Here is an post-mortem analysis / "how to" for the MT. 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
Carbocation stability.... due to (i) alkyl groups, which are weak electrons donors and (ii) resonance with pi bonds. These effects add 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.

Qu2: AB
The systems are all dienes... diene stability is such that conjugated are more stable than isolated which in turn are more stable than cumulated.  For individual C=C, alkene stability is such that a more highly substituted C=C (i.e. more alkyl groups on the C=C unit) is more stable. The more stable a system is, the less exothermic the heat of hydrogenation will be for the reduction to the alkane. i is an isolated diene, ii is conjugated with two disubstituted C=C and iii is conjugated with one trisubstituted and one disubstituted.   Overall then iii > ii > i.

Qu3: B
If you know your pKa's then this is easy : water = 15.7 terminal alkyne pKa = 26, t-butanol = 18.  Remember the lower the pKa the stronger the acid, so i > iii > ii.   What if you don't remember your pKas ?  (why not ?)  Then you'll need to deduce it.  Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-..... first compare water and the alcohol.... if you recall, alkyl groups are weak electron donors and stabilise cations.... therefore they destabilise anions.... so water is more acidic than an alcohol.  Now what about the alkyne ? Maybe the easiest way here is to think about the bases that are used in synthesis to make the acetylide ion.... sodium amide is the common choice.... that implies that alkoxides, RO-, are probably not strong enough bases so the alkyne is a weaker acid than the alcohol.

Qu4: A
Resonance energies measure stability. Benzene i has a high resonance energy (36 kcal/mol) due to the aromaticity of the cyclic pi system. Furan ii is also aromatic but the aromatic stabilisation is less than that of benzene (16 kcal/mol).  The cyclohexadiene in iii is an isolated diene so it has zero resonance energy (well, there is no resonance !).  Therefore we get i > ii > iii.

Qu5: AB
Reaction is the Diels-Alder reaction. The normal Diels-Alder is favoured by having electron withdrawing groups on the dienophile. i the oxygen substituent here  has a lone pair next to the pi system which can be donated to the pi system by resonance. In ii we have a carbonyl which is an electron withdrawing group (think of the important contributor where the C=O is drawn as +C-O-  hence the +ve C attracts electrons. In iii there are 2 carbonyl groups attached to the C=C.  Overall then we have iii > ii > i.

Qu6: A
All CC bonds, the thing that is changing is the hybridisation of the C atoms involved in the bond. Review ? In i both C are sp3, we have a simple standard sigma bond (1.54 pm)  In ii one C is sp3 the other is sp2.  Since the sp2 has greater s character, the orbital is smaller and the result is a shorter bond (1.51 pm). In iii both C are sp2, hence shorter again, plus there is the additional bonding interaction due to the adjacent p orbitals (1.46pm). Overall  i > ii > iii.

Qu7: AB
Reaction is the hydrogenation of either alkene or alkynes. The reactivity towards reduction is determined by the strength of the pi bonds so alkynes reduce more rapidly than alkenes. Aromatic C=C reduce more slowly than alkene C=C due to the aromatic stabilisation. So iii > ii > i.

Qu8: A
Reaction is the hydroboration oxidation of an alkene. This gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Since no C+ forms, there is no carbocation rearrangement.  First draw 2-methyl-1-pentene, then add the -OH to the least substituted end of the C=C.  So i is the major product and ii is a minor product (the Markovnikov product). iii is only obtained is there has been a rearrangement. Overall then i > ii > iii.

Qu9: A
Sucrose is the disaccharide of glucose and fructose, they are the component monosaccharides. 

Qu10: A
It's about making a better leaving group... protonating the oxygen in the glycosidic bond means there is a neutral leaving group.

Qu11: B
An anomeric carbon is typically attached to two oxygen atoms by single bonds. In this case there is only one C-O bond there.

Qu12: A
Reactions are second order if the rate law has depends on the product of the concentrations of two molecules, either different species A and B or the same species, A with A.

Qu13: A
Alcohols undergo dehydration, an elimination with strong acid via the E1 pathway.

Qu14: B
The reaction of alcohols with HCl in the Lucas test is an SN1 type reaction so the rate depends on the relative stability of the carbocation produced.

Qu15: B
Secondary alcohols are oxidised to ketones.  Primary alcohols are oxidised to aldehydes then carboxylic acids.

Qu16: B
2,4-DNP reacts with the carbonyl group in aldehydes and ketones only.

Qu17: A
Both Kevlar and nylon contain repeating units joined by an amide group, O=C-N unit and hence are polyamides.

Qu18: B
Terephthalic acid is 1,4-benzene dicarboxylic acid.

Qu19: A
Yes esters are hydrolysed by water under acid or base conditions. You did it under basic conditions in the laboratory and were asked to draw both the acid and the base reaction mechanisms in the report.

Qu20: A
Do you know your functional groups ?

