353 Winter 2008 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: A
The reactivity of carbonyl groups towards a hydride reagent, looking at the relative reactivity of i nitro aldehyde, ii aldehyde, and iii ketone. 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.  The reactivity of carbonyl systems is impacted by the substituents attached to the carbonyl, here that is -H and -R. The stronger that group is as an electron donor, then the less electrophilic the carbonyl carbon is. Now what about the two aldehydes ? The nitro group in i is a strong electron withdrawing group via resonance. This makes the C=O in i more +ve and more electrophilic than a simple aldehyde like ii. Hence in terms of reactivity i > ii > iii.

Qu2: D
The reaction is electrophilic aromatic substitution, a nitration, and we need to look at the substituent effects on the aromatic ring.  The ester group in i is connected via the carbonyl group and so is a strong electron withdrawing group via resonance - it's a deactivating group. The methoxy group in ii is a strongly activating group via resonance. The ester group in iii is attached to the ring via the O attached by single bonds (i.e. the alkoxy O) of this 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 but due to the adjacent C=O where there is competing resonance, it's not a activating as the methoxy group in ii. So the reactivity is ii > iii > i.

Qu3: AB
Acidity... if you know your pKa's then this is easy : ester enolate =25, alcohol =17 and dicarbonyl / active methylene = 11.  Remember the lower the pKa the stronger the acid, so iii > ii > 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 two of the systems have the acidic H are alpha to at least 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 alcohol ii is more acidic than the ester i.  In the active methylene iii, the enolate can be resonance to two C=O groups making it more acidic than a simple alcohol like ii.

Qu4: A
The easiest way to answer this, is by knowing that alkenes typically undergo electrophilic addition with hydrogen halides where the first step (also rate determining step) is the addition of the proton (the electrophile) to give the carbocation. Since iii is a nucleophile, it is quite unreactive towards alkenes. If we now look at i and ii, then it is the higher acidity of HI that makes it more reactive than HBr so overall we have, i > ii > iii.

Qu5: C
The systems are substituted alkenes.... electropilic addition of H2SO4 to an alkene.... this is governed by the stability of the carbocation produced. If you look at the three alkenes, the key change is that there is a methyl substitutent in i, a methoxy group in ii and a trichloromethyl group in iii. So i will give a simple secondary carbocation. In ii 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 ii will be more reactive than i (think of the electron donating alkoxy group making the C=C more electron rich). In contrast, the -CCl3 group is electron withdrawing (due to Cl electronegativity) and will hence destabilise an adjacent +ve charge making iii less reactive than i ... so ii > i > iii.

Qu6: AB
The reaction is electrophilic aromatic substitution, a nitration, 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 and directs meta. The methyl group in ii is a weakly activating group due to inductive effects (hyperconjugation) and is ortho / para directing. The t-butyl group in iii is like the methyl group in ii and is ortho / para directing. The steric size of the t-butyl group blocks the adjacent ortho positions to some extent and so the yield of the para isomer is increased. So the reactivity is iii > ii > i.

Qu7: C
Draw out 1-methyl-1,3-cyclopentadiene, this should help you recognise the reaction as the Diels-Alder reaction . In terms of stereoisomers we need to think about (i) regioselectivity and (ii) endo/exo selectivity. Reaction of i can give a single regio isomer but as either endo or exo. The reaction of ii can give two regio isomers each of which has an endo and exo isomer. While with iii only one isomer can be formed. So the number of isomers possible is 2, 4 and 1 respectively, hence ii > i > iii.

Qu8: B
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 with just 1 adjacent CH2 and therefore 2 enolisable H. iii is an ester, with an adjacent methyl group and hence 3 enolisable H, so i > iii > ii.

Qu9: C
Propanal is an aldehyde - they react via nucleophilic addition with nucleophiles.... so i - iii are the nucleophiles. i is a Grignard reagent and should be viewed as a carbanion - since C is not very electronegative, this is a very reactive nucleophile. ii is a orgnanolithium reagent and should be viewed as a carbanion, also a very reactive nucleophile. Since Li is more electropositive than Mg, RLi are more reactive than RMgX. Finally, we have iii, a thiol, the neutral S atom in the nucleophilic site, but it's not nearly as reactive as a carbanion. Hence ii > i > iii.

Qu10: D

A radical halogenation question about (note the reaction conditions) which is controlled by the stability of the carbon radical formed as an intermediate. i would require the formation of an unstable phenyl radical (note the Br is normally added to alkyl benzenes using FeBr3 . ii forms via a secondary and benzylic radical and iii via a primary radical, therefore is terms of radical stability and therefore rate of reaction, ii > iii > i.

