Here is an post-mortem analysis / "how to" for the '98 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: E
Acidity of carbonyl systems is governed by the nature of the electron withdrawing groups near them. The better the group can stabilise the -ve charge in the conjugate base, the enolate, the more acidic the original system. So -CHO (pKa =17) is better than -CO2R (pKa =25) since the -OR portion is also donating towards the C=O), but between two esters is best of all (pKa = 11). This means  iii > i > ii.

Qu2: B
Substituent effects on benzenes. -OMe groups are strongly activating (resonance of the O lone pairs), -Me groups are weakly electron donating (hyperconjugation) and -CN is strongly electron withdrawing (resonance onto the electronegative N), so i > iii > ii.

Qu3: E
All about the relative reactivity of carboxylic acid derivatives, so one can think of the LG ability or the effect the Substituent has
on the electrophilicity of the C=O. LG ability : -Cl > -OMe > -NMe2. Electron withdrawing ability : -Cl > -OMe > -NMe2. So iii > i > ii.

Qu4: A
Aldehyde (i) is more reactive than ketone (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 the ester, (iii) the -OR is strongly electron donating, further decreasing the electrophilicity.  So i > ii > iii.

Qu5: A
At low temperature, HBr addition occurs under kinetic control  giving 1,2- rather than 1,4- addition, i.e.  (i) > (ii). Product (iii) is not possible from this reaction. So i > ii > iii.

Qu6: AB
Less reactive (softer) systems give more conjugate addition at C than direct addition at O (hard). Enolates (iii) are more less reactive than organometallics and tend to give more conjugate addition. Grignard reagents are less reactive than organolithiums So iii > ii > i.

Qu7: AB
In addition to C=C, the alkene is the Nu, so if the substituents on the double bond donate electrons, it will be more reactive. Here the substituents electron donating ability is -OMe > -Me > -CO2Me. So iii > ii > i.

Qu8: E
Since alkyl groups are weak electron donors, t-butyl- will direct o,p but the large size will favour p over o. Methyl is also o,p directing and due its smaller size, more o will be produced. -NO2 is a m director.  Thus, iii > i > ii.

Qu9: E
Simply count the H a to the C=O. (i) has 3, (ii) has 2 and (i) has 4 ,  so iii > i > ii.

Qu10: AB
In the Diels-Alder reaction the dienophile is usually the electrophile, so electron withdrawing groups enhance the reactivity giving iii > ii > i.

Qu11: AB.
Two exchangeable H suggest the -NH2 group.

Qu12: A.
IR gives a C=O but no -OH or C-O. H-nmr more like ketone than ester.

Qu13: D
IR gives C=O and -OH. H-nmr very acidic H at 12.8 ppm.

Qu14: E
IR gives C=O more like an ester. H-nmr has deshielded peaks consistent with those in -O-CH systems.

Qu15: C
H-nmr with ArH (near 7ppm) and an exchangeable H suggest the phenol.

Qu16: C
Has ArH (7 - 8 ppm) and aliphatic H (eg. 1 ppm).

Qu17: NONE !
Esters are not very polar, nor very acidic. Only aldehydes and ketones react with 2,4-DNP. The FeCl3 test is for phenols and iodoform for methyl ketones.

Qu18: AB
Monosubstituted, aromatic ester with -OEt group.

Qu19: AC
A Gabriel synthesis of a primary amine. First radical bromination of the benzylic position to get an alkyl bromide. Substitution by the N nucleophile to alkylate on the N. Finally the hydrazine releases the amine AC.

Qu20: A
Bromobenzene will give mainly p-dibromide. Convert to the Grignard and react with CO2, acid work-up to get the p-dicarboxylic acid, A.

Qu21: C
Me group directs nitration o,p then oxidise the benzylic position to the carboxylic acid, so C.

Qu22: E
Nitration, reduction to the amine, diazotise, introduce -CN (Sandmeyer type reaction) and hydrolyse -CN to -CO2H.

