353 Winter 2009 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: AB
The reactivity of carbonyl groups towards an organometallic reagent, looking at the relative reactivity of i amide, ii ester, and iii ketone. Electronic and steric factors need to be considered. First compare i and ii, esters tend to be more reactive than amides because the N is a better electron donor than an O (think about the electronegativity) and the N group is a poorer leaving group than the similar O group. Both factors make the amide less reactive than the ester.  Now compare iii and ii. The ketone attached group is a C atom and this is only a weak e donor and so the ketone C=O is more reactive (more electrophilic) than the ester system. Hence in terms of reactivity iii > ii > i.

Qu2: E
The reaction is radical halogenation (note the reaction conditions) which is controlled by the stability of the carbon radical formed as an intermediate. In i the methyl groups allows the formation of a stable benzyl radical and iii forms a very stable doubly benzylic radical. In contrast, ii can either form a methyl or phenyl radical, neither of which are very stable (no resonance). Therefore is terms of radical stability and therefore rate of reaction, iii > i > ii.

Qu3: E
Acidity... if you know your pKa's then this is easy : ketone =20, amine =35 and dicarbonyl / active methylene = 11.  Remember the lower the pKa the stronger the acid, so iii > i >ii. 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 two of the systems have the acidic H that are alpha to at least one carbonyl group. In the active methylene iii, the enolate can be resonance to two C=O groups making it more acidic than a simple ketone like i. Now compare the ketone and the amine. The key here is knowing that the N-H bond is not very acidic and that the effect of the C=O is more significant than the effect of the N.

Qu4: C
Allylic hydrogens are those on the carbon atoms adjacent to alkene C=C bonds, i.e. H-C-C=C. A has 5, B has 6, and C has 2.

Qu5: A
The systems are substituted alkenes.... electrophilic addition of H2SO4 to an alkene.... The key change is that there is a methoxy substitutent in i, a methyl group in ii and a trifluoromethyl group in iii. So ii will give a simple secondary carbocation. In i and iii the substituents will impact the stability of the cation. An alkoxy group, RO- is able to stabilise the +ve charge by resonance so i will be more reactive than ii (think of the electron donating alkoxy group making the C=C more electron rich). In contrast, the -CF3 group is electron withdrawing (due to Cl electronegativity) and will hence destabilise an adjacent carbocation making iii less reactive than i ... so i > ii > iii.

Qu6: E
We are looking at the acidity of a series of substituted phenols hence it's the aromatic substituent effects we need to focus on. An electron withdrawing substituent will increase the acidity of the parent phenol since the electron withdrawing group will stabilise the conjugate base. In the structures : i has a methyl ketone (moderate electron withdrawing), ii has a methyl group (weak electron donor) and iii has a nitro group (strong electron withdrawing). Therefore the acidity is iii > i > ii.

Qu7: C
Here we are looking the stability of carbanions, i and iii are sp3 hybridised systems and ii is sp. The more stable the anion, the less basic it is. So since the acetylide ii system is the least basic, it's the most stable (pKa = 25). If we compare i and iii, then the difference is the number of alkyl groups attached to the -ve C. i has one and iii has three. Since alkyl groups are weakly electron donating, more alkyl substituents makes the carbanion less stable (more basic).

Qu8: E
Draw out the structures and identify the alpha positions. i is a symmetrical ketone, 2 adjacent CH2 groups means 4 enolisable H. ii is an aldehyde where the CHO group is attached to a phenyl ring, therefore no enolisable H. iii is non-symmetrical ketone, with an adjacent methyl group and a CH2 group and hence 5 enolisable H, so iii > i > ii.

Qu9: D
The reaction is electrophilic aromatic substitution, a bromination and we need to look at the substituent effects on the aromatic ring. The nitrile group in i is connected via the carbon atom group and so is a strong electron withdrawing group via resonance - it's a deactivating group. The amino group in ii is a strongly activating group due to resonance effects. The phenyl group in iii is weakly activating. So the reactivity is ii > iii > i.

