Here is an post-mortem analysis / "how to" for the 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
The reaction is an acyl substitution of  carboxylic acid derivatives... this involves the loss of a leaving group.  Either use leaving group ability (CH₃CO₂^- > CH₃O^- > NH₂^- ) which is related to the stability of the conjugate bases, or the electronic effect the group has on the electrophilicity of the C in the carbonyl group....so we get iii > i > ii.

Qu2: A
The reaction is electrophilic aromatic substitution, and we need to look at the substituent effects on the aromatic ring.  -OMe is an activating group due to electron donation by resonance, -Cl is slightly deactivating due to electronegativity and the nitrile (CºN) is like a carbonyl (C=O) and so is deactivating due to electron withdrawial by resonance. So i > ii > iii.

Qu3: B
Look at where the H is attached and think of the factors that stabilise charge such as electronegativity, resonance, inductive effects etc. i gives an O -ve that is further resonance stabilised to give a second O -ve (a carboxylic acid), both ii and iii give  C -ve that are resonance stabilised to give a O -ve. ii is an ester and iii is a ketone. In the ester ii the resonance stabilisation of the enolate has to compete with the resonance interaction with the other O atom and so is less effective.  So i > iii > ii.

Qu4: D
The reaction is the hydride reduction of carbonyls, so we need to pay attention to the electrophilicity of the C in the carbonyl. Look at the attached groups. i is an ester (alkoxy, aryl), ii is an aldehyde (H, aryl) and iii a ketone (alkyl, aryl).  Electron donating groups will reduce the electrophilicity of the carbonyl. -OR is a electron donating group by resonance, -R is weakly electron donating due to hyperconjugation and H is the reference (i.e. no effect).
So ii > iii > i.

Qu5: D
This is about the Diels-Alder reaction. Methyl propenoate is a dienophile so we are looking at the dienes, so i as only an alkene is not going to react. ii is locekd in the favoured s-cis conformation whereas iii is not, so ii > iii. Therefore, overall we have  ii > iii > i.

Qu6: A
Draw the enols and look for stabilising factors. i gives phenol, a highly favoured aromatic enol. ii gives an enol that can be stablised by a favourable intramolecular, H-bond as part of a favourable six membered ring and iii gives a simple enol.  So we have  i > ii > iii.

Qu7: C
The reaction is electrophilic addition to an alkene with an acid. The controlling factor (rate determining step) will be the stablility of the carbocation that is produced. i gives a secondary carbocation, ii gives cation that can be resonance stabilised by the O lone pairs and iii gives a comparatively unstable vinyl cation. So overall rates ii > i > iii.

Qu8: A
Another electrophilic addition to an alkene but now looking at different acids on the same alkene. The controlling factor (rate determining step) will be the strength of the acid used. First the two strong acids, but HI > HBr (ability of halide to accept charge due to size, strength of H-X bond), and water is only a weak acid.  So overall rates i > ii > iii.

Qu9: A
The reaction is electrophilic aromatic substitution, and we need to look at the directing effect of the groups. i and ii both have alkyl groups => a electron donating group => ortho/para director whereas iii is a CO₂H => a electron withdrawing group => meta director  Steric effects will block the ortho site more with the larger R group in i that ii , so yield of para i > ii > ii.

Qu10: AB
Look at how the stability of the deprotonated form changes (i.e. the conjugate base) so look at the stabilisation of the charge and the way this is influenced by the aromatic substituents. Remember simple electrostatics (like charges repel = destabilise). -OCH₃  is an electron donating group by resonance = decreased stability, -CH₃ is slightly electron donating due to hyperconjugation = slightly destabilised and H is electronically neutral (i.e. no effect). Overall then we have  iii > ii > i.

Qu11: D
Due to changes in atmospheric pressure, boiling points change at different elevations. At sea-level, the boiling point is about 1^o higher for evry 15^o above 50^o, so our boiling point will be about (193-50)/15 degrees higher = 10^o higher = 203^o C.

Qu12: E
The water solubility and the positive Lucas test indicate an alcohol. The negative ferric chloride test means it is not a phenol, and the negative DNP test means it is not an aldehyde. An amine would probably be soluble in HCl.

Qu13: BC
The DNP test on STU-2 shows the presence of a carbonyl from an aldehyde or ketone. Carboxylic acid derivatives don't react with 2,4-DNP.

Qu14: BD
The yellow precipitate due to the formation of iodoform, CHI₃, in the iodoform test indicate the presence of a methyl ketone.

Qu15: C
The Lucas test is the reaction of an alcohol ROH with HCl and ZnCl₂ to give the alkyl chloride, RCl via a substitution reaction of SN1 (via carbocation) character.

Qu16: BDE
In the dichromate oxidation test, the orange Cr VI of Cr₂O₇^2- is reduced to green Cr III. Primary and secondary alcohols are oxidised to carbonyls but tertiary alcohols don't oxidise as they lack the critical H at the carbinol C for the elimination step.

Qu17: C
Melting points are not sensitive to atmospheric pressure changes.

Qu18: E
The IR shows an -OH at about 3350 cm^-1, the H NMR shows a mono-substituted aromatic (7.2 ppm, 5H), and a deshielded CH (4.8ppm, 1H) quartet so coupled to 3H adjacent, and a methyl group (1.4ppm, 3H) coupled to a single H.

