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: C
The reaction is nucleophilic addition to carbonyl groups, looking at the relative reactivity of aldehydes (i and ii) and ketones (iii). Electronic and steric factors mean than the number of alkyl groups are important : they both make the carbonyl C less electrophilic and more hindered hence less reactive so we get ii > i > iii.

Qu2: C
The reaction is electrophilic aromatic substitution, and we need to look at the substituent effects on the aromatic ring.  The -OC(=O)CH₃ group in i is attached to the ring via the O attached by single bonds (i.e. the alcoholic O) of this ester group.  This O has then lone pairs that can be donated to the ring and hence it is an electron donating group and hence is an activator.  The -OH group in ii of course all has then lone pairs that can be donated to the ring and hence it is an electron donating group and hence is also an activator. What's the difference ? In the ester in i the O lone pair is also involved in resonance with the C=O group, which makes the lone pair less available for donation to the  aromatic ring. The -CF₃ in iii is a deactivating group due to electron withdrawal by strong inductive effects due to the electronegativity of the F atoms. So the reactivity is ii > i > iii.

Qu3: A
If you know your pKa's then this is easy : ammonia = 35, terminal alkyne pKa = 26, t-butanol = 18.  Remember the lower the pKa the stronger the acid, so i > ii > iii.   What if you don't remember your pKas ?  (why not ?)  Then you'll need to deduce it.  Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-..... first compare ammonia and the alcohol.... if you recall, N is less electronegative than O so N is less stable as an anion compared to O, hence alcohols are more acidic than ammonia.  Now what about the alkyne ? Maybe the easiest way here is to think about the bases that are used in synthesis to make the acetylide ion.... sodium amide is the common choice.... that implies that alkoxides, RO-, are probably not strong enough bases so the alkyne is a weaker acid than the alcohol.

Qu4: A
Enols of active methylenes (where there are 2 C=O groups separated by a CH₂ group) are more stable than those of simple carbonyls, one reason being the possibility of a very favourable if the 6 membered ring, as would be the case here for both i and ii.  The difference here between these two is the degree of substitution of the alkene unit.... more alkyl groups make it more stable. So the dialkyl-substituted alkene in i is more stable than the monoalkyl-substituted alkene in ii.  So we have i > ii > iii.

Qu5: E
Electrophilic addition to C=C in a series of substituted alkenes with HBr = hydrohalogenation reaction . Reaction is controlled by the stability of the intermediate carbocations formed by the addition of the H+ to the C=C.  System iii gives a tertiary carbocation where the lone pairs on the adjacent O in the ether group can give some extra resonance stabilisation.  System ii gives a secondary cation, and iii gives a secondary carbocation where the lone pairs on the adjacent O in the ether group can give some extra resonance stabilisation (an electron donating group). Remember that the more stable cation is the easiest / fastest to form, so iii > i > ii.

Qu6: C
Look at how the stability of the deprotonated form changes (i.e. the conjugate base using the simple acid equation HA <=> H+    A-   then look for factors that stabilise A- ).  So look at the stabilisation of the charge and the way this is influenced by the aromatic substituent effects. Remember simple electrostatics (like charges repel = destabilise). -OCH₃ in iii is an electron donating group by resonance = decreased stability, -NO₂ in ii is strongly electron withdrawing due to resonance and inductive effects, and H in i is electronically neutral (i.e. no effect). Overall then we have  ii > i > iii.

Qu7: B
Remember the rules for ranking resonance structures  : complete octets are most important...draw a Lewis structure complete with the lone pairs first.  If we look at i has complete octets at C, N and O despite the charges, and note it has one more pi bond than the other two.  As for the other two, the negative charge on the more electronegative O atom is more important.  So i > iii > ii. 

Qu8: A
The systems are all organometallic reagents where the C atoms react as nucleophiles. Either look at the reactivity of the metal in its elemental form or consider the electronegativity. The more reactive the metal itself, the more reactive the organometallic reagent. So overall then i > ii > iii.

Qu9: C
The reaction is electrophilic aromatic substitution, a nitration,  we need to consider the directing effects of the substituents. The two C=O systems are both meta- directing and the Br is ortho-/ para- directing, hence the major product will be at site ii.  Steric effects of the larger tBu group will make iii less reactive than i. So the yields are such that ii > i > iii

Qu11: B
The pKa is more like 12.  It's an active methylene, and you used ethoxide (pKa = 16) to deprotonate it.

Qu12: B
The balanced equation is :  2 Na + 2 CH₃CH₂OH -> 2 CH₃CH₂ONa + H₂

Qu13: B
The structure shown is ethyl acetoacetate. Diethyl malonate is the diester = CH₃CH₂O₂CCH₂CO₂CH₂CH₃

Qu14: B
The iodide reacts with the butyl bromide to make butyl iodide via an SN2 reaction.

Qu15: B
Ethoxide and methoxide have very similar basicities, the reason is to avoid transesterification.

