Chem 353 Final '99

Here is an post-mortem analysis / "how to" for the '99 Final. 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.

Qu1: B
All about the relative reactivity of carboxylic acid derivatives, so one can think of the LG ability or the effect the subtituent has on the electrophilicity of the C=O, giving i > iii > ii.

Qu2: C
Substituent effects on benzenes. -Cl groups are weakly deactivating (induction), -OH strongly electron donating (resonance with the lone pairs) and -NO2 is strongly electron withdrawing (resonance onto the electronegative O and induction due to the +ve N atom), so ii > i > iii.

Qu3: E
All H in question are a H, but in iii it is on O in a carboxylic acid so it is the most acidic (pKa =5).  Aldehydes (i) are more acidic (pKa =17) than esters (iii) (pKa =25) since in the ester the resonance donation from the -OR group reduces the stability of the enolate. So iii > i > ii.

Qu4: A
Methanal (i) is more reactive than ethanal (ii) towards Nu since it is less hindered and alkyl groups are weak electron donors which also decreases the electrophilicity of the C=O. In an ester, the -OR is strongly electron donating, further decreasing the electrophilicity.  So i > ii > iii.

Qu5: A
Nu increases down the periodic table (polarisability) so RS- is the best Nu. For the two O systems, resonance stabilisation of the carboxylate (iii) decreases the availiability of the electrons and therefore the nucleophilicity, so i > ii > iii.

Qu6: AB
The reaction conditions promote SN1. In (iii) we have a good LG with a 2o benzylic carbocation, so it will be very fast. In (ii) we have a poorer LG and a simple 2o carbocation, and (iii) is an unreactive vinylic chloride. So iii > ii > i.

Qu7: AB
Dehydration of an alcohol usually gives the more stable, morely highly substituted alkene, iii > ii. (i) is not a product of 1,2-elimination from 1-methylcyclohexanol. So iii > ii > i.

Qu8: E
Since alkyl groups are weak electron donors, t-butylbenzene will direct o,p but the large size will favour p over o. Thus, iii > i > ii.

Qu9: C
Radical bromination depends on radical stability: 3o benzylic > 1o > phenyl,  so ii > i > iii.

Qu10: D
Diels-Alder reaction requires that the diene be able to adopt an s-cis conformation. Cyclopentadiene is locked in this conformation and butadiene can achieve it by rotation about the C2-C3 bond, but (i) is locked s-trans and is therefore unreactive = ii > iii > i.



LABORATORY
Read the questions then go back to the data to see what you need to use, pull it all together to get the last question (then review based on that ?)

Qu11: D
Since the uncorrected boiling point is 223oC, the corrected value should be about 1o higher for every 15o above 50o, so about 12o higher => 235o.

Qu12: AD.
DNP requires a red precipitate (the reagent itself is red), and the iodoform requires a yellow precipitate of CHI3 be formed.

Qu13: BC

Qu14: B
1800-1600 cm-1 is the approximate range of the C=O.

Qu15: AB
A quartet integration of 2 suggests a deshielded -CH2- next to a -CH3.

Qu16: D or E
The aromatic region (7 - 8 ppm) has a doublet of doublets, integrating for a total of 4H, thus either D or E is possible.

Qu17: AC
Aromatic based on nmr. No -OH in IR so not an acid. Could be an ester, but not a ketone (DNP test) or an ether.

Qu18: BC
Para-disubstituted aromatic with -OEt group, and a partly deshielded -Me group, but cannot be a methyl ketone.



PRODUCTS OF SYNTHESIS
Work from the starting materials to the products using the reagents to "see" what product to look for.

Qu19: B
2-butene with Br2 gives 2,3-dibromobutane that can be eliminated with NaNH2 (strong base) to 2-butyne, which will undergo addition (hydration) in the presence of Hg to give the enol that will tautomerise to the ketone, 2-butanone.

Qu20: E
1-butene will undergo hydroboration / oxidation to 1-butanol. Oxidation with aq. Cr will give the butanoic acid.

Qu21: D
Propanal will undergo addition with the Grignard to give 2-butanol. After tosylation, the strong bulky base, tBuO- will cause elimination to give the less substituted alkene, 1-butene.

Qu22: C
1-bromopropane reacts with Mg to give the Grignard with will react with CO2, then work-up to give the carboxylic acid. Reaction of the acid with an alcohol gives the methyl ester.

