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 a nucleophile(here water) adding to a carbonyl of an ester i, acid chloride ii and an amide iii, hence these are all carboxylic acid derivatives, so it's nucleophilic acyl substitution and we are looking at the acid derivatives. The electronic effect of the group (the leaving group) attached to the C=O controls the electrophilicity of the C=O and obviously the leaving group ability.  In order of the groups electron donating ability (CH₃)₂N- > CH₃O- > Cl-  so the amide is the least electrophilic and the acid chloride the most electrophilic.....so we get the reactivity order of ii > i > iii

Qu2: A
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 CH₃- group in ii is weakly electron donating and the -CN in iii is a deactivating group due to electron withdrawal by resonance and inductive effects. So the reactivity is i > ii > iii

Qu3: A
If you don't know the pKa's then look at where the H is attached and think of the factors that stabilise the negative charge such as electronegativity, resonance, inductive effects etc. i gives an O -ve that is further resonance stabilised to give a second O -ve (it's a carboxylic acid, pKa about 5). In ii the anion is a C -ve - it's an enolate of an active methylene compound where the -ve charge can be  resonance stabilised onto 2 possible C=O creating 2 different O -ve (it's a beta-diester, pKa about 11). In iii the charge ends up as an O-ve no resonance stabilisation because there is no pi system (it's an alcohol, pKa about 16).  So in terms of acidity  i > ii > iii

Qu4: E
Draw 1,3-cyclopentadiene first..... The reaction is a Diels-Alder reaction (alkene plus conjugated diene). The selections are the dienophile components, and it the normal Diels-Alder reaction, the reactivity is increased if the dienophile has electron withdrawing groups (since it can be thought of as the electrophilic component of the reaction).   In i we have just ethene, in ii we have two methyl groups, these are weakly electron donating and so they decrease the reactivity. In iii we have an ester connected to the alkene via the C=O so it is an electron withdrawing group.  Therefore reactivity is iii > i > ii

Qu5: A
Electrophilic addition to C=C in a series of substituted alkenes with aq. sulfuric acid = hydration reaction . Reaction is controlled by the stability of the intermediate carbocations formed by the addition of the H+ to the C=C.  System i gives a primary carbocation where the lone pairs on the adjacent O in the ether group can give some extra resonance stabilisation.  System ii gives a simple secondary cation, and iii gives primary carbocation where there is an unfavourable electron withdrawing group next to the C+ center. Remember that the more stable cation is the easiest to form, so i > ii > iii.

Qu6: E
The systems are all amines, in fact they are all anilines (i.e. aminobenzenes), it is the N atoms in these systems that we need to consider. Bases according to the Lewis definitions at least are electron pair donors. So the difference in the structures is the aromatic substituent, so these groups must influence the electron availability - i.e. substituent effects.  In i we have an H which we consider to have zero effect.  In ii we have a nitro group, -NO₂, which is electron withdrawing through resonance. An electron withdrawing group will reduce the electron availability at the N atom making it less basic. In iii we have a methoxy group, CH₃O-, which is electron donating through resonance, which will increase the the electron availability at the N atom making it more basic.  So we have iii > i > ii

Qu7: AB
Assigning oxidation states ? Count the bonds attached to the C, each H counts +1, C counts 0 and a bond to a more electronegative atom (e.g. O or N) counts -1. Total the count and then switch the sign since the oxidation state for the C plus the groups attached must equal 0.  In i the C is attached to C (count 0), two bonds to H (count +2) and O (count -1) therefore oxidation state C = -1. In ii the C is attached to C (count 0), two bonds O (count -1 per bond) and H (count +1) therefore oxidation state C = +1.   In iii the C has four bonds to O (count -1 per bond) therefore oxidation state C = +4.  Therefore  iii > ii > i

Qu8: B
The reaction is electrophilic aromatic substitution, a nitration.  First draw methylbenzene.... the attached group is a methyl group so this will direct the nitro group to the ortho or para positions corresponding to products i and iii. There will be minimal meta.  So waht about ortho vs para. The methyl group is small and so is the incoming nitro group, there are two ortho to each para, so here the reaction will give more ortho than para. (expt. yields are 63 : 3 : 34).  So the yields are such that i > iii > ii

