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
Acidity...all the H we are interested in are located on N atoms, the difference is the number of C=O groups in proximity. Each C=O allows us  resonance stabilisation via delocalisation of the -ve charge in the conjugate bases to the electronegative O atom (in the same way as for a simple carboxylic acid).  Thus iii > i > ii.
approximate pKa's iii = 16, i = 30, ii = 35

Qu2:C
Since the +ve center is not adjacent to the benzene ring, the question is really comparing a secondary, a tertiary and a vinyl cation.  Since alkyl groups are weakly electron donating due to hyperconjugation and polarisation effects, more alkyl groups give a more stable carbocation. Hence tertiary cations are more stable than secondary cations.  Vinyl cations are quite unstable due to the nature of the empty orbital. So ii  > i > iii.

Qu3:C
i is a simple amide, ii is a simple ketone, if you know it, these are about 1660 and 1715 cm^-1 respectively (or use the tables).  iii is has the C=O with 2 NH groups, one either side. In an amide, the N donates electrons via resonance to the C=O which increases the single C-O character and so lowers the carbonyl frequency. This is effect is even more pronounced with the two N atoms in iii, therefore ii  > i > iii.

Qu4:C
The reaction of alcohols with HBr is essentially an SN1 type reaction and hence depends on the stability of the carbocation produced.  i is primary, ii is secondary and iii is an aromatic alcohol.  Remember that the C attacked in SN reactions need to be sp3 ? Hence ii > i > iii.

Qu5: A
Each type of carbon will give a single peak.  i has 5 types, ii has 4 and iii has 2 types so i > ii > iii.

Qu6: E
13C nmr spectroscopy... We are looking at 3 different methyl groups next to 3 different C=O groups. What's the effect ? The more electron withdrawing the C=O group unit, then the great the deshielding and therefore the greater the chemical shift of the methyl group.  Remember that the N of an amide is an electron donor via resonance.  The shifts are 31ppm, 22 ppm and 34 ppm respectively, so iii > i > ii.  

Qu7: E
H nmr spectroscopy.... We are looking at the effect of a CH2 group that is attached to a -Cl and another but different group... benzene, H and OCH₃ respectively.  The O in iii will deshield due to the electronegativity of the O.  Benzene rings cause a little deshielding due to the magnetic anisotropy.  H exerts no electronic effect (it's the normal situation), so iii > i > ii.

Qu8: AB
The reaction here is SN2, so it will depend on the degree of steric hindrance at the C-LG center.  i  is tertiary (hence slow), ii is secondary and iii is primary (hence fast) so iii > ii > i.

Qu9: C
A good leaving group is the conjugate base of a strong acid (i.e. stable is the displaced form).  Acidity increases down the groups of the periodic table so H2S is more acidic than H2O so HS- is a better LG than HO-.  HI is a very strong acid and hence iodide, I-, is a very good leaving group. So ii > i > iii.

Qu10: AB
Straight from the midterm ! Basicity... the strongest base will create the most of the conjugate base. Compare to O systems first... O- is a stronger base than O, so ii  > i.  Now compare N- and O- .  N is less electronegative than O so it is a better electron donor and hence a better base (Lewis definition) so iii  > ii. Therefore overall  iii > ii > i

Qu11: AB
Elimination of alkyl halides is via the E2 pathway, where the H is always removed from the C adjacent to the C-X bond. Hence, i can only give one alkene product, 1-pentene.  ii can eliminate to give 2-pentene, but this can exist as a mixture of cis and trans isomers.  iii can eliminate to give 3 alkenes.... 2-ethyl-1-butene, cis-3-methyl-2-pentene and trans-3-methyl-2-pentene. Hence iii > ii > i. 

Qu12: B
Radical stability follows the same trend as carbocation stability (since both deal with alkyl groups stabilising electron deficient centers). Here we have tertiary, primary and secondary respectively so the stability order is i > iii > ii.

Qu 13:D
Rf is the ratio of distance traveled by sample to distance traveled by solvent from the origin.  Measure these distances with a ruler. Distance traveled by sample (middle of spot)  = 17mm, distance traveled by solvent = 34mm  therefore 17/34 = 0.5

Qu14:A
If the sample shows up as two spots after development then is must contain at least two compounds, hence the sample is impure so i is true. The spot that is more polar in normal chromatography will have a lower Rf value since it will interact with the polar stationary phase more, so ii is incorrect.  The spot that elutes more rapidly is the one that moves the fastest i.e. travels the furthest, This is x so iii is false.

Qu15:BCE
A is false because benzene has an IHD of 4 due to 3 C=C and one ring.  D is false because glycine does not have a chirality center with 4 different groups attached. 

