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: D
Carbocation stability....for simple alkyl cations, more alkyl groups means a  more stable carbocation but here we also need to consider the effects of resonance. i is a phenyl cation which are less stable than primary cations. ii is a primary benzylic cation also stabilised by resonance, almost as stable as a tertiary system and iii is simple primary. So overall we have in terms of stability, ii > iii > i.

Qu2: E
All about the nucleophilicity of these groups. There are two oxygen systems and a sulfur system, all are single bonded and all are negative. Let's deal with the two oxygens first.  The possibility of resonance delocalisation in ii makes the electrons less available and therefore the carboxylate ii is less nucleophilic than the alkoxide i. Now compare the O and the S in i and iii respectively.  Since S is larger than O (one row lower in the same group of the periodic table), the sulfur is more polarisable and therefore more nucleophilic than the O.  Hence  iii > i > ii.

Qu3: AB
Leaving group ability....remember that good leaving groups need to be stable when they leave and therefore tend to be the conjugate bases of strong acids i is an amine a very, very poor leaving group (ammonia, NH3 is a very weak acid), ii chloride is a fair leaving group and iii is a tosylate, an excellent leaving group, so iii > ii > i.

Qu4: A
Basicity...either think about the availability of the electrons in the base or the stability of the bases. The stronger the bases, then the more the reaction shown moves to the right. These are all based on atoms from the first row of the periodic table. In the amide ion i, the -ve charge is on a nitrogen atom whereas in the alkoxide ii, the negative charge is localised on the O atom. In water, the system is neutral rather than negative, so it will be a weaker base. Since N is less electronegative than O, the N -ve system is more willing to donate its electrons and so is more basic. Therefore in terms of base strength and hence the amount of reaction:   i > ii > iii. In terms of the appropriate pKas: i = 35, ii = 15 and iii = -3. The lower the pKa the stronger the acid / weaker the base.

Qu5: AB
First identify the reaction.... the alcohol starting materials and reaction conditions of sulfuric acid / heat suggest alcohol dehydration via an E1 reaction. These reactions are typically controlled by the stability of the intermediate carbocations. i would give a unfavourable primary cation, ii a secondary and and iii a secondary benzylic cation (more stable since resonance stabilised). The more stable the carbocation, the more rapidly it forms and the faster the reaction....so in terms of reactivity we have iii > ii > i.

Qu6: C
First identify the reaction.... the conditions of NaI / acetone suggest SN2 (think back to the laboratory expt). A quick look at the systems shows three bromides, so we are looking at the effect of changing the alkyl group (since the leaving group is the same across the series)... since they are named the first step would be to draw the structures. i is a tertiary bromide, ii is secondary and iii is a phenyl (aromatic) bromide. Recall that SN reactions only occur on sp3 systems, so iii is not reactive. Since SN2 reactions are slower as the steric hindrance of the alkyl group increases, we have ii > i > iii.

Qu7: A
Chemical shifts of the groups in question in these systems are determined by the nature of the attached groups. All are sp3 CH systems. i is an CH2 between a phenyl group and the oxygen of a methoxy group, both groups will deshield the protons to a shift of about 5 ppm. ii is a CH3 group next to an O and is more deshielded by the electronegativity of the O to about 3.5 ppm. iii is an CH3 directly attached to a phenyl and usually about 2 ppm. Therefore i > ii > iii.

Qu8: B
Basicity...either think about the availability of the electrons in the base or the stability of the bases. There are two oxygen systems and a sulfur system, all are single bonded and all are negative. Let's deal with the two oxygens first.  The possibility of resonance delocalisation in ii makes the electrons less available and therefore the carboxylate ii is less basic than the alkoxide i (pKas are 15 and 5 respectively). Now compare the O and the S in i and iii respectively.  Since S is larger than O (one row lower in the same group of the periodic table), the sulfur is larger and therefore better able to accommodate the -ve charge. This makes the S system less basic than the O system (basicity decreases down a group in the periodic table pKas 10 and 15 respectively).  Hence  i > iii > ii.

Qu9: B
Conformational analysis so look for different types of interactions. i has an eclipsed conformation between the middle CH2 and the CH3 groups on either side as drawn (so 6 eclipsed bonds). ii is in a more favourable staggered conformation and has an eclipsed conformation on both the LHS and the RHS as drawn (for a total of 6 eclipsed interactions) and therefore is higher energy than i. iii has an eclipsed conformation on the RHS as drawn (for a total of 3 eclipsed bonds).Therefore the highest energy conformation is i and the lowest (most stable) is ii, so i > iii > ii.

