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.

RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may appear. Look for two pairs of similar systems to compare that have minimal differences in structure. If a compound is named, draw it out. If a reaction is involved, identify the type of reaction and then what the controlling factors are.

Qu1: AB
Carbocation stability.... due to (a) alkyl groups, which are weak electrons donors, and (b) resonance with a bromine atom. These effects add stability due to charge delocalisation.  i is secondary, ii is secondary with an attached bromine (further stablised by resonance with the lone pairs on the bromine) and iii is tertiary.   This means that the simple secondary cation in i is the least stable of the three. The bromine lone pair donation in ii makes it more stable than i but not to the extent that it makes it more stable than a tertiary carbocation (bromine is large, the C-Br bond is quite long and the interaction of the p orbitals is not the greatest). Hence iii > ii > i.

Qu2: D
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.   In iii both C are sp2, but we are looking at a C-C between two C=C in a conjugated diene. The double bond is stronger than either of the single bonds (there is a greater bonding interaction). Comparing the two single bonds, then the higher s character will mean a shorter stronger bond, so the sp2:sp2 in iii is stronger (in addition there is also extra bonding due to the adjacent pi systems. So in terms of strengths: ii > iii > i.

Qu3: 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.

Qu4:A
The reactions of alkenes and alkynes with aq. acid are controlled by carbocation stability and the number of alkyl groups on the pi unit (more alkyl groups makes them more electron rich and hence more nucleophilic). Alkynes are less reactive since they would require the formation of the less stable vinyl carbocations. Therefore i > ii > iii.

Qu5: AB
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. ii and iii are both conjugated dienes whereas i is an isolated diene so ii and iii are more stable than i. The difference between ii and iii is that ii has two disubstituted C=C while iii has a tri- and a disubstituted C=C, so iii is more stable than ii. The more stable system will have the least -ve heat of hydrogenation (all react to give 2-methylhexane). Therefore we get iii > ii > i in terms of heats of hydrogenation.

Qu6: A
All about enantiomeric excess and optical rotation. For i based on the masses of the two enantiomers drawn, the e.e. is 66% (use (1.25 - 0.25)/(1.25 + 0.25)). For ii a little more complex, but work out the e.e. via the rotation... [a]D (sample) = a / cl = +1.27 / (2.0/10) = 6.35. Therefore e.e. = 50%. And for iii a racemic mixture has an e.e. of 0 (by definition). So i > ii > iii.

Qu7: AB
The reaction is the hydroboration / oxidation of alkenes. First draw out the starting material, 2-methyl-2-pentene. ii and iii have correctly added the HO and H across the C=C therefore i has the lowest yield. ii is the Markovnikov product, while iii is the anti-Markovnikov product. Since hydroboration / oxidation gives the anti-Markovnikov product as the major product then we get iii > ii > i.

Qu8: A
Check the pKas, 15 vs 25 vs 35 respectively of the acids or think about the bases that are used to prepare an acetylide ion !  In terms of acid strength i > ii > iii.
Remember that the hybridisation of the C (sp) makes it difficult to compare to the O and N systems (if you did, you would have got B).

Qu9: 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 

Qu10: B
We discussed the stability of enols and ketones when we looked at the hydration of alkynes, and concluded that the ketone is the more stable form. Now we need to compare the two enols, which are really substituted alkenes.... we discussed alkene stability.... more alkyl groups gives a more stable alkene, so iii is more stable than ii. So overall, in terms of stability, i > iii > ii.

STARTING MATERIALS AND PRODUCTS:
If you are trying to find the product, then you should probably just work forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably the best way to go....
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu11: BE
Working forwards... hydration of an alkyne to give a ketone. Internal alkynes where there are two simple alkyl groups give mixtures of ketones because there is no significant difference in the cation character (equal stability of carbocations). So both B and E are formed.

