353 Winter 2011 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.

RELATIVE PROPERTIES:
Identify the controlling feature, which is not always as obvious as it may appear. Look for two pairs of similar systems to compare that have minimal differences in structure. If a compound is named, draw it out. If a reaction is involved, identify the type of reaction and then what the controlling factors are.

Qu1: D
Acidity... if you know your pKa's then this is easy : aldehyde = 17,ester = 25 and ketone =20.  Remember the lower the pKa the stronger the acid, so ii > iii >i. 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 of the systems have the acidic H that are alpha to a carbonyl group. In the aldehyde i , ester ii and the ketone iii, the conjugate bases (they are enolates) can be resonance stablised by the C=O group. The difference between each carbonyl system is the related to the difference between the effect that the substituent (-H, -OR and the -R group respectively) has. Since an alkoxy group (-OR) is strongly electron donating, it will destabilise the conjugate base the most making the ester the least acidic. An alkyl group (-R) is slightly electron donating, it will destabilise the conjugate base making the parent ketone iii less acidic than the aldehyde i. . Overall then in terms of pKa, ii > iii > i.

Qu2: C
The basicity of phenolates is determined by the electron availability of the electrons on the -ve O atom. This is affected as the substituents change. In this question we have 3 electron withdrawing substituents which stablise the -ve O to different extents, the stronger the withdrawing effect the less basic the O atom is (since the electron withdrawing group decreases the availability of the electrons and this makes the base less able to donate those electrons). The first two groups in i and ii are both halogens, which are slightly electron withdrawing group via inductive effects due to electronegativity.... so since -F is more electronegative than -Cl, the -F has the greater electron withdrawing effect. The nitro group in ii is a strongly deactivating group due to resonance and inductive effects. So the basicity is ii > i > iii.

Qu3: D
The reaction is an intramolecular aldol type reaction of a symmetrical diketone. These typically give a beta-hydroxy carbonyl (like ii or iii) and intramolecular examples typically favour 5- or 6- membered rings.... so ii > iii > i.

Qu4: C
The reactivity of carbonyl groups towards an organometallic reagent, methyl lithium, depends on the electrophilicity of the carbonyl carbon. Looking at the carbonyls, we have i an aldehyde, ii a trichloro substituted aldehyde and iii a dimethyl substituted ketone. Electronic and steric factors need to be considered. First compare the simple aldehyde i and the ketone iii. Simple aldehydes are usually more reactive than simple ketones because the extra alkyl group of a ketone electronically and sterically makes the carbonyl less reactive. The tri-Cl in ii means that there is an electron withdrawing substituent and so it will make the aldehyde C=O more +ve and hence more electrophilic and more reactive than aldehyde i. So overall, ii > i > iii.

Qu5: E
 The reaction is electrophilic aromatic substitution, a bromination, and we need to look at the substituent effects on the aromatic ring. The halide group in ii is a slightly electron withdrawing group via inductive effects - it's a deactivating group. The methyl group in i is a slightly activating group due to hyperconjugation. The amide group in iii is moderately activating due to resonance donation of the N lone pair. So the reactivity is iii > i > ii.

Qu6: C
The systems are substituted alkenes so we are looking at electrophilic addition of HCl to an alkene . A closer look at the structures shows that the three systems have similar base structures (i.e. alkenes) each with a different substituent: i has an alkyl group, ii has an alkoxy group and iii a nitro group. Each alkene will react with the HCl to give a carbocation and the substituent will affect the availability of the pi electrons: electron donating substituents will make them more available and electron withdrawing will make them less available. An alkoxy group, RO- is a strong electron donor, an alkyl group, -R, is a weak electron donor and a nitro group is a strong electron withdrawing. So ii will be more reactive than i (think of the electron donating alkoxy group making the C=C more electron rich) and iii the least reactive... so ii > i >iii.

Qu7: C
The reaction is electrophilic aromatic substitution, a nitration and we need to look at the substituent effects on the aromatic ring. The halide group in i is a slightly electron withdrawing group via inductive effects - it's a deactivating group but ortho/para directing. The isopropyl group in ii is a slightly activating group due to hyperconjugation and ortho/para directing - it's steric size will favour para since the ortho groups will be partially blocked. The ester group in iii is connected via the carbonyl and is strongly deactivating due to resonance, a meta directing group. So the para yield will be ii > i > iii.

