Here is an post-mortem analysis / "how to" for the '99 Final. 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: B
All about the relative reactivity of carboxylic acid derivatives, so
one can think of the LG ability or the effect the subtituent has on the
electrophilicity of the C=O, giving i > iii > ii.
Qu2: C
Substituent effects on benzenes. -Cl groups are weakly deactivating
(induction), -OH strongly electron donating (resonance with the lone
pairs)
and -NO2 is strongly electron withdrawing (resonance onto
the
electronegative O and induction due to the +ve N atom), so ii > i
> iii.
Qu3: E
All H in question are a H, but in iii it
is on O in a carboxylic acid so it is the most acidic (pKa =5).
Aldehydes
(i) are more acidic (pKa =17) than esters (iii) (pKa =25) since in the
ester the resonance donation from the -OR group reduces the stability
of
the enolate. So iii > i > ii.
Qu4: A
Methanal (i) is more reactive than ethanal (ii) towards Nu since it
is less hindered and alkyl groups are weak electron donors which also
decreases
the electrophilicity of the C=O. In an ester, the -OR is strongly
electron
donating, further decreasing the electrophilicity. So i > ii
> iii.
Qu5: A
Nu increases down the periodic table (polarisability) so RS-
is the best Nu. For the two O systems, resonance stabilisation of the
carboxylate
(iii) decreases the availiability of the electrons and therefore the
nucleophilicity,
so i > ii > iii.
Qu6: AB
The reaction conditions promote SN1. In (iii) we have a good LG with
a 2o benzylic carbocation, so it will be very fast. In (ii)
we have a poorer LG and a simple 2o carbocation, and (iii)
is
an unreactive vinylic chloride. So iii > ii > i.
Qu7: AB
Dehydration of an alcohol usually gives the more stable, morely highly
substituted alkene, iii > ii. (i) is not a product of
1,2-elimination from
1-methylcyclohexanol. So iii > ii > i.
Qu8: E
Since alkyl groups are weak electron donors, t-butylbenzene will direct
o,p but the large size will favour p over o. Thus, iii > i > ii.
Qu9: C
Radical bromination depends on radical stability: 3o
benzylic
> 1o > phenyl, so ii > i > iii.
Qu10: D
Diels-Alder reaction requires that the diene be able to adopt an s-cis
conformation. Cyclopentadiene is locked in this conformation and
butadiene
can achieve it by rotation about the C2-C3 bond, but (i) is locked
s-trans
and is therefore unreactive = ii > iii > i.
Qu11: D
Since the uncorrected boiling point is 223oC, the corrected
value should be about 1o higher for every 15o
above
50o, so about 12o higher => 235o.
Qu12: AD.
DNP requires a red precipitate (the reagent itself is red), and the
iodoform requires a yellow precipitate of CHI3 be formed.
Qu13: BC
Qu14: B
1800-1600 cm-1 is the approximate range of the C=O.
Qu15: AB
A quartet integration of 2 suggests a deshielded -CH2- next
to a -CH3.
Qu16: D or E
The aromatic region (7 - 8 ppm) has a doublet of doublets, integrating
for a total of 4H, thus either D or E is possible.
Qu17: AC
Aromatic based on nmr. No -OH in IR so not an acid. Could be an ester,
but not a ketone (DNP test) or an ether.
Qu18: BC
Para-disubstituted aromatic with -OEt group, and a partly deshielded
-Me group, but cannot be a methyl ketone.
Qu19: B
2-butene with Br2 gives 2,3-dibromobutane that can be eliminated with
NaNH2 (strong base) to 2-butyne, which will undergo addition
(hydration)
in the presence of Hg to give the enol that will tautomerise to the
ketone,
2-butanone.
Qu20: E
1-butene will undergo hydroboration / oxidation to 1-butanol. Oxidation
with aq. Cr will give the butanoic acid.
Qu21: D
Propanal will undergo addition with the Grignard to give 2-butanol.
After tosylation, the strong bulky base, tBuO- will cause
elimination
to give the less substituted alkene, 1-butene.