Qu9:
Reaction is the hydroboration oxidation of an alkene. This gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Since no C+ forms, there is no carbocation rearrangement.  First draw 3-methyl-1-butene, CH₂=CHCH(CH₃)CH₃, then add the -OH to the least substituted end of the C=C.  So i is the major product and ii is a minor product (the Markovnikov product). iii is only obtained if there has been a rearrangement and since there is no C+ formed, there is no way for this product to have been formed. Overall then i > ii > iii

Qu10: C
Resonance energies.  Benzene, i, has the highest resonance energy per C=C of any system at about 36 kcal/mol.  Naphthalene ii has higher resonance energy then benzene because there are more C=C units, 61 kcal/mol (but it is less that 2 x benzene) and pyrrole, iii, has slightly less resonance energy than benzene at about 22 kcal/mol.  So we have  ii > i > iii.

Qu11: E
Boiling points increase with increasing pressure so they are higher at sea level than here in Calgary at almost 4000ft above sea level. A rough rule of thumb for Calgary is 1^o for every 15^o above 50^o. Thus for 235^o we are talking about a 12^o increase, hence about 248^o

Qu12: C
Since #0001 doesn't give a colour change with ferric chloride, it can't be a phenol so A is wrong.  If it were a tertiary alcohol, it should react rapidly in a Lucas test and would not be oxidised by dichromate, so B is wrong. Since #0001 gives a precipitate with 2,4-DNP, it must be an aldehyde or a ketone, and since the unknown is oxidised by dichromate, it must be a aldehyde (ketones don't easily oxidise), so C is true and D is false. A carboxylic acid should dissolve in NaOH and NaHCO₃, would not give a precipitate with 2,4-DNP and would not be oxidised by dichromate, so E is false.

Qu13: B
An imine has a C=N unit and an oxime has a C=N-OH unit.

Qu14: ACE
A is true... look at the H-nmr peaks at abouit 4ppm and 1.4ppm. B is false, there is no -OH in the IR (about 3500cm^-1), or in the H-nmr (broad, exchangeable singlet).  C is true, the H in RCHO typically show up around 10ppm.  D is false, a monosubstituted benzene ring would have to have integration for 5H in the aromatic region. E is true, the classis C=O stretch region.

Qu15: BDE
A is false... a negative Lucas test and no dichromate oxidation rule this out. B is true, if #0001 is an aldehyde, STU-A must be a carboxylic acid so C must be false and D is true too. If #0001 contains a benzene ring (see Qu14), then so must STU-A and so E is true.

Qu16: ABD
C is false because tertiary alcohols can't be oxidised. E is false, it's chromium VI.

Qu17: AC

Qu18: B

Qu19: E or AE
If n=2, then we need a 10 pi system..... E has 5 C=C and hence 10 pi electrons as part of an aromatic pi system. AE has 4 C=C and the N lone pair, here the N is like that of pyridine with the N lone pair being its contribution to the aromatic system.

Qu20: AB
Only AB and AD have 3 double bonds, but AD is pyridine and is aromatic.

Qu21: A
4n p electron systems are A, B, AC and BC. B is non-aromatic because it is not a cyclic conjugated p system (there is an sp3 C in the ring). AC  is non-aromatic because it is not a conjugated p system. BC will turn out to be the answer to qu 25. A is an 4 p system, boron has an empty p orbital and hence creates a cyclic and conjugated p system.

Qu22: AC
All except AC have at least 4 adjacent sp2 centers and are therefore conjugated systems.

Qu23: D or AD or AE
A, C, D, AD and AE contain none C atoms in the rings and so are heterocyclic compounds. Of these C, D, AD and AE are aromatic. Protonation of C gives a non-aromatic conjugate acid because the N lone pair is part of the aromatic system and loss of this destroys the aromaticity.

