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.
Qu1: A
Carbocation stability is dictated by the substituents at the C+ center.
It may help here to note that the carbocations are related to those
encountered
during the reaction of an alkyne with HBr. i is a secondary
system,
with a Br atom which through resonance of a lone pair can further
stabilise
the +ve charge, ii is a simple secondary system and iii is a secondary
vinylic cation (which is like a primary) due to the hybridisation at
the
C (note no resonance as the pi system is perpendicular = no overlap).
Therefore
we get i > ii > iii.
Qu2: D
Radical bromination of an alkane. Draw the alkane first, then the
products.
Bromination is quite selective for the more highly substituted
positions
due to the stability of the radicals. i is primary, ii is tertiary and
iii is secondary, so ii > iii > i.
Qu3: B
i is an alkene, so it can protonate to give in this case a tertiary
carbocation. ii is an alkane which does is not a good base since all
the
electrons are in strong single bonds. iii is an alkyne that
protonates
to give an unfavourable vinyl cation. So i > iii > ii.
Qu4: C
Alkenes react via electrophilic addition where the rate determining
step in the addition of the electrophile. Here the electrophile will be
H+ and so it depends on the acidity of the reagent. Acidity
increases
down a group due to bond strength and the ability of the conjugate base
to accept charge. So HBr > HCl, we should know water is not a strong
acid,
or can think about willingness of O to accept charge. Overall then ii
>
i > iii.
Qu5: C
Need to work out which of the 3 stereoisomers is (S,S). i is
meso (note the vertical mirror plane that reflects left to right) so it
is (R,S) and hence is an "internal racemate" and is optically inactive
= 0o rotation. Group priorities are CO2H > CHCO2H >
CH2 > H.
If we look at ii then the right hand carbon has the R configuration,
and
so ii must be (R,R) and will have a positive rotation. Overall then we
have ii > i > iii.
Qu6: C
Catalytic hydrogenation depends on the strength of the p
bond. The second p bond in an
alkyne
is weaker than that in an alkene, so reactivity of ii > i.
Carbonyl
groups have strong p bonds and are not
easily
reduced under these conditions. So overall we have ii > i > iii.
Qu7: A
Reaction is the hydroboration oxidation of an alkene. This gives the
anti-Markovnikov alcohol via a syn addition due to the concerted
addition
of the B and H across the C=C. Since no C+ forms, there is no
carbocation
rearrangement. i is the major product with the -OH at the less
substituted
end and ii is a minor product (the Markovnikov product). iii is only
obtained
is there has been a carbocation rearrangement. Overall then, i > ii
> iii.
Qu8: AB
A Diels-Alder reaction and we are looking at the diene component, which
is the Nu in a normal Diels-Alder reaction. This means electron
donating
groups on the diene make it more electron rich = stronger, more
reactive
nucleophile. Plus the diene needs to be s-cis (or cisoid) for the
reaction
to occur. In i the diene cannot be s-cis and will not react. ii
can
rotate to be s-cis and iii is already locked s-cis and also has an
alkyl
substituent (weakly electron donating). So iii > ii > i
Qu9: ACE
In B the equation is wrong, it should be [a]
= a / c l . In D, the pKa of HCl =
-7.
Qu10: B
Milk sugar is lactose. Benedicts solution gives a red precipitate with
a reducing sugar. The precipitate is mainly casein. Proline gives a
yellow
spot with ninhydrin because it's a secondary amine.
Qu11: BC
A steam distillation is based on immiscible liquids while Raoults law
describes the vapour pressure for miscible liquids. For D, the organic
layer can either be top or bottom depending on the densities of the
layers.
For E, IR gives the functional groups, nmr spectroscopy gives the
hydrocarbon
skeleton.
Qu12: ABCD
E Sodium sulphate is soluble in water (as are most Na+ salts), the
precipitate was the carboxylic acid..
Qu13: A
B reflux apparatus is for heating liquids at their boiling points.
