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: B
Remember the rules for ranking resonance structures  : complete octets are most important.... i has complete octets at C,N and O despite the charges, and note it has one more pi bond than the other two.  As for the other two, the negative charge on the more electronegative O atom in iii is important.  So i > iii > ii

Qu2: A
Carbocation stability.... (i) alkyl goes are weak electrons donors and (ii) resonance with pi bonds adds stability due to charge delocalisation.  i is secondary and benzylic. ii is tertiary and iii is a phenyl cation (which are very unstable C+).  A simple primary benzylic cation is almost as stable as a tertiary so a secondary benzylic cation is more stable than a tertiary cation, hence i > ii > iii.

Qu3: C
Acidity .... Know your pKa's or work it out ?  If you need to work it out, then consider the general acidity equation HA <=> H+  A-.   Look at the factors that stabilise the conjugate base, A-.   The lower the pKa, the stronger the acid.  H on electronegative atoms tend to be more acidic, here we have a phenol i and a carboxylic acid iii.  Carboxylic acids have pKa's about 5, phenols 10.  The resonance delocalisation in a carboxylate ion of an acid to a second electronegative O atom stabilises the conjugate base. For ii the H is on a C adjacent to a C=O group, hence the carbanion is resonance stabilised allowing the negative charge to be delocalised to the O.... but the pKa is around 20. So ii > i > iii.

Qu4: D
The general rule is that the number of lines in a coupling pattern is the number of  neighbours plus 1, however we also need to remember that we do not see the coupling of equivalent H systems.  So i has 2 neighbours hence a triplet (3 lines), ii has 4 neighbours hence a quintet (5 lines) and iii has only 3 neighbours hence a quartet (4 lines : note since the 2 CH2 groups are equivalent i.e. of the same type, we do not see them coupling). Hence ii > iii > i.

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
Chemical shifts are determined by shielding - this is affected by electronegative atoms and by pi systems.  In iii the H are on a C that is connected to an electronegative O, hence the shift will be about 4 ppm (actually 3.76 ppm).   In i the H are on a C that had a Br attached. The effect of the Br is less than that of an O, the H are seen at 2.68 ppm. In ii the H are on a C that is next to an aromatic ring.... typically shift just above 2 ppm (actually 2.34 ppm). So iii > i > ii.

Qu7: D
Recognise the connection as the radical halogenation of alkanes or relate bond length to bond strength to help. In the reaction, the heat or light is used to promote the reaction, which it does by cleaving the Br-Br bond because it's the weakest.  Then the C-H bonds.... methyl C-H versus a tertiary C-H.... the reaction selectivity favours the tertiary C-H cleavage, again because it's the weaker bond. So in terms of bond strengths we have ii > iii > i.

Qu8: D
The reaction is an elimination of an alcohol to give an alkene (dehydration due to strong acid and heat) and it will be E1 hence the rate of reaction depends on carbocation stability - the more stable the carbocation, the faster it will be formed.  i is a primary, ii is tertiary and iii is secondary.   Hence ii > iii > i.

Qu9: C
Chapter 3 on conformational analysis.  i has 2 staggered conformations of different energies where the C-Cl torsional angles are 60 or 180 degrees.  ii has 3 staggered conformations of different energies where C-Cl bond at carbon two is between the Br and Cl on carbon one (60 degree torsional angle to both C-Cl and C-Br bonds) or anti to the C-Br or anti to the C-Cl.  iii has only 1 staggered conformations of different energy where the C-Cl is at 60 degrees to 2 C-H bonds.

Qu 10: B
Really about alkene stability. Notice they are isomeric.  The more stable the alkene the less exothermic it's heat of combustion.  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 2 and iii has 3, so i is more stable than iii which is more stable than ii.  Hence for heats of combustion, the least exothermic is i, ii is the most exothermic.  Hence i > iii > ii.