Qu24: B
Non-aromatic as drawn means we are considering B, AB, AC and BC.  Loss of a proton to give the conjugate base means we will be looking at the anions. AB would then be an 8 pi electron systems and hence not aromatic. AC still would not have a cyclic, conjugated pi system, and BC will turn out to be the answer to qu 25.

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

Qu26: A
The alkene reacts then the HBr under these conditions in an anti-Markovnikov fashion to give 1-bromobutane, CH₃CH₂CH₂CH₂Br. Mg/THF gives the Grignard reagent, CH₃CH₂CH₂CH₂MgBr which in then reacted with ethanal, an aldehyde to give, after the normal acid work-up, a secondary alcohol, 2-hexanol, A.

Qu27: E
First, the methyl group in the methylbenzene will direct the Friedel-Crafts acylation to the ortho / para positions, with para the major for steric reasons. The second step is vigourous oxidation that will convert both the methyl group and the acyl group to carboxylic acid groups. Hence the product is E.

Qu28: B
Mg/THF gives the Grignard reagent, PhMgBr which in then reacted with CO₂ to give the carboxylic acid after a normal acidic work-up. LiAlH₄ will then reduce the acid to the primary alcohol.

Qu29: AD
The alkene will react with the peracid to give an epoxide. Reaction of the epoxide with the acetylide nucleophile (SN2 like fashion) will add the acetylene at the least hindered end and put the alkyne group trans to the O. Work-up (step 3) gives the trans tertiary alcohol. Finally, reduce the alkyne to the alkyl group. Both enantiomers of the product will be formed. So look for the -OH on the same C as the methyl group and so that the ethyl group is cis to the methyl but trans to the -OH.

Qu30: C
Classic Diels-Alder with cyclopentadiene favouring the endo product followed by hydrolysis of the ester to the carboxylic acid.  Count carbon atoms.

Qu31: C
The Grignard will react with the epoxide in a SN2 like fashion (so it attacks the least hindered end), normal work-up gives 2-phenylethanol. Then a Williamson ether synthesis, sodium to make the alkoxide, the Nu which then reacts with methyl iodide via an SN2 to give the ether, C.

Qu32: A
The beta-ketoester will react with the 1,2-diol to give a cyclic acetal, a protecting group for aldehydes and ketones.  This leaves the ester to be reduced to a primary alcohol by the LiAlH₄, with a normal acid work-up.  Heating with aq. acid then remove the acetal, so we have the keto-alcohol, A.

Qu33: C
The product is a 1,2-diol, made by the reaction of permanganate with an alkene formed in a Wittig reaction. In this example, the cyclohexyl unit came from the ylid side of the Wittig reaction so our starting material is an alkyl halide, C.

Qu34: D
Diazonium chemistry of an aromatic system.... track back to the reduction of the nitro group to give an aminobenzene. The starting material needed is meta-nitromethylbenzene.

Qu35: C
The product is an aldehyde, made by a PDC oxidation of a primary alcohol. The alcohol has been prepared by the anti-Markownikov hydration via hydroboration / oxidation of an alkene, C. (note there are no C+ formed and hence no rearrangements in hydroboration / oxidation).  Count C atoms and make sure the C=C is in the correct location.

Qu36: D
The product is an ether that has been made by the addition of the alcohol to the C=C unit. The alkene has been formed via an E2 elimination (note the strong base in an aprotic solvent and heat) of an alkyl halide. The required alkene has the C=C outside the ring (exocyclic) and so the halogen must be the primary. B has too few C atoms.

Qu37: B
An oxime, these are made by the reaction of hydroxyl amine with the carbonyl, in this case a ketone. If you look at the required ketone, 3-hydroxycyclohexanone, the 1,3-realtionship of the C=O and the -OH suggests an aldol reaction. If we dissconnect the aldol product we are going to need 5-oxohexanal..... CH₃C(=O)CH₂CH₂CH₂CHO formed by the ozonolysis with a reductive work-up of 1-methylcyclopentene.  Note that A has too few C atoms to work and D will ozonise to give a dialdehyde.

