Download CHM412 June 2013 paper

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CHM412 June 2013 paper
Question 4
a) Cleavage in chemistry means to break/split. Homolytic cleavage involves breaking of a bond between two
atoms of the same type, e.g. a Br and another Br atom, so that one of the two e- in the breaking bond goes
onto one atom that made up bond.
Some examples
Br
Br
H
.
2 Br
C
H
H
H
H
C
H
.
C
.
C
H
H
This can happen in the manufacture of poly(ethene),
although a substance called an 'initiator' us usually used
instead. The initiator forms radicals which last a bit longer
than the ethane diradical - which quickly reforms the
pi bond.
Heterolytic cleavage is when a bond between atoms of different types break and both e- in the breaking bond
go onto one of the atoms only. They will go on the most electronegative atom in the bond.
e.g.
H
H
C
O
H
+
C
.. _
O
H
Another important difference is that homolytic cleavage (or homolytic fission) produces RADICALS
whereas heterolytic fission produces IONS. Also note the different arrows. The single hooked arrow shows the
movement of just one electron. The double hooked arrow shows the movement of two electrons.
H
H
Cl
Cl
2 Cl
.
Homolytic cleavage
H
C
Cl
H
C+
+
.. _
Cl
H Heterolytic cleavage H
So in the answer explain about which typrs of bonds break homo or heterolyically, the distribution of leectrons
and what is produced (ions or radicals)
b)
C6H14 We must get 6 CO2’s to use up all six carbons in the C6H14. This requires 6 O2’s
We must get 7 H2O’s to use up the fourteen H in the C6H14. This requires 3½ O2’s
Total O2’s used was therefore 9 ½
C6H14 + 9 ½ O2  6CO2 + 7 H2O
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c) In general, H-bond species have the highest boiling points (need more energy put into them to overcome
the strong hydrogen bonds), then molecuels with dipoles which have lower bpt’s followed by (non-polar)
molecules with LDF’s only.
Within the H-bonding species, Carboxylic acids form dimers (as part d) mentions), that is, two sets of Hbonding to one other molecule, so their H-bonds are stronger than the hydrogen bonding in water.
Withing the LDF categories, the species with the more open / straight chain structure and the more electrons
will have the higher LDF’s
We need to be careful. The order required is by INCREASING boiling poitn, so the species with the lowest bpt
should be written first. Thereoire
Propane < (is less than) < propanol < propanoic acid.
ii) All these species have the same number of electrons, so we then look at which has the greatest surface area
over which the LDF’s can act. The more ‘straight’ the chain is the higher the bpt. The more
‘branched/rounded/closed’ the structute is, the less surface area the molecule has hence has lower bpt.
2,2-dimethylpropane < (is less than) 2-methylbutane < n-pentane
d)
Ethanoic acid
Ethanoic acid
O
H3C
H
O
C
C
O
H
CH3
O
Hydrogen bonds
Question 5
a)
Benzene to 4-chlorobenzoic acid. Note: this is a 5 mark question. So there’s quite a bit of work to do!
COOH
Step one is to draw the structures:
This arrow
means
"multiple
steps"
Cl
Break down the problem. We obviously can see that we need to do at least two steps… put a Cl on and put a
COOH on. We should be familiar with Cl2 and AlCl3 in dry ether to clorinate a benzene ring.
Next step: Putting on a COOH geoup. We do not know of any way to do this directly. Thinking deeper, we can
see a C species has been added to the ring. So we should think about Friedel Crafts alkylation or acylation.
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So maybe take chlorobenzene and add methanoyl chloride + AlCl 3 in dry ether. That would give us something
H
O
like…
O
+
AlCl3
H C
Cl
Cl
Cl
Then we know from our
done using K2Cr2O7 + H2SO4 (aq)
H
O
OH
O
+ K2Cr2O7
+ excess
H2SO4(aq)
Cl
diagram to realise we could oxidise the aldehyde to a carboxylic acid. This is
We could also do friedel crafts alkylation. And afterwards use hot acidified
(using H2SO4) KMnO4 solution. Which will cut off all C atoms in the alkyl
chain except the one attached to the ring. And turn the ‘benzylic’ C into a
COOH group straight away.
Cl
So there you have it. Two ways to do the reaction.
When we need to put TWO GROUPS on a benzene ring it is usually VERY IMPORTANT to consider which one
should be done first, as groups on the benzene have the directing effect and an activating/deavtivating effect.
If we put Cl on first, then Cl is o,p directing so position of the carbon based sidegroup at the ‘top’ of the
molecule is OK. But the benzene ring will be deactivated slightly. However, Friedel Crafts (FC) reactions are
usually quite reactive etc, so there is no problem with putting Cl on first.
What if the HCOCl group was put on first? Then we’d have a major problem as it is meta directing so we would
get 3-chloro and not 4-chloro. So we cannot acylate the ring first!
If we put the alkyl group on first (from FC alkylation) then the alkyl chain would be o,p directing and also
activating of the ring. So we can do this and it may give a better result to do so – especially if we use a longish
chain which would help block the 2-position, and favour pushing the Cl into the 4-position.
