353 FIN Winter 2006

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: C
The reactivity of carbonyl groups towards hydride reagents, looking at the relative reactivity of i ketones, ii aldehydes, and iii esters. 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. The reactivity of carbonyl systems is impacted by the substituents attached to the carbonyl, here that is -R, -H and -OR. The stronger that group is as an electron donor, then the less electrophilic the carbonyl carbon is. We treat -H as electronically neutral, -R are weak electron donors and -OR are strong electron donors. Hence in terms of reactivity ii > i > iii.

Qu2: B
The reaction is electrophilic aromatic substitution, a Friedel-Crafts alkylation, and we need to look at the substituent effects on the aromatic ring.  The methoxy group in i is a strong electron donor via resonance - it's an activating 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 i > iii > ii.

Qu3: A
Acidity... if you know your pKa's then this is easy : carboxylic acid = 5, aldehyde enolate = 17 and ester enolate = 25.  Remember the lower the pKa the stronger the acid, so i > ii > 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 ii and iii it's C and i 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 aldehyde 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: AB
Oxidation state.... more C-O bonds means a higher oxidation state so we get iii > ii > i.

Qu5: A
Reaction is the electrophilic addition to alkenes and we are looking at the alkenes reacting with HBr in the dark, hence the stability of the intermediate carbocations is the critical issue.  The carbocation produced from i will be tertiary from ii will be secondary and that from iii destabilised due to the electron withdrawing group. Hence i > ii > iii.

Qu6: C
Acidity... Think of the simple acid equation HA <=> H+    A-   then look for factors that stabilise A-.....note that two are N bases, the other is an O base. N is a better e- donor than O. i is pyridine which is aromatic and has an aromatic conjugate acid. ii is pyrrole which is aromatic but has a non-aromatic conjugate acid (which means that the conjugate acid is quite acidic) and iii is a simple alcohol to alkoxide. Therefore in terms of acidity, then ii > i > iii.

Qu7: C
Draw out 1-methylcyclohexene... hydroboration / oxidation forms the anti-Markovnikov alcohol via an overall syn addition. ii is the major product, i is the Markovnikov product and will be formed in minor amounts, whereas iii will not be formed since it would require that the alkene was originally exocyclic.

Qu8: AB
More C=C in conjugation = more resonance stabilisation (greater resonance energy). If a compound is aromatic, then there is even greater resonance. i is a non-aromatic, conjugated diene = 4 kcal / mol ii is aromatic (benzene = 36 kcal / mol) and iii has two aromatic rings...(naphthalene = 61 kcal / mol).

Qu9: C
These reagents react as nucleophiles in the nucleophilic addition to aldehydes. I and ii are both organometallic reagents so we think of them as carbanions which will be much more reactive than the alcohol, a weak nucleophile in comparison. In terms of I and ii, the more electropositive the metal (or the more reactive the metal) the more anionic the carbon in the organometallic reagent so RLi > RMgX.

Qu10: D

Check the pKas, 9 vs 25 vs 16 of the acids or work it out.... the systems are a diketone (an example of an active methylene compound, two electron withdrawing groups), a terminal alkyne and cyclopentadiene which has an aromatic conjugate base.


LABORATORY:
Based on the general principles cover in the laboratory so far. Need to know the principles and details of the steps in the experiments.

Qu11: E
Acetophenone
is an aromatic ketone (total 8C) and therefore is too non-polar to dissolve in aqueous solvents.

Qu12: D
Aniline is aminobenzene, C6H5NH2, and therefore will dissolve in aqueous acid when the polar ammonium ion, C6H5NH3+ is formed.

Qu13: C
Phenol is an aromatic alcohol, C6H5OH, and will dissolve in aqueous base when the polar phenoxide ion C6H5O- is formed. NaOH is a strong enough base to do this but NaHCO3 is not.

Qu14: B
The positive test is the formation of a silver mirror due to the oxidation of an aldehyde.

Qu15: AE
The test is positive for methyl ketones.

Qu16: BC
The test is positive for phenols.

Qu17: A or BC
-OH are need for this reactions so either alcohols or phenols.