Qu10: E
The multiplicity (i.e. coupling) of the signals for each of the positions indicated is determined by the number of neighbours that are of a different type (since H of the same type do not show coupling). i is a CH2 between a phenyl group and a methyl group so it has 3 neighbours and therefore appears as 4 lines. ii is a CH2 group in cyclohexane where there is only one type of H and therefore we see a singlet. iii is an CH between two methyl groups so it has 6 neighbours and therefore appears as 7 lines.. Therefore iii > i > ii.

Qu11: C
The reaction is the E2 elimination of an alkyl halide (draw it from the name) with different strong bases. The key here is that the larger the base, the lower the yield of the more highly substituted, more stable Zaitsev product because the larger base is hindered from abstracting the more hindered proton. So since the base sizes are iii > i > ii it means the yields of the Zaitsev products is ii > i > iii.

Qu12: AB
A question about alkene stability. Notice they are isomeric.  In terms of alkene stability, the general rule is the more alkyl groups on the C=C unit, the more stable it is. i has 1 alkyl groups, ii has 3 alkyl groups and iii has 4, so iii is the most stable then ii then i is the least stable. So in terms of stability iii > ii >i.

Qu 13: AE
Since the SN1 goes via a carbocation, the reaction with the most stable carbocation will react fastest...but in order for the carbocation to form, there must be a good leaving group (this rules out the alcohols under these reaction conditions), so we are looking at the SN1 of alkyl halides. Of the alkyl bromides, look at the carbocations that would be formed. The most favourable carbocation is the one from AE, a tertiary and tribenzylic system.

Qu14: B
Alcohols do not react under these conditions (they lack a good leaving group), so we are looking at the SN2 of an alkyl halides (think back to the laboratory expt) with less steric hindrance being the controlling issue. So the primary benzylic system reacts fastest, B and aromatic halides (e.g. A) don't undergo SN2 reactions.

Qu15: CE
The reaction is the E2 elimination of an alkyl halide with a strong base. Of the alkyl halides to select from, only CE will react fastest because the substituent pattern favours the -Br being in the reactive axial position with 2 C-H bonds in the required 1,2-position that can eliminate.

Qu16: E
An application of acidity. Look at the functional groups (alkyl halides, aromatic alcohols, alcohols and amines), think about pKas etc. The -OH in E is a phenol which makes it the most acidic (pKa about 10).

Qu17: AD
An application of basicity...either think about the availability of the electrons in the base or the stability of the bases. Look at the functional groups (alkyl halides, aromatic alcohols, alcohols and amines), think about pKas etc. The -NH groups in AC and AD (amines) have lone pairs that are more available than similar O systems (due to electronegativity) but in AC there is also some delocalisation of the lone pair to the aromatic.... this makes the lone pair less available.

Qu18: D
Bond dipoles are caused by atoms of different electronegativities at each end of a bond.

Qu19: D
All about types of H and coupling. Since D only has one type of H, the spectrum will be one singlet.

Qu20: BD
Look at the substituted cyclohexanes. You need a system that will give two chair conformations of the same energy. This is most easily achieved if there are two substituents of the same type that are trans with one axial and one equatorial.

Qu21: C
Need to work backwards.... the nitrile / cyanide has been introduced via anSN2 reaction of a tosylate. Since the SN2 causes inversion, the original alcohol needs to be trans to the methyl group as in C. Note that A is the enantiomer of C. B and D have the wrong stereochemistry.

Qu22: C
Need to work backwards.... the product is an ester. The reagent in step 2 shows the source of the acid group which is being used in an SN2 reaction as the nucleophile with (based on step 1) an alkyl bromide. The bromide has been made from the corresponding alcohol.

Qu23: B
We should work forwards ....alcohol to alkyl bromide, but requires a rearrangement, so need to make sure it's SN1. A gives bromides via SN2. C would cause elimination....D would give a rearranged alcohol but the SN2 reaction of PBr3 is not efficient due to the tertiary structure. E would not involve a carbocation so no rearrangement.