Qu16: B
The chemical shift for the CH unit at 4.2 ppm and the CH₂O units at 2.9 ppm are incorrect for diethyl n-butylmalonate.

Qu17: B
2,4-DNP only reacts with aldehydes and ketones not acids or esters.

Qu18: A

Qu19: B
The Lucas test depends on SN1 type reactivity. Phenols do not react via the SN1 pathway due to the unfavourable phenyl cation that would be required.

Qu20: A

Qu21: C
B, D and E are all amines. A is really an imide.

Qu22: ABCDE
Count the attached groups including lone pairs but remember to consider the impact of resonance.  If the N is adjacent to a pi system then it will be involved in resonance and hence the lone pair will be in a p orbital.  In E the N is part of a double bond.

Qu23: BDE
Review aromaticity ?

Qu31: D
The first step is ozone with and a reductive work-up overall an ozonolysis reaction, cleaving the alkene to give 2 molecules of the ketone acetone.   The Grignard reagent is then added to the ketone carbonyl followed by a normal dilute acid work-up to give the tertiary alcohol.
Note : only D is a tertiary alcohol. Count C atoms !

Qu32: D
The starting material is a ketone. Reagent 1 is lithium diisopropyl amide (LDA), a strong non-nucleophilic base - this will react with the ketone to form the enolate.  Reagent 2, methyl iodide will react with the enolate in an SN2 reaction to give the alkylation product where a methyl group has been added adjacent to the carbonyl group i.e. alkylation of an enolate.  Reagent 3, sodium borohydride, NaBH₄, will reduce agent the ketone carbonyl to a secondary alcohol.  C is a tertiary alcohol and has the methyl group at the wrong location.

Qu33: A
The first reaction will the electrophilic addition of the HBr to the alkene in accord with Markovnikov's rule, so giving 2-bromopentane.  Reaction of this alkyl bromide with triphenyl phosphine and a strong base will form an ylid for a Wittig reaction - this is completed when the carbonyl is added ... the product will be the alkene where the new C=C is formed between the original carbonyl C and the C-Br center. B has the wrong C=C location (moved in ketone portion), C has the C=C to the wrong carbon of the alkyl bromide. D and E have the wrong transformation.

Qu34: C
The benzene ring will react with the anhydride via a Freidel-Crafts acylation reaction to give the ketone. Treatment of the ketone with Zn / Hg / HCl is a Clemmensen reduction which converts the C=O to a CH₂.  i.e. ethylbenzene in this case.  The final step is a Freidel-Crafts alkylation reaction to give para-ethylmethylbenzene, C, directed by the o/p directing effect of the ethyl group.

Qu35: E
This is a Diels-Alder reaction.... hints are the cyclohexadiene, the conjugated diene, and a substituted alkene, the dienophile.  Remember the reaction is concerted and the stereochemistry of the diene and the dienophile is preserved in the reaction.  Only A, D and E correspond to DA reactions of cyclohexadiene.  B and C are from reactions of cyclopentadiene (count the C atoms). A has trans-acid groups so it has the wrong stereochemistry from the cis-dienophile. D has the wrong functional groups.... aq. acid will hydrolyse the anhydride from the dienophile not reduce it.

Qu36: C
The Grignard reagent will add to the epoxide at the least hindered end (since we have a strong Nu, the reaction is SN2 in character). This reaction is completed via a normal acid work-up giving an alcohol.  Step 2 is a Williamson ether synthesis - conversion of the alcohol to an ether : Na removes the proton generating the alkoxide that reacts in an SN2 reaction with the alkyl iodide.  The only products that are ethers are C and E. E has the wrong regiochemistry based on the Grignard step.

Qu37: E
The starting material is an aldehyde and an ester. The 1,2-diol in the first reagents will react with the aldehyde to form a cyclic acetal, a common protecting group for aldehydes and ketones.  Step 2 is lithium aluminium hydride, a strong hydride reducing agent - this will reduce the ester group to a primary alcohol.  Step 3 removes the protecting group to reveal the aldehyde.  Need a C4 system or only E is an aldehyde !

Qu38: B
The product is a cyclic acetal formed by the reaction of a 1,2-diol with a ketone.  : see the ketone, acetone in step 2.  In step 1, osmium tetroxide reacts with an alkene to give 1,2-diols.  Hence we need an alkene.  C is a diene, the extra C=C would either react or should still be there.

Qu39: A
Really need to work backwards with this one. Step 2 will add Br to aromatic rings : need to check the regiochemistry. The right hand aromatic ring has an alkyl group (o/p) and a nitro group (m) - so this is consistent. The left hand ring also has a alkyl group and an alkoxy group (both o/p), again consistent. So the 2 Br atoms could be added in this step, or one could be in place already, if that's the case, then the relative reactivity of the two aromatic rings will come into play = substituent effects.  The right had has the nitro, a deactivating group and the left has an alkoxy group, an activator.  Step 1 is a Williamson ether synthesis : base (K₂CO₃)  then alkyl iodide - this converted a phenol to a methyl ether.  Only A,C and D are alcohols for step 1. D lacks the methyl group meta to the ether in the product, so that can't be right. In C the problem will be further bromination of the top ring.