Qu10: C
 Resonance energy indicates the stability of the conjugated system. Aromatic systems have high stablisation and therefore high resonance energies. Since ii is a conjugated aromatic alkene, it the longest conjugated system and it will have the highest resonance energy. If we compare iii and i which both have 3 C=C, because i is aromatic while iii is a conjugated triene, i is more stable and has the higher resonance energy, hence ii > i > iii.

STRUCTURES and PROPERTIES:
You need to know about functional groups and reactions, and how to apply concepts related to structure such as hybridisation, aromaticity acidity, and reactivity.

Qu11: A
The set has two carboxylic acids (typical pKa = 5) and three phenols (typical pKa = 10). The benzoic acid with the electron withdrawing group will be the most acidic of this set since the withdrawing group will further stabilise the carboxylate conjugate base.

Qu12: E
There are phenols in B, C and E. E will be the most basic phenol since it has the weaker electron withdrawing group.

Qu13: BC
The SnCl2/HCl will reduce the nitro groups (electron withdrawing) to amines (electron donating). These means the nitrophenols, B and C will become less acidic once they have been reduced.

Qu14: A
This requires we identify a substituent that doesn't interact with the pi system and hence has only sigma bonds. Nitro groups (B and C) have a pi system that interacts through resonance. The methoxy group (D) is a resonance donor due to the O lone pair. The methyl ketone in E is resonance withdrawing.

Qu15: CD
A and B have 2 each in the CH2 group of the ethyl substituent. C has 6, 2 from each in the CH2 group of the two ethyl substituents and 2 more from the central CH2 group. D has 5, 2 from each in the CH2 group of the two ethyl substituents and 1 from central CH group. E has 2 from the CH2 group of the ethyl substituent not attached to the oxygen.

Qu16: D
What factors stabilise an enolate? 1,3-dicarbonyls give more stable enols due to the presence of an intramolecular hydrogen bond as part of a favourable 6 membered ring. Therefore in this set we are looking at C and D. So what differences are there in these two? D will give an enolate with more alkyl substituents on the C=C and hence it will be more stable.

Qu17: CDE
 The dicarbonyls will react with two equivalents of Grignard reagents to give diols and esters also react with two equivalents to give tertiary alcohols.

Qu18: BCD
Sodium borohydride will reduce aldehydes and ketones but not acid derivatives such as esters (E) and amides (A).

Qu19: ABDE
Carboxylic acid derivatives are all hydrolysed back to the parent carboxylic acid when heated with aq. acid. Since C is a ketone and therefore not a carboxylic acid derivative, it's the only exception.

Qu20: C
Sodium borohydride will reduce aldehydes and ketones (C) but not acid derivatives such as esters (D) and amides (E). The more reactive anhydride A and acyl chloride Bwill reduce first to the aldehyde and then benzyl alcohol.

Qu21: E
All about carboxylic acid derivative reactivity. Amides are the least reactive of the acid derivatives present.

Qu22: ABD
The ketone C will reduce to a secondary alcohol (see qu20). The amide E will reduce to an amine.

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: E
A and C are non-aromatic due to the sp3 C atom in the rings. B is a cyclic, conjugated pi system with 4 pi electrons and is therefore antiaromatic. D is non-aromatic because it does not contain a cyclic pi system.

Qu24: BCD
Heteroaromatic systems require a non-C atom to be present within the aromatic system. That means E is not a heteroaromatic system. A is non-aromatic due to the sp3 C atom in the ring. B is a cyclic, conjugated pi system with 2 pi electrons - the boron contributes an empty p orbital to the pi system - and is therefore aromatic. D aromatic because and includes a pyrrole unit.

Qu25: DE
A is aromatic - it contains a benzene unit. B is a cyclic, conjugated pi system with 6 pi electrons and is therefore aromatic. C is not a conjugated system, it only contains a single C=C pi system. D and E are conjugated trienes and non-aromatic because they do not contain a cyclic pi systems.

Qu26: D
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. B is a n=1, 6 pi electron system. E is cycloocatetraene and is non-planar and therefore non-aromatic.

Qu27: AE
B, C and D are all non-aromatic as drawn, but the sp3 C center can not be changed in a resonance contributor, hence these 3 structures don't have aromatic resonance contributors.

