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
Qu8: AB
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:
Acetophenone
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
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
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
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
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
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
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).
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.
Hexachloroacetone is a very reactive carbonyl due to the inductive withdrawal of the six adjacent electronegative Cl atoms.
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