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 is dictated by the substituents at the C+ center. It may help here to note that the carbocations are related to those encountered during the reaction of an alkyne with HBr.  i is a secondary system, with a Br atom which through resonance of a lone pair can further stabilise the +ve charge, ii is a simple secondary system and iii is a secondary vinylic cation (which is like a primary) due to the hybridisation at the C (note no resonance as the pi system is perpendicular = no overlap). Therefore we get i > ii > iii.

Qu2: D
Radical bromination of an alkane. Draw the alkane first, then the products. Bromination is quite selective for the more highly substituted positions due to the stability of the radicals. i is primary, ii is tertiary and iii is secondary,  so ii > iii > i.

Qu3: B
i is an alkene, so it can protonate to give in this case a tertiary carbocation. ii is an alkane which does is not a good base since all the electrons are in strong single bonds.  iii is an alkyne that protonates to give an unfavourable vinyl cation.  So i > iii > ii.

Qu4: C
Alkenes react via electrophilic addition where the rate determining step in the addition of the electrophile. Here the electrophile will be H+ and so it depends on the acidity of the reagent.  Acidity increases down a group due to bond strength and the ability of the conjugate base to accept charge. So HBr > HCl, we should know water is not a strong acid, or can think about willingness of O to accept charge. Overall then ii > i > iii.

Qu5: C
Need to work out which of the 3 stereoisomers is (S,S).  i is meso (note the vertical mirror plane that reflects left to right) so it is (R,S) and hence is an "internal racemate" and is optically inactive = 0^o rotation. Group priorities are CO2H > CHCO2H > CH2 > H. If we look at ii then the right hand carbon has the R configuration, and so ii must be (R,R) and will have a positive rotation. Overall then we have ii > i > iii.

Qu6: C
Catalytic hydrogenation depends on the strength of the p bond. The second p bond in an alkyne is weaker than that in an alkene, so reactivity of ii > i.  Carbonyl groups have strong p bonds and are not easily reduced under these conditions. So overall we have ii > i > iii.

Qu7: 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.  i is the major product with the -OH at the less substituted end and ii is a minor product (the Markovnikov product). iii is only obtained is there has been a carbocation rearrangement. Overall then, i > ii > iii.

Qu8: AB
A Diels-Alder reaction and we are looking at the diene component, which is the Nu in a normal Diels-Alder reaction. This means electron donating groups on the diene make it more electron rich = stronger, more reactive nucleophile. Plus the diene needs to be s-cis (or cisoid) for the reaction to occur.  In i the diene cannot be s-cis and will not react. ii can rotate to be s-cis and iii is already locked s-cis and also has an alkyl substituent (weakly electron donating).  So iii > ii > i

Qu9: ACE
In B the equation is wrong, it should be [a] = a / c l .  In D, the pKa of HCl = -7.

Qu10: B
Milk sugar is lactose. Benedicts solution gives a red precipitate with a reducing sugar. The precipitate is mainly casein. Proline gives a yellow spot with ninhydrin because it's a secondary amine.

Qu11: BC
A steam distillation is based on immiscible liquids while Raoults law describes the vapour pressure for miscible liquids. For D, the organic layer can either be top or bottom depending on the densities of the layers.  For E, IR gives the functional groups, nmr spectroscopy gives the hydrocarbon skeleton.

Qu12: ABCD
E Sodium sulphate is soluble in water (as are most Na+ salts), the precipitate was the carboxylic acid..

Qu13: A
B reflux apparatus is for heating liquids at their boiling points. C charcoal is for removing coloured impurities. D waste needs to go in the appropriate drum. E what No safety glasses ???

Qu14: D
Reaction is the oxymercuration-demercuration of an addition reaction of an alkene. This gives the Markovnikov alcohol due to the addition of the Hg species across the C=C giving a cyclic mercurinium ion. This is then opened by water to give the alcohol at the more highly substituted position (SN1 like) with the Hg at the less hindered end. The NaBH4 removes the Hg. Since no C+ forms, there is no carbocation rearrangement.  A is not an alcohol. B is an elimination product. C is only obtained is there has been a carbocation rearrangement and E is the anti-Markovnikov alcohol.

Qu15: C or D
H2SO4 causes an elimination of an alcohol to give an alkene that is then subjected to ozonolysis with a reductive work-up. Looking at the dicarbonyl product, we can work out the required alkene by "snapping" together the two C=O units to give the C=C. This gives us 1-methylcyclopentene which is C (draw carefully, count carbon atoms). The fact that one end of the dicarbonyl product is an aldehyde and the other is a ketone tells us that the two ends of the alkenes are different. Since C would not react with the acid to do anything, it is an acceptable answer. A would give the wrong symmetrical alkene, which would have given 1,6-hexanedial on ozonolysis. B and E would not eliminate under these conditions, alkyl halides need a base to undergo elimination reactions. D eliminates to give the desired alkene.

Qu16: C
Addition of a hypohalous acid to a C=C puts the -OH at the more highly substituted position and the -Cl trans to it... but treating this with base allows the formation of the epoxide. So C is the correct answer.

Qu17: C
Terminal alkynes are deprotonated with sodium amide, a strong base to give the acetyilide ion, a good Nu. This reacts SN2 fashion with the primary alkyl bromide to give Ph-CH₂-C#C-CH₂-Ph.  Alkynes undergo ozonolysis without the need for a special work up to give carboxylic acids.  Counting carbon atoms gives C as the correct product.