Qu23: C
Diels-Alder reaction, to the substituted cyclohexene. Reduction of the C=C then the Wittig reaction to convert the aldehyde group to C=CH2.

Qu24: B
Cyclohexanol will be oxidised to cyclohexanone. LDA will form the enolate which can be methylated.

Qu25: B
Alkenes plus peracid to epoxide, which will open SN2 fashion at the less hindered end with the strong nucleophile like a Grignard to give the alcohol.



STARTING MATERIALS
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.

Qu26: D
Diazonium chemistry.... the diazo compound being prepared from the amine.

Qu27: D
Benzyl alcohol was formed by reduction of a C=O system, which was obtained by the oxidation of an alkyl aromatic substituent with a benzylic H.

Qu28: D
A long one to plug back through.... the maide was formed by the reaction of the amine with the acid chloride prepared from the acid. The acid was obtained by the reaction of CO2 with the Grignard of the bromide. So the methyl group was already in place.

Qu29: C
Cyclic acetal formed by the reaction of the ketone with the 1,2-diol. The 1,2-diol must have been prepared by addition to the alkene.

Qu30: E
Overall this seems to be an enolate condensation type reaction with decarboxylation (it can't de dehydration since there is no C=C), so an ester system must be involved. The ozonolysis with the oxidative work-up followed by MeOH also suggests this.
If we have 5 C in the product and we lost 1 in CO2, then we needed 6 in the alkene. Only E can give a C6 diester....

Qu31: A
Another condensation reaction, this time an Aldol, since the C=C has been removed by the reduction. System has 7 C atoms, so needs to be a di-methyl ketone.... this is "seen" by breaking open the product at the C2-C3 bond where the C=C would be located.

Qu32: C
Williamson ether synthesis, preceded via an anti-Markovnikiv addtion of HBr to a C5 alkene with a 2o alcohol.



REAGENTS
Look at the functional groups in the starting material and products to try to determine what may have happened. Look at the
reagents in each option to see what effect they would have on the SM....

Qu33: BE
Acid to ester = esterification, usually alcohol, H+ and heat.

Qu34:  CD or CE
Ester to keto-ester = Claisen condensation of ester enolate, so need the alkoxide base or the stronger non-nucleophilic base like LDA.

Qu35: CD or CE
Alkylation of the enolate... need the base again

Qu36: AD
Oxidation of alcohol to ketone, PCC.

Qu37: E
a-mono-chlorination of ketone uses HCl. Basic conditions will cause poly chlorination.

Qu38: BC
Hydrolysis and decarboxylation, aqueous acid or base and heat. Here base would cause problems will Aldol reactions.

Qu39: AC
Reduction of the ketones to alcohols, use hydride reagents.

Qu40: AB
Alcohol to chloride, PCl3 would avoid acidic conditions and potential rearrangements.

Qu41:  ABC
Need to add the two ester groups and form the ring via the enolate of diethyl malonate.



STEREOCHEMISTRY
Need to know the mechanisms or stereochemistry and regiochemistry of your reactions. Identify the reaction then go from there.... Watch out for the enantiomers, achiral materials will give racemic products.

Qu42: AB
Diels-Alder reaction, syn addition, stereospecific, cis dienophile gives cis products, A and B are enantiomers.

Qu43: C
E2 type elimination, needs to be anti-periplanar H and Cl. -OH converted to -Cl via SN2 reaction with inversion, so here it has to eliminate away from the methyl group.

Qu44: D
Epoxide opens under acidic conditions to put the -Cl on the same C as the methyl group (more stable carbocation), -OH and -Cl will be trans.

Qu45: E
Syn reduction with poisoned catalyst to the cis-alkene, then anti-addition to the 2,3-dibromide. Need the RR and SS stereoisomers.

Qu46: CE
Anti-Markovnikov hydration via hydroboration, -OH and -H added syn.



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: D
In amides the strong electron donating effect of the N system inhibits the resonance stabilistion of the enolate.

Qu48: D
Acid catalysed hydration of alkenes via carbocation, so rearrangement needs to be considered.

Qu49: B
It is the substitutent already on the benzene that dictates the location of the incoming substituent.

Qu50: C
Grignards are strong bases, so protonation will occur destroying the carbanion.