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
The reaction is a nucleophile adding to a carbonyl of an ester i, aldehyde ii and ketone iii.  The electronic effect of the group attached to the C=O controls the electrophilicity of the C=O.  In order of the groups electron donating ability CH₃O- > CH₃- > H- so the ester is the least electrophilic and the aldehyde the most electrophilic.....so we get the reactivity order of ii> i > iii.

Qu2: AB
The reaction is electrophilic aromatic substitution, and we need to look at the substituent effects on the aromatic ring.  -NO₂ is a deactivating group due to electron withdrawal by resonance and inductive effects, -Cl is slightly deactivating due to electronegativity and the N atom of an amide slightly activating due to the possibility of electron donation by resonance. So iii > ii > i.

Qu3: A
Look at where the H is attached and think of the factors that stabilise charge such as electronegativity, resonance, inductive effects etc. i gives an O -ve that is further resonance stabilised to give a second O -ve (a carboxylic acid), both ii and iii give  C -ve. In ii we have an active methylene where the -ve charge can be  resonance stabilised onto 2 possible C=O creating O -ve. iii is a terminal acetylene but still a C- is formed. FYI : pKa values.... acid about 5, diketone about 9 and an acetylene about 25.  So in terms of acidity  i > ii > iii.

Qu4: A
A set of similar organometallic reagents.... reactivity correlates to reactivity of the metal, so Li > Mg > Cu.... therefore i > ii > iii.

Qu5: A
Nucleophiles are electron donors so we have to look at factors that affect electron availability. Compare i and ii, they only differ in the nature of the charged atom O vs S.  Nucleophilicity increase down a group due to polarisability factors so i > ii.  ii and iii are both O systems. The phenoxide iii has some resonance stabilisation of the charge making it less nucleophilic than the simple alkoxide, So overall we have  i > ii > iii.

Qu6: A
Benzene, i, has the highest resonance energy per C=C of any system at about 36 kcal/mol.  Pyridine ii has slightly less at about 28 kcal/mol and furan iii is a lot lower around 16kcal/mol (more diene character).  So we have  i > ii > iii.

Qu7: D
Compare O and S systems first (same group).  Acidity increases down a group (e.g. HI > HBr > HCl >HF) due to size effects (i) length and strength of HX bond and (ii) the ability of the larger atom to accept the charge, so RSH > ROH (acidity) therefore RO- > RS- in terms of basicity. Note electronegativity should not be used when comparing atoms in different rows of the periodic table.  Now compare O and C systems (same row).  O is more electronegative than C so is less willing to share electrons and therefore it is less basic... so C- > O- (basicity). Overall then we have C- > O- > S- or  ii > iii > i.

Qu8: C
Electrophilic addition to CC pi bonds in alkenes and alkynes with sulphuric acid. Reaction is controlled by the stability of the intermediate carbocations. System i gives a secondary cation, ii gives a secondary cation where the lone pairs on the adjacent O can give some resonance stabilisation and iii gives less favourable vinyl cation. Most stable cation is the easiest to form, so ii > i > iii.

Qu9: A
The reaction is electrophilic aromatic substitution, and we need to look at the directing effect of the groups. i and ii both have alkyl groups => a electron donating group => ortho/para director whereas iii is a C=O of a ketone => a electron withdrawing group => meta director  Steric effects will block the ortho site more with the larger R group in i that ii , so yield of para i > ii > ii.

Qu10: C
Comparing i and ii we have an isolated and a conjugated diene, so the conjugated diene, ii, is more stable. The cyclobutadiene in iii is also a conjugated system, but it is antiaromatic and quite destabilised. Overall then we have ii > i > iii.

A = IR gives C=O, H-NMR gives -OCH₃ at 4ppm and a monosubstituted aromatic at 7-8ppm... it's an ester, methyl benzoate,
B = IR gives an alcohol, H-NMR gives a monosubstituted aromatic at 7-8ppm, a CH₂ and the -OH so we have benzyl alcohol
C = IR gives an alcohol, H-NMR gives a disubstituted aromatic, a CH₃ and the -OH so we have a phenol, p-methylphenol
D = IR gives C=O, H-NMR gives an aldehyde at 10ppm, and a monosubstituted aromatic at 7-8ppm... its an aromatic aldehyde, benzaldehyde
E = IR gives no C=O, H-NMR gives a monosubstituted aromatic at 7-8ppm and alkene at 5-7ppm : phenylethene
AB = IR gives C=O, H-NMR gives a monosubstituted aromatic at 7-8ppm and a CH₃ at 2.5ppm (next to a C=O ?) its an aromatic ketone, acetophenone

Qu11: B
Lucas test is for alcohols but not phenols.

