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: B
Allylic hydrogens are found on the sp3 C immediately adjacent to a C=C unit. i has 7 (CH3 and 2 x CH2), ii has only 1 (a CH) and iii has 3 (CH2 and a CH) hence i > iii > ii.

Qu2: A
Alkenes and alkynes react with aq. sulfuric acid to give carbocations that then react with water as the Nu. Alkenes give alcohols, alkynes give enols that tautomerise to ketones. The rate determining step is the addition of the H+ to the pi bond and is controlled by the stability of the resultant carbocation, so alkenes react faster than alkynes. Alkanes are weak bases (so they do not protonate easily - where would the electrons come from ?) so we have i > ii > iii

Qu3: C
If you know your pKa's then this will help,  remember that the lower the pKa the stronger the acid.   What if you don't remember your pKas ?  Then you'll need to deduce it.  Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-.... here we are looking at a hydrocarbon, i corresponds to a simple sp2 vinylic CH (pKa about 45) ii is an allylic position and iii is a simple sp3 alkane type CH (pKa > 60).  The key issue here is that the allylic anion would be resonance stabilised, hence ii  > i > iii 

Qu4: A
Resonance energies measure stability due to conjugation. If we look for conjugation we see that i is a conjugated triene, ii has a conjugated diene and iii has only isolated C=C (so it has zero resonance energy as there is no resonance !).  Therefore we get i > ii > iii.

Qu5: D
Reaction is the Diels-Alder reaction and we are looking at the dienes reacting with CH2=CHCO2CH3 ....  In order for the reaction to occur, the diene needs to be s-cis.  i  is locked s-trans (i.e. no free rotation), ii is locked s-cis so that's fine. iii can easily rotate around the sigma bond to become the required s-cis. So ii > iii > i.

Qu6: E
All the CC bonds are between sp2 C atoms, the thing that is changing is the type of bond between the C atoms involved in the bond. In i we have an aromatic bonds in benzene, so it's about half way between a C=C and a C-C (about 140 pm). In ii we have a standard 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 iii > i > ii.

Qu7: C
Epoxidation of an alkene.... more alkyl groups makes the alkene more nucleophilic because alkyl groups are weak electron donors. i is disubstituted, ii is trisubstituted and iii is monosubstituted so reactivity is ii  > i > iii 

Qu8: C
Reaction is the addition of HCl to an alkene. This gives the Markovnikov alkyl chloride via the most stable carbocation intermediate.  First draw phenylethene C6H5CH=CH2, then add the H+ to the least substituted end of the C=C to give the more stable conjugated and hence resonance stabilised carbocation, C6H5CH(+)-CH3.  So ii is the major product and i is a minor product (the anti-Markovnikov product). iii not formed... it destroys the aromaticity. Overall then ii  > i > iii 

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

Qu10: A

Qu11: A
An anomeric carbon is typically attached to two oxygen atoms by single bonds, here a OH group and a OC group

Qu12: B
Reactions are first order if the rate law has depends on the product of the concentrations of a single molecules.  If a reaction depends two molecules of A (i.e. rate = k [A] ² ) then that reaction is second order.

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

Qu14: A
The reaction of alcohols with HCl in the Lucas test is an SN1 type reaction and forms the alkyl chloride. These are less polar than alcohols and tend not to be soluble in the aqueous medium.

Qu15: A
Tertiary alcohols are not easily oxidised to carbonyl compounds.

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

Qu17: B
Nylon [6.6] is made from a diacid system with 6 carbons and a diamine containing 6 carbons.

Qu18: A
For example, the an ester synthesis : RCO2H + R'OH-->  RCOOR' + H2O.

Qu19: A
Yes - this causes the solvent to condense and run back in to the flask so that it does not boil dry.

Qu20: A
Do you know your functional groups ?

Qu21: A
Grignard reagents contain Mg.

Qu22: B
No the iodine actually reacts with the Mg. 

Qu23: A
This is typical of work-up... neutralise the reaction conditions to allow the easy isolation of a neutral organic molecule.

Qu24: A
Grignard reagents contain Mg.

Qu25: C
Working forwards....the first reaction is the addition of hydrogen bromide to a conjugated diene. At -80^oC, this will be under kinetic control and so the initial product will be the direct or 1,2-addition product and will have the Br derived from the more stable C+ location.  A has added the HBr 1,2- but in an anti-Markovnikov fashion,  B is the thermodynamic product from 1,4-addition.   D and E don't make any sense....they would require vinyl and primary carbocations be formed.

Qu26: D
Working backwards.... look at the product (note it's C7) and the reagents... ozonolysis with a oxidative work up... reconnect the two carbonyl groups to get the require alkene, ethylcyclopentene.  This would be formed by the dehydration of an alcohol.  The only alcohols are A, D and E. However A and E will not give an ethyl substituted system and E is only C6.

Qu27: D
Working forwards....the reaction is the oxymercuration - demercuration of an alkene.... gives an alcohol following Markovnikov's rule, no carbocation rearrangements.  Probably a good idea to draw out the alkene starting material in the same type of line diagram as the products...
The only alcohols are C, D and E.  C requires a rearrangement via a 1,2-alkyl shift of a carbocation intermediate and E is the anti-Markovnikov product.

