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
Carbocation stability is dictated by the substituents, i is the very poor phenyl cation between primary and methyl in stability (no resonance stabilisation of the +ve charge, the empty orbital is perpendicular to the pi system), ii is a simple secondary system and iii is dibenzylic, so lots of resonance to spread out and stabilise the charge. Therefore we get iii > ii > i.

Qu2: C
Resonance energy measures the extra stability of conjugated systems compared to the same number of isolated C=C. You don't need to know the numbers to work out the answer, they are here for reference.  i is just two separate butadiene units (care ! it is not the aromatic compound naphthalene), each one about 4 kcal/mol so a total of 8 kcal/mol. ii is benzene itself an aromatic compound with significant resonance energy (cyclic conjugated system) of 36 kcal/mol. iii is a single butadiene system. So ii > i > iii.

Qu3: C
Like carbocations, radicals are stabilised by alkyl substituents, but more importantly for this question by resonance. i is a primary system with 2 other resonance contributors, ii is tertiary with 2 other resonance contributors and iii is a secondary system (not conjugated = no resonance here). So ii > i > iii.

Qu4: B
The reactions are electrophilic additions, so we need to look at the intermediates that are produced, the carbocations. in accordance with Markovnikov's rule (H+ adds the the end with most H already present).  i will give a tertiary carbocation, ii a vinyl cation (worse than primary due to hybridisation being sp) and iii gives a secondary carbocation,  so i > iii > ii.

Qu5: D
Maximum number of stereoisomers is given by 2ⁿ where n = number of chiral centers.  i has one chiral center so it can be R or S, the double bond section has no stereoisomerism. ii has two chiral centers, so RR, RS, SR and SS are possible. iii also has two chiral centers but they each have the same set of four groups attached so we have the situation where a meso compound arises, so we only have RR, RS = SR and SS (3) permutations. Overall then we have ii > iii > i.

Qu6: C
Catalytic hydrogenation is controlled by the nucleophilicity of the pi system. Alkynes ii are more electron rich therefore more nucleophilic than alkenes i. Benzene pi systems are delocalised and this makes them less nucleophilic. So ii > i > iii.

Qu7: D
Could be regard as a question about the radicals that result (remember that homolytic means one electron goes each way) or you could look at the bond strengths based on the hybridisation of the atoms involved in the bond.... all the bonds are C-H. The radicals method is probably easiest (if you know radical stabilities). i is allylic so is resonance stabilised (very stable, weaker bond)  ii is vinyl (not very stable / strong bond) and iii is secondary (moderate stability / typical sp3 C-H bond). Overall then, ii > iii > i.

Qu8: AB
A Diels-Alder reaction and we are looking at the dienophile component, which is the E+ is a normal Diels-Alder reaction. This means electron withdrawing groups on the alkene make is more electron poor = stronger, more reactive electrophile. In i the -OMe group is a electron donor by virtue of the lone pairs on the O, ii has a carbonyl group next to the alkene and so is electron withdrawing and iii has two carbonyl groups (even more electron withdrawing) so iii > ii > i

Qu9: A
Lactose is milk sugar.

Qu10: B
No the equation is wrong, it should be [a] = a / c l .

Qu11: A
The recycling part of the polymer experiment.

Qu12: A
Nylon was made by the TA in the polymer experiment and Kevlar was a write-up question.

Qu13: A
More C=C means it will react with more iodine = higher iodine number.

Qu14: A
Anhydrous salts are used to dry solvents / solutions.

Qu15: A
The amino acid proline stains yellow with ninhydrin (from the milk experiment)

Qu16: A
Yes this is true, from the detergent experiment, part of the write-up.

Qu17: B
Saponification refers to making soap where you hydrolyse esters with NaOH to prepare the carboxylate salt.

Qu18: A
Yes is the soap experiment you hydrolysed the ester bond, creating to prepare the carboxylic acid (in its salt form) and the alcohol, glycerol.

Qu19: B
Cyclopentadiene, B, is non-aromatic because there is not a cyclic,  conjugated, p system. The two C=C give 4p electrons.  AD and BC are both 6p electron systems due to the inclusion of lone pairs on S or N into the give p system.

Qu20: AE
If n=1, then we need a 6p system..... if it is ionic we need AE, CD or CE... CD is 2p electrons and CE is 8p electrons.

Qu21: B
Cyclopentadiene, B, is a very common diene in the Diels-Alder reaction. When deprotonated it gives a 6p electron aromatic system.

Qu22: AC (or AD)
Aromatic as drawn means we are considering AB, AC, AD, AE, BC, and CD.  It also needs to be basic, so look for lone pairs that are NOT part of the p system this could be AC (all the N lone pairs are in sp2 type orbitals and are not part of the aromatic system) or maybe AD (one S lone pair is in an sp2 hybrid type orbital and is not part of the aromatic system).