Qu12: AB
DNP implies an aldehyde or ketone, the chromate tells us it is not oxidisable, so it must be the ketone.

Qu13: C
Ferric chloride test is for phenols.

Qu14: B
DNP implies it is not an aldehyde or ketone (no precipitate) and the chromate tells us it is is oxidisable, so it must be an oxidisable alcohol (and therefore not a phenol).

Qu15: AB
The yellow precipitate due to the formation of iodoform, CHI₃, in the iodoform test indicate the presence of a methyl ketone.

Qu16: E
Tests for the presence of C=C in alkenes (and alkynes).... this system is an alkene.

Qu17: D
DNP implies an aldehyde or ketone, but the iodoform says it is not a methyl ketone, so must be the aldehyde.

Qu18: A
DNP implies it is NOT an aldehyde or ketone, therefore, given the C=O system is required, from the options it must be the ester.

Qu19: AB
The heterocycles (i.e. systems with non-C atoms in rings) are E, AB, AC, AE and BC.... but only E, AB and AE contain nitrogen. The N atoms in E aren't basic (no lone pairs to donate).  In AE the lone pair on N is part of the aromatic system so it is not readily donated and therefore the N is not very basic. In contrast in AB, the lone pair is NOT part of the aromatic system (there are already 6 p electrons from the 3 p bonds) so it is quite basic.

Qu20: A
If n=2, then we need a 10 p system..... only A has 10 p electrons.

Qu21: AD
4n p electron systems are B, C and AD. B is non-aromatic because it is not a cyclic conjugated p system. C will turn out to be the answer to qu 25. AD is an 8 p system, it is cyclic and conjugated.... if planar it would be anti-aromatic and therefore destabilised. In order to avoid this destabilisation, the ring twists out of planarity and so is "non-aromatic".

Qu22: A or AB or AC
Aromatic as drawn means we are considering A, E, AB, AC, AE and BC. Now we are looking for a basic system that also has an aromatic conjugate acid, so the lone pair used in the acid/base reaction cannot be part of the initial aromatic unit.   Protonation of E, AE and BC will give non-aromatic conjugate acids

Qu23: B
Non-aromatic as drawn means we are considering B, C and D.  Now we are looking for a system that becomes aromatic when a proton is removed. B is a 4 pelectron system, so the extra lone pair formed when H+ is removed from the -CH2- makes an aromatic 6 p electron system. C and D are discussed in Qu24 and 25 below.

Qu24: D
Non-aromatic as drawn means we are considering B, C and D.  The example of tautomerism we met was enol / ketone (during addition of water to
alkynes). If we look at D and convert each of the H-C-C=O to C=C-OH we get a tri-enol, 1,3,5-trihydroxybenzene, which is aromatic.

Qu25: C
Non-aromatic as drawn means we are considering B, C and D.  In either B or D resonance cannot alter the CH2 units to become part of C=C and so allow a cyclic conjugated system to form.  In C if we make the C=C become -C-C+ this gives a 6p electron system in the five membered ring and a 2p electron system in the three membered ring, both of which are aromatic.

Qu26: D
Esters react with 2 equivalents of Grignards to give tertiary alcohols after work up so in this case we will get a diol since we have a cyclic ester to start with. Excess PBr₃ will convert the alcohols to bromides which are then eliminated  giving a diene. Counting C atoms : SM = 5 plus 2 from 2 eq. methyl magnesium bromide = 7 C.

Qu27: C
Amines react with acid chlorides to form amides. Friedel-Crafts alkylation of the aromatic ring will be para- . The alkyl group is then oxidised to an acid -CO₂H and finally the amide is hydrolysed back to the amine. So we are looking for p-aminobenzoic acid, C.

Qu28: C
The acid reacts with thionyl chloride to form the acid chloride which in turn reacts with the amine to form the amide. Reduction of the amide gives an amine so must be B or C. It maybe useful to count C atoms.... remember the C from the original carbonyl will now be in the amine.

Qu29: B
The peracid will react with the C=C to form an epoxide. The amide ion is a good nucleophile and will attack the epoxide from the least hindered end giving the 2-amino-3-alcohol product, B.

Qu30: D
Diels-Alder reaction followed by reduction of the ketone to a secondary alcohol but no reduction of the C=C. Count C atoms.....

Qu31: B
Diazotise the amine, then convert it to the nitrile, which is then hydrolysed to the acid.... so we are looking for m-hydroxybenzoic acid, B.