Qu22: C
1-bromopropane reacts with Mg to give the Grignard with will react
with CO2, then work-up to give the carboxylic acid. Reaction
of the acid with an alcohol gives the methyl ester.
Qu23: C
Diels-Alder reaction, to the substituted cyclohexene. Reduction of
the C=C then the Wittig reaction to convert the aldehyde group to C=CH2.
Qu24: B
Cyclohexanol will be oxidised to cyclohexanone. LDA will form the
enolate
which can be methylated.
Qu25: B
Alkenes plus peracid to epoxide, which will open SN2 fashion at the
less hindered end with the strong nucleophile like a Grignard to give
the
alcohol.
Qu26: D
Diazonium chemistry.... the diazo compound being prepared from the
amine.
Qu27: D
Benzyl alcohol was formed by reduction of a C=O system, which was
obtained
by the oxidation of an alkyl aromatic substituent with a benzylic H.
Qu28: D
A long one to plug back through.... the maide was formed by the
reaction
of the amine with the acid chloride prepared from the acid. The acid
was
obtained by the reaction of CO2 with the Grignard of the
bromide.
So the methyl group was already in place.
Qu29: C
Cyclic acetal formed by the reaction of the ketone with the 1,2-diol.
The 1,2-diol must have been prepared by addition to the alkene.
Qu30: E
Overall this seems to be an enolate condensation type reaction with
decarboxylation (it can't de dehydration since there is no C=C), so an
ester system must be involved. The ozonolysis with the oxidative
work-up
followed by MeOH also suggests this.
If we have 5 C in the product and we lost 1 in CO2, then
we needed 6 in the alkene. Only E can give a C6 diester....
Qu31: A
Another condensation reaction, this time an Aldol, since the C=C has
been removed by the reduction. System has 7 C atoms, so needs to be a
di-methyl
ketone.... this is "seen" by breaking open the product at the C2-C3
bond
where the C=C would be located.
Qu32: C
Williamson ether synthesis, preceded via an anti-Markovnikiv addtion
of HBr to a C5 alkene with a 2o alcohol.
Qu33: BE
Acid to ester = esterification, usually alcohol, H+ and
heat.
Qu34: CD or CE
Ester to keto-ester = Claisen condensation of ester enolate, so need
the alkoxide base or the stronger non-nucleophilic base like LDA.
Qu35: CD or CE
Alkylation of the enolate... need the base again
Qu36: AD
Oxidation of alcohol to ketone, PCC.
Qu37: E
a-mono-chlorination of ketone uses HCl.
Basic conditions will cause poly chlorination.
Qu38: BC
Hydrolysis and decarboxylation, aqueous acid or base and heat. Here
base would cause problems will Aldol reactions.
Qu39: AC
Reduction of the ketones to alcohols, use hydride reagents.
Qu40: AB
Alcohol to chloride, PCl3 would avoid acidic conditions and potential
rearrangements.
Qu41: ABC
Need to add the two ester groups and form the ring via the enolate
of diethyl malonate.
Qu42: AB
Diels-Alder reaction, syn addition, stereospecific, cis dienophile
gives cis products, A and B are enantiomers.
Qu43: C
E2 type elimination, needs to be anti-periplanar H and Cl. -OH
converted
to -Cl via SN2 reaction with inversion, so here it has to eliminate
away
from the methyl group.
Qu44: D
Epoxide opens under acidic conditions to put the -Cl on the same C
as the methyl group (more stable carbocation), -OH and -Cl will be
trans.
Qu45: E
Syn reduction with poisoned catalyst to the cis-alkene, then
anti-addition
to the 2,3-dibromide. Need the RR and SS stereoisomers.
Qu46: CE
Anti-Markovnikov hydration via hydroboration, -OH and -H added syn.
Qu47: D
In amides the strong electron donating effect of the N system inhibits
the resonance stabilistion of the enolate.
Qu48: D
Acid catalysed hydration of alkenes via carbocation, so rearrangement
needs to be considered.
Qu49: B
It is the substitutent already on the benzene that dictates the
location
of the incoming substituent.
Qu50: C
Grignards are strong bases, so protonation will occur destroying the
carbanion.