Qu21: B
Since organometallic compounds contain carbon - metal bonds and metals are more electropositive than carbon, C-M systems are polarised such that the C is negative and the metal is positive.  Hence the C atoms of C-M systems are nucleophilic.

Qu22: A
Review the experimental for this experiment.... the acid product was obtained in the final step as a white precipitate by the acidification of a solution containing the carboxylate salt.

Qu23: B
The dry glassware is needed because Grignard reagents react with water.

Qu24: B
Grignard reagents contain Mg.

Qu25: A
The first reaction is the addition of bromine to a conjugated diene. At 50^oC, this will be under thermodynamic control and so the initial product will be the 1,4-addition product, 1,4-dibromo-2-butene.  The resultant alkyl halide will then react with excess CH3S- as a nucleophile (S is the same group as O, and is a better Nu) to give the 1,4-dithioether, A.  B would be the product of kinetically controlled 1,2-addition of bromine in step one. All the other products would require that more than 1 mole equivalent of bromine were added to the diene.

Qu26: A
The most obvious structural feature in the product is the fused cyclopropane ring - this has been formed in step 3, a Simmons-Smith reaction - which adds a carbene type system to an alkene.  Step 2 would be consistent with that if it were an elimination reaction, these strongly basic conditions would apply if we were eliminating a halide.... this is fine since step 1 is a radical bromination.  Put this together, we need to take away the one C atom added in the Simmons-Smith reaction to reveal the starting alkene for that reaction = cyclopentene. Cyclopentene would be formed by the elimination of bromocyclopentane which in turned would be prepared from cyclopentane, A.

Qu27: C
Osmium tetroxide reacts with alkenes to give 1,2-diols. Alcohols, when heated with sulfuric acid react by dehydration to give alkenes via an E1 type of mechanism which favour the most stable products.  A diol will give a diene, conjugated dienes are often favoured because of their extra stability. Although could be formed, there is a more stable and hence more likely product. Epoxides as in B are not usually prepared from diols. C is a conjugated diene with two tri-substituted alkene units. D and E are more like the products of alkene ozonolysis.

Qu28: B
The product is a cis-alkene, the last set of reagents indicate the catalytic hydrogenation of an alkyne.  The preceding steps look like a terminal alkynes has been  deprotonated with sodium amide, a strong base to give the acetylide ion, a good Nu and then reacted in an SN2 fashion with the benzyl bromide to give the product. In order to get the require alkyne for the reduction, the original alkyne needs to be propyne, B.  If A were the starting material, then the product of the sequence would have one less C atom, Ph-CH₂CH=CH₂.  C, D and E are not terminal alkynes and so would not react with step 1 and hence 2.

Qu29: D
The starting material is an alcohol, first set of reagents, sulfuric acid and heat suggest dehydration to give the alkene.  Step 2 is ozone  and step 3 corresponds to a reductive work-up overall an ozonolysis reaction. The C=C has to form to give 1,1,3-triphenylpropene.  We are given the ketone product from the disubstituted end of the alkene, the monosubstituted end will give an aldehyde under the reducing work-up conditions, corresponding to phenylethanal, D.

Qu30: C
The product is an epoxide and it has been formed by the reaction of a halohydrin. Reaction of an alkene with a hypohalous acid gives a 1,2-halohydrin followed by treatment with a base to allow an intramolecular SN2 reaction (also an example of a intramolecular Williamson ether synthesis) to give an epoxide.  The starting material needs to be an alkene, so A, B or C.   In order to put the epoxide in the correct location, the C=C needs to be in the ring and be methyl substituted so 1-methylcyclopentene, C.  A would give an epoxide but the methyl group would be in the wrong location and B would also give an epoxide but it would be external to the cyclopentane unit rather than on the cyclopentane.

Qu31: A
The starting material is an alkyne, the first set of reagents indicate a dissolving metal reduction of an alkyne to the trans-alkene. The second reagent is a peracid and this will cause epoxidation of the alkene. This is a concerted reaction and the stereochemistry of the alkene is preserved in the epoxide. The final step, aqueous acid will cause the epoxide to undergo acidic ring-opening in an SN1like fashion to give a 1,2-diol. If you know it gives a diol, A is the only diol in the answers.

Qu32: AB
Reaction of an alkene with a halogen in the presence of water gives a 1,2-halohydrin followed by treatment with a base (sodium carbonate) allows an intramolecular SN2 reaction (also an example of a intramolecular Williamson ether synthesis) to give an epoxide (hence answer must be A, B or C). The reaction will lead to the halohydrin via the cyclic bromonium ion, the regiochemistry will put the -OH at the more substituted position with anti stereochemistry to the Br atom.  For the SN2 the nucleophilic HO must attack at 180 degrees (i.e. also anti) to the C-Br bond, therefore the two methyl groups will still be trans in the product epoxide.  So A and B both have the correct stereochemistry (they are the enantiomers that correspond to the addition of the Br to opposite faces of the alkene unit). C has the wrong stereochemistry. D and E are diols not epoxides.