Qu32: C.
Protect the more reactive ketone as a cyclic ketal. React the ester with 2 equivalents of the Grignard reagent, work-up then removal of the ketal protecting group reforms the ketone.... hence we have a tertiary alcohol / ketone system, C.

Qu33: B
The cyclopropane is formed by the Simmons-Smith reaction of the Zn system with an alkene, which has in turn been made by a Wittig reaction (note the P species) from cyclohexanone, B.

Qu34: E
The ketone has been formed by the oxidation of a secondary alcohol.  The secondary alcohol must have been formed by the reaction of a Grignard reagent (E) with the methanoate ester.

Qu35: D
The alkyl group has been added to the benzene via an acylation then a Wolff-Kischner reduction. We are looking at the preparation of the acyl chloride for the acylation, so we need the C4 acid, D.

Qu36: C
The sequence looks like an aldol (note the enone product) preceded by an ozonolysis with a reductive work-up.  The product is C6, so the original cycloalkene is also C6. The required intermediate dicarbonyl can be visualised by "breaking" the product at the C=C, where the C=O would have been on the beta-C of the C=C. Hence we have an aldehyde there with a methyl ketone at the other end of the chain.  A gives a dialdehyde on ozonolysis, B is C7, C is good, D and E don't react with ozone as they lack C=C.

Qu37: E
Sequence involves the methylation of a ketone enolate, where the ketone has been formed by the PCC oxidation of a secondary alcohol itself formed by the hydroboration of an alkene (anti-Markovnikov). Working with the product, remove one methyl group, convert the C=O to C-OH then to the trisubstituted alkene reveals E.

Qu38: E
Tough one despite its apparent simplicity. The only material with the correct oxidation state is E. Reaction involves loss of water followed by a carbocation rearrangement.  A, B and C would give alcohols via hydration reactions. D would react to via a rearrangement to also give an alcohol, 2,3-dimethyl-2-butanol.

Qu39: C
The sequence looks like an intramolecular Claisen (i.e. a Dieckmann) condensation achieved via the esterification of a dicarboxylic acid from an ozonolysis with an oxidative work-up.  The product is C6, so the original cycloalkene is also C6. The required intermediate diester can be recreated by "breaking" the product at the ring C-C between the -OH and -CH₂OH.  To get the diacid from the ozonolysis, the alkene needs C6, symmetrical and only 1,2-disubstituted... so it has to be cyclohexene.

Qu40: C
A Sandmeyer reaction of a diazonium salt to give the bromide : use CuBr.

Qu41: BC
Convert the Br to a carboxylic acid with one new C atom. This suggests using a Grignard reagent and carbon dioxide.

Qu 42: D
Acid to acid chloride, an acid derivative..... need to use thionyl chloride. The triethyl amine is used to react with the HCl by-product.

Qu43: AB
Not easy to spot, but this is a Friedel-Crafts reaction on benzene.

Qu44: B
Convert the -CH₃ to a -CH₂Br.  Since this is a reaction that converts a simple C-H to a C-Br, need radical substitution.

Qu 45: AD
Bromide to a nitrile... looks like a nucleophile substitution of a benzylic halide using a cyanide as the nucleophile.

Qu46: CD
Need to add an ester group next to the nitrile which is an electronwithdrawing group.... this means we could use a strong base to form an enolate then add an acylating agent.

Qu47: ABC
Now need to add a methyl group next to a nitrile and an ester, both of which are electronwithdrawing groups. Use a base to form an enolate then add an alkylating agent.

Qu 48: BE
Another tricky one.... how to get rid of a C group and get a carboxylic acid ? Decarboxylation ? So hydrolyse both the nitrile and ester to a dicarboxylic acid which will then undergo the loss of carbon dioxide.

Qu49: D
Draw the product, you've added -OH and -Cl to an alkene. Hydration is addition of water, there is no rearrangement. You need to know the mechanism by which this reaction occurs.... HOCl is equivalent to HO- and Cl+, so the chlorine is the electrophile that adds first to the C=C.

Qu50: D
Aromaticity...the stability of the product conjugate base is what makes the system so acidic for a seemingly simple C-H bond.

Qu51: B
High T suggests thermodynamic conditions when equilibria and hence reversible reactions are present.

Qu52: B
Need to know the process, the oxidation is really an 1,2-elimination of a chromate ester... the critical "item" is the presence of the H- at the carbinol carbon... i.e. we need to have had a H-C-OH unit to start with.... clearly not possible in a tertiary alcohol where all the groups attached to the carbinol C are alkyl groups.

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