Here is an post-mortem analysis / "how to" for the FINAL. 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
An application of acidity and basicity. First identify the acidic H, probably draw out the conjugate base, A-. The functional groups are i a carboxylic acid, ii a ketone and iii an amine. Carboxylic acids are the most acidic since the carboxylate ion puts the negative charge on an electronegative O where there is also resonance delocalisation. In the ketone, the enolate has the negative charge on a carbon but there is resonance delocalisation that puts the negative charge on an electronegative O. In the amine, the negative charge is localised on the N atom. Or we could use the pKas...carboxylic acid = 5, ketone = 20 and amine = 35. The lower the pKa the stronger the acid and the weaker the conjugate base. So we have in terms of acidity i > ii > iii.

Qu2: B
Since radical substitution using chlorine will give product from each type of hydrogen, this is a question about counting H types. First draw the structure for each name. Then count H types.  i has 4 types, ii has 1 type and iii has 3 types. Hence  i > iii > ii.

Qu3: E
Really a question about alkene stability. Notice they are isomeric.  The more stable the alkene the less exothermic it's heat of combustion.  In terms of alkene stability, the general rule is the more alkyl groups on the C=C unit, the more stable it is. i has 2 alkyl groups, ii has 3 alkyl groups and iii has 1, so ii is the most stable then i then iii is the least stable. So iii has the most exothermic heat of combustion.  Hence, in terms of the heats of combustion (most exothermic to least) iii > i > ii.

Qu4: D
First identify the reaction.... the conditions of NaI / acetone suggest SN2 (think back to the laboratory expt). A quick look at the systems shows three alkyl bromides, so we are looking at the effect of changing the alkyl group (since the leaving group is the same across the series). i is tertiary, ii is primary and allylic and iii is secondary. Since SN2 reactions are slower as the steric hindrance of the alkyl group increases, we have ii > iii > i.

Qu5: D
Basicity...either think about the availability of the electrons in the base or the stability of the bases. The stronger the bases, then the more the reaction shown moves to the right. These are all oxygen systems, water, an alkoxide and a carboxylate respectively. In the carboxylate iii, the -ve charge is delocalised to a second electronegative oxygen atom whereas in the alkoxide ii, the negative charge is localised on the O atom. In water, the system is neutral rather than negative, so it will be a weaker base. Therefore in terms of base strength and hence the amount of reaction:   ii > iii > i

Qu6: AB
An application of acidity. The functional groups are i a terminal alkyne, ii an alcohol and iii a carboxylic acid. Carboxylic acids are the most acidic since the carboxylate ion outs the negative charge on an electronegative O where there is resonance delocalisation. In an alcohol, the negative charge is localised on the O atom. In an alkyne, the negative charge is localised on the sp C atom (remembering what bases are used to deprotonate alkynes is also useful here - alkoxides don't work). Or we could use the pKas... alcohol =15, alkyne = 25 and carboxylic acid = 5: the lower the pKa the stronger the acid and the weaker the conjugate base. So we have in terms of acidity iii > ii > i.

Qu7: B
Radical stability: More alkyl substitutents implies a more stable radical but resonance can also stabilise radicals. i is a secondary allylic radical, ii is a secondary radical and iii is a primary allylic radical so i > iii > ii.

Qu8: D
Chemical shifts of the C in question in these systems are determined by the hybridisation and the proximity of the electronegative O. All are sp2 C atoms. i is an aromatic C with a simple methyl group attached, and so it has a shift of about 140ppm. ii is a ketone carbonyl C and is therefore next to an O and is more deshielded by the electronegativity of the O than the aromatic C in i. Ketone C=O are usually > 190ppm. iii is an ester carbonyl C however, the resonance donation of the alkoxy group in the ester shields the carbonyl compared to a ketone, so ester C=O are usually < 190ppm. Therefore ii > iii > i.

Qu9: A
The reaction is the E2 elimination of an alkyl halide with a strong base. The key here is to count up how many different 1,2-H atoms are present to give different alkene products and don't forget to count for cis and trans alkenes. i can give 3 alkenes, ii can give 2 and iii can only give1 so i > ii > iii.

