353 Winter 2010 FINAL

Here is an post-mortem analysis / "how to" for this exam. 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.

STRUCTURES and PROPERTIES:
You need to know about functional groups and reactions, and how to apply concepts related to structure such as stereochemistry, acidity/basicity and reactivity etc.

Qu11: AD
Achiral molecules don't rotate plane polarised light.... but all the examples have chirality centers, so we need to identify the meso compounds. Assign the configurations as R or S at the chirality centers....The group priority order is Ph > CPh(H)CH₃ > CH₃ > H. Note that A can be identified as meso by virtue of the fact that each of the substituents are opposite in 3D space across the center C-C to the like group.

Qu12: BC or BE
Enantiomers are non-superimposable mirror images and therefore have opposite configurations are all chirality centers.

Qu13: BCE
The required trans-alkene (Ph(CH₃)C=C(CH₃)Ph) reacts with H₂ / metal catalyst via a syn addition to give two enantiomers.

Qu14:CE
In order for two of these structures to rotate light in the same direction, they will need to have the same configurations.

Qu15: B
All the structures contain N atoms, but only 3 are amines... the lone pairs on N in amines makes them basic. The primary amine in B is the most basic because there is no resonance of the N sp3 lone pair with the pi system (as there is in A). In D the lone pair is in an sp2 hybrid and is therefore closer to the N nucleus and less available.

Qu16: CE
The formal charge on the N in the neutral amines in A, B and D is zero. In a nitro group, C, and an ammonium group, E the N has a +1 formal charge.

Qu17: ABE
Primary and secondary amines react with nitrous acid (formed from NaNO₂ and H+) to give diazonium salts.

Qu18: B
Acidity... of carbonyl (enolate) systems. In general terms, aldehydes > ketones > esters in terms of their ability to stabilise enolates so the beta-ketoaldehyde B is the most acidic of these systems.

Qu19: E
Across the set of 5 compounds most of the H are alpha-hydrogens next two carbonyl groups except E where they are only next to one carbonyl (hence less resonance stabilisation of the conjugate base).

Qu20: D
Sodium borohydride is a milder hydride reagent that reduces aldehydes and ketones or more reactive carbonyls but not ester or less reactive carbonyls. Hence the diester, D, is the only system that would not be reduced.

Qu21: CD
Systems where a carboxylic acid group is beta to another carbonyl group can be decarboxylated. Carboxylic acid derivatives (e.g. esters) are readily hydrolysed to carboxylic acids under these conditions.

---

AROMATICITY and RESONANCE:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria (cyclic, planar, conjugated pi system with 4n+2 pi electrons).

Qu23: ACD
B and E do not have cyclic pi systems (note the sp3 C in the ring) and are therefore nonaromatic. D is aromatic because the boron atom has an empty p orbital available to contribute to the pi system and the carbanion adds 2 e to the 4 from the 2 C=C to make a 6 pi system.

Qu24: AD
Antiaromatic systems satisfy the first three aromaticity criteria but are 4n systems and therefore do not meet the Huckel rule. B is aromatic (one S lone pair is in the 6e pi system), C is aromatic because the boron atom has an empty p orbital contributed to the conjugation of to the 2e pi system) and E is aromatic because the carbocation center has an empty p orbital contributed to the conjugation of to the 2e pi system).

Qu25: A
B is a cyclic, conjugated pi system with 4 pi electrons and is therefore antiaromatic. C is a cyclic, conjugated pi system with 6 pi electrons and is therefore aromatic. D and E are not conjugated systems.

Qu26: C
In order to be aromatic hydrocarbons, they can only contain C and H atoms, no heteroatoms such as N. If n=2 in the Huckel rule it means we need 10 pi electrons. A is 10 pi electrons, but non planar and therefore is non-aromatic. B is not a conjugated system, D is not a hydrocarbon and E has 14 pi electrons (i.e. an n=3 aromatic system).

Qu27: ACD
B has an aromatic tautomer but not an aromatic resonance structure. Remember that atoms can not move in resonance contributors. The sp3 carbon in E prevents the aromatic character. In A, C and D the charge separation in the resonance contributor can create an aromatic system. In A and D these are 5 membered anionic rings and in C its a 7 membered cationic ring.

Qu28: AD
B and E are all non-aromatic as drawn due to the sp3 C centers in the cyclic systems. The lone pair on the N in C is part of the aromatic system and therefore has a non-aromatic conjugate acid.

Qu29: AC
B, D and E are aromatic as drawn. Removal of H+ from the structure A and C gives a anionic and neutral aromatic system respectively.

---

STARTING MATERIALS AND PRODUCTS:

If you are trying to find the product, then you should probably just work forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably the best way to go....
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu30: C
Working forwards, the peracid reacts with the alkene to give an epoxide. The Grignard reagent will then react with the epoxide at the least hindered end to form a tertiary alcohol (B or C). B has the -OH on the wrong carbon atom, and D has the wrong regiochemistry in the epoxide opening step.

