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: D
Carbocation stability....for simple alkyl cations, more alkyl groups means a  more stable carbocation but here we also need to consider the effects of resonance. i is the relatively unstable phenyl cation. ii and iii are both secondary but ii has resonance stabilisation... so ii > iii > i.

Qu2: E
All about the nucleophilicity of these groups. These are all oxygen systems and all are negative. The possibility of resonance delocalisation in i and ii makes the electrons less available and the O is less nucleophilic. However there is a difference in the resonance. In the phenol i the charge can only be delocalised to C, in the carboxylate ii it can be delocalised to O which is more electronegative. Therefore the phenolate i is more nucleophilic than ii. Now compare the O in i and iii respectively.  There is no resonance in iii, this means the -ve charge is fixed on the single oxygen. Hence i is less nucleophilic and overall in terms of nucleophilicity, iii > i > ii.

Qu3: C
An application of acidity and basicity. All the species are neutral, and all are in the same row. First we need to consider electron availability.... O and N both have lone pairs but C only has bonded pairs. Therefore the C is the weakest base. In the same row, basicity increases as an atom gets less electronegative, so N is more basic than O. So we have in terms of basicity that ii > i > iii.

Qu4: AB
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 systems are from the same row of the periodic table, water, an alkoxide and amide ion. respectively. In the alkoxide ii, the negative charge is on the O atom, in iii the negative charge is on the less electronegative N atom which means the electrons are more available. 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:   iii > ii > i

Qu5:E
Another application of acidity and basicity. This time acidity.... The functional groups are i an alcohol, ii an amine and iii a carboxylic acid. Carboxylic acids are the most acidic since the carboxylate ion puts the negative charge on an electronegative O where there is resonance delocalisation to a second electronegative O. In the alcohol, the negative charge is on an electronegative O while for the amine, the negative charge would be on a less electronegative N and hence is less favourable. Or we could use the pKas... alcohol =15, amine = 35 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 > i > ii.

Qu6: A
First identify the reaction.... the conditions of AgNO3 / aq, ethanol acetone suggest SN1 (think back to the laboratory expt). A quick look at the systems shows bromides, so we are looking at the effect of changing the alkyl (R) group (since the leaving group is the same across the series), first draw the structures from the names. i is tertiary, ii is secondary and iii is phenyl (aromatic). Since SN1 reactions are slower as carbocation gets less stable, we have i > ii > iii.

Qu7: E
Chemical shifts of the H in structures are determined by the hybridisation and the proximity of the electronegative atoms and the benzene ring. All are sp3 C systems. i is an CH2 between an aromatic and an electronegative O atom, predict a shift about 4.5 ppm. ii is a methyl group next to an O (only) and is only deshielded by the electronegativity of the O to about 3.5 ppm. iii is a CH2 between an electronegative O and Cl and will be the most deshielded, about 5.5 ppm. Therefore iii > i > ii.

Qu8: B
i is a C-Br bond involving an sp3 C atom. Since Br 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 in 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 sp3 C atoms. So overall, in terms of lengths, i > iii > ii.

Qu9: 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.

Qu10: C
First draw the starting material out and then identify the H leading to the product as primary, secondary or tertiary. Note that iii is impossible! ii is secondary and i is primary, so in terms of yields, because Br is much more selective for the more stable radical ( where primary < secondary < tertiary), then ii > i > iii.

Qu11: D
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 just one alkene, pent-1-ene, ii can give three, pent-1-ene and cis- and trans-pent-2-ene, and iii can only give3-methylbut-1-ene and 2-methylbut-2-ene, so ii > iii > i.

Qu12: AB
Really a question about alkene stability. Notice they are isomeric.  The more stable the alkene the more exothermic it's heat of formation.  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 1 alkyl groups, ii has 2 alkyl groups and iii has 3, so iii is the most stable then ii then i is the least stable. So iii has the most exothermic heat of formation.  Hence, in terms of the heats of formation (most exothermic to least) iii > ii >i.

LABORATORY:
Need to know the basis of your experiments and / or be able to relate it the concepts in the course as a whole.

Qu 13: BE
Since the spot due to "a" is higher than that for "b", "a" has eluted further than "b". Rf is the ratio of distance traveled by sample to distance traveled by solvent from the origin. Since "a" is above half way on the plate, it's Rf > 0.5, and since it is below the solvent front, it's Rf <1.

Qu14: B

Qu15: B
Since methoxybenzene is not acidic or basic and lacks polar groups, it will be soluble in the organic solvent diethyl ether.

Qu16: DE
The aq. ethanolic silver nitrate conditions are for SN1 of alkyl halides. A is false because ethanol is a polar solvent. B is false since the ethanol would be a nucleophile. C is false since the precipitate is the silver halide.

Qu17: C
Boiling points are lower at altitude (lower pressure) so A is false. Silica gel is the stationary phase in TLC so B is false. Charcoal is used to remove coloured impurities so D is false. Separating funnels are used to immiscible liquids so E is false.

Qu18: D
An application of  acidity and basicity.  It's a question about relative acidity or pKa. The most acidic of the organic functional groups are alcohols, D, pKa = 15.

Qu19: AB or D
Alcohols are eliminated using acid and heat in an E1 type of reaction giving pent-1-ene. Amines can be eliminated in a similar manor. Since the alkyl chains are the same, the same alkenes are possible.

