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 question about stability carbocations. Remember that vinyl carbocations are quite unstable while resonance stabilises carbocations (e.g. allylic systems are resonance stabilised).
Qu 2:	All about the nucleophilicity of these molecules. All the systems are neutral, but it is the lone pairs on the O and S that make those molecules nucleophilic. There are two oxygen systems and a sulfur system, all are single bonded and two are negative. Let's deal with the two oxygens first.  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 oxygen and S-ve more nucleophilic than O-ve. There are no available electrons on the alkane, it's a very poor nucleophile.
Qu 3:	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. Here we have a tosylate, an alcohol and an alkane. Alkyl groups are terrible leaving groups and tosylates are a method of making -OH into a better leaving group.
Qu 4:	Basicity...either think about the availability of the electrons in the base or the stability of the bases (look for the lone pairs) or relate to the pKa's of the appropriate systems. 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.
Qu 5:	First identify the reaction.... the conditions of AgNO3 / aq. ethanol suggest an SN1 reaction (think back to the laboratory expt). A quick look at the systems shows three chlorides, so we are looking at the effect of changing the alkyl group (since the leaving group is the same across the series). SN1 reactions (where the leaving group is the same) are typically controlled by the stability of the intermediate carbocations.
Qu 6:	First identify the reaction.... the alcohol starting materials and reaction conditions of HBr suggest formation of an alkyl bromide via an SN1 reaction. These reactions are typically controlled by the stability of the intermediate carbocations. .
Qu 7:	Chemical shifts of the groups in question in these systems are determined by the nature of the groups attached to the methyl group (that is what is changing). Chemical shifts are affected by electronegative atoms (especially such as O) and magnetic anisotropy (due to pi systems).
Qu 8:	Acidity.... Need to look at the stability of the conjugate bases and consider the factors that stablise that. H attached to more electronegative atoms tend to be more acidic (the electronegative atom stabilises the conjugate base) F is the most electronegative atom in the periodic table. H on larger atoms tend to be more acidic, S is in the second row. A knowledge of the pKa values of common acids really helps!
Qu 9:	This is a radical substitution reaction. With bromination, the reaction is much more selective for the more stable radicals (i.e. tertiary > secondary). For the two secondary systems, the more H there are of that type, the greater the reaction of that site. The starting material has been drawn and the H atoms of the relevant types are colour coded.
Qu 10:	The number of lines = 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 e.g. the CH2 groups in butane.)
Qu 11:	The question is about elimination reactions. Alkyl halides eliminate with strong base / heat, alcohols with strong acid / heat... therefore it's only the alkyl halides that eliminate under the conditions of the question. The Zaitsev product is favoured the most, the better the leaving group.
Qu 12:	Counting types of hydrogen

Qu 13:	First identify the reaction.... the alcohol starting materials and reaction conditions of HCl suggest formation of an alkyl chloride via an SN1 reaction where the leaving group is the same) are typically controlled by the stability of the intermediate carbocations. The benzyl system II forms a resonance stabilised carbocation.
Qu14:	Acidity of alkynes (relates to a MT topic). Terminal alkynes I are the most acidic because the H attached to the sp C gives a conjugate base where the -ve charge is stabilised by the proximity of the +ve nucleus.
Qu15:	The question is about elimination reactions. Alkyl halides eliminate with strong base / heat usually via E2 pathways, and alcohols with strong acid / heat usually via E1 pathway. Here, an E1 reaction via a carbocation would rearrange to give 1,2-dimethylcyclohexene. Therefore it's only the alkyl bromide that eliminate under the conditions of the question to give the product shown via an E2 where there is no carbocation intermediate and therefore no rearrangement.
Qu16:	Easiest to answer by pushing the curly arrows from X to derive the resonance contributor.
Qu17:	The reaction is the conversion of an alcohol to its tosylate. In this reaction, the alcohol O reacts as a nucleophile and the O becomes part of the tosylate. This means that the C-O does not break or change in anyway and therefore the product tosylate has the same stereochemistry as the starting material alcohol.
Qu18:	Maximising the number of bonds / complete octets is the most important factor. The atoms like C,N,O can't expand their octets, so they can only form a maximum of 4 bonds.

Qu19: E
Need to work backwards ....looking at the product, it's an amine produced by a nucleophile substitution of an alkyl bromide. The alkyl bromide was formed with HBr from an alcohol via an SN1 reaction. Count C atoms!

Qu20: B
We should work forwards....looking at the starting material and the product, we need to convert an alcohol to an ether but there also needs to be a rearrangement meaning SN1 reactions are involved. One common way to make an ether is the reaction of an alkyl halide and an alkoxide. HBr will reaction with the alcohol to give the rearranged bromide that can then undergo an SN2 reaction with the nucleophilic methoxide to make the secondary methyl ether. A and E would give the primary methyl ether. D would give a secondary alcohol.

Qu21: D
We should work forwards.... starting from an alcohol and going to an alkene - and it's the anti-Zaitsev product, which means we can not just use a straight forward E1 elimination in a alcohol dehydration using a strong acid / heat which we would expect to give the Zaitsev product. So, need to use E2 chemistry and that means make the -OH into a better leaving group, then use a bulky base to drive the reaction to the less hindered anti-Zaitsev product. A would give the Zaitsev product, B, C and E are all basic an d would not cause an elimination since alcohols eliminate under acidic conditions.

