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 reactivity of carbonyl groups towards Grignard reagents, looking at the relative reactivity of i esters, ii aldehydes, and iii ketones. Electronic and steric factors need to be considered. First compare ii and iii, aldehydes tend to be more reactive than ketones because (1) they are less hindered and (2) alkyl groups are weak electron donors. Both factors make the carbonyl C less electrophilic and more hindered hence less reactive.  In esters, the RO- group of the carbonyl is a strong electron donor, making them even less reactive than ketones.  So we get ii > iii > i.

Qu2: AB
The reaction is electrophilic aromatic substitution, and we need to look at the substituent effects on the aromatic ring.  The ketone group in i is connected to the ring at the carbonyl carbon : this C has positive character due to the electronegative O and there is resonance withdrawal by the C=O group - it's a deactivating group. The alkyl group in ii is a weakly activating group due to an inductive effect. The -OC(=O)CH₃ group in iii 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.  So the reactivity is iii > ii > i.

Qu3:C
Acidity... if you know your pKa's then this is easy : ketone enolate = 20, carboxylic acid = 5 and ester enolate = 25.  Remember the lower the pKa the stronger the acid, so ii > i > 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-.....Note that all the acidic H are alpha to one carbonyl group. Now look at the atom the H is attached to.  In i and iii it's C and ii it's O. Recall, that C is less electronegative than O so C is less stable as an anion compared to O, hence the carboxylic acid is more acidic than the ketone or the ester.  In the ester the RO- group has lone pairs on the O that can be donated via resonance to the C=O. This competes against the donation from the C lone pair in the enolate making the ester enolate less stable so the ester is less acidic.

Qu4: D
The factors involved here are aromaticity in ii and iii and ring strain in i. Cubane, i, has 6 cyclobutane rings.... lots of ring strain... this makes is unstable.  Styrene (aka ethyenylbenzene), ii, is a conjugated aromatic system.... this makes it quite stabel.  Cycloctatetraene iii is non-aromatic =  "typical stability". Therefore ii > iii > i.

Qu5: A
Electrophilic addition to C=C in a series of substituted alkenes with aq. sulfuric acid = hydration reaction . The reaction is controlled by the stability of the carbocations formed by the addition of the H+ to the C=C.  System i gives a secondary 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 vinyl carbocation - vinyl cations are quite difficult to form.  Remember that the more stable cation is the easiest / fastest to form, so i > ii > iii.

Qu6: B
Basicity of substituted aromatic amines means look at the substituent effects on the aromatic ring.  An electron donating group on the ring will make the lone pair on the N more available, so making the amine more basic. The methoxy group, CH3O- in i is a strong electron donor, the Cl- in ii is slightly electron withdrawing and the H- in iii is the "reference" - i.e. no effect.  Therefore in terms of basicity, i > iii > ii.

Qu7:AB
Catalytic hydrogenation is easier the weaker the pi bond.  This means alkynes reduce more readily than alkenes and that carbonyl groups are difficult to reduce in this way (least reactive)..... iii > ii > i.

Qu8: A
Enolisable H are adjacent to carbonyl groups.  In i there are two (one each side), ii has one (on the left, also adjacent to the methyl group) and iii has none. So i > ii > iii.

Qu9: E
The reaction is electrophilic aromatic substitution, a nitration,  we need to consider the directing effects of the substituents. The Cl is ortho-/ para- directing.  The C=O system is deactivating and meta- directing. Alkyl groups are ortho-/ para- directing and the steric effects of the larger tBu group make it favour para over ortho.  So iii > i > ii

Qu11: E
The Lucas test is for alcohols that undergo SN1 reactions. Only the IR spectra A, D, E and AB show -OH groups. D is an aromatic alcohol (phenol) which don't react in the Lucas test and AB H-NMR suggests a carboxylic acid. The spectra of E suggests it's benzyl alcohol, C6H5CH2OH which reacts rapidly under SN1 due to the favourable carbocation intermediate.

Qu12: C
The DNP test gives an orange precipitate with aldehydes and ketones.  Only the IR spectra C and AB show C=O groups.  The spectra of AB suggest a carboxylic acid (e.g. IR has an -OH and H-NMR peak at 12ppm).

Qu13: D
The ferric chloride test is for phenols.  Only the IR spectra A, D, E and AB show -OH groups. Of these only D and AB have aromatic H in the H-NMR. D is the aromatic alcohol (phenol), the spectra of AB suggest a carboxylic acid (e.g. IR has an -OH and H-NMR peak at 12ppm).