Qu19: AD
Oxidation of #0001 gives a methyl ketone, still need the aromatic unit, so has to be AD.

Qu20: D
Peracid reacts with the alkene to give the epoxide which then reacts with the nucleophilic Grignard reagent undergoing ring opening from the least hindered end giving the alcohol product.

Qu21: B
Friedel-Crafts acylation para to the strong activator (alkoxy group) then reduction of the aromatic ketone to give the secondary alcohol.

Qu22: E
Diols react with ketones to give cyclic acetals as protecting groups.  Excess Grignard reagent will then add twice at the ester group giving a tertiary alcohol with 2 methyl groups, aq. acid and heat will remove the acetal deprotecting the ketone.

Qu23: B
Hydroboration/oxidation of an alkene occurs with anti-Markovnikov regiochemistry so the -OH will be introduced at the least hindered end. PCC (pyridinium chlorochromate) oxidises the primary alcohol to the aldehyde.

Qu24: B
Diene and dienophile = Diels-Alder reaction making a substituted cyclohexene system. The hydride reduces the ketone to a secondary alcohol but doesn't reduce the C=C.

Qu25: A
Diene and dienophile = Diels-Alder reaction making a substituted cyclohexene system. Here the diene is furan so we have a O atom in the bridge and the endo product with the carbonyl groups fold under should be preferred.  The ethanol reacts with the cyclic anhydride, a carboxylic acid derivative to give a diester and then the C=C is reduced by a catalytic hydrogenation.

Qu26: B
A Wittig reaction of a carbonyl of a ketone. Here is gives a simple exocyclic C=C system that then undergoes a Simmons-Smith cyclopropanation reaction to give a cyclopropane unit. Here the Simmons-Smith reagent has two methyl groups attached which will be geminal to each other in the product.

Qu27: C
The carboxylic acid in the product looks to have arisen from the hydrolysis of a nitrile (view as a carboxylic acid derivative) that was introduced via diazonium chemistry. So the starting material needed to be the appropriately substituted amine (i.e. look at the location of the phenol).

Qu28: E
The ethyl ester of the product was formed from the parent carboxylic acid with ethanol.  The parent acid was in turn obtained by reaction of carbon dioxide with the aromatic Grignard reagent.

Qu29: C
The amide product was formed by the reaction of the amine with a more reactive carboxylic acid derivative, an acid chloride that had been prepared from the parent carboxylic acid.

Qu30: B
The product contains a cyclic acetal that had been formed using the 1,3-propanediol and the ketone cyclopentanone  which was formed by the PCC oxidation of the corresponding secondary alochol.

Qu31: B
Not an easy one... partly due to those seemingly easy reagents, aq. acid and aq. base. However there is a hint in the options and the first reagent. Look for alkylation of an enolate, twice, to give the cyclic unit.... counting C atoms and considering the b-acid arrangement needed for decarboxylation to get the single substitutent, we need to start with the active methylene system of a malonate as in B.

Qu32: D
The b-hydroxy ketone suggest an aldol type process that arose from an intramolecular reaction. Use the intact carbonyl to spot the enolate of a methyl ketone that had reacted with an aldehyde to give the secondary alcohol, so break that bond to reveal 5-oxohexanal (count C) that arose from a ozonolysis with a reductive work-up of 1-methylcyclopentene.

Qu33: D
The simple diimine system comes from two carbonyls as methyl ketones 1,4- with respect to each other, generated from a C6 system by ozonolysis.

Qu34: DE
Need to add a methyl group a to an electron withdrawing group.... use an enolate type approach = strong base then alkylating agent.

Qu35: BC
Hydrolysis of the nitrile to the carboxylic acid.

Qu36: A
Reduction of the carboxylic acid, CO₂H to a primary alcohol, CH₂OH requires the stronger hydride reagent, LiAlH₄.

Qu37: AB
Now oxidise the primary alcohol to an aldehyde with the selective reagents, here PCC.

Qu38: BE
Track the previous aldehyde and recognise the conversion of the aldehyde to an alkene via a Wittig type reaction.

Qu39: A
Now reduce the ester to the primary alcohol, CH₂OH which requires the stronger hydride reagent, LiAlH₄.

Qu40: D
Make the ester by reaction of the alcohol with acid derivative, here a reactive acyl bromide.

Qu41: BD
Convert the C=C to an alcohol with anti-Markovnikov slectivity... so hydroboration sequence is needed.

Qu 42: CE
Alcohol to an ether.... Williamson.... need a base to make the alkoxide Nu then the alkyl halide as the alkylating agent.

Qu43: D
The ester.

Qu44: E
Component alcohol.

Qu45: B
Acid by oxidation.

Qu46: A
Parent carboxylic acid.

Qu47: D
Anti-Markovnikov hydration of an alkene because the B of borane has only 6 electrons around it and so is the electrophile. The B is replaced by O by attack of HOO- during the oxidation steps.

Qu48: D
Polyalkylation is a problem because the product is more reactive than the starting material since alkyl groups are weak electron donors.

Qu49: B
The aldehyde group, CHO, is electron withdrawing due to resonance and so directs meta.

Qu50: D
Resonance is the key.... try drawing the resonance structures to see.

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