353 Final Winter 2000

Here is an post-mortem analysis / "how to" for the 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. 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: E
The reaction is an acyl substitution of a carboxylic acid derivatives, this involves the loss of a leaving group.  Either use leaving group ability ( Cl- > CH3CO2- > NH2-) which is related to the stability of the conjugate bases, or the electronic effect the group has on the electrophilicity of the C in the carbonyl group....so we get iii > i > ii.

Qu2: D
The reaction is electrophilic aromatic substitution, and we need to look at the substituent effects on the aromatic ring.  -OMe is an electron donating group by resonance, -Me is weakly electron donating due to hyperconjugation and the carbonyl of an ester is electron withdrawing by resonance. So ii > iii > i.

Qu3: AB
Look at where the H is attached and think of the factors that stabilise charge such as electronegativity, resonance, inductive effects etc. iii gives a resonance stabilised O-ve, ii gives a simple O-ve and i gives a C- species (no stabilising factors : not an enolate !). So iii > ii > i.

Qu4: C
The reaction is the hydride reduction of carbonyls, so we need to pay attention to the electrophilicity of the C in the carbonyl. Look at the attached groups. i is a ketone with two alkyl groups, ii is an aldehyde (H, alkyl) and iii an ester (alkyl, alkoxy).  Electron donating groups will reduce the electrophilicity of the carbonyl. -OR is an electron donating group by resonance, -R is weakly electron donating due to hyperconjugation and H is the reference (i.e. no effect).
So ii > i > iii.

Qu5: AB
Bases are electron pair donors, so look at the electron availability.  First compare O and C systems (same row) O is more electronegative which means it is a poorer donor, so C is more basic than O (i.e. iii > ii). Now compare O and S (same group) S is larger than O so charge is more spread out making it more stable so O is more basic then S, (i.e. ii > i). Overall then we have  iii > ii > i.

Qu6: AB
The enolisable H are those adjacent to the C=O group. i has none, ii has two and iii has three, so iii > ii > i.

Qu7: E
Aldehyde H (CHO) are usually 9-10 ppm. Vinyl H (C=CH) are usually 5-7 ppm.  The resonance effects of the C=O make the beta position more +ve (so more deshielded) in character than the alpha position.... overall iii > i > ii.

Qu8: AB
The greater the double bond character of the C=O the greater the frequency (thinking of the resonance contributors to the C=O group). Acid chlorides (Cl = electron withdrawing group, gives more double bond character) are higher than esters (-OR = electron donating => more single bond character) i.e. iii > ii. Conjugation also gives more single bond character (draw the resonance contributors) so ii > i.  Put this together : so iii > ii > i.

Qu9: E
The reaction is electrophilic aromatic substitution, and we need to look at the directing effect of the ester which is PhOC(=O)CH3... so the group attached to the aromatic ring is an O => a electron donating group => ortho/para director. Steric effects will block the ortho site to some degree, so iii > i > ii

Qu10: B
Bases are electron pair donors, (here the N is the basic site) so look at the electron availability and the way this is influenced by the aromatic substituents. -OCH3  is an electron donating group by resonance = increased electron availability, -CN is strongly electron withdrawing by resonance =  decrease electron availability and -Cl is electron withdrawing (electronegativity = induction). Overall then we have  i > iii > ii


LABORATORY:
Based on the unknowns experiment that in itself covers lots of material...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: C
Melting points are not particularly sensitive to pressure changes.

Qu12: E
Boiling points increase with pressure so the corrected value for sea-level are higher by  about 1o higher for every 15o above 50o, so about 14o higher => 274o.

Qu13: ABC

Qu14: BC
1800-1600 cm-1 is the approximate range of the C=O and 1200-1000 cm-1 is the approximate range of the C-O.

Qu15: B

Qu16: BC
Peak at 3 ppm integrates for 2H => CH2 group, appears as a quartet so has 3 neighbours (e.g. -CH2CH3) and is slightly deshielded, enough to be next to a C=O group but not just an -O- (nearer to 4 ppm).

Qu17: DE
The pair of doublets each for 2H represents two types of aromatic H where there are two of each type. If there are two different substituents on the aromatic ring then it must be a para substituted system.

Qu18: ADE

Qu19: E
Pull it all together from the previous questions and pay attention to the H-nmr spectra .... para disubstituted benzene narrows it to E, AC, AE, BC, BD, BE.
The methyl group singlet at 4 ppm indicates an -OCH3 group.... leaving us E and AE. In AE there would also be a -OCH2- group near 4 ppm (not seen) so leaves E.


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

Qu20: A
The alkene reacts with the peracid to give an epoxide which is then ring opened by the methoxide nucleophile SN2 fashion (least hindered end) to give A.
C and E have the wrong regiochemistry.

Qu21: B
Brominate para to the methoxy group (an activator, o/p director), make the Grignard, and make the carboxylic acid then the ethyl ester.....B.