Qu16:BCD
A is false because acetone is a polar, aprotic solvent.  E is false because this reaction reflects SN2 reactivity and that means the less hindered system will react the most rapidly. 

Qu17: BD
A is false because it reacts to give an alkyl bromide, alkenes react to give 1,2-dibromides.  C is false because methylbenzene reacts to give a resonance stabilised secondary radical rather than a simple secondary radical.  E is false because the low reactivity is due to the lack of a benzylic H and hence the low radical stability. 

Qu18: B
The heat of formation... more exothermic means a more stable compound, -13.5 kcal/mol is the most exothermic. So if we are looking for the most stable isomers, this will be the one with the two substituents in the more favourable equatorial positions, hence B.

Qu20: D
The isomer with the most Van der Waals strain will be the one with the most 1,3-diaxial strain, that will be the one with the substituents axial, D.

Qu21: B or D
A constitutional isomer has different connectivity due to different branching and / or different functional groups... but of course they have to have the same molecular formulae to be isomers.  A is the same as the question, C is stereoisomer and E is not an isomer (C5).

Qu22: B or D
A conformational isomer has the same connectivity but is different due to rotation about single bonds... but of course they have to have the same molecular formulae to be isomers. A and C are constitutional isomers and E is not an isomer (C5).

Qu23: D
There are two chirality centers, one at C6 and one at C7. Hence there are 2² optical isomers.

Qu24: C
The are 4 types of C in the aromatic ring.  There are 2 equivalent ethyl groups attached to the N, so we are up to 6 total. Then there is the C=O, plus 3 more in the cyclopropane ring and a the C attached to the amine. That is a total of 11.

Qu25: B
O5 is part of a double bond, C=O, so it is sp2.  C7 is sp3 because it has 3 C and 1 H attached, hence 4 single bonds.   N9 is sp3 because it has a C atoms attached, 2 H atoms and the N lone pair but there is not resonance since there is not adjacent pi system, it's an amine N.

Qu 26: B
Oxidation states
C4 has the following bonds : 1 x C (count 0) and 2 x O (count -2) and 1 x N (count -1) so the sum = -3 therefore the oxidation state of C4 = +3.
C7 has the following bonds : 3 x C (count 0) and 1 x H (count +1) so the sum = +1 therefore the oxidation state of C7 = -1.

Qu 27: C
A ethyl group requires a triplet and a quartet. The quartet is for the CH2 coupled to the CH3, but because the CH2 is also attached to the electronegative N atom, is will have the greater chemical shift.  Since other H on the molecule only has one of that type, the integral for the ethyl group will be 4:6 (note we also gave full marks for AB, but C is the best answer).

Qu28: D
Working backwards, the second reaction is an SN2 to replace the a leaving group with a  -CN.  The leaving group turns out to be a Br, put in place by a  radical halogenation of alkane type CH at the most stable radical site (bromine selectivity).  Alcohols A and C may react in both steps but the -OH group (poor LG) would still be present in the product. Alkene B would react with the Br₂ to give a dibromide. So this leaves D or E. Counting carbons shows that we need a C2 side chain.

Qu29: D
The first reagent is tosyl chloride, TsCl, so we are making a tosylate of an alcohol, ROTs. That narrows the choices to B, C or D. This is followed by an SN2 reaction to give the iodide. Since stereochemistry is implied in the product, we need to pay attention to that.  When an alcohol reacts with TsCl the C-O bond in the original alcohol is unchanged and therefore the stereochemistry in unchanged in that step.  However, SN2 reaction usually occur with inversion so the stereochemistry of the final iodide is the opposite to that of the original tosylate and hence alcohol.  B will give a racemic product, C will give the enantiomer of the required product, whereas D gives the highest yield of the required product.

Qu30: D
PCl₃ reacts with alcohols to give alkyl chlorides via an SN2 type reaction.  This chloride would then react with the ethoxide to give a ether via a Williamson ether type synthesis, another SN2 reaction.  Ethoxide will not substitute a aryl chloride (since the C is not sp3) - this rules out C and E.
A and B require an oxidation, but we do not have an oxidising agent here.

Qu31: E
A simple alkyl bromide being heated with a sterically hindered base under aprotic conditions suggests an E2 reaction to give an alkene rather than a substitution.  That limits the answer to C, D or E. Since E2 reactions occur in such a way that the LG and H are lost from adjacent carbon atoms (i.e. they are 1,2-elimination reactions), then E is the only viable answer.