Qu8: B
Resonance energy indicates the stability of the conjugated system. Aromatic systems have high stablisation and therefore high resonance energies. Let's compare benzene i with the pyridine system iii (note even though the N is +ve, this cation is an aromatic system) then we need to recall that benzene is more aromatic than pyridine and therefore benzene has the higher resonance energy, hence i >iii. Now look at ii. In this pyrrole type case the +ve N and this means that the lone pair from the pyrrole type N is not available to complete the aromatic system and hence iii is non aromatic. Therefore it has a low resonance energy more like a diene. So i > iii > ii.

Qu9: D
First identify the alpha positions. i is an alkyl bromide meaning it lacks a carbonyl and therefore has no enolisable H. ii is a cyclic ketone nd the C=O group is between 2 CH2 groups meaning a total of 4 enolisable H. iii is an aromatic ketone with an phenyl group and a CH3 group and hence 3 enolisable H, so ii > iii > i.

Qu10: B
The reactivity of carboxylic acid derivative carbonyl groups towards hydrolysis looking at the relative reactivity of i acyl halide, ii ester, and iii acyl anhydride. The electronic factors of each of the substituents / leaving group on each of the carbonyl groups need to be considered. In i the Br is a weak electron withdrawing group and a good leaving group. In ii the methoxy group is a strong electron donor and a poor leaving group. In iii the carboxyl group is a moderate electron donor and a reasonable leaving group. Electron donors on the carbonyl make the carbon less electrophilic and less reactive, a better leaving group makes the acid derivative more reactive. Hence in terms of reactivity i > iii > ii.

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
The equatorial positions are those around the plane of the ring rather than pointing perpendicular to the ring (axial).

Qu12: CE
The easiest way to solve this problem is assign the configurations as R or S at each of the two chirality centers in each molecule. A = (1S, 2S), B = (1R, 2R), C = (1S, 2R), D = (1R, 2S) and E = (1S, 2R).

Qu13: AB
1-methylcyclohexene will react with Br2 to give the 1,2-dibromide via an anti-addition - this means that the two Br atoms are on opposite sides of the ring i.e. trans. The relative positions of the Br atoms are trans in A and B, and cis in C, D and E.

Qu14:AB or CD or DE
Enantiomers are non-superimposable mirror images and therefore have opposite configurations are all chirality centers. Use the configurations from Qu12 to identify the structures with opposite configurations.

Qu15: A
N atoms where the N atoms and / or their lone pairs are not part of aromatic systems are more basic since they are more available (i.e. the lone pairs are not extensively resonance delocalised).

Qu16: B or C or E
Apply the four criteria for aromaticity (cyclic, planar, conjugated pi system with 4n+2 pi electrons). A and D are not aromatic because they do not have cyclic pi systems due to the sp3 C at the top of the drawn structures.

Qu17: D
Each peak in a normal 13C NMR represents a type of C... hence count the number of C types in each structure...A = 3, B = 3, C = 1, D = 4 and E = 1.

Qu18: A
First identify the alpha positions adjacent to each of the carbonyl groups then sum the number of H attached to each adjacent C....A = 8, B = 5, C = 5, D = 2 and E = 6.

Qu19: B
The most deshielded peak (highest chemical shift) will be the aldehyde H in B (9-10 ppm).

Qu20: E
The reaction involved is a Friedel-Crafts acylation which uses either an acyl halide or an anhydride. A = diketone, B = ketone / aldehyde, C = ketone / ester , D = diester and E = anhydride.

Qu21: B
Since A-D all have alpha-hydrogens between two carbonyl groups (this means more resonance), E is the least acidic of these systems where the alpha-H are only adjacent to one carbonyl group. In terms of their ability to stablise -ve charge, aldehydes > ketones > esters, therefore the ketone / aldehyde B has the most acidic H.

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).

Qu22: E
The aromatic compounds here are C, E, AC and AE. For n = 2 in the Huckel rule, we require 10 pi electrons. C, AC and AE are all 6 pi electron systems.

Qu23: AB or D
The non-aromatic compounds here are A, B, D, AB and BC. A and B each have two C=C and are dienes. BC is non-planar (the anti-aromatic character of the planar conformation destabilises it) and has 4 C=C and is therefore a tetraene.