Qu12: E
Working backwards...reactions look like intramolecular alkoxymercuration / demercuration of an alkene (note that no water is being added) - this would explain the ether as the product. Therefore the starting material we are looking for should have both a C=C and -OH.

Qu13: D
Working backwards... it looks like the alkylation of a nucleophile (SN2) obtained from alkene ozonolysis with an oxidative work-up. This would account for the dimethyl ester products. Since the products are dicarbonyls, we need to consider reconnecting the pieces to make a ring.... in this case they reconnect to make the six membered ring with an isolated diene. Since the ozonolysis gives a diacid, the alkenes involved can only be 1,2-disubstituted. This means the answer can not be A, or B since they would give ketone-acids. E would give a diketone and a diacid.

Qu14: ABD
Working forwards... The ozonolysis will cleave the alkyne to carboxylic acids and the alkene to aldehydes / ketones (because the work up is reductive). If we look at the alkene, we see at least one H atom at each end of the C=C, this means we will get aldehydes. Alkynes always give carboxylic acids.

Qu15: E
Working backwards... The ozonolysis with an oxidative work up......will cleave an alkene to ketones / carboxylic acids. If we number the carbon chain starting at the acid carbonyl group, then the two carbonyls are on C1 and C6. This means we need a cyclohexene. Since we have a methyl ketone, there must be a methyl group on the alkene, i.e. 1-methylcyclohexene. The alkene has been formed by E2 elimination of the alkyl halide.

Qu16: E
Working backwards.... The target is a 1,1-dibromocyclopropane, this can be made via the carbenoid addition of 1,1-dibromocarbene to the alkene using bromoform and base (i.e. CHBr₃) . Given the cis stereochemistry of the cyclopropane, this indicates the need to start with the cis-alkene, which would be made via the catalytic hydrogenation reduction of alkyne to an alkene using Lindlar's catalyst of the alkyne 2-pentyne. This alkyne comes from the methylation of 1-butyne via an SN2 with the nucleophilic terminal acetylide. Note that the Simmons-Smith using organic zinc reagents from dibromomethane would give the cyclopropane with a -CH₂- added not a -CBr₂- added.

Qu17: C
Working forwards.... the first reaction is the addition of HX to an alkene. This gives the Markovnikov product via the most stable carbocation intermediate, in this case the secondary carbocation will form, but this will then undergo a 1,2-hydride shift to a more stable tertiary carbocation and hence the tertiary bromide forms (this is actually answer option E). The second step then is an elimination of the alkyl halide to give the Zaitsev alkene (more stable, more highly substituted), C.

Qu18: CD
Working backwards... the product is an alkyne, reagent a strong base, so it looks like the elimination to form an alkyne. This could be either a double loss of HCl from a vicinal or geminal dihalide or the loss from HCl from a vinyl halide. Simple alcohols do not eliminate using basic reactions conditions.

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:C
The first reaction is the catalytic hydrogenation reduction of alkyne to an alkene which is a syn addition giving the cis-alkene. The alkene is then reacted with cold alkaline potassium permanganate to give a 1,2-diols via a syn addition - hence not A or B (wrong functional group). The syn addition of the diols to the cis-alkene will give a meso compound (C) rather than the pair of enantiomers D and E. Darn those Fischer projections!

Qu20: DE
The first reaction uses a peracid to epoxidise an alkene followed by basic ring opening using methoxide (note the -OH and -OMe groups in the product). This means that the reaction is SN2 like and the methoxide attacks the least hindered C in the epoxide.... hence we end up with 1,2-dimethoxy groups.. via an anti type ring opening. Looking at the Fischer projections the regiochemistry of the reactions means we need the -OH and -OMe adjacent to the ethyl group (this rules out B and C). Careful inspection of the 3D diagrams reveals the stereochemistry..... with the alkyl groups set up to match the Fischer diagrams, the two methoxy groups need to be on opposite sides.