Qu14: A
Need an alcohol that can dehydrate to give an alkene. Only the IR spectra A, D, E and AB show -OH groups. D is an aromatic alcohol (phenol) which don't react in the Lucas test and AB H-NMR suggests a carboxylic acid. The spectra of E suggests it's benzyl alcohol, C6H5CH2OH which reacts rapidly under SN1 due to the favourable carbocation intermediate.  A is cyclohexanol which can eliminate to give cyclohexene.

Qu15: AB
The DNP test gives an orange precipitate with aldehydes and ketones.  Only the IR spectra C and AB show C=O groups.  The spectra of AB suggest a carboxylic acid (e.g. IR has an -OH and H-NMR peak at 12ppm) which would not react with DNP.

Qu16: B
Alkenes react with bromine in chloroform. The C=C of an alkene can be seen in the IR of B near 1600cm-1 and the peaks in the H-NMR near 5.6ppm.

Qu17: AC
Basic compounds like amines dissolve in 10% HCl.  The IR of AC suggests an NH2 (near 3400cm-1).

Qu18: D
The tests indicate we need a moderate acid... a phenol not a carboxylic acid.  Only the IR spectra A, D, E and AB show -OH groups. D is an aromatic alcohol (phenol) which don't react in the Lucas test and AB H-NMR suggests a carboxylic acid.

Set 1 : Assign the configurations to A-E. A = RS, B =SS, C = RR, D = RS and E = RS (note the different alkyl group in E compared to the rest).

Qu19: AD
Meso compounds require RS where the two chirality centers have the same set of 4 groups attached.

Qu20: BC
Enantiomers are non-superimposable mirror images - they have the opposite configurations at chirality centers.

Qu21: AB or AC or BD or CD
Diastereomers are stereoisomers that are not enantiomers - the configuration at least one but not all centers are the same.

Qu22: B

Set 2: All about the reactivity of carbonyl compounds, ketones and esters.

Qu23: B
The most reactive of these ketones is the ring strained cyclopropanone.

Qu30: E
Step 1 is LDA, a strong base... this will create the carboanion adjacent to the nitrile group (compare with a C=O), it's a type of enoalte really. Step 2 will add a methyl group via an SN2 reaction of the carboanion (it's an enolate alkylation) then step 3 will hydrolyse the nitrile via the amide to the carboxylic acid.

Qu31: D
Step 1, the 1,3-diol is used to convert the aldehyde to a cyclic acetal. Step 2... the SN2 reaction of the acetylide anion on the alkyl bromide. Step 3 reduces the alkyne to the cis-alkene. Step 4 converts the alkene to the epoxide where the expoxide has the same stereochemistry as the alkene. B and C have the wrong stereochemistry. A and B have one too few carbons in the cyclic acetal. 

Qu32: C
Step 1 is the radical halogenation of the cyclohexane to give bromocyclohexane. Step 2 - SN2 with the PPh3 then the strong base forms the ylid ready for the Wittig reaction in step 3 with the ketone to give the alkene.  A and B have the wrong alkene.

Qu33: E
Step 1 is the electrophilic aromatic substitution, in this case chlorination and so it adds a Cl to a benzene ring - in this case it will add mainly para to the large aryl group in the ring without the nitro group since the nitro group deactivates the ring it is attached to. Step 2 is a Friedel-Crafts alkylation, this will also occur on the more activated ring, ortho to the aryl group.  A has the wrong regiochemistry as both the Cl and the methyl have added meta. B-D neither of the substitutions occur on the deactivated ring.

Qu34: D
Step 1 is the dissolving metal reduction of the alkyne to the trans-alkene. Step 2 converts the alkene to the epoxide where the expoxide has the same stereochemistry as the alkene. (i.e. trans). Step 3 - the Grignard reagent opens the epoxide in an SN2 type fashion (least hindered end attacked). 

Qu35: B
Step 1 is the electrophilic aromatic substitution, in this case bromination and so it adds a Br to a benzene ring para to the large t-butyl group.  In step 2, the aromatic bromide is converted to the Grignard reagent then in step 3, this is added to the methanal to give a primary alcohol. Finally, step 4, a controlled oxidation by PCC to give the aldehyde. A has too many carbon atoms in the aldehyde substitutent (we only added one C), C has too few carbon atoms and is a ketone. D is the alcohol obtained after step 3.  E ... PCC is not a strong enough oxidant to form this product.

Qu36: D
The starting material is a lactone... a cyclic ester. In step 1 the excess Grignard reagent will react with the ester to add two methyl groups giving after step 2, the work-up, a tertiary alcohol from the original carbonyl and a primary alcohol from the cleavage of the ester.  In step 3, an SN2 reaction will convert the alcohols to alkyl bromides and then step 4 is the E2 elimination to give the alkenes in this case a diene and in accord with Zaitsev's rule.  Only D and E are dienes, E has the wrong regiochemistry. 