 

LABORATORY:
Based on the general principles covered in the laboratory. In this case, primarily spectroscopic skills and the unknowns laboratory for functional group and solubility tests.

Qu11: E
If a compound dissolves in 10% HCl, it implies that it is basic, so we are probably looking for an amine. Amine N-H stretches in the IR are around 3500cm-1 and they are weaker and less broad than O-H. The IR in E suggests an NH2, which the H-NMR supports. IR C, AB and AC have O-H bands.

Qu12: B
An orange precipitate with 2,4-DNP implies an aldehyde or a ketone. The orange solution with dichromate implies that there is no oxidation, so the unknown is a ketone. The IR spectra A, B, D and AB all show C=O (near 1700cm-1). The H-NMR of A suggests an aldehyde (peak at 9-10 ppm), D suggests an ester (alkoxy group due to peak at 4 ppm) and AB suggests a carboxylic acid (peak at 10-12 ppm).

Qu13: A
An orange precipitate with 2,4-DNP implies an aldehyde or a ketone. The green solution with dichromate implies that there is oxidation, so the unknown is an aldehyde. The IR spectra A, B, D and AB all show C=O (near 1700cm-1). The H-NMR of A suggests an aldehyde (peak at 9-10 ppm).

Qu14: C
The Br2 in chloroform decolourisation detects the presence of an alkene or alkyne. Alcohols react via dehydration when heated with H2SO4 to give alkenes. So we need to identify the alcohols that can eliminate. The IR spectra C, AB and AC suggest O-H, but AB is a carboxylic acid not an alcohol. The H-NMR of AC implies we have benzyl alcohol, C6H5CH2OH, which can not eliminate.

Qu15: D
An precipitate with 2,4-DNP implies an aldehyde or a ketone. Since we know there is a C=O, no precipitate means it's not an aldehyde or ketone. The lack of colour with indicator paper means we need a neutral compound. The IR spectra A, B, D and AB all show C=O (near 1700cm-1). The H-NMR of A suggests an aldehyde (peak at 9-10 ppm), B suggests a ketone (peaks near 2 ppm), D suggests an ester (alkoxy group due to peak at 4 ppm) and AB suggests a carboxylic acid (peak at 10-12 ppm).

Qu16: AC
The rapid reaction in the Lucas test (HCl / ZnCl2) implies that we are looking got an alcohol that favours SN1 reaction due to the formation of a stable carbocation. The IR spectra C, AB and AC suggest O-H, but AB is a carboxylic acid not an alcohol. The H-NMR of AC implies we have benzyl alcohol, C6H5CH2OH, which can form a resonance stabilised carbocation. The H-NMR of C implies we have butan-2-ol, CH3CH(OH)CH2CH3, which forms a less stable secondary carbocation.

Qu17: C
The moderate reaction in the Lucas test (HCl / ZnCl2) implies that we are looking got an alcohol that undergoes SN1 reaction due to the formation of a moderately stable carbocation. The IR spectra C, AB and AC suggest O-H, but AB is a carboxylic acid not an alcohol. The H-NMR of AC implies we have benzyl alcohol, C6H5CH2OH, which can form a resonance stabilised carbocation (hence fast Lucas reaction). The H-NMR of C implies we have butan-2-ol, CH3CH(OH)CH2CH3, which forms a less stable secondary carbocation.

Qu18: AB
If a compound dissolves in 5% NaOH, it implies that it is acidic, so we are probably looking for a carboxylic acid or a phenol. The red colour with indicator paper means we need an acidic compound. The IR spectra C, AB and AC suggest O-H (broad around 3600cm-1). The IR of AB also shows a carbonyl (near 1700cm-1) and the H-NMR AB suggests a carboxylic acid (peak at 10-12 ppm). None of the O-H systems are phenols. AC is a monosubstituted benzene and therefore not a phenol.

 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.

Qu19: ABD
 Assign the configurations as R or S at the chirality centers.... note that since B has an easily seen vertical mirror planes then it can't be (R,R) or (S,S), it must be (R,S) (i.e. they are meso compounds). The group priority order is OH > CH(OH)Et > Et > H. The absence of a mirror planes in C (vertical) and E (horizontal) means they can't be meso,

Qu20: AC, AE, BC, BE, CD, or CE
If you have the configurations from qu 19 worked out, then select the systems that are stereoisomers but not enatntiomers (i.e. different configurations at one but not at all the chirality centers).

Qu21: CE
Cis-hex-3-ene will react with peracid via an syn addition to give a cis-epoxide with will then react with aqueous acid to give a diol via a trans ring opening. This process is equivalent to an anti addition to the alkene and so the cis-alkene gives (R,R)- and (S,S)-3,4-hexanediol.