Qu23: BC
Aromatic as drawn means we are considering AB, AC, AD, AE, BC, and CD. Now we are looking for a basic system that becomes non-aromatic, so look for lone pairs that are part of the p system that will make the protonated form non-aromatic. The lone pair on N in BC is part of the 6p electron system but involving the electrons in an N-H bond removes them from the 6p system.

Qu24: E
Which compounds are non-aromatic as drawn ? A, B, C, D, and E.  The example of tautomerism we met was enol / ketone (during addition of water to alkynes). If we look at E and convert the H-C-C=O to C=C-OH we get the enol "phenol", which is aromatic.

Qu25: A or C.
Which compounds are non-aromatic as drawn ? A, B, C, D, and E. Resonance in A (make the C=O become +C-O-) gives a 6p electron system, and in C (make the C=C become -C-C+) gives a  6p electron system in the 5 membered ring and a 2p electron system in the 3 membered ring, both of which are aromatic.

Qu26: AC
Isoelectronic means the same number of electrons as benzene. Look for the same number of non-H atoms as a quick start.

Qu27: C
The alkene starting material will undergo hydration to give the Markovnikov alcohol via the carbocation.... but that will rapidly rearrange to before the water adds, giving C.  (D would result if the carbocation didn't rearrange).

Qu28: E
KOH/heat looks like elimination of an alkyl bromide to make a C=C which then gets ozonolysis with an oxidative work-up. B would eliminate to give cyclohexene then O3 / H2O2 gives hexanedioic acid. To get the ketone/acid, need E. Try redrawing the product and "joining" the two C=O groups back together to see where the C=C was.

Qu29: B
Bromination of a C=C followed by a very strong base) suggests elimination.  This method typically gives alkynes (count C to get the right one).

Qu30: D
Deprotonate the terminal alkyne with the strong base, then add one more C atom by alkylating with methyl iodide (an SN2 reaction), finally an anti reduction to give the trans-alkene, D.

Qu31: B
Hydroboration / oxidation gives an anti-Markovnikov alcohol which when heated is giving a cyclic ether.... sort of a Williamson type reaction... need to make a 5 membered ring, so there needs to be a 4 C chain, with a C=C and a good leaving group. Has to be B. C doesn't work because the hydroboration reaction will add the OH at the wrong carbon.

Qu32: D
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 cyclohexene, using the curly arrows allows you to see where that methyl group needs to be.

Qu33: C
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) then the diene is a methylated cyclohexene, using the curly arrows allows you to see where that the dienophile is an alkyne system with two ester groups, C.

Qu34: D
Reaction of an epoxide under acidic conditions : protonation of the epoxide O gives cation character so the Nu (here the alcohol) adds to the more highly substituted C of the epoxide (has more cation character = SN1 like), but still form the opposite side to the O atom of the epoxide. So the OCH3 group needs to be on the same C and the CH3 group and inverted compared to the original epoxide. A / B / E  have the wrong regiochemistry, C the wrong stereochemistry.

Qu35: B
Product will be a 1,2-bromohydrin : Br-C-C-OH. Bromine adds to form the cyclic bromonium ion which is than attacked at the more substituted C atom by the water as a Nu and displacing the -Br from that C (compare with the above opening of the protonated epoxide). This gives the -OH on the same C as the CH3 group and anti to the -Br. A and D have the wrong stereochemistry (-OH and Br are cis), C has the wrong product, E has the wrong regiochemistry.

Qu36: CD
Dissolving metal reduction of the alkyne gives the trans alkene then the peracid gives the epoxide (a syn addition) so look for the trans epoxides. C and D are enantiomers, since alkyne and alkene are achiral, product will be racemic, same amount of C and D. A has the wrong substitution pattern, B the wrong stereochemistry based on the reduction and E is the wrong product.

Qu37: D
Catalytic hydrogenation of the alkyne gives the cis alkene then the permanganate gives a 1,2-diol HO-C-C-OH (via a syn addition). Syn addition to the cis alkene means a meso product. A,B,C and E are not meso !

Qu38: BD
Hydroboration / oxidation gives the anti-Markovnikov alcohol (regiochemistry : -OH not at the same C as the -CH3 group) and overall syn addition of the -H and -OH (stereochemistry : -OH and -CH3 need to be trans). A has the wrong regiochemistry, C and E have the wrong stereochemistry. B and D are enantiomers and both are produced in equal amounts.

Qu39: C
Hydration of the terminal alkyne will give the enol then tautomerise rapidly to the methyl ketone C. A and B are at the wrong oxidation state (from hydration of C=C, D has the wrong regiochemistry and E is the enol intermediate.

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