Qu18: B or AE
Oximes are derivatives of aldehydes and ketones.


 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: AD
An E2 reaction…this requires that the H and the Br be axial in order to be at 180o to each other.  Remember that the larger the substituent, the greater it’s preference to be equatorial. This means that a tBu group has a very strong preference to be equatorial.  Look for structures where the Br is already axial while the tBu is equatorial as this means no ring flip will be required.

Qu20: AD
Assign the configurations as R or S at the chriality centers.  B and C are not chiral. A = (S,S), D = (R,R) and E = (R,S). Note that since E is a cis isomer while A and D are trans isomers and therefore E can’t be an enantiomer of either A or D.

Qu21: BC
Since A, D and E are configurational isomers (see qu 20), they can’t be conformational isomers. B and C are 1,4- and both have trans substituents – they can be interconverted by a ring flip.

Qu22: A
See Qu 20.

Qu23: B
Acetals are formed by aldehydes and ketones and aldehydes are more reactive than ketones, hence it’s the aldehyde B.

Qu24: A
Look for H atoms alpha to carbonyls – only the ketone A has enolisable H.

Qu25: C
Lithium aluminium hydride reduces carbonyls… so it will not reduce the alkene C.

Qu26: B
Schiff's reagent reacts with aldehydes so B.

Qu27: BD
A is non-aromatic, C is anti-aromatic (4 pi electrons) and E is anti-aromatic (8 pi electrons).

Qu28: BD
Both B and D can be protonated on (B on the S and D on the N) giving aromatic conjugate acids.

Qu29: A
A has an aromatic resonance structure where the is a -ve charge in the five membered ring and a +ve charge on the three membered ring meaning that both rings are aromatic (one 6 pi and one 2 pi system).


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.

Qu30: C
The transformation is electrophilic aromatic substitution introducing bromine. This will occur on the most reactive aromatic ring (so you need to know which substituent activates the benzene ring the most) - this will be the ring with the electron donating methoxy group (which directs o,p).  The second step is the reduction of the nitro group to the amine which requires a basic work-up (step 3). Hence C is the answer.

Qu31: E
Step 1 converts the aldehyde into a cyclic acetal which acts as a protecting group. Subsequent reaction of the ester with 2 equivalents of the Grignard reagent followed by aqueous acid work-up gives the tertiary alcohol and then the protecting group is removed to reveal the original aldehyde group, E.

Qu32: A
Step 1 converts the alkene to the epoxide which is then reduced with lithium aluminium hydride in an opens the epoxide in an SN2 type fashion from the least substituted end to give the secondary benzylic alcohol. This is followed by a Williamson ether synthesis to give the methyl ether, A.

Qu33: A
Step 1 is a dissolving metal reduction of the alkyne to the trans-alkene and this is followed by the addition of HBr (Markovnikov's rule) so the Br ends up at the site of the more stable benzylic carbocation intermediate, A.

Qu34: A
Step 1 is the selective sodium borohydride reduction of the ketone over the ester. The ketone gives a secondary alcohol which is then converted to a tosylate which then undergoes an SN2 reaction to give the bromide, A.

Qu35: D
Step 1 forms the aromatic Grignard reagent with is then reacted with H+ to give the overall effect of -Br to -H. The last step removes the cyclic acetal protecting group to reveal an aldehyde, D.

Qu36: E
Step 1 converts the carboxylic acid into the more reactive acid chloride ready for the reaction with the amine to give an amide. The amide in then reduced with the lithium aluminium hydride to convert the C=O into a CH2 group, E. Note the aniline is aminobenzene.


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).

Qu37: B
A long one...the last two steps look like -CN added and then hydrolysed to carboxylic acids. Note that the CN results in extra C atoms being added. The nitrile has been made by the SN2 on a bromide. The bromide has been formed from an alcohol generated by the sodium borohydride reduction of an aldehyde or ketone. The carbonyls have been formed by ozonolysis with a reductive work up. If both of the carboxylic acid groups in the final product have been produced via nitrile hydrolysis, then two extra C atoms were added... if you work with this, it means that the ozonolysis gave a ketone and an aldehyde which defines the original alkene structure as B.