C charcoal is for removing coloured impurities. D waste needs to go in
the appropriate drum. E what No safety glasses ???
Qu14: D
Reaction is the oxymercuration-demercuration of an addition reaction
of an alkene. This gives the Markovnikov alcohol due to the addition of
the Hg species across the C=C giving a cyclic mercurinium ion. This is
then opened by water to give the alcohol at the more highly substituted
position (SN1 like) with the Hg at the less hindered end. The NaBH4
removes
the Hg. Since no C+ forms, there is no carbocation rearrangement.
A is not an alcohol. B is an elimination product. C is only obtained is
there has been a carbocation rearrangement and E is the
anti-Markovnikov
alcohol.
Qu15: C or D
H2SO4 causes an elimination of an alcohol to give an alkene
that is then subjected to ozonolysis with a reductive work-up. Looking
at the dicarbonyl product, we can work out the required alkene by
"snapping"
together the two C=O units to give the C=C. This gives us
1-methylcyclopentene
which is C (draw carefully, count carbon atoms). The fact that
one
end of the dicarbonyl product is an aldehyde and the other is a ketone
tells us that the two ends of the alkenes are different. Since C
would not react with the acid to do anything, it is an acceptable
answer.
A would give the wrong symmetrical alkene, which would have
given
1,6-hexanedial on ozonolysis. B and E would not eliminate under these
conditions,
alkyl halides need a base to undergo elimination reactions. D
eliminates
to give the desired alkene.
Qu16: C
Addition of a hypohalous acid to a C=C puts the -OH at the more highly
substituted position and the -Cl trans to it... but treating this with
base allows the formation of the epoxide. So C is the correct answer.
Qu17: C
Terminal alkynes are deprotonated with sodium amide, a strong base
to give the acetyilide ion, a good Nu. This reacts SN2 fashion with the
primary alkyl bromide to give Ph-CH2-C#C-CH2-Ph.
Alkynes undergo ozonolysis without the need for a special work up to
give
carboxylic acids. Counting carbon atoms gives C as the correct
product.
Qu18: D
Radical bromination at the more highly substituted position followed
by E2 elimination to give the alkene, probably the anti-Zatseff, less
highly
substituted alkene due to the bulk of the base and then form the
cyclopropane
ring via the carbenoid reaction (with :CH2 as the reactive species).
The
position of the cyclopropane ring confirms the location of the C=C as
being
exocyclic. Only C and D have the right number of carbon atoms to be
possible
solutions. The alkene in C would probably give a diene after
elimination....
Qu19: B
The catalytic hydrogenation has removed the C=C. But you should be
able to recognise the two esters as being from the dienophile and
therefore
that the C=C should be on the opposite side of the product ring. So we
are looking for the diene of the Diels-Alder reaction. If you draw the
curly arrows for the retro-Diels-Alder reaction, (start at a C=C and
push
around the 6-membered ring) then the diene is a methylated
1,3-butadiene,
using the curly arrows allows you to see where that methyl group needs
to be on C2 of the diene.
Qu20: D
Now looking for the dienophile of the Diels-Alder reaction. If you
draw the curly arrows for the retro-Diels-Alder reaction, (start at a
C=C
and push around the 6-membered ring), this allows you to see where that
the dienophile has two ester groups. The only way they could be trans
in
the product is if the trans alkene system is used. Note that it can't
be
the alkyne because the esters would be cis in the product after the
hydrogenation.
Qu21: A
Reaction of an epoxide under basic conditions, so the nucleophile is
the alkoxide ion = a strong Nu. Hence the reaction is SN2 in
character and the Nu adds to the less highly substituted carbon of the
epoxide
(less hindered = SN2 like), but still from the opposite side to the O
atom
of the epoxide. So the OCH3 group needs to be on the opposite carbon of
the epoxide to the CH3 group.