Qu31: B
Working forwards, hydroboration and oxidation of the alkene gives the anti-Markovnikov primary alcohol. PDC is an oxidising agent giving the aldehyde which is then treated with an ylid in a Wittig reaction to give the alkene. A has too many carbon atoms and C comes from the wrong regiochemistry in the first step.

Qu32: A
Working forwards, thionyl chloride reacts with the carboxylic acid to make the acid chloride. This then reacts with the secondary amine to give the N,N-dimethyl amide.

Qu33:B
Working forwards, the aromatic ester undergoes an electrophilic aromatic substitution, specifically Friedel-Crafts alkylation on the more reactive ring (left RO- substituted aromatic) giving a para methyl group. This is followed by another electrophilic aromatic substitution, specifically nitration again on the more activated ring. A and D have the wrong regiochemistry for both the electrophilic aromatic substitutions since the carbonyl group deactivates that ring. C has the wrong regiochemistry for nitration since both the methyl and the alkoxy group activates the other ring. E has the wrong regiochemistry since the first reaction is directed para by the alkoxy group.

Qu34: D
Working backwards, the last step is hot acidic permanganate that has caused the oxidation of a benzylic substituent to give the benzoic acid. Step 1 is an electrophilic aromatic substitution, specifically Friedel-Crafts alkylation giving the t-butyl group. Remember that these alkylations only work on activated aromatics and that t-butyl groups can't be oxidised by permanganate.

Qu35: C
Working backwards, the product is a 1,2-diol. Looking at the reagents, this has come from the ring opening of an epoxide via what amounts to an anti-addition. Analysis of the diol product indicate the regio and stereochemistry on the required alkene.

Qu36: A
Working backwards, LDA is lithium isopropyl amide, a strong base, so it looks like we are making an anion, probably an enolate and then methylating it via an SN2 reaction. This methyl group is attached to the carbon alpha to the ketone. The 2nd step, acidic hydrolysis is not easy to make sense of on its own, but if we look at the first step, we should be able to make some sense of it. The dicarbonyl is being treated with base, so again it's enolate chemistry and it is being used to make the 5 membered ring via a double alkylation. This means the 1,4-dibromide A is what we are looking for.

Qu37: D
Working backwards, looking at the product and the 3rd set of reagents, it looks like we made the two ethyl esters probably from the carboxylic acids (it looks like a Fischer esterification). Based on the 2nd step, this was the ozonolysis of an with an oxidative work-up to give the dicarboxylic acid, so we can reveal the starting alkene by reconnecting the carbonyl groups. Looking at step 1, the cyclopentadiene and heat suggests we did a Diels-Alder reaction, so reverse the arrows in the product to reveal the starting material, D.

Qu38: A
Working forwards, the aldehyde ester will be reduced by the excess hydride to the 1,3-diol. Reaction of the 1,3-diol with the ketone will form a cyclic ketal. The key here is getting the right number of atoms in each ring. B would be formed by reaction with cyclohexanone with 1,2-ethanediol. D has too many carbon atoms (it would be formed by reaction with cyclohexanone), and E has too few carbon atoms (it would be formed by reaction with cyclopentanone with 1,2-ethanediol).

Qu39: B
In order to convert benzene into the m-nitrobenzoic acid we need to introduce a carbon substituent and a nitro group. The key issues are the Friedel-Crafts don't work on deactivated aromatics and nitro groups and carboxylic acid groups interfere with Grignard reactions. These factors rule out A, C and D. E does not work because aryl bromides don't react with carbon dioxide. The route that works is a Friedel-Crafts acylation to methyl ketone (m-director) then permanganate oxidation, under these conditions (acidic and heated) will oxidise the substituent to a carboxylic acid group. Then an electrophilic aromatic substitution, specifically nitration using nitric and sulfuric acids.

Qu40: A
In order to convert 2-pentyne into the appropriate pentan-2,3-diol, we first need to interpret the Fischer projection. The implication is that we either make the cis-alkene (catalytic hydrogenation using Lindlar's catalyst) and then do a anti addition to give the diol or make the trans-alkene (Na / NH₃ is a dissolving metal reduction) and do an syn-addition to get the diol. Cold alkali permanganate or osmium tetroxide / aq NaOH gives a diol via syn-addition. B gives an alcohol not a diol, C and D give the wrong stereochemistry, and E has the wrong nucleophile in the last step (it needs water not methanol).

Qu41: D
We need to convert the carboxylic acid group into a primary alcohol via a reduction, but in order to do this, we will need to protect the aldehyde first and then deprotect it after the reduction since the aldehyde is more reactive than the acid. A and B have the steps in the wrong order. C is a reduction followed by oxidation so it would give the dialdehyde. E does not give the desired product because NaBH₄ is not a strong enough reducing agent to reduce the acid.