Qu24: AD
Antiaromatic systems satisfy the first three aromaticity criteria but are 4n systems and therefore do not meet the Huckel rule. The aromatic compounds here are C, E, AC and AE. The non-aromatic compounds here are A, B, D, AB and BC.

Qu25: B
Conjugated pi systems have "linked" pi systems (i.e. pi systems that extend over 3 or more atoms).

Qu26: E
AC, AD and AE are not neutral. The system with the highest resonance energy will be the aromatic system with the most contributing C=C.

Qu27: C or E or AB.
The aromatic compounds here are A, C, E and AB. For n = 2 in the Huckel rule, we require 10 pi electrons. C, AC and AE are all 6 pi electron systems. Of these, the conjugate acid of A (formed when H+ adds to the N) is non-aromatic since the lone pair on the N in A was part of the aromatic system.

Qu28: AE
The non-aromatic compounds here are D, AC, AD and AE. AC and AD have aromatic resonance structures. Remember that atoms can not move in resonance contributors. The tautomer of AE is aniline (i.e. aminobenzene, C6H5NH2)

Qu29: AC or AD
The non-aromatic compounds here are D, AC, AD and AE. AC and AD have aromatic resonance structures. Remember that atoms can not move in resonance contributors.

Qu30: B
The aromatic compounds here are A, C, E and AB. The non-aromatic compounds here are D, AC, AD and AE.

Qu31: E
The aromatic compounds here are A, C, E and AB. For n = 2 in the Huckel rule, we require 10 pi electrons. A, C and AB are all 6 pi electron systems.

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.

Qu32: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 since the -OH ends up and the end opposite to where the nucleophilic Grignard reagent attacked.

Qu33:A
Working forwards, hydroboration and oxidation of the alkene gives the anti-Markovnikov primary alcohol. PCC is an oxidising agent giving the aldehyde which is then treated with an ylid in a Wittig reaction to give the C6 alkene (i.e. A, B or C). B has the C=C in the wrong location and C has too few carbon atoms.

Qu34:B
Working backwards, the product is a 1,2-diol. Looking at the reagents, this has come from the syn-addition to an alkene. This reveals a substituted cyclohexene formed via a Diels-Alder reaction, where the dienophile is a reagent in step 1, so we are looking for the diene. Note that since the two methyl substituents in the product are on opposite side (i.e. trans), then the two methyl groups in the diene need to have opposite stereochemistries in the diene.

Qu35:E
Working forwards, the first step is an electrophilic aromatic substitution, specifically Friedel-Crafts acylation adding a methyl ketone (i.e. an acetyl group) to the N substituted ring (since it is the more activated ring). The second step is hydride reduction by excess lithium aluminium hydride that reduces both the amide to an amine and the ketone to a secondary alcohol, so E.

Qu36:C
Working backwards, the product is an n-alkyl substituted aromatic ether. The second step is the Clemmensen reduction using Zn/Hg and HCl to reduce a C=O to a CH2. Step 1 is an electrophilic aromatic substitution, specifically Friedel-Crafts acylation giving the aromatic ketone para to the strongly activating methoxy group. To get the 3C n-alkyl chain it means we needed to start with the 3C acyl chloride, C. A, B and D would all give alkylation reactions to give branched alkyl substituents.

Qu37:C
Working backwards, the 1,2-diol has been formed via acidic hydrolysis of the epoxide from step 1, the peracid reacts with the alkene to give an epoxide. The formation of a diol via these reactions is an overall anti addition to the original alkene. Therefore, we need to manipulate the Fischer diagrams into wedge-hash diagrams and then rotate those in to the conformation with the -OH groups anti and hence reveal the alkene stereochemistry. D has the wrong alkene structure and E is the wrong stereochemistry.

Qu38:B
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 a carbonyl group, 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 ketoester is being treated with a base (ethoxide) so again it's enolate chemistry and it is being used to make the 6 membered ring via a double alkylation. This means the 1,5-dibromide B is what we are looking for.