Qu38: D
Periodic acid cleaves 1,2-diols to carbonyls - so reconnect the two aldehyde groups to give the 1,2-diol. The diol comes from the reaction of
osmium tetroxide with an alkene
this should reveal a classic Diels-Alder product.... the cyclopentadiene reaction should also suggest that. So reverse the Diels-Alder reaction to reveal the dienophile and pay attention to the stereochemistry.

Qu39: A
The product is a hemi-ketal, which has been formed by an intramolecular reaction of a methyl ketone and a primary alcohol. The primary alcohol has been formed from a Grignard reagent reacting with methanal... this also added 1 C atom, so we needed a benzylic halide to give us the right Grignard reagent... the first step protected the ketone as a cyclic ketal.

Qu40: A
Looks deceptively easy... but H+ can be tough to work out....In this case the structure is a ketal... spot the ketone and the two -OH groups to reveal the answer.

Qu41: C
The product contains a cyclopropane that has been made by the Simmonds-Smith reaction - of the carbenoid with the alkene... the alkene has been made by a Wittig reaction of a phosphorous ylide with cyclohexanone. Therefore we are looking to the right alkyl halide, C, count C to avoid mistakes!

Qu42: E
A double Grignard reaction had been used to make a cyclic structure.... the Grignard reagent would have had to react with an ester in order to do this..... in order for the OR part of the ester to still be present, then the ester must have been cyclic (i.e. a lactone) counting up the C atoms means the ester contained 4 C atoms, so with the -O- this means a 5 membered ring, E.

Qu43: D
Simple addition of bromine across an alkene... looking at the positions of the 2 Br atoms tells us where the alkene was which tells us it is either


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).  

Qu44: ABD
Need to form a ketone enolate is 100% yield so need a strong base such as LDA.

Qu45: BE
Now need to add a single carbon with an -OH.... an aldol reaction...need methanal.

Qu46: ABE
Oxidation of the primary alcohol to an aldehyde.

Qu47: D
Aldehydes react with secondary amines to give enamines.

Qu48: AB
Reduction of an aromatic nitro group to an amine - due to the H+ present, need a basic work up to get the free amine product rather than the ammonium salt.

Qu49: A
Nitration of the aromatic system using nitric acid.

Qu50: AE
A tough one.... looks like an addition of N-H across a CN triple bond....

Qu51: AB
Reduction of an aromatic nitro group to an amine - due to the H+ present, need a basic work up to get the free amine product rather than the ammonium salt.

Qu52: BC
Amide preparation from an acid chloride and an amine using a base such as pyridine or triethyl amine to avoid the reaction of the key amine with the HCl by-product.


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.

Qu53: C
If the two methyl groups show as two peaks in the H NMR then it tells you that they are different to each other. This is the case if there is C=N character. A similar situation would be observed in an alkene such as (CH3)2C=CH(R).

Qu54: A
C=O is about 179 kcal/mol and a C-O is about 86 kcal/mol hence this tends to favour the carbonyl side of the equilibrium. Note that if there is an excess of water (e.g. when dissolving a little methanal in water) the Le Chatielier's principle will shift the equilibrium towards the hydrate side.

Qu55: E
Carbonyls and primary amines react to give imines....pH 10 is basic, therefore, the ketone, the amine or the -OH leaving group will be protonated (i.e. +ve) at this pH. The pKa for a ketone is about 20, therefore the enolate concentration will be very low at pH 10.

Qu56: B
Esters are hydrolysed under basic conditions via a tetrahedral intermediate. The tetrahedral intermediate for the six membered ring system will be analogous to a chair cyclohexane where all the bonds are staggered - this means that this intermediate is quite stable and therefore easily obtained.

Qu57: D
This is an E2 reaction.... need to look at the reactive conformation with the Br and H at 180 degrees to each other. In this conformation, look at the positions of the large phenyl groups. If the are close to each other, then steric effects will destablise both the product and the transition state leding to that product.

Qu58: D
Hexachloroacetone is a very reactive carbonyl due to the inductive withdrawal of the six adjacent electronegative Cl atoms.


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