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: E
All about the nucleophilicity of these groups. There are two oxygen systems and a sulfur system, all are single bonded and all are negative. Let's deal with the two oxygens first.  The possibility of resonance delocalisation in ii makes the electrons less available and therefore the phenolate ii is less nucleophilic than the alkoxide i. Now compare the O and the S in i and iii respectively.  Since S is larger than O (one row lower in the same group of the periodic table), the sulfur is more polarisable and therefore more nucleophilic than the O.  Hence  iii > i > ii.

Qu2: A
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 a tertiary cation as drawn with a secondary contributor. ii is a secondary cation also with a secondary contributor and iii is secondary, no resonance. So overall we have in terms of stability, i > ii > iii.

Qu3: C
IR stretching frequencies are predicted by Hooke's Law. Here the atoms involved are CX bonds so it's the strength of these bonds that is the critical factor. i is a sp3 C-Cl, whereas ii is a sp2 C-Cl bond. The greater the s character in the C hybrid, the shorter and the stronger the bond, therefore the sp2 C-Cl bond is stronger and therefore higher frequency. In iii it is a sp3 C-Br bond - compare this to the sp3 C-Cl. Br is heavier than Cl (therefore it impacts the reduced mass in Hooke's law) and hence the C-Br bond is lower frequency than the C-Cl bond. Therefore IR stretching frequency, ii > i > iii.

Qu4: B
The reaction here is SN2 (note the strong nucleophile and the aprotic solvent) so it will depend on the degree of steric hindrance at the C-LG center and the nature of the leaving group. i is primary and has a very good leaving group, ii is secondary with a poor leaving group under these conditions and iii secondary with a good leaving group so i > iii > ii.

Qu5: A
Each type of carbon will give a single peak so it's counting types of carbon.  i has 5 types since the two methyl groups are equivalent. Ii has 4 types due to symmetry through the C=O unit across the middle of the ring and iii has 2 types since the molecule can be divided in half in two perpendicular directions.  Hence  i > ii > iii.

Qu6: B
Alkene stability and conformational analysis. Recognise these from the midterm ? First, the 5C atoms in cyclopentenes are planar (cyclopentanes have an envelope conformation but the C=C in the ring prevents that). First, we can use the fact that a more highly substituted alkene is more stable, therefore, since i is trisubstituted, it is more stable than either ii or iii which are both disubstituted. Ii is less stable then iii because there are more eclipsing interactions due to the two adjacent sp3 C. Hence in terms of stability  i > iii > ii.

Qu7: A
H-nmr chemical shifts of the H in question in this system are influenced by the aromatic system and / or the electronegative O. i is an aromatic H and so will have a shift of 7-8ppm due to the deshielding caused by the magnetic anisotropy. iii is on an sp3 C and is next to an O - it is deshielded by the electronegativity of the O to about 3-4ppm. Ii is on an sp3 C and is next to both an O and the aromatic ring, so it's deshielded by both and it's shift is between 5 and 6 ppm. Therefore i > ii > iii.

Qu8: B
Elimination of alkyl halides is via the E2 pathway, (KOH = strong base plus heat) where the H is always removed from the C adjacent to the C-X bond. The rates of elimination reactions (both E1 and E2) are such that tertiary > secondary > primary due to the opening of the bond angles that occurs on going from sp3 to sp2. i is tertiary, ii is primary and iii is secondary, so overall we have i > iii > ii.

Qu9: C
Elimination of alcohols is via the E1 pathway, (H₂SO₄ = strong acid plus heat) where the H is always removed from the C adjacent to the C-OH bond. The rates of elimination reactions (both E1 and E2) are such that tertiary > secondary > primary due to the opening of the bond angles that occurs on going from sp3 to sp2. i is secondary, ii is tertiary and iii is primary, so overall we have ii > i > iii.

Qu10: E
An application of  acidity and basicity.  First recognise that we are looking at the reaction of sodium methoxide (NaOCH3) which is a base with a series of acids. The strongest acid will be the one that forms the most conjugate base. So either think about their relative acidities or use the pKas (if you know them.... if not why not?). i is a carboxylic acid (typical pKa = 5),  ii is a thiol (typical pKa = 10) and iii is hydrogen fluoride (pKa = 3). The lower the pKa the stronger the acid.....If you don't know the pKas then you need to employ the acidity factors (atom, electronegativity, resonance, size etc). Overall iii > i > ii

Qu11: D
The number of lines in H-NMR coupling patterns is given by the n+1 rule when n = number of equivalnet neighbours. In i the only neighbouring H are of the same type, therefore no coupling is observed = singlet (1 line). In ii the H in question has 8 neighbours (of approx. the same type) and therefore appears as 9 lines while in iii the H has two neighbours and therefore appears as a triplet (3 lines). Hence ii > iii > i.

Qu12: D
Really 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 4 alkyl groups and iii has 3, so i is the least stable then iii then ii is the most stable. So ii has the least exothermic heat of combustion.  Hence, in terms of the heats of combustion ii > iii > i.