Qu33: C
The product is a vicinal-dibromide (a 1,2- arrangement) so this must come from an alkene rather than an alkyne.  Br₂ adds across C=C to add one Br atom at each end of the pi bond in an anti fashion.  You should redraw the product Fischer diagram first as a wedge-dash diagram viewed from the side then in a new wedge-dash diagram after rotating about the central C-C bond to get the two Br anti (best to draw the Br atoms in the plane of the page). This means that the two alkyl groups need to be cis in the alkene as in C.  A would give a tetrabromide if there were excess bromine or a vinyldihalide (Br-C=C-Br).  B has the wrong stereochemistry. D has too many C atoms, an ethyl instead of a methyl group.  E has the C=C in the wrong location to give the required product.

Qu34: DE
The starting material is an alkyne, the first set of reagents indicate the catalytic hydrogenation of an alkyne to the cis-alkene since  poisoned catalyst is being used. The second reagent is a peracid and this will cause epoxidation of the alkene. This is a concerted reaction and the stereochemistry of the alkene is preserved in the epoxide. So because the alkene has cis methyl and phenyl groups, so will the epoxide. A has one C atom too few, the methyl group is missing. B and C have the groups in the wrong stereochemistry, they are trans. D and E are enantiomers with the correct cis-stereochemistry formed by the O being added to opposite faces of the C=C.

Qu35: B
This is a Diels-Alder reaction.... hints are a substituted cyclopentadiene, the diene, and a disubstituted alkene, the dienophile.  Remember the reaction is concerted and the stereochemistry of the diene and the dienophile is preserved in the reaction. So because the dienophile is cis here, the ester groups must also be cis in the product. The methyl group on the diene needs to be tracked too. A has the cis ester groups but has the less favourable exo product, B is the correct answer with the cis ester groups in the favoured endo orientation and hence is the major product.  C has the ester groups trans and is therefore incorrect since the dieneophile is cis. D has the methyl group in the wrong location and E has lost the C=C that needs to be present from the diene portion.

Qu36: C
The reaction is the hydroboration-oxidation of an alkene. This gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Note that the -OH is formed with retention of stereochemistry when the B atom is replaced. So for methylcyclohexene, the -OH ends up on the C next to the C-CH3 unit.  Since the -OH and -H add syn, the -OH and the -CH3 must be trans, C.  A and B have the wrong stereochemistry. D has the wrong regiochemistry (this is the Markovnikov product) and E has the wrong functional group transformation.

Qu37: D
The starting material is an alkyne, the first set of reagents indicate the hydration of the alkyne to give the ketone via the enol and the tautomerisation. A and B are the enols that form, but enols are very unstable and very rapidly equilibrate to the ketone, D. C has the wrong regiochemistry since the +ve charge is most stable on the benzylic C where it is resonance stabilised. E has one C atom too few, a methyl group is missing. 

Qu38: AE
Which systems are non-aromatic as drawn ?  B, E, AB and AE. B, AB and AE are non-aromatic because the pi system are not cyclic pi systems.  B, E and AB are all 6 pi electron systems.  Note that AC is anti-aromatic, the empty p orbitals completes the cyclic conjugated unit and it is a 4 pi electron system.

Qu39: A or D or AD or BC or CD
Which systems are aromatic as drawn ? A, C, D, AD, BC, CD and CE.  Now consider adding H+ to make the conjugate acid.... the conjugate acids of C and CE are all still aromatic because the protonated lone pair is not part of the pi system.

Qu40: B or AE
Which systems are non-aromatic as drawn ?  B, E, AB and AE.  Now consider removing H+ to make the conjugate base.... the conjugate base of AB would be a 8 pi electron system.... tending to make it anti-aromatic. 

Qu41: B
Which systems are non-aromatic as drawn ?  B, E, AB and AE.  Tautomerism allows us to move H atoms typically from heteroatoms to C trading pi bonds (as in the ketone-enol tautomerism).  This applies for B.

Qu42: C
Heteroaromatic means the ring contains a non-carbon atom.  So that limits us to B, C, CD, or CE.  Of these only C has the same number of electrons as benzene.

Qu43: AD
Which systems are aromatic as drawn ? A, C, D, AD, BC, CD and CE.   Which ones of these are ionic i.e. charged ? AD or BC. If n=1 in the Huckel rule we need 4n+2 or 4x1 + 2 = 6 pi electrons. That's AD since BC is a 2 pi electron system.

Qu44: C or CE
Which systems are aromatic as drawn ? A, C, D, AD, BC, CD and CE.  Now consider adding H+ to make the conjugate acid.... the conjugate acids of  C and CE are all still aromatic because the protonated lone pair is not part of the pi system.

Qu45: AC
The only anti-aromatic system here is AC.   Which systems are aromatic as drawn ? A, C, D, AD, BC, CD and CE. Which systems are non-aromatic as drawn ?  B, E, AB and AE.

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