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.
Qu1: AB
Catalytic hydrogenation reactivity depends on the strength of the p
bonds.
This means that alkynes reduce more readily than alkenes since the
second
p
bond
is weaker, therefore iii > ii. The extra stability of aromatic
C=C
means they are not as reactive as simple alkenes, so ii > i. Overall
then
iii > ii > i.
Qu2: D
Alkenes react via electrophilic addition where the rate determining
step in the addition of the electrophile. Here the electrophile will is
H+ and so the rate of reaction depends on the stability of the
intermediate
carbocation, the more stable the C+, the more readily (faster) it will
form. Alkanes like i don't have electrons available to react with the
H+
so they are essentially unreactive. Cyclohexene ii will react to
give a secondary C+ and the alkyne iii will give a vinyl cation which
are
only about as stable as primary cations. Overall then ii > iii
>
i.
Qu3: C
If you know your pKas this is easy : terminal alkyne pKa = 26, amide
ion -NH2 about 35 and t-butoxide, (CH3)3CO-
= 18. Remember the higher the pKa the stronger the base, so ii
>
i > iii. What if you don't remember your pKas ?
(why
not ?) Do you know your reactions ? Recall that terminal alkynes
i.e. R-CºC-H can be deprotonated
and
the most common base for this is NaNH2... so the amide
ion
-NH2 must be a stronger base than the acetylide ion R-CºC-
. Alkoxides are weaker bases than amides and acetylides (or we
would
use it for this reaction !). You can rationalise the N vs O
system
by thinking about electronegativity.
Qu4: E
Carbocation stability is dictated by the substituents at the C+
center.
i is a simple secondary cation, ii is a secondary vinylic cation (which
is like a primary) due to the hybridisation at the C (note no resonance
as the pi system is perpendicular = no overlap) and iii is secondary
and
allylic - the C=C means we should recognise the possibility of
resonance
stabilisation. Therefore we get iii > i > ii.
Qu5: E
i has a chirality center (the C with the Br attached) and so can be
R or S. ii is achiral. iii has a chirality center (the C with the
Br attached) and the C=C can be either E or Z. So iii has 4 isomers, i
has 2 and ii has only 1. Overall then we have iii > i > ii.
Qu6: C
Reaction is the hydroboration oxidation of an alkene. This gives the
anti-Markovnikov alcohol via a syn addition due to the concerted
addition
of the B and H across the C=C. Since no C+ forms, there is no
carbocation
rearrangement. First draw 1-methylcyclopentene, then add the -OH
to the least substituted end of the C=C. So ii is the major
product
and ii is a minor product (the Markovnikov product). iii is only
obtained
is there has been a rearrangement. Overall then, ii > i > iii.
Qu7: A
Observed rotation is given by a = [a]D.c.l
We are told the pathlength l is a constant.
The diagram in i is the S enantiomer (so +ve) and the concentration
of the sample is 2/20 = 0.1 g/ml, so a =
+23.1
x 0.1
The diagram in ii is also the S enantiomer (so +ve) and the
concentration
of the sample is 1.5/30 = 0.05 g/ml, so a =
+23.1 x 0.05
In iii, we have more R than S, the observed rotation will be produced
by 1g of R in 20ml i.e. a = -23.1 x 0.05
So i > ii > iii.
Qu8: D
The more stable the alkene, the less exothermic the heat of
hydrogenation.
The more highly substituted the alkene, the more stable it is. i is
tri-alkyl
substituted, ii is mono- and iii is di, so i is the least exothermic
and
ii is the most, i.e. ii > iii > i.
Qu9: BC
Sucrose is the disaccharide of glucose and fructose, the pKa of HCl
= -7 and acetals are formed by the reaction of aldehydes or ketones
with
alcohols (they are important elements of carbohydrate structure).
Qu10: ABC
Cystine acid is a very minor component of casein (did you see it on
your chromatogram ?). Less polar materials interact less with the
paper (the stationary phase) of the chromatogram and so they run
faster,
a higher Rf value.
Qu11: ACE
Steam distillation relies on the fact that for immiscible liquids,
the vapour pressure of the mixture is based on the sum of the vapour
pressures
of the components. In an extraction in a separatory funnel, the density
of the solutions determines which layer is on top and which is at the
bottom,
for example water and dichloromethane, the water is typically the top
layer
whereas for ether and water, the water is usually the bottom layer.
Qu12: AD
Cross-linking makes the polymer less flexible and hence more rigid.
Kevlar is a polyamide (like nylon). PETE is the polymer of terephthalic
acid and ethylene glycol.
Qu13: CE
Buchner funnels are used for vacuum filtrations (i.e. for separation
of solids from solutions or liquids). Boiling chips are used to provide
a surface on which to initiate boiling and prevent super heating.
Aqueous
waste should be poured in the appropriate waste container and not
washed
away in a sink.
