is. BUT unsaturated fats are better for you, as saturated are more likely to
clog up your arteries (heart attacks are more likely)
Industrial alcohol can be made by the addition of steam to ethane with a catalyst.
11.3.7
Alkenes can be polymerised, here the molecules add on to each other to produce a
macromolecule called a polymer…the original molecule is called a monomer. Polythene, the
commonest plastic is a polymer of ethene, while PVC is a polymer of chloroethene. The whole
process is started by adding an initiator molecule.
I
I
I
I
I I I I
I
C = C Polymerisation------. C – C – C – C – C – C – C etc.
I
I
I
I I
I I
I I
X
X
X
X
X
X can be many different things and it will affect the properties of the plastic. Example if X is C6H5
you have styrofoam
11.3.8
Esterification is the reaction of an alcohol and a carboxylic acid in the presence of a strong acid
catalyst
C–C–O–H
Ethanol +
+ C – C – C - COOH --( C – C – O – C – C – C – C + H2O
II
O
butanoic acid -------------( ethyl butanoate + water
Esters are fruit flavours….ie can be used as artificial flavours in foods and as scents in perfumes.
11.3.9
Ethanol can be oxidised to ethanoic acid by powerful oxidising agents such as Cr2O72- The
dichromate ion is orange/yellow and is reduced to green Cr3+ . (the intermediate is ethanal, which is
usually not isolated)……so wine can go off if exposed to air…..it will turn to vinegar!
C2H5OH-----( CH3CHO + 2H+ + 2e --(further oxidation + H2O ---( CH3COOH + 2H+ + 2 e
The ethanal could be easily distilled off, since it has no H bonds and thus its boiling point is much
lower than that of either ethanol or ethanoic acid. Ethanoic acid would be the only product if lots of
oxidising agent was present and the mix was refluxed for a while.
11.3.10
If the molecules used in 11.3.8 had had active groups at both ends, then the reaction could have
continued to produce a polymer…this is not an addition polymer as made by alkanes, but a
CONDENSATION polymer, so called because a small molecule (water) is ejected. It requires
monomers to have an active group at each end.
Di acid + di amine ---------( NYLON
Di acid + di alkanol --------( POLYESTER
The monomers used in the creation of classic nylon are hexan 1,6-dioic acid and 1,6-diamino
hexane while normal polyester would be made from ethan1,2 -diol and benzene 1,4-dicarboxyllic
acid
11.3.11
Amino acids are molecules that have an amine group up one end and an alkanoic acid
group up the other. . There are lots of possibilities but the simplest is :
H
I
H – N – C – C –O - H
I I II
H H O
This is GLYCINE or 2-amino propanoic acid to me. The bold H can be swapped for other things to
create a whole heap of “natural” amino acids found in living tissues.
THESE ARE THE BUILDING BLOCKS OF PROTEINS. Note they have two active ends….although
the ends are different…they can be combined intp condensation polymers as before and a water is
ejected when each amino acid is added…..two amino acids = a dipeptide…..a few more = a
polypetpide and lots = protein.
H–N–C–C–N–C–C–N–C–C -N–C–C–N–C–C–O-H
I
II I
II I
II I
II I
II
H
O
H
O H
O H
O H
O
Note the difference between nylon and protein…..nylon goes A – A – B – B – A – A – B – B while
protein goes A – B – A – B – A – B – A – B - where A = acid remnants ( C=O) and B = base
remnants ( N - H ).
Organic Test
1) Which of the following is an ester? A) CH3OCH3 B) CH3CHO C) CH3CH2OOCCH3 d)
CH3OCH2COOH
2) The addition of Bromine to ethene will produce a) 1,2 dibromo ethene b) 1,2 dibromoethane c)
Bromoethane d) 1,1 dibromoethane
3) Which of the following chemicals is a secondary alcohol? A) CH3OH b) C2H5OH c) CH3OCH3 d)
CH3CHOHCH3
4) A free radical that could attack a methane molecule is
i) CH4 ii) Cl2
iii) Cl
a) I and iii only b) ii only c) iii only d) ii and iii only
5) The correct name for the organic drawn below is a) hexane b) 1 methyl pentane c) 2-methyl
pentane d) 1,3-dimethyl butane
C
C
I
I
C-C-C-C
6) An addition polymer a) has only one molecular monomer b) ejects a small molecule, such as
water or a hydrogen halide, per monomer added onto the chain c) relies on breaking a C=C bond in
the monomer.
d) Adds a small molecule such as water into the chain as each monomer is added on to the chain.