Qu24: E
We should work forwards ....looking at the product, it's an alkene - but it's the anti-Zaitsev (Hofmann) product. Since the starting material is an alcohol, a simple dehydration with acid and heat is not going to work, as it would give the Zaitsev product. Instead, convert the -OH to a halogen and then do an E2 elimination using a large base to favour the anti-Zaitsev product. A would give the Zaitsev alkene. B, C and D would not cause an elimination reaction of the alcohol.

Qu25: E
We should work forwards.... The first step is radical halogenation of a benzylic position followed an SN1 using silver nitrate and methanol (think about the lab expt). In the SN1, the nucleophile will be the O of the methanol so the product is a benzylic methyl ether.

Qu26: B
Need to work backwards ....looking at the product, it's an alkene. Since we are looking at KOH / ethanol / heat, we are looking at an E2 elimination of an alkyl halide. To work out which bromide to use, look at the E2 stereochemical requirement of the H-C and C-Br bonds needing to be at 180 degrees.

Qu27: C
We should work forwards .... halogentaion of an alcohol using thionyl chloride followed an SN2 using a thiolate ion will give a methyl thioether.

Qu28: A
We should work forwards ....looking at the product, it's an alkene formed by the dehydration of an alcohol... so you need a strong acid.

Qu29: C
A is a pentane, B is missing a methyl group, D is a 2,2-dimethylbutane and E is a pentane.

Qu30: B
The torsional angle between the chlorine atoms is 60 degrees for the gauche conformation.

Qu31: BC
A is trans-1,4-, D is cis-1,3- and E is trans-1,4-.

Qu32: D

Qu33: D
Enantiomers are non-superimposable mirror images. Test for that (use models if needed) or assign the configurations at the chirality centers. The Newman projection is R and the wedge-hash is S.

Qu34: A
The interaction between the large t-butyl group and the methyl group is an example of a Van der Waals strain.

Qu35: D
It will be most stable dimethylcyclohexane. The cis-1,3-dimethyl will have the least strain.

Qu36: A
The smallest ring will have the most angle strain per CH2 since the bond angles will be the most different from the sp3 optimal of 109.5 degrees.

SPECTROSCOPY:
Use any IR information to get the functional groups. Use the H-NMR to get the number of types of H, how many of each type from the integral and what they are next to from the coupling patterns. Chemical shifts should tell you if the group is near -O- or maybe C=O groups etc.

Qu37: B
IR shows a carbonyl (i.e. C=O) at 1718 cm^-1 which is a typical ketone. The 13C-nmr peak at 209ppm supports the C=O and also suggests an aldehyde or a ketone.  This means it could be B, C, or AB. The H-nmr doesn't have a peak at 9-10ppm, so it is not an aldehyde. The H-nmr has 3 types of H and the C-nmr has 4 types of C so it has to be B. The H-nmr coupling shows an ethyl group and a methyl group singlet.

Qu38: AD
IR shows C-H at 2974 cm^-1. The lack of 13C-nmr peaks above 50 ppm suggests no sp, sp2 C or C attached to more electronegative atoms.  This means it could be AD or BE. The H-nmr has 2 types of H and the C-nmr has only 2 types of C. The H-nmr coupling shows an ethyl group but no methyl group.

Qu39: BE
IR shows C-H at 2974 cm^-1. The lack of 13C-nmr peaks above 50 ppm suggests no sp, sp2 C or C attached to more electronegative atoms. This means it could be AD or BE. The H-nmr has 3 types of H and the C-nmr has 4 types of C. The H-nmr coupling shows an ethyl group and a methyl group.

Qu40: AE
IR shows a carbonyl (i.e. C=O) at 1792 cm^-1 which quite high for typical ketone. The H-nmr has 2 types of H in an ethyl group. The 13C nmr peak at 175 ppm suggests a carboxylic acid derivative. This means we could have either AE or BC. Since anhydrides give two C=O bands in the IR, it has to be the acyl chloride, AE.

Qu41: A
IR shows a carbonyl (i.e. C=O) at 1716 cm^-1 which is a typical for a ketone. The IR also shows an -OH at 3400 cm^-1. The 13C-nmr peak at 181 ppm supports the C=O being a carboxylic acid derivative and the H nmr peak at 11.7 ppm that exchanges with D2O suggests a carboxylic acid.

Qu42:E
IR does shows an -OH at 3300cm^-1. The H-nmr exchangeable peak at 2.6 ppm supports the -OH of an alcohol. The H-nmr shows an ethyl group coupling pattern.