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: AB
Reaction is the hydrogenation of either alkene or alkynes or aromatics. 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.

Qu2: D
Carbocation stability.... due to (a) alkyl groups, which are weak electrons donors and (b) resonance with pi bonds. These effects add stability due to charge delocalisation.  i is secondary. ii is secondary and can be stablised by the lone pairs on the oxygen (note that the resonance contributor has complete octets on both C and O) and iii is an allylic cation, with primary and secondary contributors.   This means that the simple secondary cation in i is the least stable of the three. The lone pair donation in ii is better than the bonded pair donation in iii, hence ii > iii > i.

Qu3: B
All the bonds are CC bonds, but the hybridisation of the two C is changing. In i we have an sp3 : sp2 C. In ii we have a standard CC double bond (134 pm).   In iii both C are sp2, but we are looking at a C-C between two C=C in a conjugated diene (about 146 pm).....so overall i > iii > ii.

Qu4: A
Heats of reaction such as heats of hydrogenation can be used to measure the relative stability of systems. In this case we are looking at polyenes and the degree of substitution. i and ii are both conjugated dienes whereas iii is an isolated diene so i and ii are more stable than iii. The difference between i and ii is that i has two trisubstituted C=C while ii has a tri- and a disubstituted C=C, so i is more stable than ii. The more stable system will have the least -ve heat of hydrogenation (all react to give 1,2-dimethylcyclohexane). Therefore we get i > ii > iii in terms of heats of hydrogenation.

Qu5: AB
Reaction is the Diels-Alder reaction and we are looking at the dienophiles reacting with cyclopentadiene....  The reactivity increases in the normal Diels-Alder reaction with electron withdrawing groups on the dienophile.  i has an alkyl group (a weak electron donor), ii has an electron withdrawing carbonyl in the form of an ester and iii has two electron withdrawing carbonyl groups in the form of a cyclic anhydride. Therefore the reactivity is iii > ii > i.

Qu6: C
All about enantiomeric excess and optical rotation. For i based on the masses of the two enantiomers drawn, the e.e. is 50% (use (1.5 - 0.5)/(1.5 + 0.5)). For ii a little more complex, but work out the e.e. via the rotation... [a]D (sample) = a / cl = +1.613 / (1.27/10) = 12.7. Therefore e.e. = 100%. And for iii a racemic mixture has an e.e. of 0 (by definition). So ii > i > iii.

Qu7: A
Reaction is about additions of HX to an alkene. This gives the Markovnikov product via the most stable carbocation intermediate.  The rate of the reaction is controlled by the formation of the carbocation and the variable here is the acid. The stronger the acid the faster the rate of carbocation formation. In terms of acidity HBr > HCl > CH3CO2H (acidity increases down a group in the periodic table and comparing HX acids to a simple carboxylic acids (pKas are -8, -7 and 5 respectively). So i  > ii > iii 

Qu8: B
The question is about additions of HCl to alkenes and alkynes. The rate of the reaction is controlled by the formation of the carbocation and the variable here is the carbocation structure. The more stable the carbocation then the faster the rate of carbocation formation : carbocation stability. i gives a tertiary carbocation and iii gives a secondary carbocation. The alkyne ii would require the formation of an unfavourable vinyl cation and therefore goes via a statistically slower termolecular reaction. So i  > iii > ii 

Qu9: AB
The question relates to laboratory material and the Lucas test for alcohols which is really an SN1 type reaction and it thus controlled by carbocation stability. The implication is that the tertiary alcohol reacts faster than the secondary alcohol which in turn reacts faster than the primary alcohol so iii > ii > i.

Qu10: B
Bond strength can be determined by analysis of the components. We have an aromatic CH (sp2 C) and two aliphatic CH (sp3 C). The aromatic CH is shorter and therefore stronger than the sp3CH bonds due to the increased s character in the sp2 hybrid compared to the sp3 hybrid. When we compare the sp3 CH bonds, we can use the fact that ii is secondary and iii is primary and that secondary CH are weaker than primary. We can also think about radical halogenation (which preferentially reacts with the weaker CH bonds) and that ii is also benzylic. Answer in terms of bond strength : i  > iii > ii 

Qu11: D
This question brings some laboratory aspects in with lectures...Working forwards...... the bromobenzene reacts with the Mg to form a Grignard reagent which then reacts in an SN2 fashion with the epoxide, i.e. at the least substituted end, followed by a normal dilute acid work up to give an alcohol. Note that the -OH group ends up on the C adjacent to the C attacked by the Nu in epoxide reactions. In this case the product is the secondary alcohol, D.