Qu22: CE
Trans-hex-3-ene will react with cold alkaline permanganate via a syn addition to give (R,R)- and (S,S)-3,4-hexanediol.

Qu23: BDE
Amines contain C-N units where the N is connected to C or H. A is a nitro group, C is an amide (due to the adjacent C=O).

Qu24: E
The hybridisation of N atoms can be impacted by resonance interactions. This occurs in amides such as C and aromatic systems such as B. In these cases the involvement of the N lone pair with the adjacent pi systems means that the N is infact sp2 hydridised.

Qu25: ABD
 Aromatic compounds need cyclic pi systems with an odd number of pi electrons.

Qu26: E
The most basic N will be one where the N lone pair is the most available for donation. The absence of a resonance interaction in the simple tertiary amine in E makes it the most basic.

Qu27: C
Acidity... if you know your pKa's then this is easy : carboxylic acid = 5, ketone enolate = 20 and dicarbonyl / active methylenene = 9 to 13 (diketone to diester) Remember the lower the pKa the stronger the acid.   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 at least one carbonyl group. That's why we have ignored A which has no alpha H. In the active methylenes, the enolate can be resonance to two C=O groups making it more acidic than a simple ketone like ii. Since a ketone is a better stabliser than an ester group, the diketone, C, is the most acidic.

Qu28: C
Enolisable H are all the H in alpha positions. The counts of enolisable H for the series are 0, 6, 8, 5 and 2 respectively.

Qu29: CDE
Methyl magnesium bromide is a Grignard reagent which react with C=O groups to give alcohols. Ketones react to give tertiary alcohols and so do esters.

Qu30: CDE
Active methylenes typically contain CH2 groups between at least two carbonyl groups.

PRODUCTS OF SYNTHESIS:
If you are trying to find the product, then you should probably just work forwards through the sequence of reactions. 
Basically depends on the need to know and identify the reactions, this is often triggered by looking at the functional groups in the molecules.

Qu31: C
The alkene undergoes ozonolysis with an oxidative work-up to give the methyl ketone / carboxylic acid system. The second step is a hydride reduction. Since NaBH4 is a milder reducing agent it will only reduce both the ketone to a secondary alcohol, in this case forming 5-hydroxyhexanoic acid. In the presence of the acid, this alcohol / acid will easily form a cyclic ester via an intramolecular Fischer esterification type of process.

Qu32:B
The diol will then react with the ketone to form a cyclic ketal. The acetylide then undergoes an SN2 type reaction with the alkyl bromide. Na / NH3 is a dissolving metal reduction of the alkyne to give the trans-alkene, which then reacts with the peracid to give an epoxide (i.e. A or B) with the same geometry of the alkyl groups (i.e. trans).

Qu33:A
The cycloalkene will under go allylic radical bromination to give the secondary allylic bromide. Then a Wittig reaction is used to form a new C=C by first preparing the ylid (step 2) then adding the carbonyl (step 3) so we will have a diene.

Qu34:A
The starting material is really just an phenol, an aromatic alcohol. The first step, alkyl halide and base, will convert the phenol -OH into a methyl ether via a Williamson type ether synthesis. This is followed by an electrophilic aromatic substitution, specifically Friedel-Crafts alkylation.... this will occur on the most activated ring which is the -OR substituted ring. Since the -OR is activating, it is an ortho / para director.

Qu35:E
The Grignard reagent will react with the ester to form a tertiary alcohol (it's a di Grignard so it can react twice with the same carbonyl). This will break the existing ring, but make a new 5 membered ring. Acid work-up and then oxidation of the primary O-H formed when the ester reacted will give an aldehyde group.

Qu36:E
The first step is an electrophilic aromatic substitution, specifically halogenation. Since the substituent is an alkyl group, it is a weak activator, and will direct the bromine substitution to the para position. Reaction of the new aryl bromide with Mg gives the aryl Grignard reagent which is then reacted with methanal to give the primary alcohol. Permanganate, under these conditions (acidic and heated) will oxidise the primary alcohol and the alkyl substituent to carboxylic acid groups, giving the para-dicarboxylic acid.

Qu37:E
The carboxylic acid will react with the acid chloride to form an acid anhydride. Subsequent reaction of the anhydride with ammonia gives the corresponding aromatic, primary amide. The amide is then reduced with LiAlH4 to the primary amine.

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).

Qu38: C
Step 3 uses tosic acid and a ketone to give a cyclic ketal, so it looks like we needed to start with a diol. Step 2 is the strong reducing agent LiAlH4 reducing carbonyl groups to alcohols. In this case one primary and one secondary alcohol. Step 1 LDA is lithium diisopropyl amide, a strong base, often used to form enolates quantitatively (i.e. 100% enolate). In step one, we are forming an enolate and then methylating it. In this case the position of the methyl group in the product is important, note that it is not at the ring junction. This means the scheme is the cyclic ester C undergoing alpha alkylation, then reduction to give the two primary alcohols.