Qu22: D
We should work forwards ....Sodium metal (Na) reacts with alcohols (ROH) to prepare the alkoxide ion (RO-). This will react with the primary alkyl bromide to give the ethyl ether. Since the C-O bonds not change, the product ether has the same stereochemistry as the alcohol starting material.

Qu23: B
We should work forwards....halogenation of an alcohol using thionyl chloride converts an alcohol to an alkyl chloride via an SN2 reaction (i.e. inversion of stereochemistry).

Qu24: C
We should work forwards.... In cyclohexane 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, the starting material is drawn to show the two groups equatorial so we need to draw the ring flip version where both groups become axial. This means the axial position of the methyl group prevents the E2 elimination to give the Zaitsev product (A) and therefore the elimination occurs away from the methyl group to give the anti-Zaitsev product, C. E is the enantiomer of the actual product.

Qu25: A
We should work forwards ....starting from an alkane, the radical chlorination will give 2-chloro-3-methylpentane as the major product. Heating this alkyl chloride with the bulky strong base will cause an anti-Zaitsev E2 elimination.

Qu 26: B
We should work forwards ....starting with a ketone, we are making a C-C at the adjacent position.... this suggests we are using an enolate nucleophile to react with an alkyl halide to make the new C-C bond. We can prepare the enolate by reacting the ketone with a suitable base.

Qu 27: C
Draw out the 2,2-dimethylpentane (C7H16).... this reveals the presence of a t-butyl unit which you can look for. On the other hand, only C is C7.

Qu 28: E
The torsional angle between the chlorine atoms is 120 degrees in this eclipsed conformation.

Qu 29: ACD
All the structures drawn are 1,2-dimethylcyclohexane. Trans requires that the two methyl groups are on the opposite face of the ring. B and E are both cis-1,2-dimethylcyclohexane.

Qu30: C
The Newman projection shows a staggered conformation and shows the two indicated bonds at 180 degrees and hence anti (this is the best, i.e. most complete term)

Qu31:D
The structures are isomeric (same MF) but what type of isomer are they? Comparing the drawings or picking up one model and comparing allows one to they are enantiomers.

Qu32: B
All the structures show 1,4-dimethylcyclohexane. The most stable conformation will be i. chair (A, B, C) and ii. have both methyl groups equatorial therefore B.

Qu33: A
Small rings have the most deviation from the optimal sp3 C bond angle of 109.5 degrees and therefore they have the highest angle strain.

Qu34: D
Need to assign the configurations based on the Cahn-Ingold-Prelog rules for the alkene stereochemistry and the chirality center.
For the alkene, consider the two ends separately. At the top, the C with the O attached out ranks the C with 3 H attached. At the bottom, the N outranks the C. So the two higher priority groups are on the same side, and hence this is a Z isomer (German; zusammen = together).
For the chirality center, the groups in priority order are OH > C(C)=C > C(C)CH > H. Given that the lowest priority group is towards us as drawn, this means that we are looking at the S configuration.

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.

Qu35: BC
IR shows a carbonyl (i.e. C=O) at 1737 cm^-1 which is high for a typical ketone. The IR band at 1200 cm^-1 suggests C-O. The C NMR peak at 170 ppm suggests the C=O is a carboxylic acid or derivative. The H-nmr has 3 types of H, the singlet at 3.8 ppm, suggests 2 x -O-CH₃ groups. The peaks at 1.4 ppm and 3.3 ppm tells us that we have a CH₃ next to a CH. Overall the pieces are CHCH₃, and 2 x -OCH₃, and C=O likely in esters.

Qu36: A
The C NMR (C 59 and 72 ppm only) indicate only 2 C types and both sp3 (no C=O). H NMR shows only 2 types of H and the peaks esp. integrals suggests only -CH₃ and -CH₂- groups. The peaks are singlets = no coupling.

Qu37: AB
IR shows a carbonyl (i.e. C=O) at 1713 cm^-1 which is about normal for a typical ketone. This is supported by the C NMR peak at 207 ppm. Since there is no H-nmr peak at 9-10ppm, it is not an aldehyde. H NMR shows only 2 types of H and the peaks esp. integrals suggests only -CH₃ and -CH₂- groups. The peaks are singlets = no coupling.

Qu38: B
The C NMR (C 19, 52 and 102 ppm) indicates 3 C types, no C=O and likely no ArC. H NMR shows 3 H types. The H NMR peaks and the peaks esp. integrals suggests only -CH₃ and -CH groups. Overall the pieces are CHCH₃, and 2 x -OCH₃. If there is no C=O, it can't be an ester, but ether is fine.

Qu39: C
IR shows an -OH at 3350 cm^-1 and consistent with the exchangeable broad singlet (1.8 ppm 1H) in the H NMR. The C NMR (C 25 and 64 ppm only) indicate only 2 C types and both sp3 (no C=O). H NMR shows only 2 types of H and the coupling and the integrals suggests a (CH₃)₂CH- group.

Qu40: E
IR shows an -OH at 3340 cm^-1 and consistent with the exchangeable broad singlet (2.3 ppm 1H) in the H NMR. The C NMR (C 10, 26 and 64 ppm only) indicate 3 C types and all sp3 (no C=O). H NMR shows 3 types of H and the coupling and the integrals suggests a CH₃CH₂CH₂- group.