Qu21: BC
Reaction of an alkene with a halogen in the presence of water gives a 1,2-halohydrin via a cyclic bromonium ion, the regiochemistry will put the -OH at the more substituted position with anti stereochemistry to the Br atom and the -OH group. Treatment with sodium carbonate as a base forms the epoxide with the same stereochemistry as the original alkene. This means the two methyl groups need to be trans in the epoxide. Hence, A has the wrong stereochemistry, D and E have the wrong functional groups.

Qu22: B
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 via a cyclic bromonium ion to give a 1,2-dibromide via an anti addition. The best approach (other than using models) is to derive the answer using wedge-hash diagrams and then convert to a Newman projection then rotate each of the potential answers to see if it matches your answer.

Qu23: CD
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 cyanide in DMSO.... therefore the nucleophilic cyanide will undergo an SN2 reaction with the Br resulting in an inversion of the stereochemistry at that position to give cis-1,2-cyanoalcohol. This means A and B have the wrong functional groups. E has the wrong regiochemistry from the first step.

Qu24: A
The reaction is the hydroboration-oxidation of an alkene followed by a Williamson ether synthesis. The hydroboration reaction gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. The product ether will have the same stereochemistry as the alcohol because the oxygen of the alcohol functions as the nucleophile. In terms of regiochemistry, the -OH and hence the -OR group are added at the least hindered end of the alkene (i.e. the methyl group end). The best approach (other than using models) is to derive the answer using wedge-hash diagrams and then convert to a Newman projection then rotate each of the potential answers to see if it matches your answer. D and E have the wrong regiochemistry, B and C the wrong stereochemistry.

Qu25: D
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. 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 using aqueous acid to give a 1,2-diol. This means we can ignore A and E. The challenge here is to track the stereochemistry. The HOCl adds via an anti addition, then SN2 so Nu and LG at 180 degrees (inversion) to form the epoxide then opening the epoxide with another inversion : i.e. opening the epoxide gives a trans-1,2-diol. B and C have the wrong diol stereochemistry.

PI SYSTEMS:
All about knowing properties of pi systems such as resonance, structure, bonding, shapes, and terminologies etc.

Qu26: CDE
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. A has 2 isolated C=C. B is a hydrocarbon with just one C=C. C has is a conjugated diene. D has a C=O inS conjugation with a C=C. E is an allylic carbocation, so the carbocation is conjugated.

Qu27: E
Resonance contributors are derived via the delocalisation of the pi electrons with the pi system and can be derived by pushing curly arrows. A B C and D can be ruled out because the +ve charge can not be delocalised to atoms that were sp3 hybridised in the original structure.

Qu28: ACD
Tautomers (remember alkyne hydration ?) of ketones are the enols. B is a resonance contributor. E is a conjugate base of the enol.

Qu29: AC
Cumulated systems have an sp atom shared by two double bonds. B D and E are conjugated systems.

Qu30: ADE
sp2 hybridisation is typical of alkenes and other double bonds (e.g. carbonyl groups). It is also typical of simple alkyl carbocations. B contains sp C and N.

Qu31: B
sp hybridisation is typical of alkynes and other triple bonds (e.g. nitrile groups).

Qu32: A
Allylic hydrogens are those on the carbon atoms adjacent to alkene C=C bonds, i.e. H-C-C=C. A has 7, B none, C none (but it has 9 benzylic), D has 6 and E has 2.

Qu33: A
Polyene stability.... Here we are comparing different diene types and we have small ring cycloalkene. A is a conjugated dienes which are more stable than isolated or cumulated dienes. Alkynes (B and C) are similar in stability to cumulated systems and so an alkyne is less stable than isolated or conjugated dienes. Small ring cycloalkenes like D or E are destablised due to high ring strain.

Qu34: C
All the bonds are CC bonds, but the hybridisation of the two C is changing. In A, B and D we have sp3 : sp2 C. In E we have a standard CC single bond (154 pm).   In C both C are sp2, but we are looking at a C-C between two C=C in a conjugated diene (about 146 pm).

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