Qu11: E
The reaction is an elimination of an alkyl halide to give an alkene (dehydrohalogenation due to strong base and heat). Best strategy is the draw the starting material and then consider removing an H from each of the C adjacent to the Cl leaving group.  i can give 2-methyl-1-butene and 2-methyl-2-butene, ii can give only 1-butene and iii can give 1-butene, cis-2-butene and trans-2-butene.  So iii > i > ii. 

Qu12: C
Assigning oxidation states ?  i is attached to O, C and 2H so -1 + 0 + 2 = +1 therefore oxidation state = -1.  ii is attached to N, O twice and C so -1 -2 + 0 = -3 therefore oxidation state = +3.  iii is attached to 3H and C so +3 + 0 = +3 therefore oxidation state = -3.  Hence ii > i > iii.

Qu13: D
Either use the formula or draw an example and then count rings and pi bonds. i has 1, ii has 5 and iii has 4 so i > ii > iii.

Qu14: A
Leaving group ability.... i is a tosylate, an excellent leaving group, ii chloride is a fair leaving group and iii has OH, a very poor leaving group. So i > ii > iii.

Qu15: B
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)  = 26mm, distance traveled by solvent = 34.5mm  therefore 26/34.5 = 0.75

Qu16: ACE
B is wrong.... the precipitates are the silver halides therefore they are insoluble. Most silver salts are insoluble, even in water. D is wrong, bromide is a better leaving group than chloride, so bromides are more reactive than chlorides (to both SN1 and SN2 reactions).

Qu17: ACE
Bromine reacts with methylbenzene under these conditions to give (bromomethyl)benzene.  t-butyl bromide reacts slowly because it only has primary H.

Qu18: A
NaI / acetone is the reagent .... it's a solution so it must be soluble.  The reagents tests SN2 reactivity (strong Nu, polar aprotic solvent). Bromide is a better leaving group than chloride, so bromides are more reactive than chlorides (to both SN1 and SN2 reactions).  Since this is an SN2 reaction, the less hindered systems react faster so primary faster than tertiary. 

Qu19: BC
Separatory funnels are used for immiscible liquids (not mixed = layers). Distillation is best with slow careful patient heating. A fluted filter paper would not be good for a vacuum filtration which is when you would use a Buchner funnel.

Qu20: A
The alkane will undergo radical halogenation with bromine via the most stable radical to give 2-bromo-2-methylpropane via reaction of the tertiary H atom. This tertiary bromide with undergo elimination (dehydrohalogenation) when heated with a strong base like NaOEt via an E2 reaction to give 2-methylpropene.

Qu21: E
Alcohols react with thionyl chloride to give chlorides, the resulting alkyl chloride reacts via an SN2 reaction with the nucleophilic acetylide ion to give E. Count C atoms.

Qu22: C
Convert the alcohol to the tosylate in step one, the water reacts as the Nu to hydrolyse the tosylate to give the alcohol via an SN2 process so inversion occurs C.

Qu23: D
We have tertiary alkyl bromide = favourable carbocation and good leaving, with a weak nucleophile, water.  Tertiary systems don't usually react via SN2. Water in acetone is reasonably polar....the factors point towards SN1, so the product should be the racemic alcohol, D.

Qu24: A
The reagents and the product are consistent with an E2 elimination of an alkyl halide. E2 require the C-H and C-LG bonds to be coplanar, most often at 180 degrees (torsional angle).  Since the C=C in the product is not the Zaitsev product, but a disubstituted alkene the elimination to the trisubstituted system must be prevented. This means the -Br and the -CH3 group must be trans-1,2-. D would give the enantiomer of the required product.

Qu25: D
The product looks to have been formed by 2 nucleophilic substitution reactions - likely SN2. So look for a system with 2 different leaving groups with the right stereochemistry (remember to account to the inversions) and the better LG must be in the right location to put the CN group in the correct location this is D.  has the wrong stereochemistry at C2, B would be difficult to control with 2 equivalent leaving groups, C would give the CN and the OH switched  and E would not react in either of these reactions.