Qu39:C
Working forwards, the first step is the Grignard reagent reacting with the cyclic ester to give a diol, a tertiary -OH and a primary -OH after the acidic work up (step two). This is followed by conversion to the diol to a dibromide, via SN2 with PBr3, followed in the last step by double E2 elimination of the alkyl bromides to the diene.

Qu40:D
Working forwards, the first step is the reduction of the diester 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. A would be formed by reaction with cyclopentanone with 1,3-propanediol. B would be formed by reaction with cyclohexanone with 1,2-ethanediol. C would be formed from an intramolecular system (1,8-dihydroxyoctan-4-one), and E has too few carbon atoms (it would be formed by reaction with cyclopentanone with 1,2-ethanediol).

REAGENTS FOR SYNTHESIS:
Need to be able to look at reactions, looking at the functional groups in the starting materials and products to think about how you have got there.
 How well do you know your reagents ? Look at what has actually happened in terms of the reaction functional group transformation and then first look for any regiochemical issues then finally the stereochemistry last (it's the hardest to sort out).

Qu41:E
Overall we need to achieve the alkylation of two benzenes... the only methods we considered were based on Friedel-Crafts type chemistry. Note that the reaction of Grignard reagents with halides is not a good way to make new C-C bonds.

Qu42:C
The product has a 1,3-dicarbonyl and therefore is suggestive of enolate chemistry using condensation reactions. Given the starting material being an alkene, we are going to need to generate two carbonyls and use a base to generate the enolates. So ozonolysis will generate the carbonyls. B gives an aldehyde / ketone which will undergo an aldol to give a hydroxy ketone... i.e. gets us to the wrong oxidation level. C gives the carboxylic acid / ketone which is first converted to the ester / ketone before doing the condensation reaction.

Qu43:B
Notice that we need to add one C atom to the system via a new C-C bond. Note that the reaction of Grignard reagents with halides is not a good way to make new C-C bonds, Grignards react well with aldehydes, ketones, esters and epoxides to give alcohols.

Qu44:E
It looks like oxidation of the amine group to a nitro group... but the oxidation conditions in each option would also oxidise the benzylic position. So the best option is to remove the amine via diazonium reactions and then nitrate to add the nitro group.

Qu45:B
The product is a cyclic hemi-acetal which would be formed by the intramolecular reaction of a 4-hydroxybutanal. In order to get 4-hydroxybutanal from the starting material, we need to reduce the ester, but since the ester is less reactive than the aldehyde we need to protect the aldehyde first.... so convert the aldehyde to a cyclic acetal, then reduce the ester, then deprotect and catalyse the hemi-acetal formation (acid).

Qu46:C
All about the stereochemistry. Looking at the starting material and the products, the approach looks to be based on alkyne to alkene to 1,2-dibromide. Since the bromination of alkenes goes via anti addition, we should redraw the Newman projection of the product to reveal the required alkene stereochemistry... this means rotating the Newman projection with the Br at 180 degrees - which reveals the trans relationship of the methyl and ethyl groups and hence the need for the dissolving metal reduction.

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.

Qu47:B
Methyl benzoate has an ester group, -CO2R, on the benzene ring and this group directs the nitro group meta.

Qu48:C
Draw the two named structures and / or look at the functional groups and reagents to reveal the reaction as hydration of an alkene. Looking at the name structures, we can deduce there has been a rearrangement of the intermediate carbocation, a 1,2-methyl shift.

Qu49:D
The two N systems have quite different acidities / basicities - a factor of 1,000,000,000 difference. If we use the typical approach to acidity problems of looking at the stabilities of the conjugate acids (or bases) we should recognise the aromaticity implications of these pi systems. We can come to similar conclusions by considering the availability of the basic N lone pairs which relates to how they are involved in the aromatic pi system.

Qu50:D
The deactivated ammonium ion would direct meta, but the much more activated free amine would direct ortho / para.

Qu51:D
In order for Hb to be removed, it needs to be acidic and this requires that there needs to be a resonance interaction of the conjugate base with the carbonyl but the geometry of the ring system means that the -ve charge and the C=O are not aligned and hence there is no resonance stabilisation.

Qu52:C
The reactivity of carbonyl groups towards nucleophiles depends on the electrophilicity of the carbonyl carbon. . Electronic and steric factors need to be considered, which for an aldehyde and a ketone means comparing -H and an -R group.

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