Qu28: C
A, B, D and E are all non-aromatic as drawn. This is because A, Band D all have sp3 C centers in the cyclic systems. E is non-aromatic as drawn because the protonated N center means there is no p orbital on the N creating the cyclic pi system. The N in C is part of a C=N system and therefore does have a system pi system. Loss of a proton from C creates the heteroaromatic compound pyridine.

Qu29: ABE
C is aromatic as drawn (see qu 28 above) and D would give an antiaromatic conjugate base when a proton is lost.

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: D
Working forwards, the bromide will react with the Mg to form the Grignard reagent which will then react with the epoxide at the least hindered end to form a secondary alcohol (B, C, or D). B is missing a C atom, and C has the wrong regiochemistry in the epoxide opening step.

Qu31: B
Working forwards, the ketone will react with LDA (a.k.a. lithium diisopropylamide), a strong base to form an enolate which is then methylated via an SN2 reaction. Reaction with the peracid will cause the Baeyer-Villager oxidation to give the cyclic ester (a lactone) where the oxygen has inserted on the more substituted side. The aqueous acid step hydrolyses the ester, breaking the ring to the carboxylic acid and the alcohol.

Qu32: C
Working forwards, the NBS (N-bromosuccinimde) results in a radical bromination at the benzylic position. Reaction with PPh3 (triphenylphosphine) then nBuLi results in the formation of the ylide ready for a Wittig reaction with the ketone to give an alkene. The key to separating answers C-E is the number of methyl groups... one from the bromide side of the Wittig and one from the methyl ketone.

Qu33: C
Working backwards, the product is a cyclic ester, the reagent sequence looks like borohydride reduction of an aldehyde or ketone produced by ozonolysis of an alkene or alkyne. Look at the ester product to reveal the parent carboxylic acid and secondary alcohol.... the secondary alcohol is formed by the reduction of a methyl ketone, in this case forming 5-hydroxyhexanoic acid, the carboxylic acid resulted from the preceding ozonolysis with an oxidative work-up. The required methyl ketone / acid, 5-hydroxyhexanoic acid, is the product of the ozonolysis of alkene C.

Qu34: AE
Working backwards, the last two steps are the formation of a ketone enolate using the strong base LDA (a.k.a. lithium diisopropylamide), which is then methylated via an SN2 reaction. PCC is an oxidising agent.... so a secondary alcohol has been oxidised to give the ketone. The first 2 steps in the sequence are benzaldehyde then an acidic work up.... this implies that the starting material was probably an n-propyl organometallic reagent to give the right chain system.

Qu35: A
Working backwards, the product is an alkene that could have been formed by an elimination of an alcohol probably formed by the reaction of an ester with an organometallic reagent. Esters react with 2 equivalents of Grignard reagents.

Qu36: A
Working backwards, step 3 is an electrophilic aromatic substitution, specifically Friedel-Crafts alkylation giving the t-butyl group. Steps 1 and 2 are diazotisation and subsequent chlorination. This means that the starting material is an unsubstituted amine.

Qu37: E
Working backwards, step 5 is the strong reducing agent LiAlH4 reducing carbonyl groups to alcohols, in this case a ketone to a secondary alcohol. The preceding sequence 1-4 is the nitration / reduction / diazotisation and subsequent substitution with cyanide. This means that the starting material is a monosubstituted aromatic methyl ketone.

Qu38: C
Working backwards.... The product is a conjugated aldehyde and the reaction conditions also suggest that we are looking at an aldol condensation here. To reveal the starting material, reverse the process, by breaking the C=C and adding in a "lost" carbonyl at the end of the C=C furthest from the product ketone. Count carbons carefully.... we are looking for a alpha substituted cyclohexanone, the substituent a butanal chain.

STARTING MATERIALS FOR SYNTHESIS:
Need to be able to work backwards.... but again look at the functional groups in the products to think about how you may 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).

Qu39: C
In order to convert ethylbenzene into the m-nitrobenzoic acid we need to oxidise the ethyl group (an o,p-director) to the carboxylic acid (m-director) then nitrate. Permanganate, under these conditions (acidic and heated) will oxidise an alkyl substituent to carboxylic acid group. Then an electrophilic aromatic substitution, specifically nitration using nitric and sulfuric acids. A gives the para-isomer. NaNO2 does not add the nitro group (it's used to diazotise primary amines). B and D give benzoic acid and E gives no reaction with steps 1-3 so yields 1-ethyl-4-nitrobenzene.