Qu 26: D
Double elimination of HBr to give a triple bond.  A would eliminate to an alkene, B would not eliminate to an alkyne, C OH are not eliminated under basic conditions and E can only undergo substitution to give benzyl alcohol.

Qu 27: E
Product is an ether.... think Williamson.... hence the introduction of the bromide in step i.  If the methoxy group and the methyl group are trans in the product and since the methoxy group is introduced via an SN2 reaction (hence with inversion) then the alkyl bromide must have the -Br and methyl groups cis. In order to get the Br in via substitution reaction, we need a good leaving group.... so it can't be one of the -OH systems, must be one of the tosylates.  Since the Br is being introduced via an SN2 reaction (hence with inversion) then the tosylate and the methyl group must be trans.... E. Note that A would give the enantiomer of the required product.

Qu28: D
Need to add a leaving group before we can eliminate. For an alkane starting material that means use radical halogenation then a strong base / heat to eliminate. Bromination is more selective for the tertiary position than chlorination and a better leaving group in the elimination.

Qu29: E
For the -OH to be substituted, you need to convert to a better leaving group hence A, B and C fail.  To get the nitrile CN group in, we need to use SN2 type conditions.....SN1 would give a C+ that would rearrange to the more stable tertiary cation and hence the CN would end up in the wrong location. D will not work because the alcohol will react with the HBr via an SN1 type pathway and so a C+ rearrangement occurs. Tosylates are normally only used on secondary and primary alcohols and tend to react via SN2 pathways.

Qu30: D
To make the alcohol with the opposite stereochemistry, need and SN2 inversion. Make a better leaving group then hydrolyse via an SN2. A will cause a dehydration to an alkene, B will not react, C will give the correct functional groups but the wrong stereochemistry as there will be 2 SN2 inversions (so back to original stereochemistry).

Qu31: D
Making a non-symmetrical ether from an alcohol... Williamson ether synthesis.... hence base to make the alkoxide then add the alkyl halide = D.

Qu32: B
Alcohol to bromide where functional group moves primary to tertiary.... sounds like a substitution via an cation rearrangement so SN1... use strong acid so HBr to get the cation to form. A would add a Br to the tertiary C but we would still have the OH. C  would give the primary bromide via an SN2. Neither D nor E would give any reaction.

Qu33: D
Alcohol to alkene - a dehydration, an elimination. Alcohols are eliminated using strong acid and heat.

Qu34: A or B
Two types of H / compound. Both singlets therefore no coupling. Chemical shift of 3.7 ppm implies an H on a C with an O attached i.e. a H-C-O unit. Chemical shift of 7.0 ppm implies an H on a C=C probably deshielded by the resonance electron withdrawing effect of a C=O (or something similar). Note that D would have the vinyl H at about 3 ppm (yes that's right !) since OR groups are resonance donors and hence shield the vinyl H.

Qu35: AB
Four types of H in H-nmr / compound. IR of 1745 cm^-1 implies a carbonyl, a little high for a ketone.  The presence of a peak at 4 ppm in the H-nmr suggests a O-CH system and therefore an ester (AB) rather than a ketone (E) where the CH2 would connected to the C=O would be at about 2ppm.

Qu36: AE
Three types of H in the H-nmr / compound.  Peak at 3.4 ppm implies an H on a C with an O attached i.e. a H-C-O unit.  No suggestion of an -OH, only AE fits these criteria. 

Qu37: C
Four types of H in H-nmr / compound. IR and 1725 cm^-1 and 1675 suggest 2 C=O possibly conjugated systems and 1625 cm^-1 implies C=C which is supported the H-nmr peaks at 6.9 and 7.3 ppm. These two peaks integrate for 1H each and so we are looking at a disubstituted alkene, where the two substituents are different.  Only C fits these requirements.

Qu38: AC
Four types of H in H-nmr / compound. The broad exchangeable singlet at 2.3 ppm and the broad IR band at 3400 cm^-1 implies an -OH of an alcohol so either AC or AD.   But we can rule out AD (i) only has 3 types of H and (ii) wrong coupling patterns.

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