Qu22: D
Add the Grignard to the ketone C=O to get a tertiary alcohol, remove the cyclic acetal deprotecting an aldehyde then oxidise to the carboxylic acid.... count C atoms..... D

Qu23: C
Oxidise the primary alcohol to the aldehyde (PCC can be used to stop there) then make a cyclic acetal..... count C atoms !   C

Qu24: D
Diels-Alder reaction to give a acyl cyclohexene system, Wittig on the ketone, then reduce both the C=C count C atoms !  D

Qu25: E.
Baeyer-Villager of the ketone on the more substituted side to give the ester, then reduction of the ester.... carbonyl side in the acid unit to primary alcohol plus the secondary alcohol from the alcoholic portion of the ester... count C atoms....E

Qu26: E
Friedel-Crafts acylation para to the methyl group then vigourous oxidation of both groups to carboxylic acids.... E



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.

Qu27: C
Reagents suggest the processes based on diazonium reactions (need to recognise this)... reduce the nitro to the amine, diazotise and substitute....the only SM that gets to the meta dibromide is C.

Qu28: D.
Only D and E have enough C atoms.... the primary alcohol has been formed by a reduction of a carbonyl system.  Oxidation of aromatic alkyl groups with benzylic H (so not E) gives aromatic carboxylic acids.

Qu29: D
The final amide was formed by the reaction of the amine with a carboxylic acid derivative, the acid chloride which was obtained from the carboxylic acid. This was in turn made by the carboxylation of the Grignard from the bromide... so we must have started with methylbenzene.

Qu30: C
Recognise the acetal and its formation from the 1,2-diol prepared from the alkene...C

Qu31: B
Working backwards, the primary alcohol is obtained by the hydride reduction of a carbonyl group maybe an aldehyde, an acid or an ester. But then the two aq. steps are too difficult to unwrap.  So go to the beginning. Since all the starting materials are dicarbonyls, maybe we should be thinking of alkylating enolates especially as step one uses a base (NaOMe) and an alkyl halide. A and D can't form enolates. Only B or E are active methylenes (easier enolate formation)... What happens is that the enolate forms, this gets alkylated, then the base forms a new enolate and a second, now intramolecular alkylation occurs forming a five membered ring (4C from the dihalide and 1C from the enolate). The NaOH causes both esters to hydrolyse and forms a dicarboxylic acid that then with acid / heat is decarboxylated (loss of CO2).

Qu32: B
H2/Pt suggests a reduction of an alkene or alkyne.  Given the possible starting materials and the use of NaOH, a base, in step 1, an intramolecular aldol would be a possibility. This would mean that the aldol product would need to be 3-hydroxycyclohexanone in order to get to cyclohexanone after elimination then reduction. In order to get 3-hydroxycyclohexanone, the starting material would have to be a methyl ketone / aldehyde system with 6 carbons....

Qu33: D
Periodate is used to cleave 1,2-diols... which are made from alkenes with KMnO4  to get the methyl ketone / aldehyde, the alkene needs to be a methyl substituted cycloalkene with 6 carbons....



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....

Qu34: AE
Make the cyclic acetal, you have the diol, so you need the C=O.

Qu35: B
Convert the aldehyde to a secondary alcohol.... adding a methyl group : looks like a Grignard reaction using MeMgBr.

Qu36: AB
Oxidise the secondary alcohol to the ketone without deprotecting the acetal (i.e. stay away from aq. acid).

Qu37: BE
Ketone to alkene looks like a Wittig reaction, so recognise the P system...

Qu38: BD
Anti-Markovnikov hydration of the alkene requires hydroboration / oxidation.

Qu39: C
Substitute the -OH to get the -Br without rearrangement.

Qu40: CE
Prepare an enolate, use a base. LDA is the best choice to give 100% deprotonation.

Qu41:  BC
Deprotection (removal of the acetal) uses aq. acid.



MECHANISMS
Need to identify the type of reaction based on classic functional group descriptions.

Qu42: BD

Qu43: B

Qu44: CE

Qu45: AC

Qu46: DE

Qu47: CE



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.

Qu48: C
The stability of the conjugate base due to its aromaticity makes its formation favourable and so lowers the pKa. A isn't true, B neither does cyclopentane, D so does cyclopentane and E cyclopentane is non-aromatic too.

Qu49: E
Friedel-Crafts alkylations are prone to carbocation rearrangements (to more stable cations).  A and C are true but not explanations, B and D are false.

Qu50: A
This type of ester where it is connected to the aromatic ring via a C=O is deactivating, electron withdrawing and meta directing. B is true but irrelevant to the question since it is a ketone type group, C and D : the incoming group does not control the regioselectivity, E is false, they are the same statistically.

Qu51: D
Remember that bases and nucleophiles are quite similar, many reagents can do both. A isn't true they react with other groups such as epoxides, nitriles... B is false (an ester is more hindered but reacts), C false (its the same as an ester which does react). E is true but not relevant here.


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