Qu 13: AC
Since naphthalene and aniline are both organic systems, they will dissolve in the diethyl ether. Since no acid or base has been added, no ionic species will have been formed.

Qu14: A
Extracting with acid (HCl) will protonate the basic amine - the ionic ammonium ion D will be in the aqueous layer, leaving the naphthalene A in the organic ether layer. 

Qu15: D
Extracting with acid (HCl) will protonate the basic amine - the ionic ammonium ion D will be in the aqueous layer, leaving the naphthalene A in the organic ether layer. 

Qu16: C
In order to neutralise the 3 x 15mL of 5%HCl, you need to add an equivalent amount of base (NaOH). Neither A nor B add enough moles of base to neutralise 45mL of 5% HCl.

Qu17: C
Once the acidic ammonium salt D has been neutralised with base, the aqueous layer will contain the neutral amine, C.

Qu18: C
The neutral amine C will be extracted back from the neutral aqueous layer into the organic layer.

Qu19: D
A is used for decolourising (see acetaminophen experiment), B and C are reagent in the nucleophilic substitution experiment, E would not be suitable since it would react as an acid with the basic amine and you used AB as a stain in the chromotography experiment.

Qu20: DE
In ABC, the N is part of an amide where the N is involved in resonance with the carbonyl C=O group and therefore the N is sp2 hybridised. DE are amines.

Qu21: DE
Chirality centers (in their simplest case) require a atom with four different groups attached. So look for an sp3 C with 4 different groups attached - this only occurs in D and E.

Qu22: D
First draw out the name (i.e. nomenclature) then work out the IHD (faster method) or calculate the molecular formula (slower). In each case also check the number of CNO. N,N-diethylaminoethanal has C6H13NO and an IHD = 1. ABC also have IHD =1 (1 C=O pi bond) and E has an IHD =1 due to one ring. D has an IHD = 2 (1 pi bond plus 1 ring).

Qu23: DE
An application of  acidity and basicity.  It's a question about nitrogen basicity. In order for the N to be protonated to more than 50% when treated with one equivalent of ethanoic acid, the conjugate acid formed must be a weaker acid than ethanoic acid. This requires that the N be an amine (D and E) and not an amide (A, B and C). The N in an amide is less basic due to the resonance with the carbonyl group.

Qu24: D
The IHD here is 5 (1 C=O pi bonds, 3 C=C pi bonds and 1 ring).

Qu25:B
There are 6 types of H here.... 1 OH (position 7), 1 NH (position 11), one CH₂ (position 8), one sp3 CH (position 10), and two types of aromatic CH (positions 1/5 and 2/4).

Qu 26: D
Hybridisations.  C10 has four groups attached (C8, C12, N11 and an H) so it's sp3. N11 is part of an ammonium ion - there are four groups (C10 and 3H) so it's sp3. O13 is part of a resonance system that makes it resemble a carbonyl group so it's sp2.

Qu 27: E
Oxidation states are calculated by looking at the distribution of the electrons in the bonds due to electronegativity. C6 is attached to 3 x C atoms (equally electronegative = 3 x 0 = 0) and 1 x O (more electronegative = -1), sum these up = -1 therefore C6 oxidation state +1.

Qu 28: A
Oxidation states are calculated by looking at the distribution of the electrons in the bonds due to electronegativity. N11 is attached to 1 x C atoms (less electronegative = 1 x +1 = +1) and 3 x H (less electronegative = 3 x +1 = +3), sum these up = +4 .... but there is a charge of +1, therefore since N oxidation state + plus of attached groups = N formal charge, N11 oxidation state -3.

Qu 29: C
An application of  acidity and basicity.  It's a question about NH acidity. The pKa for the loss of a proton from an ammonium ion is about 10.

Qu30: C
Need to work backwards : reagents 2) look like a nucleophilic substitution (SN2) of a ditosylate forming a ring containing an alkyne this means the nucleophile is probably an acetylide - this is consistent with reagents 1) - amide ion used as a base to convert a terminal alkyne into an acetylide. The only terminal alkyne is C

Qu31: C
Need to work backwards : reagents 2) could either be a nucleophilic substitution or an elimination (probably since we heat it too). There is a methoxy group in the product but it is on an aromatic ring - so it is NOT from a nucleophilic substitution. There is also an alkene, so we are probably looking at the elimination of an alkyl halide to give an alkene. Reagents 1) look like radical halogenation of an sp3 C atom (look for the most stable radical). The only answer that makes sense is C since that is the only way to end up with the methoxy as an aromatic substituent. The process forms the benzylic radical and hence the benzylic bromide which is then eliminated to give the conjugated alkene.

Qu32: A
We should work forwards : reagent 1) is N-bromosuccinimide = NBS = source of Br radicals for radical bromination. This will occur at the site of the most stable radical so we will get the tertiary bromide formed. Reagent 2) is a nucleophilic substitution of the Br for the cyanide.

