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: A
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 4 alkyl groups, ii has 3 alkyl groups and iii has 2, so i is the most stable then ii then iii is the least stable. So in terms of stability i > ii >iii.

Qu2: E
The number if lines = multiplicity (i.e. coupling) of the signals for each of the positions indicated and 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.

Qu3: B
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 iodide is a very good leaving group, ii is an amine a very, very poor leaving group (ammonia, NH3 is a very weak acid) and iii is an alcohol where HO- is a poor leaving group, so i > iii > ii.

Qu4: E
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 alcohol i, the lone pairs are on an oxygen atom, in the alkane ii, there are no lone pairs and in the amine iii, the lone pair is on a nitrogen atom. Since N is less electronegative than O, the N system is more willing to donate its electrons and so is more basic. Therefore in terms of base strength:   iii > i > ii.

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: D
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 CH3 attached to a phenyl group which will deshield the protons to a shift of about 2.2 ppm. ii is a CH2 group next to an O and a phenyl group which will both cause deshielding to about 5ppm and iii is an CH3 directly attached to an O and is deshielded by the electronegativity of the O to about 3.5 ppm. Therefore ii > iii > i.

Qu8: B
Basicity...either think about the availability of the electrons in the base or the stability of the bases. Here, they are all neutral, but the nature of the basic atom is changing. First notice that all 3 systems have lone pairs, all are negative and all from the same row of the periodic table. In the case, the less electronegative atom will be the most basic, so since O is more electronegative than N, O holds on the it's electrons more than N so it is the N that is more basic than O and C > N So in terms of basicity i > iii > ii.... the more basic the anion, the greater the reaction. 

Qu9: 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.

Qu10: A
Acidity.... First let's compare i and ii (because they are very similar, both are carboxylic acids) : the difference between them is the halogen atom... the presence of the electronegative atom will stablise the conjugate base due to inductive electron withdrawal through the sigma bonds, so in terms of acidity, the more electronegative F will stabilise more than the Cl so in terms of acidity, i > ii. Now let's compare iii, it's a thiol... In terms of general principles, -SH systems are more acidic than OH systems due to the size of the S compared to the O. But that would only explain an alcohol compared to a thiol. In the carboxylic acid, there is resonance delocalisation in the conjugate base.... carboxylic acids are stronger acids that thiols. Other things that would help answer this question are knowing the pKa's (carboxylic acid are about 5, thiol about10). Therefore in terms of acidity, i > ii >iii.

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: 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 secondary. So overall we have in terms of stability, ii > iii > i.

Qu 13: E
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, so we are looking at the SN1 of alkyl halides. Look at the carbocations that would be formed. The most favourable carbocation is the one from E, a primary allylic system with a tertiary allylic resonance contributor.

Qu14: D
We are looking at the SN2 of an alkyl halides (think back to the laboratory expt) with less steric hindrance being the controlling issue for sp3C-Br centers. Vinyl halides (e.g. D) don't undergo SN2 reactions.

Qu15: B
The reaction is the E2 elimination of an alkyl halide with a strong base. We need to look at the nature of the diene that would be formed : A, D and E give cumulated, B conjugated, C isolated. Since conjugated dienes are more stable than isolated than cumulated, then B is the answer.

Qu16: E
The most deshielded chemical shift means higher chemical shift. Deshielding is caused by electronegative atoms and proximity to C=O in these cases. Notice that E is an aldehyde which has a chemical shift of 9-10ppm.

Qu17: A
An application of acidity. The most acidic H will be one of the H on the CH between the 2 C=O groups (since they provide resonance stabilisation) i.e. A or C. The presence of the electronegative F atom in A makes it the most acidic since the F can further stabilise the conjugate base.

Qu18: A
E1cb reactions are favoured when there is a stablised conjugate base (see qu17) and a weaker leaving group (e.g. F)

Qu19: AB
Enantiomers are required to have the same melting points.

Qu20: C
Methoxycyclohexane is C7H14O, IHD =1. ABE all had IHD = 2 (1 ring + 1 pi bond). D has a different molecular formula and therefore is not an isomer.

Qu21: E
We should work forwards .... halogenation of an alcohol using HCl. These reactions tend to be SN1 in character (protonation of the OH gives a good leaving group) and proceed via a planar carbocation and here we will get the racemic product.

Qu22: C
We should work forwards.... The first step is radical halogenation of a tertiary 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 t-butyl methyl ether, C.

Qu23: D
Need to work backwards ....looking at the product, it's an alkene - but it's the anti-Zaitsev product, which means we are looking at an E2 elimination which are typical of alkyl halides which could be made via a radical halogenation of methylcyclohexane to give 1-bromo-1-methylcyclohexane. This is followed by an E2 elimination using a large base to favour the anti-Zaitsev product.