Qu28: A
Working backwards.... look at the product : it's C7 and a methyl ketone, 2-heptanone, formed by the hydration of an alkyne.  The options A - E allow for heptyne to be made with the triple bond at C1 to C5 respectively.  The most efficient synthesis will be from 1-heptyne so that the methyl ketone is the major product since the alkyne would be terminal.  2-heptyne (from B will give a 50:50 mixture of the required product and 3-heptanone.

Qu29: B
The transformation is alkene to alkyne, to make pi bonds, we usually use eliminations... so we need to get some leaving groups in and then take them out.... if we remove the C=C we need to then make two new pi bonds... hence two leaving groups are needed..... two halogens is probably the best choice then need a strong base for the E2 eliminations of the halides.  A will give an alkyl bromide then an alkene, C will give an alkane via addition, D will give an alcohol then an alkene and E will not cause an elimination as there is no -OH leaving group.

Qu30: E
Working backwards.... look at the product...this is a Diels-Alder reaction.... hints are the cyclohexane units. The key is to recognise that the diol coming from an alkene (the cyclohexene unit) by the permanganate reaction. If you draw that in, then you can push the curly arrows to "reverse" the Diels-Alder and reveal the diene and the dienophile.

Qu31: D
See Qu 30.

Qu32: E
The starting material is an alkyne, the first set of reagents indicate a dissolving metal reduction of an alkyne to the trans-alkene. Alkenes react with aq. acid to undergo additions to alcohols in accord with Markovnikov's rule, based on the resonance stabilisation of the carbocation intermediate, E is the correct answer. A, B and C could only come from an alkyne, D has the wrong regiochemistry.

Qu33: C
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. A has the wrong regio- and stereochemistry,
B has the wrong regiochemistry, D has the wrong stereochemistry, and E has the wrong regiochemistry.

Qu34: B
Radical addition of two equivalents of HBr across an alkyne gives a 1,2-dibromide. Look at the mechanism and remember that (i) Br can stablise a radical (like it does a C+) and (ii) that it is the Br radical that adds first to the alkyne and then to the alkene in the second reaction.

Qu35: B
This is a Diels-Alder reaction.... hints are the dienophile, and the bridged structure of the product.   The challenge here is to realise that the C=C has been removed by the catalytic hydrogenation... the only way to recognise where it was it to recall that it is opposite in the cyclohexane unit to the location of the dienophile which is easily recognised by looking at the ester groups. With the C=C back in place, then you can push the curly arrows to "reverse" the Diels-Alder and reveal the diene.  A lacks the O atom, C has the methyl group in the wrong location, D lacks the methyl group altogether and E is not a diene.

Qu36: C
Osmium tetroxide reacts with alkenes to give 1,2-diols. The product is a geminal-diol so this must come from an alkene rather than an alkyne. The -OH groups are added across the C=C in a syn 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 -OH syn (best to draw the OH groups in the plane of the page). This means that the two alkyl groups (one methyl and one ethyl) need to be cis in the alkene as in C.  A is an alkyne.  B has the wrong stereochemistry. D and E have too few C atoms, an methyl instead of an ethyl group.

Qu37: E
The reaction is the hydroboration-oxidation of an alkene to give 3-methyl-2-pentanol. 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. Note that the -OH is formed with retention of stereochemistry when the B atom is replaced.  Redraw one of the enantiomers so the -OH and the H that added are syn then remove them to reveal the C=C unit.  A and B would give 2-ethyl-1-butanol.  C would give 3-methyl-1-pentanol. D has the wrong stereochemistry.

Qu38: C
Which systems are non-aromatic as drawn ?  A, C, E, AB, CD and DE. Of these A, C, E, AB and DE are non-aromatic because the pi system are not cyclic pi systems.  Only C is a 4 pi electron systems.

Qu39: A
In order to be non-conjugated, need either an alkene or an isolated polyene.

Qu40: E
Which systems are non-aromatic as drawn ?  A, C, E, AB, CD and DE.  To be a triene we need 3 x C=C as in E. CD and DE are not trienes, they are dieneones.

Qu41: E or DE
Which systems are non-aromatic as drawn ?  A, C, E, AB, CD and DE.  Tautomerism allows us to move H atoms typically from heteroatoms to C trading pi bonds (as in the ketone-enol tautomerism).  This applies for DE.  But it could also apply occur in E in a hydrocarbon variation.

Qu42: AC or AE
Which systems are aromatic as drawn ? D, AC, AD, AE, BC and CE. If n=2 in the Huckel rule we need 10 pi electrons. D, AD, BC and CE have 6 pi electrons in their aromatic systems.

Qu43: C
Look for a non-aromatic system with an aromatic conjugate base....Which systems are non-aromatic as drawn ?  A, C, E, AB, CD and DE.
Which are hydrocarbons ? A, C, E and AB. Which will have aromatic conjugate bases ? C and E. Both A will have anti-aromatic conjugate bases.  AB will have a nonaromatic conjugate base. 

Qu44: AD, BC or CE
Which systems are aromatic as drawn ? D, AC, AD, AE, BC and CE. If n=1 in the Huckel rule we need 6 pi electrons = D, AD, BC and CE.   Now consider adding H+ to make the conjugate acid.... the conjugate acids of D and CE are all still aromatic because the protonated lone pair is not part of the pi system.

Qu45: AB
Which systems are non-aromatic as drawn ?  A, C, E, AB, CD and DE.  Now consider adding H+ to make the conjugate acid.... the conjugate acids of AB would be aromatic because the C+ would form at the tertiary C in conjugation with the 3 C=C in the 7 membered ring, where n=1 in the Huckel rule.

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