Qu10: C
Conformational analysis so look for different types of interactions. i has an eclipsed conformation on the RHS as drawn (so 3 eclipsed interactions). ii has an eclipsed conformation on both the LHS and the RHS as drawn (for a total of 6 eclipsed interactions) and therefore is higher energy than i. iii is in a more favourable staggered conformation. Therefore the highest energy conformation is ii and the lowest (most stable) is iii, so ii > i > iii.

Qu11: A
Leaving group ability.... i is a tosylate, an excellent leaving group, ii chloride is a fair leaving group and iii has OH, a very poor leaving group. So i > ii > iii.

Qu12: B
i is a C-I bond involving an aromatic C atom. Since I is a large atom, this bond going to be long. The other two are CC bonds and will therefore be shorter. ii is a CC bond is a benzene ring so the resonance in the aromatic system means that it is half way between a C-C and C=C. iii is a C=C between 2 sp2 C atoms. So overall, in terms of lengths, i > iii > ii.

Qu 13: E
An application of acidity. Look at the functional groups (alcohols, alkyl halides, aromatics, ketones and an ammonium salt), think about pKas etc. The ammonium group in E is positive which makes it easier to loose a proton (pKa about 10).

Qu14: AD
Alcohols are eliminated using acid and heat in an E1 type of reaction. Since the E1 goes via a carbocation, the reaction with the most stable carbocation will react fastest... so look at the carbocations that would be formed. The most favourable carbocation is the one from AD, a tertiary and benzylic system.

Qu15: C
The conditions of NaI / acetone suggest SN2 of an alkyl halide (think back to the laboratory expt) with less steric hindrance being the controlling issue. So the primary benzylic reacts fastest. Alcohols do not react under these conditions (they lack a good leaving group) and aromatic halides don't undergo SN2 reactions.

Qu16: B
The reaction is the E2 elimination of an alkyl halide with a strong base. Of the alkyl halides to select from, only B has a 1,2-H that can eliminate.

Qu17: B
The aq. ethanolic silver nitrate conditions are for SN1 of alkyl halides. Since the SN1 goes via a carbocation, the reaction with the most stable carbocation will react fastest... so look at the carbocations that would be formed. The most favourable carbocation is the one from B, a secondary and benzylic system. Alcohols do not react under these conditions (they lack a good leaving group).

Qu18: D
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. Here there are 4 rings and 10 pi bonds to give a total of 14.

Qu19: A
Each type of carbon will give a unique signal. There is symmetry in the system in (a) the para substituted aromatic ring so 4 aromatic C and (b) the two equivalent ethyl groups in the ammonium salt.

Qu20: B
The lowest chemical shift means the least deshielded i.e. towards the zero end of the spectral scale. Deshielding is caused by electronegative atoms and proximity to C=C. C is less electronegative than N. Hence the answer is neither A or D. C3 is near deshielding groups and C9 is part of an aromatic ring.

Qu21: D
The lone pairs on O5 will be such that one is in a p orbital to allow for the resonance interaction with the double bond and the other lone pairs will be in sp2 hybrid orbitals.

Qu22: B
C3 is a chirality center. The groups in priority order are : N₂ > CS(CH₃)₂ > CO₂- > H .... assign the sense of direction, but make sure you evaluate that for when the lowest priority group is away from you.

Qu23: C
We should work forwards .... Alkylation of a terminal alkyne. The NaNH₂ removes the acidic terminal H to create an acetylide that reacts as a C nucleophile which is then reacted in an SN2 reaction with an alkyl halide, benzyl bromide, C₆H₅CH₂Br. A and B have too few C atoms, A has lost the methyl from the alkyne and B the methylene from the benzyl system. D and E have the wrong functional group (the Br has been substituted).