B and C are different conformations of the same molecule with the wrong
stereochemistry at the -OH center (that stereochemistry is defined by
the epoxide). These compounds would be formed if the enantiomer of the
original epoxide were used. D and
E have the wrong regiochemistry.
Qu22: D
Br2 adds across C=C or C#C to add one Br atom at each end
of the p bond. A would give a tetra-bromide
under these conditions not a dibromide (where do the H come from
?).
B would put the two Br atoms on C1 and C2 not C2 and C3 (i.e.
regiochemistry).
E would not react under these conditions. So is it C or D. Need to pay
attention to the stereochemistry. The required product is a meso system
as shown by the Fischer diagram. Since the addition of Br is an
anti
addition process, this requires that the original alkene is trans, i.e.
D. The cis alkene C would give the pair of enantiomers with the
(R,R)
and (S,S) configurations.
Qu23: D
Catalytic hydrogenation of an alkyne with Lindlars catalyst gives the
cis alkene. Generation of a carbene (here loss of HCl gives :CHCl)
which
then adds to the C=C to give a cyclopropane in a syn fashion so the
stereochemistry
of the C=C is preserved. A has the wrong carbene, B has the wrong
original alkene stereochemistry, C has wrong carbene and wrong alkene
stereochemistry.
D and E have right carbene and right original alkene
stereochemistry.....
the reaction will occur preferentially to put the larger Cl atom away
from
the other two groups (sterics), so D is the right answer.
Qu24: C
Dissolving metal reduction of the alkyne gives the trans alkene then
the peracid gives the epoxide with the same stereochemistry.
Treatment
of the epoxide with aq. NaOH opens the epoxide to give the 1,2-diol
HO-C-C-OH
in an SN2 like fashion so the 2 -OH are trans, so in effect completing
an anti addition of the diol system to the C=C in the two step process.
Overall anti addition to a trans alkene means a meso product. A,B,D and
E are not meso !
Qu25: B
Permanganate oxidation of an alkene gives the syn 1,2-diol. A is not
a diol. C and D have the wrong stereochemistry. E has the wrong
regiochemistry,
the -OH groups need to be where the C=C was !
Qu26: A
Probably the toughest of the bunch. Reaction is the
hydroboration-oxidation
of an alkene. This 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. Only A and B have the C=C in the right location, so it's down
to the stereochemistry in these two. Careful drawing of the syn
addition
process confirms that A is the correct stereoisomer.
Qu27: AD
Reaction is the hydroboration-oxidation of an alkene. This gives the
anti-Markovnikov alcohol so we are looking for Ph-CH2CH2OH.
The two connected CH2 units will give two triplets as in AD.
Qu28: AB
Reaction is ozonolysis with an oxidative work up. This will give the
carboxylic acid and CO2 in this case. The acid is
characterised
by a singlet at high chemical shift around 12 ppm as in AB. Note that
10
ppm as in AC is a little low for an acid and is more likely an
aldehyde.
Qu29: A
Reaction is the catalytic hydrogenation of the C=C to the alkyl group,
so we are looking for Ph-CH2CH3. The ethyl
group will have a characteristic quartet and triplet with integrals in
the 2:3 ratio as in A. Note that the chemical shift of the
quartet
in B is too high for it just to be attached to the aromatic ring (more
like it is attached to an O atom).
Qu30: D
Reaction is hydration of the C=C to give the Markovnikov alcohol, so
we are looking for Ph-CH(OH)CH3. This will be
characterised
by the 1H quartet and a 3H doublet plus OH singlet.
Qu31: C
Reaction is the catalytic hydrogenation of the C#C to the C=C, so we
are looking for Ph-CH=CH2. The vinyl group will have a
characteristic complex coupling patterns in the region of 5-7 ppm as in
C.
Qu32: E
Reaction is the mercury assisted hydration of a terminal alkyne to
give a methyl ketone i.e. Ph-C(=O)CH3. This will be
characterised
by a 3H singlet at about 2.2 ppm.