7) Which has the highest boiling point? A) ethanol, b) ethanal c) ethanoic acid d) ethane
8) How many different structural isomers are there for C4H9 Br?
a) 2 b) 3 c)4 d) 5
9) The catalyst added in order to speed up oxidation of alcohols is a) Cr2O72- b) H2O c) H+ d) None
is used
10) Incomplete combustion of a hydrocarbon could produce which of the following
I) CO
II) C
III CO2
IV) Water
a)
I and II only b) I,II and IV only c) II and IV only d) All of them are possible products
11a) Define the term SN1(1)
b) What type of bromoalkane undergoes SN1 reactions easiest(1)
c)
12)
Show the mechanism for the SN1 reaction between a halogenoalkane, C4H9Br and OH . (3)
The monomer chloroethene can be polymerised to produce the hard plastic PVC.
a)Draw out the monomer and a 6 carbon section of the polymer PVC (2)
13)a)Name the two organic products that could be produced when ethanol is oxidised.(2)
a) In the oxidation of ethanol, the process of REFLUX is carried out. Give TWO reasons
why this process is carried out(2)
14a) Carboxylic acids have the general formula Cx H2xO2 . Using the value where x = 3 draw out a
structure that is a carboxylic acid and name the molecule you have drawn. (2)
Draw another structure that is NOT carboxylic acid but has the same formula as that in 14a above
and name one functional group that this molecule possesses (2)
TEST : Organic chem
1) Combustion of an alkane normally produces a) soot, CO2 and water b) only CO and CO2 c)
only CO2 and H2
d)Alkenes and H2
2) The molecule
CHClCHCl is named a) chloroethene b) 1,2 dichloroethene c)
dichloroethene d)1,1 dichloroethene
3)The formula of the molecule known as propanal is : a) C3H7CHO b) C3H7OH c) C2H5CHO d)
CH3COCH3
4) The chemical propanal is produced by the partial oxidation of a) Propanone b) propanoic acid c)
Propanamide d) propan-1-ol
5) The chemical ethylamine , C2H5NH2 when added to water is a) Basic and insoluble b) acidic and
insoluble c) Basic and soluble d) basic and insoluble …
6) Which of the following is NOT a member of the same homologous series? A)CH4 b)C2H4 c) C3H8
d)C4H10
7)The product of the reaction between CH3COOH and CH3CH2OH is a) CH3COOCH2CH3
b) CH3CH2COOCH2CH3 c) CH3CH2COOCH3 d)CH3COOCH3
8) When bromine is added to propene, what is formed ? a) 1,2-dibromopropene b)
1,1-dibromopropane c) 1,2- dibromopropane d) 1-bromopropane
SECTION B
9) Account for the fact that the boiling point of alkanes a)increases as the length of the carbon
chain increases and is lower than the boiling point of alcohols with a similar # of carbon atoms (4)
10) Several chemicals exist with the formula C2H4O2 . Draw out any two that contain different
functional groups and suggest, with reasons, which would be more soluble in water. (3)
11)
Styrene has the formula : H
H
C=C
H
C6H5
This molecule can be polymerised to form polystyrene (styrofoam) a packing material. What type of
polymerisation is this? Draw out a 6 carbon section of the polymer, substituting X for C6H5 if you
wish. (2)
11) Amino acids react together by a condensation reaction. Using your data booklet draw out an
organic product produced by the reaction of alanine and cysteine. Name the non-organic product.
(3)
b) Alanine exists in optical isomeric forms. Explain why this is so. (1)
c) Diethyl ether, C2H5 0 C2H5 is a functional isomer of butan-1-ol. Explain the term” functional
isomer” using these two chemicals as examples(2)
d) Draw out and name a position isomer of butan-1-ol (2)
e) Draw out and name a chain isomer of butan-1-ol (2)