Qu20: C
The conditions of NaCN / DMSO suggest SN2 of an alkyl halide ( with less steric hindrance being the controlling issue. So the primary allylic reacts fastest. Alcohols and amines do not react under these conditions (they lack a good leaving group) and vinyl halides don't undergo SN2 reactions.

Qu21: AC
Constitutional isomers are where there is a different arrangements of the groups. Isomers require the same molecular formula.

Qu22: D
The index of hydrogen deficiency (IHD) is a count of the number of pi bonds and rings. Here there is 1 ring and 3 pi bonds (two C=O in the amide and ester groups) and one C=C) to give a total of 4.

Qu23: B
An application of nmr spectroscopy.  The most downfield signal is due to the most deshielded carbon with have the highest chemical shift... we should expect this to be a carbonyl (due to the double bond and the electronegative O atom).

Qu24: E
Each of the carbons is a different type.

Qu25:A
O14 is part of a simple ether so the O is sp3 (4 attached groups, 2C and 2 lone pairs).

Qu 26: E
C12 is connected to N, O x 2 and C... adding up the attached groups e count gives -3 (due to the attachment to the three more electronegative atoms) and hence the oxidation state is +3.

Qu 27: C
An application of ir spectroscopy.  The highest frequency will be the N-H rather than the C-H.

Qu28: B
We should work forwards .... reaction of the alcohol with tosyl chloride will form the tosylate which is then undergoing SN2 with HO- to give the alcohol with inversion of stereochemistry.

Qu29: D
We should work forwards ....looking at the starting materials, it's an alkyl bromide and is reacting with a strong base / heat and undergoing an E2 elimination using a small base to favour the Zaitsev product. Draw the alkyl bromide with the H and Br bonds at 180 degrees to see the formation of the product with the methyl groups trans.

Qu30: C
Need to work backwards ........looking at the product, it's an ether and the reaction conditions look like a Williamson ether synthesis, (nucleophilic substitution) so we need an alcohol and a halide so we need an alcohol. It will need to be the alcohol with the same stereochemistry as the ether (since that C-O is not changed during the reaction). A is the enantiomer of the required alcohol. B has the wrong stereochemistry at the -OH. D and E don't have the right functional groups.

Qu31: A
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.

Qu32: D
Need to work backwards ........looking at the product and the last set of reagents it looks like an allylic radical halogenation. The first step looks like alcohol dehydration, an E1 elimination.

Qu33: C
We should work forwards ....looking at the product, it's an anti-Zaitsev alkene and we started from an alkyl halide. We then need E2 elimination of the alkyl halide (so base / heat) to give the anti-Zaitsev alkene (less highly substituted = use a bulky base) to give the alkene

Qu34: E
We should work forwards .... looking at the starting material it's an alcohol, we need to convert this to an alkyl chloride without rearrangement. This means we should use an SN2 reaction. Use thionyl chloride to give the alkyl chloride. A would give a tertiary chloride since the SN1 path would cause a rearrangement.

Qu35: A
We should work forwards .... looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - we have the halide, we need the alcohol, ethanol.... the bromide is tertiary, so SN1 will work, no rearrangement to worry about. C would likely result in an elimination and so would E because we have a strong base reacting with a tertiary bromide.

Qu36: C
We should work forwards .... looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - we have the alcohol which we need to react with a base and then ethyl halide (need a leaving group).

SPECTROSCOPY:
Use any IR information to get the functional groups. Use the H-NMR to get the number of types of H, how many of each type from the integral and what they are next to from the coupling patterns. Chemical shifts should tell you if the group is near -O- or maybe C=O groups etc.

Qu37: BD
IR shows C-O but no C=C or C=O or OH. The 13C-nmr peaks are too low for ArC or C=C. This rules out all except A, BC and BD. The H-nmr only has two H types, singlets (no coupling = no neighbours), ratio 1:1, the second deshielded by O. The H nmr details rule out A and BC.

Qu38: A
IIR shows C-O but no C=C or C=O or OH. The 13C-nmr peaks are too low for ArC or C=C. This rules out all except A, BC and BD. The H-nmr only has two H types, singlets (no coupling = no neighbours), ratio 3:2 both deshielded by O. The H nmr details rule out BC and BD.

Qu39: B
IR shows C=O at 1715 cm^-1. The 13C-nmr peak at 207ppm implies an aldehyde or ketone. The H-nmr only has two H types, singlets, ratio 3:2 (note that there is no CH2 coupling since the neighbours are of the same type).

Qu40: BC
IR shows C-O but no C=C or C=O or OH. The 13C-nmr peaks are too low for ArC or C=C. This rules out all except A, BC and BD. The H-nmr has three H types, ratio 3:6:1 and the coupling shows a CH3CH unit. ), the third type is deshielded by O. The H nmr details rule out A and BD.

Qu41: AB
IR shows an OH carbonyl at 3400 cm^-1 this is supported by the H nmr exchangeable peak at 1.8 ppm and implies we probably have an alcohol. So it looks like E or AB. The 13C-nmr peak at 64 ppm supports the C-O. The H-nmr has three H types, ratio 6:1:1 and the coupling shows a (CH3)2CH unit.), the third type is deshielded by O.

Qu42:BE
IR does shows C=C at 1606 cm^-1 and C-O near 1150 cm^-1 so this means D, AC, AD, or BE. The 13C-nmr peaks at 100-161 ppm suggest 4 types of ArC consistent with the meta isomer.

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