Qu37: B
The product is an ester - these are typically formed from carboxylic acids and alcohols with an acid catalyst. The alcohol could come from the reduction of a ketone (since it would be a secondary alcohol).  The first step is ozonolysis with an oxidative work up ... this accounts for the formation of the carboxylic acid and the ketone from an alkene .... need to get the alkene in the right place and need a system with 6 carbon atoms.  Only B has the right number of C atoms and gives a keto-acid after ozonolysis.

Qu38: D
The product is 1,3,5-tribromobenzene. A quick glance at the reactions suggests we are looking at diazonium chemistry to remove an amine group.  So the first step in the formation of the amine by reduction of a nitro group.  Aniline is very activated and reacts with bromine to give the 2,4,6-tribromoaniline.

Qu39: A
PDC is used to oxidise alcohols, usually to aldehydes or ketones.  Here the product is an aldehyde (remember tertiary alcohols don't oxidise easily).  THe only system that could react to give a tertiary alcohol here is A by hydration of the alkene. The first step would reduce the ester to the primary alcohol.  Neither B nor C with undergo hydration to give a tertiary alcohol and D and E will be reduced by LiAlH4 to give diols (one secondary -OH and one primary -OH).

Qu40: D
The product is a conjugated enone, a typical aldol product.... the last two reaction steps are consistent with that... in this case it's an intramolecular aldol of 5-oxohexanal... disconnect the enone to reveal the carbonyl groups and then reconnect to see which alkene needed to be ozonised (reductive work-up, step 1). 

Qu41: B
The product is a highly substituted cycloehexane with specific stereochemistry. This suggests a Diels-Alder reaction.  Step 2 would convert an alkene into a 1,2-diol with cis stereochemistry - this information can be used to "reveal" the alkene. Reverse the Diels-Alder to reveal the required diene, 2,4-hexadiene. The opposite stereochemistry of the methyl groups in the product shows the stereochemistry required of the methyl groups in the diene - only B has the right stereochemistry (one cis and one trans).

Qu42: D
The product is an aromatic dicarboxylic acid... step 2 in the reactions suggests that the acid groups have been formed by the oxidation of aromatic alkyl group.  Step 1 is a Friedel-Crafts alkylation, these only work on activated aromatics - A, C and E are deactivated aromatics so no reaction.  B has a t-butyl group which does not oxidise to a carboxylic acid (it doesn't have a benzylic H). 

Qu43: A
Straight from the midterm. The reaction is the hydroboration-oxidation of an alkene to give 3-methyl-2-pentanol. The reaction gives the anti-Markovnikov alcohol via a syn addition due to the concerted addition of the B and H across the C=C. Note that the -OH is formed with retention of stereochemistry when the B atom is replaced.  Redraw one of the enantiomers so the -OH and the H that added are syn then remove them to reveal the C=C unit.  D and E would give 2-ethyl-1-butanol.  C would give 3-methyl-1-pentanol. B has the wrong stereochemistry.

Qu44: ABD
Need to add a t-buytl group via a Friedel-Crafts alkylation type reaction, here the t-butyl carbocation is formed by the reaction of the alcohol with the acid catalyst.

Qu45: AE
Now need to add a methyl group via a Friedel-Crafts alkylation reaction using RCl and AlCl3.

Qu46: CD
Radical chlorination of the benzylic methyl group via a stable benzylic radical.

Qu47: B
Alkylation of the enolate of an active methylene needs a base to prepare the enolate.

Qu48: A
Hydrolysis of the diester and loss of one acid group via decarboxylation - requires aq. base and heat (see section directions... heat can be assumed if needed).

Qu49: D
Convert the carboxylic acid to the ester : typical reaction is ROH plus an acid catalyst.

Qu50: BD
Convert the ethyl ester into a tertiary alcohol with two new methyl groups by using an organometallic reagent such as a methyl Grignard.

Qu51: AC
A little more difficult... an intramolecular alkylation... a Friedel-Crafts alkylation type reaction, here the t-butyl carbocation is formed by the reaction of the alcohol with the acid catalyst., here the carbocation is formed by the reaction of the alcohol with the acid catalyst.

Qu52: BC
And finally we need to add an acyl group via a Friedel-Crafts acylation type reaction, RCOCl plus AlCl3. 

Qu53: D
Hydroboration / oxidation of alkenes. The reaction is concerted (no carbocation, carbanion or radical intermediate), where the B is the electrophilic atom.

Qu55: C
Addition of HBr to alkynes... during the addtion of the second equivalent of HBr, the lone pairs on the Br already added helps stabilise the developing cation by resonance.

Qu56: B
Resonance of the N lone pairs of Nb and Nc with the adjacent C=O make them less nucleophilic.

Qu57: C
In this case, the enol is aromatic and hence very favourable.

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