353 MT Winter 2007
Here
is an post-mortem analysis / "how to" for
the MT. 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.
Acidity
and pKas. For simple alkanes, alkenes and alkynes, terminal alkynes are
the most acidic (due to hybridisation of orbital associated with the -ve charge),
then vinyl H. The allylic H in ii give a resonance stablised conjugate
base and have a pKa near 45, very similar to the vinyl H pKa. So in terms
of pKa, ii > iii. What about i (cyclopentadiene) ?
The conjugate base of i is aromatic
and hence very stable, which therefore makes cyclopentadiene quite acidic.
Since the stronger the acid, the lower the pKa, the order is ii >
iii > i (actual pKas approx 40, 25, 16 respectively).
Qu2: C
Carbocation stability...look at the degree and the number of resonance contributors
i is 2o and benzylic, ii is 1o as drawn but the first contributor
gives a 2o benzylic (so ii has more contributors than i). iii
is a simple 2o carbocation.
Qu3: C
i is an aromatic CC - which has a length between a C=C and C-C due
to the resonance. ii is a C-C single bond between 2 sp2 hybridised
C atoms and there is also some conjugation of the adjacent C=C systems so
the bond is a little shorter than a normal C-C. iii is a normal C=C.
Double bonds are shorter than single bonds (stronger interaction = shorter
bond). Therefore, in terms of bond length, ii > i > iii.
Qu4: B
We have 3 dienes. i and iii are conjugated, ii is isolated
so it will be the least stable and therefore is at the highest initial energy
and will therefore will have the most exothermic heat of combustion. Since
i contains a trisubstituted C=C, it will be more stable than iii.
Hence it terms of heats of combustion (least -ve to most -ve), i >
iii > ii.
The reaction here is a Diels-Alder
reaction and we are looking at the diene components. The critical issue
is the ability to adopt the s-cis conformation. ii is the most reactive
as it is locked in that reactive conformation. i and iii must
equilibrate into that conformation. In iii the terminal methyl group
will undergo a steric interaction with a terminal H at the other end of the
diene system in the s-cis conformation and this destablises this form and
so restricts the equilibrium compared to that for i. Hence ii
> i > iii.
Qu6: C
Number of products from ozonolysis of alkenes. The cyclic alkene in i will
give a single dialdehyde, ii is a diene and will cleave to give methanal,
1,3-propandial and propanal. iii is an alkene and will give propanone
and 3-methylpentan-2-one. So ii > iii > i.
Qu8: B
The reactions of alkenes and alkynes with aq. acid are controlled by carbocation
stability and the number of alkyl groups on the pi unit (more alkyl groups
makes them more electron rich and hence more nucleophilic). Alkynes are less
reactive since they require the formation of the less stable vinyl carbocations.
Therefore i > iii > ii.
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.
A hemi acetal has an HO-C-OR unit... formed from ROH + R'CHO.
Qu10: A
Qu11: A
Qu12: A
Secondary
alcohols are oxidised to ketones.
Qu14: B
Alcohol
dehydration is catalysed by acids. The bromine in chloroform is a test
for the presence of C=C in the products of the dehydration.
Qu15: B
The reagent used is not a phenol...(i.e. Ar-OH) the test reagent is 2,4-dinitrophenylhydrazine
(Ar-NHNH2).
Qu16: B
Acid waste needs to be diluted and thrown into the aqueous waste.
Qu17: B
The statement describes condensation polymers.
Qu18: B
Nylon is a polyamide.
Qu19: A
Qu20: B
Reflux occurs at the normal boiling point - the apparatus prevents evaporation.
Qu21: A
Qu22: B
A cryogen causes burns due to extreme cold.
Qu23: A
Qu24: B
The structure shown is phenyl magnesium bromide.
STARTING
MATERIALS AND PRODUCTS OF REACTIONS:
If you are trying to find the product, then you should probably just work
forwards through the sequence of reactions.
If you are looking for the starting material, then working backwards is probably
the best way to go....
Basically depends on the need to know and identify the reactions,
this is often triggered by looking at the functional groups in the molecules.
Working backwards,
ozonolysis with a oxidative work up... up giving a single
keto-acid - this would need to originate from a
cyclic trisubstituted alkene which seems to have
been formed by an elimination
of an alkyl halide rather than an alcohol (since base was used).
Qu26: C
Working forwards....the
reaction is the addition
of hydrogen bromide to a conjugated diene. The phenyl substituted systems should be treated with care as the kinetic product from the most stable C+ contributor also leads to the thermodynamic product, the conjugated system. So the product
will result from protonation at C4 to give the resonance stabilised carbocation Ph-CH=CH-CH+-CH3 via 1,2-addition product across the C3-C4 C=C. Addition of the Br- to this carbocation has been followed
by a nucleophilic substitution (methanol as the nuclephile to give the ether).