Qu40: C
Step 3, PCC = pyridinium chlorochromate, an oxidising agent for alcohols.  Step 2.... oxymercuration ..... looks like of an alkyne (peek at step 1) to give a ketone most likely, and step 1 looks like the acetylide has been added to a carbonyl giving an alcohol.  If we look at the product, then the (O=C)CH₃ unit on the right has come from the alkyne, so we need the C₆H₅CH₂CHO starting material = C.  A  would react with 2 eq. of the acetylide or give a mixture of products. B the acidity of the -OH would just protonate the acetylide. D would give a tertiary alcohols that would not be oxidised and E lacks a carbonyl for the acetylide to react with.

Qu41: A
Not an easy one.... need to spot the condensation - the "hint" is that step 1 is an alkoxide base and all the possible starting materials are carbonyl compounds. Counting C atoms may help.... It needs to be an ester then to get to the right oxidation state - so a Claisen condensation followed by decarboxylation. B and E would undergo aldol reactions to give a beta-hydroxy carbonyls - and the product would have too many C atoms. C would just be deprotonated by the base, D lacks an alpha-H so it can't form an enolate.

Qu42: D
The product is a cyclic ester..... step 4 probably formed this from the corresponding acid and alcohol - break the ester to see that. Sodium borohydride, NaBH₄, is a reducing agent for aldehydes and ketones. Since the alcohol needed is secondary, it requires that it was a ketone that was reduced.  The ketone and the acid were formed by the ozonolysis with an oxidative work-up (hydrogen peroxide) of an alkene.  Counting C atoms eliminates B and C as answers.  Since A is a diene the ozonolysis will remove 3C atoms as a dicarbonyl by product.  E would give an isomeric (regioisomer) product.

Qu43: C or D
Step 2 is the oxidation of an aromatic alkyl group to give a carboxylic acid.  Step 1 could be either a Friedel-Crafts alkylation or an acylation using either an alkyl halide or an acyl halide respectively.  This narrows the choices to B, C or D. However, it can't be B since the tert-butyl group does not oxidise to the carboxylic acid since it lacks a benzylic H. Both in fact would work since aromatic ketones can be oxidised to carbooxylic acids (infact the oxidation of alkyl aromatics to aromatic acids probably occurs via alcohols and aldehydes / ketones).

Qu44: B
Step 3 is a chromium oxidation of a primary alcohol or an aldehyde to give the carboxylic acid.  Steps 2 and 1 correspond to a hydroboration / oxidation to give the primary, anti-Markovnikov alcohol from the alkene. D is the first to go - it's not an alkene. E has too many C atoms. A and C have the wrong alkene regiochemistry.

Qu45:D
Protect the ketone as this cyclic acetal needs the 1,2-diol :  HOCH₂CH₂OH / H+

Qu46: BC
Need  to convert the alkene to an anti-Markovnikov alcohol : hydroboration / oxidation

Qu47: AB
Convert the alcohol to a bromide - remember -OH is a poor leaving group so E will not work, use PBr₃.

Qu48: ABC
Carboxylic acid to an ester : need the right alcohol and an acid catalyst.

Qu49: DE
The ester has been acylated i.e. the CH₃C=O group has been added.... a Claisen condensation of an ester with a second molecule of itself - so we need a base, the corresponding alkoxide being the best choice - it's an ethyl ester so use ethoxide.

Qu50: DE
The active methylene has been alkylated, looks like an enolate question so we need need a base, the corresponding alkoxide being the best choice - it's an ethyl ester so use ethoxide (again !)

Qu51: CD
We have removed the ester group.... remember beta-ketoesters can be hydrolysed and decarboxylated : here best to choose basic conditions to avoid deprotecting the acetal system at the same time.

Qu52: AC
The ketone has been converted to an alkene with one C added, a Wittig reaction.

Qu53: C
Now convert the alkene to an epoxide = epoxidation = use a peracid

Qu54: BE
Open the epoxide to make a 1,2-diol and deprotect the cyclic acetal = aq. acid.

Qu55: B
Le Chatilier's principle - shift the equilibrium by removing the by-product. C is incorrect, it's equilibrium (reaction requires heat). D is wrong, check the oxidation states and E is wrong as there are no rearrangements.

Qu56: C
Looking at an SN1 type reaction, so look at factors affecting the carbocation intermediate, build a model if needed. A isn't true, the alkyl groups would stabilise the carbocation.  B isn't correct..... sterics is important in SN2, but in SN1 once the Br has left, the one face of the carbocation would be very accessible.  D is wrong ! SN2 are fastest for methyl and primary due to low steric hindrance.  E although water is a weak Nu, the rate determining step of an SN1 is the carbocation formation.

Qu57: E
Cyclopentadiene is non-aromatic so A and C are wrong.  B  is true, but fails to explain the pKa difference. D the anion of cyclopentane is non-aromatic.

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