Qu12: D
Working backwards.... look at the product and the third set of reagents : it looks like we have reduced an alkene or alkyne. Benzyl bromide is C6H5CH2Br - this must have undergone a substitution reaction to make a new C-C bond and judging by the first set of reagents it was with a acetylide so the answer must be a terminal alkyne... D or E.... but E has too many C atoms.

Qu13: A
Working backwards.... look at the product (note it's a substituted cyclohexane with quite specific stereochemistry... may be a Diels-Alder reaction? The other starting material hints at this too as it is a common dienophile). The second set of reagents indicate the catalytic hydrogenation reduction of alkenes or alkynes, so that explains where the C=C from the cyclohexene has gone. Once you have the cyclohexene, push the curly arrows backwards to reveal the diene. Only dienes A and B make sense. C can't be s-cis and D would put the methyl groups in the wrong location and E has too many C atoms. To decide on A or B then look at the stereochemistry, since the two methyl groups are on the face side in the product, the stereochemistry of the two double bonds in the diene must be the same i.e. A.

Qu14: E
Working forwards....the first reaction is the catalytic hydrogenation reduction of alkyne to an alkene. This is followed by the addition of hydrogen bromide to the alkene. A is the Markovnikov product of an addition of HBr to an alkyne not an alkene, and B is the anti-Markovnikov product. C is the product of addition of one equiv. of Br2 to an alkyne and D the addition of Br2 to an alkene, so E is the correct answer.

Qu15: E
Working forwards....the first reaction is the anti-Markovnikov addition of hydrogen bromide to the alkene. Now we bring in some laboratory aspects.. the alkyl bromide reacts with the Mg to form a Grignard reagent which is then reacted with the carbon dioxide followed by a dilute acid work-up.... this will give a carboxylic acid with one more carbon than the original alkyl bromide. This limits the choices to A or E but A would result from the bromide formed by Markovnikov addition to the alkene, so E is the correct answer.

Qu16: D
Working forwards....the reaction is a radical halogenation... the conditions imply of an sp3C-H bond. In this case, the weakest bonds and therefore the preference will be for the allylic position, hence D.

Qu17: A
Working forwards.... look at the starting material and the first set of reagents : preparation of a terminal acetylide followed by a substitution reaction to make a new C-C ...This is followed by the hydration of the alkyne to give a ketone (which means A or B). The C=O will be formed nearest to the phenyl ring due to the ability of the ring to help stabilise the positive character of the intermediate due to resonance, hence A.

Qu18: A
This is a Diels-Alder reaction.... hints are the reaction conditions (i.e. no reagents), and the bridged structure of the product.   The challenge here is to reveal the starting material. The easiest way is push the curly arrows to "reverse" the Diels-Alder and reveal the diene: start at the C=C and work around the cyclohexene unit.  B has the dienophile C=C in the wrong location, and C has a methyl group in the wrong location, D and E have the wrong diene unit.

REGIO- and STEREOCHEMISTRY:
How well do you know your reagents ? Look at what has actually happened in terms of the reaction functional group transformation and then first look for any regiochemical issues then finally the stereochemistry last (it's the hardest to sort out).  In cases where more than one product is formed in equal amounts (e.g. the enantiomers), then both must be selected for full marks, part marks are given when only one of the pair is selected.

Advice : in each case draw the starting material in the conformation in which is reacts or the product in the conformation in which it is initially formed using wedge-hash diagrams. It is a good idea to draw the materials in such a way that the new bonds are in the plane of the page.  Once you have drawn the materials It may also be good for you to use model kits for these questions too.  Once you have drawn the materials in this way, you may need to consider rotations around sigma bonds to make your answer match the options.  An alternative approach could be to assign configurations to your drawn answer to compare them with the options - this can be slow and prone to error.

Qu19: E
Reaction of an alkene with a halogen in the presence of water gives a 1,2-halohydrin via a cyclic chloronium ion, the regiochemistry will put the -OH at the more substituted position with anti stereochemistry to the Cl atom and the -OH group. A and B have the wrong functional group, C has the wrong regiochemistry and wrong stereochemistry, D has the wrong regiochemistry.

Qu20: A
Reaction of an alkene with a hypohalous gives a 1,2-halohydrin via a cyclic halonium ion, the regiochemistry will put the -OH at the more substituted position with anti stereochemistry to the Br atom and the -OH group. In this question, the halohydrin is then treated with a base to form the epoxide via an intramolecular SN2 (also a Williamson ether synthesis) and then the epoxide is opened to give a 1,2-diol. So only A and B have the right functional groups. The challenge here is to track the stereochemistry. The reactions are anti addition, then SN2 so Nu and LG at 180 degrees then another SN reaction where Nu and LG are at 180 degrees. This leads to A. You may need to look at models or drawings to see this.