Qu40: D
In order to conver 3-hexyne into the appropriate hexan-3,4-diol, we first need to interpret the Fischer projection - this reveals that we are making the meso diol. The implication then is that we either make the cis-alkene (catalytic hydrogenation using Lindlar's catalyst) and then do a syn addition to give the diol or make the trans-alkene (Na / NH3 is a dissolving metal reduction) and do an anti-addition to get the diol. Cold alkali permanganate gives a diol via syn-addition. Aq. sulfuric acid is simple hydration to give an alcohol. A gives the wrong stereochemistry, B gives an alcohol not a diol, C would over reduce the alkyne to the alkane and E is missing the reduction step.

Qu41: D
In order to convert benzene into isobutylbenzene, we need to avoid the potential for a carbocation rearrangement. We can do this via Friedel-Crafts acylation followed by a reduction of the carbonyl group to a methylene (CH2). A and C will give t-butylbenzene, B gives no reaction, E would give an alcohol as the product.

Qu42: E
In order to convert 1-methylcyclohexene into 2-methylcyclohexene we are trying to convert a more stable alkene into a less stable one. This means we can't just use a carbocation rearrangement, we will need to add a potential leaving group via an anti-Markovnikov addition then use an E2 process and pay attention to the stereochemistry to make sure we get an anti-Zaitsev elimination.

Qu43: E
 In order to convert the two carbonyls into CH2 groups, different strategies can be used. Hydride reagents reduce carbonyls in aldehydes, ketones, esters and acid to alcohols, except that LiAlH4 reduces amides to amines. The ketone carbonyl can be reduced to a methylene using a Clemmensen or Wolff-Kischner reductions (but they do not reduce carboxylic acid derivative carbonyl groups). A will only reduce the ketone, B will reduce the amide to the amine and the ketone to a secondary alcohol, C will reduce the ketone to a secondary alcohol, D will protect the ketone as a cyclic ketal, then reduce the amide to the amine but then deprotection restores the ketone.

Qu44: D
In order to conver 3-hexyne into the appropriate hexan-3,4-diol, we first need to interpret the Newman projection - this reveals that we are making the meso diol. The implication then is that we either make the cis-alkene (catalytic hydrogenation using Lindlar's catalyst) and then do a syn addition to give the diol or make the trans-alkene (Na / NH3 is a dissolving metal reduction) and do an anti-addition to get the diol. Cold alkali permanganate gives a diol via syn-addition. Aq. sulfuric acid is simple hydration to give an alcohol. A gives the wrong stereochemistry, B gives an alcohol not a diol, C would over reduce the alkyne to the alkane and E is missing the reduction step.

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: E
Acidity...   Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-.....the anti-aromaticity of the the conjugate base of cyclopropene is the key issue here. In terms of the other answers, cyclopropene does have high ring strain, but that does not affect the acidity. The stability of the carbocation has nothing to do with the acidity of cyclopropene, it's the carbanion that's important.

Qu46: D
Since we have a weak nucleophile and an acid catalyst, the reaction proceeds via a cationic mechanism that is SN1 like. The reactive nucleophile is the neutral alcohol. It reacts with the benzyl end of the system due to the extra stability of the benzylic cation character. Note that the system does not have a more substituted end.

Qu47: D
The double bond character of the CN bond due to the resonance contributor from the donation of the N lone pair electrons to the C=O means that one methyl group is cis to the oxygen while the other is trans. Since there is restricted rotation about that C=N, the two methyl groups have different environments and hence different chemical shifts.

Qu48: E
The variation of the product of the reaction when changing the reaction conditions including temperature shows an example of kinetic and thermodynamic control. The weaker base triethylamine gives the more highly substituted product at the higher temperature and so reflects thermodynamic control.

Qu49: B
Remember that it is the substituent on the ring that controls the site of the substitution. -OR groups are strong electron donors and so direct o/p.

Qu50: E
LiAlH4 reduces esters to alcohols. Since this is a cyclic ester, the product will be a diol (the leaving group is still attached to the rest of the system).

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