Qu33: C
Need to work backwards but it's only one step.... the reagents are typical of an elimination of an alkyl halide to give an alkene. This rules out BDE, since alcohols do not eliminate with base. If you use A then the major product will the methylcyclohexene (a trisubstituted alkene = more stable). C can only give the required product in the 1,2-elimination

Qu34:D
We should work forwards but it's only one step.... the reagents are typical of an SN1 reaction (recall the laboratory experiment?).... the bromide is left to give the secondary carbocation... but that will rearrange by a 1,2-hydride shift to a more stable benzylic carbocation and then the alcohol functions as the nucleophile to give the ether D.

Qu35: C
We should work forwards .... looking at the product it's an ether synthesis (nucleophilic substitution) so we need an alcohol and a halide. SInce the O is attached to a benzene ring, the alcohol needs to be a phenol (therefore either A or C. Draw out the required halide and then name it.....(i.e. nomenclature) 

Qu36: 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, we should see that we need two benzylic systems... draw out the structures of the names (i.e. nomenclature).... these facts limit it to either D or E. But D wil convert the initial benzyl alcohol to benzyl bromide and therefore don't allow an ether to form.

Qu37: A
Need to work backwards .... looking at the product it's a cyclic ether synthesis (nucleophilic substitution) so we need an alcohol and a halide - we are starting from a dihalide, so substitute one of the halides (the better leaving group for an -OH). A gives the product... the Br is substituted to give the 5-chloro-1-pentanol which will easily cyclise to give the favoured 6 membered ring. There is no reaction with B. D would first make a thiol, C and E are not effective.

Qu38: A
Need to work backwards .... looking at the product it's an amine, starting from an alochol so think make a better leaving group and then use a nitrogen nucleophile such as ammonia.  A is the correct answer, giving the bromide via an SN2 then replace it with ammonia. B would a tertiary bromide via radical substitution and then the problems would start (form a cyclic ether and then get a amino alcohol... i.e. wrong). C would give no reaction in the first step since there would be no good leaving group. D would probably rearrange in an SN1 type process, and NH3 would be protonated, hence no N nucleohile would be present and E is similar to D.

Qu39: E
IR shows a carbonyl (i.e. C=O) at 1716cm^-1 which is a typical ketone. The 13C-nmr peak at 212ppm supports the C=O and suggests an aldehyde or a ketone.  This means it could be E, or AB. The H-nmr has 2 types of H and the C-nmr has only 3 types of C so it has to be E. The H-nmr coupling shows an ethyl group, it does not have a methyl group singlet or the typical coupling of an isopropyl group.

Qu40: BC
IR shows a band at 1650cm^-1 which is low for a typical ketone, since the 13C-nmr does not have a peak above 160ppm it isn't a C=O and it is more likely a C=C. If it's a C=C it could be AE, BC, BD or BE. The H-nmr has 4 types of H, 10H in total. The lack of H-nmr peaks in the region 7-8ppm implies it is not aromatic and so rules out BD and BE. Since there is coupling in the region 1-2ppm, it is not just methyl groups, infact it's an ethyl group, it rules out AE.

Qu41: D
IR shows C=C at 1735cm^-1, and probable C-O at 1202 and 1122cm^-1. The H-nmr has 5 types of H, 10H in total and has some complex coupling. The 13C-nmr has 4 types of C, but neither nmr show aromatic signals. So, ruling out C=O and -OH and aromatics, this leaves D, AE or BC. But the IR suggests a C-O which best fits D. The H-nmr peaks at 1.2 and 4.1 suggest the isopropyl group with the CH connected to the O. The other three H-nmr peaks are consistent with a monosubstituted alkene.

Qu42: A
IR does shows an -OH at 3100-3500cm^-1 but no band near 1700 so no C=O.  This means it should be A or B. The H-nmr exchangeable peak at 2.4ppm supports the -OH. Since the H-nmr peak at 3.6ppm is a triplet for 2H it suggests a CH₂CH₂OH unit rather than CH₃CH(OH)CH₂ and hence it's A not B.

Qu43: BD
IR does not show a carbonyl (i.e. C=O) near 1700cm^-1 or an -OH near 3100-3500. This limits the choices to AE, BC, BD or BE.   Since the 13C-nmr shows that there are 2 types of ArC (127-142ppm), then it must be BD since BE has 3types of aromatic carbon.

Qu44: AC
IR shows a carbonyl (i.e. C=O) at 1738cm^-1 which is high for a typical ketone. The IR peak at 1248cm^-1 could be a C-O. The 13C-nmr peak at 170ppm supports the C=O and suggests a carboxylic acid or derivative.  This means it could be C, AC or AD. The H-nmr has 3types of H, 10H in total. No H-nmr peak near 12ppm rules out the acid C. There is a methyl singlet in the H-nmr at 2ppm and the peaks at 1.2 and 5 ppm suggest an isopropyl group = CH(CH₃)₂ - these things together suggest an ester with an isopropyl as the OR, i.e. AC. 

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