Qu18: D
Radical bromination at the more highly substituted position followed by E2 elimination to give the alkene, probably the anti-Zatseff, less highly substituted alkene due to the bulk of the base and then form the cyclopropane ring via the carbenoid reaction (with :CH2 as the reactive species). The position of the cyclopropane ring confirms the location of the C=C as being exocyclic. Only C and D have the right number of carbon atoms to be possible solutions. The alkene in C would probably give a diene after elimination....

Qu19: B
The catalytic hydrogenation has removed the C=C. But you should be able to recognise the two esters as being from the dienophile and therefore that the C=C should be on the opposite side of the product ring. So we are looking for the diene of the Diels-Alder reaction. If you draw the curly arrows for the retro-Diels-Alder reaction, (start at a C=C and push around the 6-membered ring) then the diene is a methylated 1,3-butadiene, using the curly arrows allows you to see where that methyl group needs to be on C2 of the diene.

Qu20: D
Now looking for the dienophile of the Diels-Alder reaction. If you draw the curly arrows for the retro-Diels-Alder reaction, (start at a C=C and push around the 6-membered ring), this allows you to see where that the dienophile has two ester groups. The only way they could be trans in the product is if the trans alkene system is used. Note that it can't be the alkyne because the esters would be cis in the product after the hydrogenation.

Qu21: A
Reaction of an epoxide under basic conditions, so the nucleophile is the alkoxide ion = a strong Nu.  Hence the reaction is SN2 in character and the Nu adds to the less highly substituted carbon of the epoxide (less hindered = SN2 like), but still from the opposite side to the O atom of the epoxide. So the OCH3 group needs to be on the opposite carbon of the epoxide to the CH3 group.  B and C are different conformations of the same molecule with the wrong stereochemistry at the -OH center (that stereochemistry is defined by the epoxide). These compounds would be formed if the enantiomer of the original epoxide were used.  D and E have the wrong regiochemistry.

Qu22: D
Br2 adds across C=C or C#C to add one Br atom at each end of the p bond. A would give a tetra-bromide under these conditions not a dibromide (where do the H come from ?).  B would put the two Br atoms on C1 and C2 not C2 and C3 (i.e. regiochemistry). E would not react under these conditions. So is it C or D. Need to pay attention to the stereochemistry. The required product is a meso system as shown by the Fischer diagram. Since the addition of  Br is an anti addition process, this requires that the original alkene is trans, i.e. D.  The cis alkene C would give the pair of enantiomers with the (R,R) and (S,S) configurations.

Qu23: D
Catalytic hydrogenation of an alkyne with Lindlars catalyst gives the cis alkene. Generation of a carbene (here loss of HCl gives :CHCl) which then adds to the C=C to give a cyclopropane in a syn fashion so the stereochemistry of the C=C is preserved.  A has the wrong carbene, B has the wrong original alkene stereochemistry, C has wrong carbene and wrong alkene stereochemistry. D and E have right carbene and right original alkene stereochemistry..... the reaction will occur preferentially to put the larger Cl atom away from the other two groups (sterics), so D is the right answer.

Qu24: C
Dissolving metal reduction of the alkyne gives the trans alkene then the peracid gives the epoxide with the same stereochemistry.  Treatment of the epoxide with aq. NaOH opens the epoxide to give the 1,2-diol HO-C-C-OH in an SN2 like fashion so the 2 -OH are trans, so in effect completing an anti addition of the diol system to the C=C in the two step process. Overall anti addition to a trans alkene means a meso product. A,B,D and E are not meso !

Qu25: B
Permanganate oxidation of an alkene gives the syn 1,2-diol. A is not a diol. C and D have the wrong stereochemistry. E has the wrong regiochemistry, the -OH groups need to be where the C=C was !

Qu26: A
Probably the toughest of the bunch.  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. Only A and B have the C=C in the right location, so it's down to the stereochemistry in these two. Careful drawing of the syn addition process confirms that A is the correct stereoisomer.

Qu27: AD
Reaction is the hydroboration-oxidation of an alkene. This gives the anti-Markovnikov alcohol  so we are looking for Ph-CH₂CH₂OH. The two connected CH₂ units will give two triplets as in AD.

Qu28: AB
Reaction is ozonolysis with an oxidative work up. This will give the carboxylic acid and CO₂ in this case. The acid is characterised by a singlet at high chemical shift around 12 ppm as in AB. Note that 10 ppm as in AC is a little low for an acid and is more likely an aldehyde.

Qu29: A
Reaction is the catalytic hydrogenation of the C=C to the alkyl group, so we are looking for Ph-CH₂CH₃.  The ethyl group will have a characteristic quartet and triplet with integrals in the 2:3 ratio as in A.  Note that the chemical shift of the quartet in B is too high for it just to be attached to the aromatic ring (more like it is attached to an O atom).

Qu30: D
Reaction is hydration of the C=C to give the Markovnikov alcohol, so we are looking for Ph-CH(OH)CH₃.  This will be characterised by the 1H quartet and a 3H doublet plus OH singlet.

Qu31: C
Reaction is the catalytic hydrogenation of the C#C to the C=C, so we are looking for Ph-CH=CH₂.  The vinyl group will have a characteristic complex coupling patterns in the region of 5-7 ppm as in C.

Qu32: E
Reaction is the mercury assisted hydration of a terminal alkyne to give a methyl ketone i.e. Ph-C(=O)CH₃.  This will be characterised by a 3H singlet at about 2.2 ppm. 

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