Qu23: CE.
Robinson annulation sequence: enolate, conjugate addition, aldol reaction, dehydration giving CE.

Qu24: AE
Cyclohexanol will be oxidised to cyclohexanone. Wittig reaction creates the exocyclic C=C which is converted to the cyclopropane via the Simmonds-Smith reaction.

Qu25: CD
Alkene plus bromine to 1,2-dibromide, which can be eliminated to the conjugated diene, 1,3-cyclohexadiene. This undergoes a Diels-Alder reaction with ethene and is then hydrogenated to give C8 system CD.

Qu26: B
Not easy ! If you can identify the location of the C4 from the dibromide, then one should be able to see an enolate reaction of diethyl malonate followed hydrolysis and decarboxylation.

Qu27: D or BD
Hoffman elimination of an amine produced by the reaction of ammonia with an alkyl halide. HBr could have reacted with C=C or -OH, but need the methylcyclopentane framework. Both D and BD would react with HBr to give 1-bromo-1-methylcyclopentane.

Qu28: AD
Hydroxylamine reacts with C=O to give the oxime and the first step is an oxidation, so it must be AD.

Qu29: E
Trans 1,2-diols are formed by the reaction of epoxides with water, MCPBA is a peracid which reacts with a C=C to generate the epoxide. So we would have needed the C=C to be produced by elimination of an alkyl halide since we have basic E2 reaction conditions. E2 requires the antiperiplanar arrangement of -H and -LG. AB would eliminate to give the C=C in the wrong location.

Qu30: AE
CO2H produced via the hydrolysis of the -CN, which would have added to cyclopentanone to give the cyanohydrin.

Qu31: BD
Final 2 steps are the Wittig reaction to create the C=C. Steps 1-3 create the anti-Markovnikov alcohol and convert it to bromide which undergoes the substitution by the PPh3.

Qu32: D
Enone product from the basic reaction conditions should suggest the intramolecular Aldol reaction so we needed a dicarbonyl (here the aldehyde - methylketone is needed) from the ozonolysis of the appropriate cycloalkene.

Qu33: AD
Oxidation of 1^o -OH to an aldehyde, so need PDC or PCC.

Qu34:  C
C=O to C=C suggests a Wittig reaction.

Qu35: AB
Allylic bromination requires radical bromination.

Qu36: AE
Conversion of -Br to -CO₂H .... use Grignard then CO₂ with acidic work-up.

Qu37: B
Reduce  -CO₂H to the -CH₂OH unit using hydride reagent.

Qu38: BE
Make the tosylate from the alcohol.

Qu39: CE
Preparation of enolate using a strong base.

Qu40: E
C=C to epoxide, use peracid.

Qu41:  BC
Epoxide to diol requires water to act as the Nu.

Qu42: BC
Diels-Alder reaction, syn addition, stereospecific, cis dienophile gives cis products.

Qu43: AE
Tosylate formed with retention at the -OH, but then cyanide displaces the -OTs in SN2 fashion with inversion so the -CH₃ and -CN need to be 1,2- and cis.

Qu44: AB
Permanganate reacts with C=C to give cis-1,2-diols.

Qu45: BC
Trans reduction using the dissolving metalt to the trans-alkene, then syn addition to the epoxide.

Qu46: DE
Reacts as HO^- and Br⁺. Br⁺ adds first to get the bromonium ion then the HO^-  adds at the more stable cationic end with inversion.

Qu47: B
Friedel-Crafts acylation gives a para-disubstituted benzene with a methyl singlet at about 2.2 ppm for the methyl ketone.

Qu48: E
Reduce the C=O to an alcohol, look for the O-CH-CH₃  peak (quartet at about 4 ppm) and the methyl group becomes a doublet. -OH singlet.

Qu49: AB
-OH to -Br, so similar to 48, but without the singlet for the -OH.

Qu50: D
-Br to -CN, very similar to 49 but the CH-CN is more deshielded.

Qu51: A
Hydrolysis of -CN to -CO₂H, look for acid -OH at 12 ppm.