353 MT Winter 2006
Here is an post-mortem
analysis / "how to" for the MT. The questions are split by the sections.
At the start of each section are a few suggestions of what to look for or
how to tackle the question type.
RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may
appear. Look for two pairs of similar systems to compare that have minimal
differences in structure. If a compound is named, draw it out. If a reaction is
involved, identify the type of reaction and then what the controlling factors
are.
Qu1: D
Vinylic hydrogens are found
on the sp2 C within alkene C=C units. i has 1, ii has 3
and iii has 2 hence ii > iii > i.
Qu2: B
Potassium t-butoxide is a strong base but a poor nucleophile (due to it’s
size). Alkyl halides react with strong bases to undergo elimination reactions
via an E2 pathway. In cyclohexyl systems, the E2 can
only occur when the leaving group is axial since the H that needs to be removed
needs to be at 180 degrees. In I the larger t-butyl group prefers to be equatorial and this
means that the Br would be axial and hence E2 elimination would be fast. In ii, with the t-butyl group equatorial, the
Br is also equatorial, therefore a ring flip to put
the Br axial would be required before the E2 can occur. However, the t-butyl
group is big enough, that it locks the ring a restricts
the ring flip…. Therefore ii is quite unreactive
under these conditions. Structure iii would have the Br equatorial
preferentially, but it can still ring flip to the reactive conformation.
Therefore i > iii > ii
Qu3: B
The 3 bonds are all
Qu4: AB
This involves knowing that conjugated dienes are more
stable than isolated dienes and that benzene is more
stable than a hexadiene – this means the
substituted benzene i has the least exothermic heat of
hydrogenation and the isolated diene iii has the most exothermic, so we get iii
> ii > i.
Qu5:
Reaction is the Diels-Alder reaction and we are looking at the dienophiles reacting with 1,3-cyclohexadiene. Since the dienophile
is the electrophile, electron withdrawing substituents increase the reaction rate,
therefore the carbonyl group in iii
makes it the most reactive.
Qu6: E
Check the pKas, 25 vs 16 vs 35 of the conjugate acids or think about the bases that
are used to prepare an acetylide ion
! In terms of
base strength iii > i > ii.
Qu7: C
i has two, ii has four and iii has
just one, so ii > i
> iii
Qu8: B
Maybe best to think of the electron availability in the conjugate bases. N is
more basic than O since N is less electronegative. The R group in iii makes the ether more basic that the alcohol in i. Therefore in terms of basic structures the
amine ii is more basic than the
ether iii then the alcohol i, so in terms
of the acidity of the structures in the question we get the reverse, i.e. i > iii > ii
LABORATORY:
Based on the general principles cover in the laboratory so far. Need to know
the principles and details of the steps in the experiments.
Qu9: A
Sucrose is the disaccharide of glucose and fructose, linked by a glycosidic bond.
Qu10: B
Not all molecules that contain chirality centers are optically active, e.g. meso compounds.
Qu11: B
Glucose contains a hemi-acetal
Qu12: A
Qu13: B
Alcohols
undergo dehydration, an elimination with strong acid via the E1 pathway
via C+ intermediates.
Qu14: B
The positive test is the red-brown colour of the bromine solution going colourless.
Qu15: A
Draw out the structure… this
secondary alcohol will be oxidised to a ketone.
Qu16: A
The Lucas test in based on the SN1 reaction that converts alcohols to alkyl
chlorides using HCl.
Qu17: B
PVC is made from vinyl chloride and there is only one Cl
per monomer unit.
Qu18: A
It comes from a tree that originated in
Qu19: A
You did this experiment, it's a radical polymerisation.
Qu20: B
Where would the sodium come from ? The product
would be HCN.
Qu21: A
Qu22: A
Qu23: A
Benzoic acid is not polar enough to dissolve in
water. Under acidic conditions, the benzoic acid exists as PhCO2H. In contrast
the benzoate ion, PhCO2- would exist in a basic aqueous layer.
Qu24: B
It is used to remove the solvent to create a
saturated solution.
STARTING MATERIALS AND
PRODUCTS OF REACTIONS:
If you are trying to find the product, then you should probably just work
forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is
probably the best way to go....
Basically depends on the need to know and identify the reactions,
this is often triggered by looking at the functional groups in the molecules.
Qu25: C
Working backwards, ozonolysis with a oxidative work up... up giving a
single diketone – this would need to originate
from a cyclic diene which seems to have been formed
by a double dehydration
of an alcohol i.e. a diol.
Qu26: E
Working forwards....the reaction is the addition
of hydrogen bromide to a conjugated diene. When
heated, this will be under thermodynamic control and so the product will be the
conjugate or 1,4-addition product and will have the Br
added. A has added the HBr 1,2- in a Markovnikov
fashion, B and C
correspond to added the HBr 1,2- in an anti-Markovnikov fashion (to different C=C) and D is an
addition of Br2 rather than HBr as the question asks.