Qu42: B
In order to convert methylbenzene into (2-methyl-1-propenyl)benzene a quick analysis shows that we need to make a new CC bond and add a total of 3 new C atoms. Two possible routes spring to mind, via a Grignard reagent / elimination route or via a Wittig reaction : B radically brominates the benzylic position then prepares the ylid ready for the Wittig reaction. A fails because the first reaction would not give the required benzyl halide (HCl does not undergo radical reactions so this is not a good route to the required halide). C does not work because Grignards don't react well with halides and it certainly would not give an alkene. Similarly for D.... for the last step to cause an elimination to give an alkene, we would need an alcohol from step 3. Again for E oxidation to the acid then reduction to the primary alcohol followed by conversion to the bromide is fine, but again Grignards don't react well with halides and it certainly would not give an alkene at the end.

Qu43: A
If we look at the starting material, a substituted cyclohexene and the product, two important things should be noted (1) the ring has been broken and (2) we have gone from a 10 carbon system to a 6 carbon system. Alkenes can be cleaved in two common ways, ozonolysis and periodate oxidative cleavage of a diol formed from the alkene (e.g. cold alkali permanganate gives a diol via syn-addition). Both ozonolysis and periodate appear in the possible answers, what it less obvious is how 4 carbon atoms are lost. In A we see the formation of the diol and then the periodate oxidative cleavage, this will give a diketone. MCPBA will oxidise the ketone in a Baeyer-Villager oxidation to give a diester by inserting on the more substituted side (it gives a diethanoate system, i.e. CH₃CO₂R) which then undergoes an acid catalysed ester hydrolysis leaving the required diol after loss to 2 molecules of ethanoic acid. B : the ozonolysis with the reductive work up gives a diketone which is then reduced by borohydride to a diol.... but no carbons are lost. C is similar, the lithium aluminium hydride reduces the diketone to the secondary diol then acid catalysed dehydration would give a ten carbon isolated diene. D is ozonolysis with an oxidative work up the diketone again, the lithium aluminium hydride reduces the diketone to the secondary diol but we still have ten carbons. E alkenes react with MCPBA to give epoxides, which is converted to a diol then cleaved by periodate to give a diketone which is then reduced by borohydride to a diol.... but still ten carbon atoms.

Qu44:C
In order to convert methylbenzene into triphenylmethanol, a quick analysis shows that we need to add two more phenyl groups, probably via Grignard chemistry with a carbonyl group (since it's common). This is achieved in C where the starting material is oxidised to benzoic acid, converted to the methyl ester and then reacted with excess phenyl magnesium bromide to give the required tertiary alcohol.

---

EXPLANATION OF PHENOMENA

Need to be able to work out not only which statements are true, but also if it is the correct rationale for the issue under consideration.

Qu45: D
When temperature affects product distribution is this way, we typically have a case of kinetic and thermodynamic control. At the lower temperature, the reaction is under kinetic control where it is the rate of formation of the carbocation intermediate that is important and this is affected by the stability of the carbocation.

Qu46: D
There are four structural criteria for aromaticity. [14]-annulene implies that is has 14 pi electrons which means that it satisfies the Huckel rule (where n=3). An inspection of the structure, shows that is does contain a cycle of conjugated pi systems. So it appears that the planarity is the issue. In [14]-annulene, it is the H atoms in the interior of the ring that are too close to each other (i.e. steric strain) that destabilises the planar conformation.

Qu47: B
When ranking basicity, looking at the availability of the lone pair electrons is the best approach. For N1, the pyridine N, the lone pair is in an sp2 hybrid orbital that is perpendicular to the aromatic pi system. For N2, the amino group N, the lone pair is in a p orbital that is involved in a resonance interaction with the aromatic pi system. Therefore, N1 lone pair electrons are more available than those on N2 and so N1 is more basic than N2. This means that on protonation of N1 to give it's conjugate acid, the amine group can stabilise the positive charge due to its electron donating character.

Qu48: C
The reaction shown is a hydride reduction of the ketone to a secondary alcohol via a nucleophilic addition. The key thing is to remember that the attacking species is an H- (not the HO- group), this H is the one attached to the same carbon atom as the -OH group. In the major product, this H has added to the bottom face due to the steric hindrance caused by the methyl substituent on the bridging carbon that partially blocks the top face.

Qu49: C
The key difference between X and Y is the alkene C=C, a question all about substituent effects. This means that the substituent in X is really just an alkyl group while that it Y is a conjugated nitrile. Alkyl groups are weak electron donors and direct ortho / para while nitriles are strong electron withdrawers and direct meta.

Qu50: E
LiAlH₄ reduces amides to amines. Since this is a cyclic amide, the product will be the cyclic amine.

Return to Homepage