Qu27: B
Working backwards...the product is a cyclic ether.... starting thinking alcohol
and alkyl halide..... Hydroboration
/ oxidation gives an anti-Markovnikov alcohol which when heated is giving
a cyclic ether.... a sort of a Williamson type reaction... need to make a
5 membered ring, so there needs to be a 4 C chain (since there is also an
O in the ring), with a C=C and a good leaving group (Br). So it has to be
B. C doesn't work because the hydroboration reaction will add
the OH at the wrong carbon.
Qu28: D
Working
forwards... the alkene
will react with the peracid to give the epoxide
which will then open with HCl in an SN1 like fashion to give a chlorohydrin.
The Cl will add at the more substituted carbon and the -OH and -Cl need to
be trans. A is a diol (wrong FG), B and C have the wrong
regiochemistry, C and E have the wrong stereochemistry (they
are cis).
Qu29:
Working forwards, the alkene starting material will undergo hydration
to give the Markovnikov product via the carbocation.... but that will rapidly
rearrange to before the nucleophile (the EtOH) adds to give an ether, giving
B. (D would result if the carbocation didn't rearrange). C
would require water to act as the nucleophile (not present).
Qu30: D
Working forwards, the alkyne is
deprotonated by the strong base to form the acetylide
nucleophile that then undergoes an SN2 reaction withe the methyl iodide.
Then a dissolving
metal reduction gives the trans-alkene D.
Qu31: E
REGIO-
and STEREOCHEMISTRY:
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).
In cases where more than one product is formed in equal amounts (e.g. the
enantiomers), then both must be selected for full marks, part
marks are given when only one of the pair is selected.
Advice : in each case draw the starting material
in the conformation in which is reacts or the product in the conformation
in which it is initially formed using wedge-hash diagrams. It is a good idea
to draw the materials in such a way that the new bonds are in the plane of
the page. Once you have drawn the materials It
may also be good for you to use model kits for these questions too.
Once you have drawn the materials in this way, you may need to consider rotations
around sigma bonds to make your answer match the options. An alternative
approach could be to assign configurations to your drawn answer to compare
them with the options - this can be slow and prone to error.
The alkyne undergoes a dissolving
metal reduction to give the trans-alkene
which then undergoes a syn-dihydroxylation
to give a diol. That means either D or E.
Qu33: B
The alkene undergoes halohydration
with trans stereochemistry to give a 1,2-halohydrin
which then forms an epoxide on treatment with base via an SN2 (with inversion).
The overall stereochemistry is syn, i.e. the epoxide stereochemistry
is the same as that of the original alkene. That means either A or
B. A
has the wrong stereochemistry since the Ph and Et need to be trans not cis.
Qu34:
D
The alkyne undergoes a catalytic
hydrogenation to give the cis-alkene which
then undergoes a cyclopropanation
to give the cyclopropane.
Working backwards, the HBr
/ peroxides means a radical reaction... since the product is a mono-bromide
we needed to have started from an alkene, 1-methylcyclohexene. The first step
suggests (at first glance) that we started from an alcohol... but the only
alcohol in the options has too few C atoms... the only other option is C
where the carbocation formed by protonation of the alkene rearranges to form
the more stable system.
Qu37: B
AROMATICITY
and RESONANCE:
Best method is to work through the molecules and decide on the aromaticity based on the four criteria (cyclic, planar, conjugated pi system with an odd number of pairs (i.e.
4n+2) pi electrons)
Only B, C, D, AB, AD and CD
are ionic. Of these, B, C, AB, and CD are aromatic
as drawn (6 pi e- except for C which is 2 pi e-).
Qu39: CE
Only A, E, AC, AE,
BC and CE are uncharged. Of these A, E,
AC and BC are non-aromatic as drawn and AE is anti-aromatic
(4 pi e-)
Only A, C, D, AC and BC are hydrocarbons.
Of these C and D are aromatic as drawn. BC will still
be non-aromatic on deprotonation where as AC gives a 10 pi e- aromatic
system.
Qu41: E
Only A, E, AC and BC are non-aromatic as drawn,
and of these only E has a tautomer (like a ketone - enol but using N instead
of O).
Qu43: AC or BC
Only A, C, D, AC
and BC are hydrocarbons. A, C and D are
fully conjugated as all the C atoms contribute to the pi system.
Qu45: AE
Heterocycles have none C atoms in the ring i.e. B, AB,
AD, AE, CD and CE. B, AB, CD
and CE are aromatic (6, 6, 6 and 2 pi e- respectively). AD is non-aromatic
as it is not a cyclic conjugated system.
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