Qu21: C
The starting material is an alkyne, the first set of reagents indicate a dissolving metal reduction of an alkyne to the trans-alkene. This then reacts with Br₂ to give a cyclic bromonium ion, which in the presence of CH₃OH (the alcohol reacts as a nucleophile) will open to give a 1,2-alkoxybromide so only C and D have the right functional groups, but D has the wrong stereochemistry. The reactions are addition to give the trans C=C and the Br and CH₃O are added anti. You may need to look at models or drawings to see this. You might want to redraw the product Fischer diagram first as a wedge-dash diagram viewed from the side.

Qu22: AC
The starting material is an alkyne, the first set of reagents indicate a catalytic hydrogenation reduction of alkyne to an alkene to the cis-alkene. This then reacts with osmium tetroxide reacts with alkenes to give 1,2-diols via a syn addition. You might want to think of the product in a wedge-dash diagram viewed from the side first to "derive" the Fischer diagram. Since A and C are enantiomers, both will be formed in equal amounts. B and D have the wrong stereochemistry.

Qu23: D
This reactions involved are the dehydration of an alcohol followed by catalytic hydrogenation of the alkene. The key issue here is that only one product is formed. This means that neither of the C with the methyl substituents attached can be part of the C=C formed in the elimination reaction. Only alcohol D can give such an alkene. All of the others will undergo Zaitsev elimination to give the more highly substituted alkenes that will include at least one of the substituted C atoms.

Qu24: CD
The reaction is the hydroboration-oxidation of an alkene to give racemic 1-phenyl-2-propanol. The reaction gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Since the -OH is added at the least hindered end the starting alkene must be 1-phenyl-1-propene, either cis D or trans C will work since the C1 carbon becomes a CH2 group.  A is a meaningless structure (valence rules) and B would give 2-phenyl-1-propanol.

Qu25: CD
Very closely related to the previous question. Alkenes react with aq. acid to undergo additions to alcohols in accord with Markovnikov's rule, based on the resonance stabilisation of the benzylic carbocation intermediate, either C or D give the required product. A is a meaningless structure (valence rules) and B would give 2-phenyl-2-propanol. Alkyne E would give a ketone.

AROMATICITY and RESONANCE:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria (cyclic, planar, conjugated pi system with 4n+2 pi electrons)

Qu26: BCDE
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact (parallel, not perpendicular). This is not restricted to all carbon systems. Make sure you draw out all the bonds to be sure (is that why you missed out C?)

Qu27: AD
Apply the criteria for aromaticity. B, C and E do not contain cyclic pi systems and hence are non-aromatic.

Qu28: AB
Heteroaromatic systems must obey the criteria for aromaticity and have a non C atom as part of the cyclic pi system. C is not aromatic, D and E are aromatic but not heteroaromatic.

Qu29: BE
Conjugation requires an extended pi system (at least 3 atoms) where the pi systems can interact (parallel, not perpendicular). To evaluate aromaticity, apply the criteria for aromaticity. To be non-aromatic, they must fail one of the first 3 criteria. A and D are aromatic. C does not have an extended pi system so it can not be conjugated.

Qu30: A
Apply the criteria for aromaticity and use n=1 in the Huckel rule so we want 6 pi electrons. B is anti-aromatic (it's not a 4n+2 pi system) and C is aromatic but n=0 (it has 2 pi electrons). D is an n=2 aromatic system and E is non-aromatic since its flexible enough to be non-planar.

Qu31: A
Apply the criteria for aromaticity. C and D are aromatic. B and E are non-aromatic but do not have tautomers. A is the N analogue of a carbonyl.

Qu32: ACE
Apply the criteria for aromaticity. Only B is not aromatic. Now think about adding a proton, H+, to each of the others... this means thinking about how the lone pair figures in the pi system. In A the N lone pair is not part of the pi system since the N is part of a C=N. In C the N is external to the aromatic pi system. In D the N lone pair is part of the aromatic pi system which means that the conjugate acid would not be aromatic. In E while the N lone pair is part of an aromatic pi system, the presence of the benzene ring in the conjugate acid means that it is still aromatic (it's a different pi system).

Qu33: BCE
This question brings some laboratory aspects in with lectures, and some nomenclature. Do you know the structures you worked with? Note anything with phenyl in the name has a benzene ring present and so must be aromatic.

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