Qu32: A
Another alkyl bromide being heated with a base - again an E2 reaction to give an alkene rather than a substitution.  Since the base here is hydroxide (hence small and unhindered), the product should follow Zaitsev's rule and give the more stable, more highly substituted alkene. This means either A or B is our answer, but which ? C and D are substitution products, E has too many C atoms. In order to select between A and B we need to look at the stereochemistry.... remember that an E2 requires that the H and the LG are at 180 degrees. If we set up the starting material in this way then the methyl groups will end up trans as in A rather than cis as in B. 

Qu33: A
We need to convert the alcohol to an ether with 2 more carbon atoms and a nitrile group.  A is a Williamson ether synthesis... make the -OH into the alkoxide, a better Nu, then SN2 reaction where the iodide (better LG) is displaced. In the final step the -CN replaces the -Cl in another SN2 reaction. B step 1 may react to give the -CH₂OCH₂CH₂OH side chain but then would not react with the NaCN.  C step 1 would react to place a Br on the cyclohexane and step 2 may react to give a disubstituted cyclohexane. D step 1 would give the cyclohexyl tosylate, but then that would not react with the BrCH₂CH₂CN (where is the Nu ?). E step 1 alcohol to bromide, step 2 bromide to alkene... step 3 no reaction.

Qu34: B
Overall reaction alcohol to alkene - hence a dehydration, an example of an elimination.  Note that we want the less highly substituted alkene, so regular conc. acid / heat (E1) is not going to work because that would give 2-pentene and not the 1-pentene required. A HBr cause substitution to give alkyl bromide... any elimination would give 2-pentene.  B step 1 reacts to give the tosylate which is then reacted with a hindered base via an E2 to give the required product. C would react to give 2-pentene. D HBr cause substitution to give alkyl bromide, but the ethoxide / ethanol reaction would give a lot of substitution and a mix of alkenes if there is any elimination (heating with a more hindered base would have worked though). E no reaction.

Qu35: B
Need to make a thioether from an alcohol, need to get the opposite stereochemistry - hence avoid SN1 and force SN2. A HBr causes substitution to give alkyl bromide but via an SN1 so we would get racemic bromide and hence racemic thioether. B step 1 reacts to give the tosylate which is then reacted with the thiolate nucleophile via an SN2 (hence inversion) to give the required thioether. C Poor reaction, Sn1 character, hence racemisation will occur. D or E will not convert the -OH into a better leaving group hence no substitution.

Qu36: B
Need to replace the -OH with a -Br with a change in the carbon skeleton... hence via a carbocation rearrangement (an alkyl shift) hence need an SN1 type reaction.  A, C and D are SN2 reactions therefore the Br will end in the same location as the original -OH with no change in the skeleton. B Perfect !  E will add a Br via radical substitution in a different location, the OH would still be present and there would be no carbocation and hence no rearrangement.

Qu37: A
4 ArH each of an individual type implies either ortho or meta disubstituted aromatic (6.8 - 7.13 ppm). 3H singlet at 3.8ppm suggests OCH3 group.  IR says no C=O or OH.  13C also lacks a C=O and has 6 types of ArC.

Qu38: AB
4 ArH of 2 individual types suggests a para disubstituted aromatic (7.2 - 7.8 ppm). No peak above 9ppm means it's not an aldehyde or an acid (hence rules out C or AD). IR shows a C=O but no O-H or C-O. 13C shows C=O at 197.5ppm (implies ketone or aldehyde). The lack of a peak near 4ppm in the Hnmr rules out the esters D and E and the ethers A and AC.

Qu39: D
4 ArH each of an individual type implies either ortho or meta disubstituted aromatic (7.1 - 7.9 ppm).  No peak above 9ppm means it's not an aldehyde or an acid (hence rules out C or AD). IR shows a C=O and C-O but not O-H.  13C shows C=O at 167ppm (not a ketone or aldehyde).  3H singlet at 3.9ppm suggests OCH3 group suggesting the esters D and E (ruled out because it's para) or ethers A and AC (ruled out because these lack a C=O).

Qu40: AD
4 ArH of 2 individual types suggests a para disubstituted aromatic (7.3 - 7.9 ppm). Peak at 12.8 ppm that is exchangeable means a carboxylic  acid (AD is the only one). IR shows a C=O and O-H and C-O. 13C shows C=O at 167.3 ppm (implies acid or derivatives).  AE lacks the C=O.

Qu41: AC
4 ArH of 2 individual types suggests a para disubstituted aromatic (6.8 - 7.05 ppm).  No peak above 9ppm means it's not an aldehyde or an acid (hence rules out C or AD).  IR shows no C=O or O-H but indicates a C-O. 13C shows no C=O.  Absence of C=O leaves us only A, AC and AE. lack of an OH rules out AE. 

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