Qu24: D or E
We should work forwards ....looking at the product, it's an alkene - but it's the anti-Zaitsev product. Since the starting material is an alcohol, a simple dehydration with acid and heat is not going to work, there would be a rearrangement. Instead, convert the -OH to a halogen and then do an E2 elimination using a large base to favour the anti-Zaitsev product. A would cause a rearrangement. B and C would not cause an elimination reaction of the alcohol (poor LG). D and E would both work since the only H that can be removed are in the adjacent methyl group.

Qu25: B
Need to work backwards ....looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - looking at the structure of the ether and the reaction step 2, we should be able to see that we need bromocyclopentane from step 1 which can be synthesised via a radical halogenation of cyclopentane, B.

Qu 26: E
Need to work backwards ....looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - looking at the structure of the ether and the reagent (Na = base), we should be able to see that we need an difunctional system with an -OH and a halide, with 4 C in a chain.

Qu 27: D
We should work forwards ....The first step is a radical halogenation of methylcyclohexane to give 1-bromo-1-methylcyclohexane. This is followed by an E2 elimination using a small base to favour the Zaitsev product.

Qu 28: C
Need to work backwards ....looking at the product, it's an alkene - but it's the anti-Zaitsev product. Since we are looking at NaOH / heat, we are looking at an E2 elimination. In cyclic systems, the critical issue is the 180 degree arrangement of the H-C and C-LG bonds - this requires that the H and the LG in question are both axial. In this question then, the best approach is to draw the cyclohexane in the chair conformation and put the t-butyl group equatorial (large groups prefer to be equatorial). Then look at the position of the Br in relation to the t-butyl group. A and B have the Br 1,3- and cis, so the Br would be equatorial. D has the Br 1,2- and trans, so the Br would be equatorial. E has the Br 1,3- and trans, so the Br would be axial but would give the wrong configuration at the chirality center (where the t-butyl group is attached).

Qu 29: E
We should work forwards ....looking at the product, it's an alkene and an alkyl halide. The only reaction of an alkane that is going to be useful to start the process is a radical halogenation to give an alkyl halide. This narrows it down to C or E! Radical bromination will give 2-bromobutane as the major product. We then need to eliminate the alkyl halide (so base / heat) to give the anti-Zaitsev alkene (less highly substituted = use a bulky base) to give the alkene and then do a allylic radical substitution to give the required product.

Qu30: A
We should work forwards ....looking at the product, it's an alkene. This means we are looking at carrying out some sort of an elimination reaction, but if we look at the location of the C=C and the original -OH, we can see that a rearrangement is going to be needed. This means that B, D and E are incorrect as these involve SN2 and/or E2 reactions as hence no carbocation and therefore no rearrangement. C does not give us the right functional group. So reviewing A, the HBr will form the alkyl bromide via an SN1 (rearranges to more stable tertiary carbocation) followed by a Zaitsev elimination.

Qu31: A
We should work forwards .... halogentaion of an alcohol using thionyl chloride followed an SN1 using silver nitrate and ethanol (think about the lab expt). In the SN1, the nucleophile will be the O of the ethanol so the product is a benzylic ether.

Qu32: A
The torsional angle between the chlorine atoms is 0 degrees for the syn eclipsed conformation.

Qu33: D
Since the C atom in question is sp3 hybridised, the bond angle will be approx. 109.5 degrees.

Qu34: D
A is not 1,3- it's 1,4-. B, C and E are cis not trans.

Qu35: C
The two C-C bonds are at 180 degrees to each other, so they are best described as "anti"

Qu36: C
The structures are isomeric, but what type of isomer are they ? If you look carefully at the Newman projections, the rear C has been twisted by 120 degrees while the front C has not moved and has the same configuration. This means we are looking at conformational isomers.

Qu37: A
Look for the staggered conformation with maximum separation of the methyl groups. B and D are eclipsed. C and E have two gauche methyl interactions while A has only one.

Qu38: D
The highest energy conformation of chlorocyclohexane will be one of the boat conformations (i.e. C or D). The 1,4-flagpole interaction of substituents in D makes D less stable than C.

Qu39: E
Lowest energy = most stable.... so look for equatorial substituents with a little bit of separation.

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