OH
O
b)
Me
C
Me
Me
C Me
Me
We have a carbonyl AND an alcohol, so our
diagram should ‘jump’ into our mind. We start with a
ketone and end up with a 3o alcohol. Remember, the advanced version of the diagram included Grignard’s
going fromNorthEast to SouthWest, pointing from carbonyls column to the alcohols column.
This, plus the fact we are adding ONE carbon suggests a grignard reaction. With the organomagnesium
compound having just one C.
So CH3Cl + Mg (dry ether)  CH3MgCl + propanone, with addition of aq. HCl at the end gives the alcohol.
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Q6
a)
When you get questions like this, draw a map and fill it in with info as you go along, or you can make a
list of statements.
E reacts with 2,4-DNPH so it must be a carbonyl (an aldehyde or a ketone).
The carbonyl doesn’t undergo oxidation so it must be a ketone.
E can be reduced to F, and we know from our
diagram that the reduction product of ketones are 2 o
o
alcohols. So F is a 2 alcohol. Confirmation of this is is given by its reaction with sodium.
F, the 2o alcohol + propanoyl chloride (an acid chloride) will give an ester (esters are said to have fruitty or
pleasant smells!)
O
OH
O
OH
+
E
F
O
Cl
O
G
The intial ketone in the scheme could be a different one, in the scheme it’s butan-3-one. But you could have
had butan-2-one here.
b)
i. an orange coloured precipitate (called copper “one” oxide, formula = Cu2O – the one refers to the oxidation
state of the Cu and not how many Cu’s there are) is produced. Fehlings reagent is a mildly oxidising reagent.
ii. propanone is a methyl carbonyl so will undergo the iodoform (triiodomethane) test. Note also methyl
alcohold that can be oxided to methyl carbonyls will also give a positive test result for this reaction, because
the reagents used in the triiodomethane test are oxidising, i.e. I 2(aq) + NaOH. They oxidise methyl 1o alcohols
and methyl 2o alcohols to methyl carbonyls, which then undergo the usual bond splitting forming CHI 3(s)
So the observations would be a tellow precipitate forms (some people say an antiseptic smell is produced but
that antiseptic smell is presnt in the beginning. I2 is used to kill germs on the skin before operations.
iii. Just like phenol, aniline (phenylamine) reacts with aq. Br2 to give a white precipitate. 2,4,6tribromophnylaniline forms.
iv. This is the bayer test. The KMnO4 tuens into MnO2(s) So the observation is the KMnO4 purple colour
becomes decolorised and dark brown/black solids form.
.
v. Br2 in UV light gives Br radicals which attack alkanes. So the Br2 which is an orangy/brown colour will be
decolourised (the product bromohexane is colourless like all alkyl halides in this course). You might see some
fumes being evolved which is HBr(g).
Q6
Too easy
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Q7
a) Anhydride plus 20 amine. You can tell from the info given that a O-(C=O)CH3 group has been lost from the
anhydrde and a H has been lost from the amine. So H-O(C=O)CH3 is the product. This is a typical side product
of anhydride reactions.
.
b) Here we have a benzene with an alkyl chain substituent on it. Cl2 + UV means we will have Cl radials.
.
We know Cl radials react with alkyl groups to make alkyl halies. We should also note there are two position on
.
the alkyl chain where Cl radials can attack. The bonds on the benzylic C (the C bonded directly to the ring) is
are more prone to attack because of the ring pulling in their electron density slightly. So that is where the Cl
goes.
Cl
c) An alcohol turns into an alkene. The OH and a H (from the adjacent carbon, top left) are removed. This is a
dehydration (or elimination of water) reaction. Use c.H2SO4 and heat to 170oC.
d) SOCl2, just like PCl5 and PCl3 is a chlorinating agent. It swaps out for –OH groups. So cyclopentanol must
have been the starting compound.
Q8
This question is quite easy. A will be butan-2-ol. C will be 2-phenylbutane. CH3CH(C6H5)CH2CH3
D is K2Cr2O7 and excess dilute H2SO4
B can be PCl5 , SO2Cl2 , or PCl3
Part c)
To differentiate (i.e. tell the difference between) 2 o alcohols from 1o and 3o alcohols, perhaps the most
efficient test is the Lucas test. A solution of ZnCl2 in c.HCl has a small amount (say ½ cm3) of the unknown
alcohol added to it.
If the solution goes ‘cloudy’ (sometimes called ‘turbid’) almiost immediately, then the alcohol was a 3 o alcohol.
If after about 30 seconds it goes cloudy then it was a 2 o alcohol. If it doesn’t go cloudy within a few minutes
then it’s a 1o alcohol.
Note: You could use other tests like Do an oxidation test using Jones’s reagent (K 2Cr2O7 + excess H2SO4(aq))
If the orange colour of the Jones reaent remains, then the alcohol was a 3 o alcohol. Remember the
diagram. If it goes blueish/green then it was a 2o or 1o alcohol. Purify the product by distillation or liwuid-liquid
extraction and then add 2,4-DNP. If it gives an orange/yellow ppte then it was oringally a 2o alcohol, otherwise
it was initially a 3o alcolhol
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