Qu39: B
Step 4 looks like the reaction of an alcohol with acid derivative to form an ester, which is consistent with step 3 being the preparation of the acid chloride from the parent carboxylic acid. Looking at steps 1 and 2, this appears to have been formed by the hydrolysis of the benzylic cyanide which was obtained via an SN2 reaction of benzyl bromide, B.

Qu40: B
The two steps look like an epoxide is formed and then reacted with aq. acid to give a vicinal diol. D and E would give triols. C has the C=C location incorrect for the location of the product OH groups. The difference between A and B is the stereochemistry of the C=C which can de deduced by an analysis of the diol stereochemistry and knowing that epoxidation (syn) followed by ring opening (trans) is equivalent to an overall anti addition of 2 OH groups to the same C=C. You might need to draw one of the products in the conformation in which it is formed in the epoxide reaction to prove it to yourself.

Qu41: A
The product is a conjugated ketone and the reaction conditions should 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 and track the isopropyl group.

Qu42: A
The looks to be from the hydration of an alkene formed by the elimination of an alkyl halide (E2). The small base, HO- and heat implies an E2 and the antiperiplanar requirement implies that A is the required starting material. B would give the wrong alkene stereochemistry. You may need to draw A and B in the antiperiplanar geometry to convince yourself.

Qu43: D
The reaction is the hydroboration-oxidation of an alkene to give 3-methylpentan-2-ol. 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. A would give 3-methylpentan-1-ol, B and C would give 2-ethylbutan-1-ol. E has the wrong stereochemistry (it would give products that are stereoisomers of those shown).

Qu44: D
The reaction uses tosic acid and a ketone to give a cyclic ketal, via an intramolecular process... pick this apart to give 2 OH groups and a C=O (a ketone), count the total C atoms and the relative positions of the OH groups.

REAGENTS FOR SYNTHESIS
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).  

Qu45: D
Need to oxidise a primary alcohol selectively to an aldehyde.

Qu46: AC
The carbonyl group of the aldehyde is used to make an alkene with one more carbon atom looks like a Wittig reaction so look for the Ph3P derived system.

Qu47: ABC
Allylic bromination is a radical reaction.

Qu48: AB
Conversion of the bromide into a carboxylic acid with one extra carbon atom.... probably Grignard followed by carbon dioxide.

Qu49: B
Reducing the carboxylic acid to the primary alcohol requires a strong reducing agent such as LiAlH4.

Qu50: CD
Convert the primary alcohol to a primary bromide using an SN2 process rather than SN1 (avoid possible C+ rearrangements).

Qu51: AD
Forming the enolate of a ketone requires a base, here the stronger base LDA is preferred for quantitative deprotonation.

Qu52: C
Alkene to an epoxide using a peracid.

Qu53: BC
Now convert an epoxide to a 1,2-diol with aq. acid.

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.

Qu54: E
Acidity...   Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-.....the aromaticity of the the conjugate base of cyclopentadiene is the key issue here. In terms of the other answers, cyclopentadiene has typical stability for a diene, and neither cyclopentadiene or cyclopentane has lone pairs, and both are non-aromatic.

Qu55: C
The pKa data tells us that the ketone is more acidic, which means the ketone enolate is more stable due to the better resonance stabilisation. Both ester and ketone enolates have resonance stabilisation due to the carbonyl group, but the added resonance of the alkoxy O in the ester competes with the enolate stabilisation.

Qu56: D
The key reason why Grignards don't react with carboxylic acids to give secondary alcohols is that an acid base reactions occurs to give the carboxylate which is a very poor electrophile.

Qu57: C
The Hg2+/ aq. Acid looks like an alkyne hydration, where the site that the water attacks is determined by sterics and electronic effects. The O nucleophile attacks at the more hindered site so allowing the larger Hg to be at the less hindered site. The same, more substituted C, also has greater electrophilic character. In this case, it is the key ability of the phenyl group to resonance stabilise the C+ that makes all the difference.

Qu58: A
In aromatic nitration, or any other electrophilic aromatic substitution for that matter, it is primarily the substitutents present that dictate the regiochemistry not the incoming electrophile. In this case, there is an ester group as a substitutent.

Qu59: D
Since NaBH4 is a milder reducing agent it will only reduce the aldehyde group to a primary alcohol. The intermediate formed is 4-hydroxybutanoic acid, which can form a cyclic ester.

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