Qu27: A
Working backwards....the reaction sequence involves the hydration
of an alkyne to give ketone,
cyclooctanone – and we are going to have to
make this ring. In order to do this, we are going to need a diterminal
alkyne, i.e. ethyne and
therefore a 6C chain with a halogen at each end. B and E would not give a cyclic alkyne. C would give a dimethylcyclohexane
system, D would give a methylcycloheptane.
Qu28: D
Working forwards.... alkene to halohydrin
to epoxide
followed by acid
catalysed ring opening to give a trans,1-2 diol. A is a cis-diol, B , C and E are not diols,
C is the product from step 1.
Qu29: B
Working forwards.... The transformation is alkene to
ether via an oxymercuration-demercuration with an
alcohol instead of water as the nucleophile. It
follows Markovnikov’s rule but since there is
no C+, there is no rearrangement.
Qu30: A
Working backwards.... ozonolysis with a reductive
work-up - materials suggest a Diels-Alder reaction, so we can look for the
cyclohexene. Try numbering the main chain in the product and reconnecting
the aldehydes and the ketones to give the original alkene....
might need some trail and error here
to get ring systems that aren’t too strained this gives a cyclohexadiene system from a Diels-Alder reaction that had an alkyne
dienophile...then you can
push the curly arrows to "reverse" the Diels-Alder
and reveal the diene and the dienophile.
Qu31: D
See Qu 30.
REGIO- and
STEREOCHEMISTRY:
How well do you know your reagents ? Look at what has
actually happened in terms of the reaction functional group transformation and
then first look for any regiochemical issues then
finally the stereochemistry last (it's the hardest to sort out). In cases
where more than one product is formed in equal amounts (e.g. the enantiomers), then both must be selected for full marks,
part marks are given when only one of the pair is selected.
Advice : in each case draw the starting
material in the conformation in which is reacts or the product in the
conformation in which it is initially formed using wedge-hash diagrams. It is a
good idea to draw the materials in such a way that the new bonds are in the
plane of the page. Once you have drawn the materials It
may also be good for you to use model kits for these questions too. Once
you have drawn the materials in this way, you may need to consider rotations
around sigma bonds to make your answer match the options. An alternative
approach could be to assign configurations to your drawn answer to compare them
with the options - this can be slow and prone to error.
Qu32: AD
The reaction is the hydroboration-oxidation of an alkene
. The reaction gives the anti-Markovnikov alcohol via
a syn addition due to the concerted addition of the B
and H across the C=C. Note that the -OH is formed with retention of
stereochemistry when the B atom is replaced.
Qu33: A
Hydrogenation to give the cis-alkene followed by epoxide
formation (which has cis Ph and isopropyl groups)
followed by attack of a weak Nu (EtOH)
under acidic conditions in an SN1 like process : this
means the OEt is attached at the benzylic
end of the epoxide.
Qu34: C
This is a Diels-Alder reaction.... hints
are the dienophile, and the bridged structure of the
product. The challenge here is to realise
that the C=C has been removed by the dihydroxylation : osmium
tetroxide reacts with alkenes to give 1,2-diols….
and notice that the rings in the products don’t have methyl substituents.
Qu35: B
The KMnO4
reaction forms a diol via a syn addition to an alkene. The Na/NH3
will form a trans-alkene from an alkyne. The product is drawn as a Newman
projection. The key issues are that the -OH groups define the C=C and hence
the alkyne. So you need an internal alkyne with a methyl substituent and an
isopropyl group... it has to be B.
Qu36: A
Qu37: A
Alkynes
react with excess HBr to give 1,1-dibromides. In this case the initial
carbocation will be the benzyic cation hence leading to A as the major
product.
AROMATICITY and
RESONANCE:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria (cyclic, planar, conjugated pi system with an odd number of pairs (i.e.
4n+2) pi electrons)
Qu38: CE
Qu39: CD
Qu40: A
The hydrocarbons are A, C, D, AC, AE, BC
and CE.
Qu41: E
The type of tautomer we have met involves the tautomerisation of a carbonyl
and an enol. O=C-C-H <=> H-O-C=C. Similar situations are possible with
C=N and C=S types of systems....see E. It's non-aromatic as drawn due
to the sp3 C alpha to the thiocarbonyl group.
Qu42: B or AD
Heterocycles have none C atoms in the ring i.e. B, AB, AD
and CD. B is aromatic (pyridine), so are AB (furan) and
AD (all 6 pi electron systems). Generally N is more basic than O since
N is less electronegative.
Qu43: BC
Only C, D, BC and CE are ionic. C, D
and CE are fully conjugated.
Qu44: AE or AD
Qu45: AC
AC isn't aromatic because it's not planar.