Qu24: B
Need to work backwards ....looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - looking at the structure of the ether and the reagent (Na2CO3 = base), we should be able to see that we need an difunctional system with an -OH and a halide, with 4 C in a chain.

Qu25: C
Need to work backwards....looking at the product, it's an alkene and an alkyl halide, specifically an allylic halide so we could use a radical halogenation to give an allylic system. We are starting from an alkyl bromide so we need to eliminate the alkyl halide (so base / heat) to to give the alkene and then do a allylic radical substitution to give the required product.

Qu 26: A
Need to work backwards ....looking at the product, it's an alkene - but it's the anti-Zaitsev product. Since we are looking at KOH / heat, we are looking at an E2 elimination. In cyclic systems, the critical issue is the 180 degree arrangement of the H-C and C-LG bonds - this requires that the H and the LG in question are both axial. In this question then, to prevent the Zaitsev product from forming, it requires that the methyl group and the Br are both axial in the reactive conformation, i.e. they are trans-1,2. So, the best approach is to draw the cyclohexane in the chair conformation and put the methyl group equatorial (large groups prefer to be equatorial). Then look at the position of the Br in relation to the methyl group. In order to get the required product we need trans 1,2-bromomethylcyclohexane A. B is the enantiomer as it has the methyl group in the mirror image location. E would give a mixture of products.

Qu 27: C
We should work forwards .... halogentaion of an alcohol using PCl₃ followed an SN1 using ethoxide / ethanol as the nucleophile to give an ethyl ether by replacing the benzylic halogenation and not the aryl halogen (which don't undergo simple SN reactions).

Qu 28: B
Need to work backwards ....looking at the product, it's an ethyl ether, but there has been a rearrangement to get there. This means we need to go via a carbocation and hence an SN1 reaction. This can be achieved by converting the -OH into a halogen using HBr (SN1) and then an substitution using ethoxide.

Qu 29: E
2-methylpentane is C6H14. Only E is C6.

Qu30: B
The torsional angle between the methyl groups is 60 degrees for cis 1,2-dimethylcyclohexane where one methyl will be axial and the other equatorial (i.e. they are staggered wrt each other).

Qu31: B
The lowest energy conformation of trans-1,3-dimethylcyclohexane will have one methyl axial the other equatorial (this is required for trans-1,3-) and it will be a chair conformation.

Qu32: C
The conformation shown shows the two indicated bonds in the plane of the page and on opposite side of the connecting bond. This means that they are at 180 degrees and hence anti.

Qu33: B
The structures are isomeric (same MF) but what type of isomer are they ? Picking up one model and turning it around allows it to be placed like the other one so they are in fact identical.

Qu34: ABE
cis-1,2-dimethylcyclopropane has a cyclopropane ring (hence ring strain). This means there are eclipsed bonds (hence torsional strain) and the two cis methyl groups will be quite close to each other (hence Van der Waals strain). Axial terminology is not applicable to cyclopropanes and the flagpole term is specific to the boat conformation of cyclohexane.

Qu35: E
The most exothermic heat of formation implies the most stable structure, the largest ring will have the least ring strain and hence be the most stable of the C5H10 isomers.

Qu36: B
In order to have two conformations of a chair cyclohexane of the same energy, look for 2 identical substituents where one is axial and one is equatorial. In these examples this means cis-1,2-dimethyl.

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: A
IR shows a carbonyl (i.e. C=O) at 1735 cm^-1 which is high for a typical ketone.  The H-nmr has 2 types of H, the peak at 3.7 ppm, suggests a H-C-O- group and the peak at 7ppm (only 1H). Look for C=O, and only two types of uncoupled H in a 3:1 ratio with a C=C....

Qu38: AC
IR shows a carbonyl (i.e. C=O) at 1745 cm^-1 which is high for a typical ketone. The H-nmr has 4 types of H. The H-nmr coupling shows an isolated methyl group and the 4.1ppm suggests a H-C-O- group suggesting a possible ester, but where the methyl group is attached to the C=O.

Qu39: BE
The H-nmr has 3 types of H. The H-nmr coupling shows an n-propyl group and the peak at 3.4ppm suggests a H-C-O- group.

Qu40: C
IR shows a carbonyl (i.e. C=O) at 1725 cm^-1 which high for typical ketone. The H-nmr has 4 types of H and includes two isolated methyl groups. ethyl group. The H nmr peaks at 6.9 and 7.3 ppm suggest a trans substituted alkene (note there are only 2H based on the integrals).

Qu41: AE
IR shows an -OH at 3400 cm^-1 and consistent with the exchangeable broad singlet (1H) in the H nmr. The H-nmr coupling shows an n-propyl group.

Qu42:BD
IR does shows an -NH2 at 3370 and 3291 cm^-1 and consistent with the exchangeable broad singlet (2H) in the H nmr. The H-nmr coupling shows an n-propyl group.