Qu38: C
The cyclohexene in the bicyclic product and the highly functionalised, bridged structure, coupled with the simple reaction conditions hint at an intramolecular Diels-Alder reaction.  Try to push the curly arrows backwards in the cyclohexene unit, starting at the C=C to reveal the starting material. The diene is a furan unit with a methyl substituent and the dienophile is a geminal disubstitued alkene, a ketone in conjugation and a methyl group, C. 

Qu39: D
Working backwards, the amide is obtained by the reaction of an acid chloride with ammonia, so we have come via a carboxylic acid (cyclohexane carboxylic acid). How did we get there ? The first set of reagents are NaOEt (a base) and a dihaloalkane - this suggests an SN2 type reaction, the base being used to form a Nu and the alkyl halide is an alkylating agent. Since all the starting materials are dicarbonyls, maybe we should be thinking of alkylating enolates - it's a dihalide so it could happen twice. C can't form an enolate as it lacks alpha-H. A, B and D are active methylenes (easier enolate formation)... What happens is that the enolate forms, this gets alkylated, then the base forms a new enolate and a second, now intramolecular alkylation occurs forming a six membered ring (5C from the dihalide and 1C from the enolate). In order to get rid of one of the carbonyl groups, a decarboxylation is needed, so we need at least one ester unit.... in fact, since we are making an acid derivative, the amide, then two ester units are required i.e. D.  So what happens ? The aq. acid causes both esters to hydrolyse and forms a dicarboxylic acid that then decarboxylates (loss of CO₂).

Qu40: ABC
Carboxylic acid to an ethyl ester = ethanol plus acid catalyst.

Qu41: CD
The product is a dicarbonyl, a beta-ketoester. This 1,3-relationship of the C=O groups suggests ester enolate chemistry.... we have ethyl esters so ethoxide as the base ? This is a Claisen condensation.

Qu 42: BE
An alkyl chain has been added to the active methylene, so we need a base, again ethoxide makes sense for an ethyl ester system, to make the enolate and then the right alkylating agent ... count carbon atoms.

Qu43: BC
The product is a ketone and a bromide, we have lost the ester group. This suggests the decarboxylation of the ester, this requires that the ester be first hydrolysed to the carboxylic acid.... this needs aq. acid or base.

Qu44: DE
Product is a cyclic acetal fromed from the ketone. This requires a 1,2-diol and an acid catalyst.

Qu 45: AD
The bromide has been replaced by a phosphorous unit, triphenylphosphine. This looks like the start of a Wittig reaction.

Qu46: A
To make the ylid, the starting material needs to be treated with a strong base such as an alkyl lithium.

Qu47: AB
Complete the Wittig reaction.... we have the ylid, we need to add a C=O system. Counting carbon atoms and looking at the C=C unit, this shows we need propanal.

Qu 48: D
Convert the alkene to an epoxide.... this requires a peracid.

Qu49: BC
Open the epoxide to a 1,2-diol and deprotect the ketone by removing the cyclic acetal - to do both of these we need aq. acid and heat.

Qu50: C
The reaction is the electrophilic addition of HX to an alkyne via carbocations. The reaction occurs via an intermediate vinylhalide, which also reacts via a carbocation but this one is stabilised by the resonance of the lone pairs on the halogen. The reactions are via carbocations not radical reactions, there is no rearrangement. Although a cyclic halonium ion forms, this does not control the regiochemistry. Both steps proceed in accord with Marknovnikov's rule.

Qu51: D
Although pyrrole and pyridine are both aromatic, the reason why pyrrole is so much less basic than pyridine is because the conjugate acid of pyrrole is non-aromatic since the N lone pair in pyrrole is part of the aromatic system and protonation removes that lone pair.

Qu52: D
O is more electronegative than N, but that means that N would normally be more basic (compare H₂O and NH₃ for example).  Draw the resonance structures to convince yourself of this answer for both the N and O protonated products, or look at the resonance contributors in the amide and note the -ve on the O as opposed to the N.

Qu53: C
Aldehydes are more electrophilic since the carbonyl group is less sterically hindered and not slightly deactivated by the weak electron donating effect of the extra alkyl group in a ketone.

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