Qu14: E
Reaction is the hydroboration-oxidation (an addition reaction) of an
alkene. This gives the anti-Markovnikov alcohol since the B adds as the
electrophile and the H as a nucleophile (electronegativities : B = 2.0,
H = 2.1). Hence the B atom adds at the least substituted end of the
alkene.
During the oxidation step the B is replaced by OH to give the alcohol.
Since no C+ forms, there is no carbocation rearrangement. A
is an organoborane, which is formed initially but would react with the
hydrogen peroxide to give the alcohol. B and C require
a
rearrangement product. D has the wrong regiochemistry, it is
the
Markovnikov alcohol.
Qu15: B
KOH / heat causes an elimination of an alkyl halide to give
an alkene that is then subjected to ozonolysis with a oxidative
work-up.
Looking at the dicarbonyl product, we can work out the required alkene
by "snapping" together the two C=O units to give the C=C. This gives us
1-methylcyclohexene (draw carefully, count carbon atoms). The fact that
one end of the dicarbonyl product is a carboxylic acid and the other is
a ketone tells us that the two ends of the alkene are different. A
and D would not react with the KOH nor the ozone. C would
only deprotonate with the base, no alkene would be formed. E would
eliminate with KOH / heat to give 1-ethylcyclopentene with would give
5-oxoheptanoic
acid, an isomer of the desired product (6-oxoheptanoic acid).
Qu16: C
CH3CO3H is a peracid which is used to convert
an alkene to an epoxide. H2SO4 / heat is therefore being used to
form the alkene by the elimination (dehydration) of an alcohol. Recall
that these are usually E1 reactions, giving the more stable (more
highly
substituted alkene) as the major product. Hence 1-methylcyclohexanol
will
dehydrate to give 1-methylcyclohexene and this in turn gives the
epoxide
C
with the peracid. Note the epoxide unit defines the location of the
C=C.
Qu17: B
Terminal alkynes are deprotonated with sodium amide, a strong base
to give the acetylide ion, a good Nu. This reacts SN2 fashion with the
secondary alkyl bromide to give Ph-C#C-CH(CH3)2.
Alkynes react with aq. H2SO4 / Hg2+
salts
to under go hydration to an enol that will rapidly tautomerise to a
carbonyl
of a ketone. The hydration of the alkyne will proceed such that the
water
Nu attacks at the end of the alkyne that would give the more stable
carbocation....
in this case this means at the end closer to the phenyl group (which
could
stabilise the C+ by resonance). Hence the ketone B is the major
product rather than D. C would be formed
from
the alkene Ph-CH=CH-CH(CH3)2 and E would
require
Ph-(CH2)2C#CH.
Qu18: D
Radical bromination at the more highly substituted position followed
by E2 elimination to give the alkene, probably the anti-Zaitsev, less
highly
substituted alkene due to the bulk of the base and then form the
cyclopropane
ring via the Simmons-Smith reaction of a carbenoid reaction (with :CH2
as the reactive species). The position of the cyclopropane ring
confirms
the location of the C=C. A would give a bicyclic product, B
would brominate in an allylic position (most stable radical) then
eliminate
to a diene.... so cyclopropanation would give either a bicyclic alkene
or a tricyclic product. C D and E all have the
right
number of carbon atoms to be possible solutions but C would
give
a mixture of bromides and hence alkenes and would still have a t-butyl
group in the product. E would result in the cyclopropane unit
on
C3 rather than C2.
Qu19: E
The trans alkene would react with the bromine to give
2,3-dibromobutane.
Sodium amide is a strong base causing a double elimination of HBr to
give
the alkyne, 2-butyne.
Qu20: D
To get a geminal dibromide, the starting material needed to be an
alkyne.
A and B are alkenes which would react to give a
monobromide.
C,
an alkane, would not react with HBr. E would give a 50:50
mixture of 2,2- and 3,3-dibromohexane (equal stability of the
intermediates).
Qu21: C
Reaction of an alkene with a hypohalous acid to give a 1,2-halohydrin
followed by treatment with a base to allow an intramolecular SN2
reaction
(also an example of a intramolecular Williamson ether synthesis) to
give
an epoxide (hence answer must be A, B or C).
HOBr reacts as HO- and Br+, the reaction will lead to the halohydrin
via
the cyclic bromonium ion, the regiochemistry will put the -OH at the
more
substituted position with anti stereochemistry. For the SN2 the
nucleophilic
HO must attack at 180 degrees to the C-Br bond, therefore the two
methyl
groups will still be cis in the product epoxide. So A and
B
have the wrong regiochemistry and D and E are diols not
epoxides.
Qu22: D
The product is a 1,2-dichloride, so this must come from a C=C rather
than a CºC. Cl2 adds
across
C=C to add one Cl atom at each end of the p
bond in an anti fashion. Analysis of the Fisher diagram,
remembering
to redraw with the two Cl anti .... this means the two methyl groups
need
to be cis in the alkene. A would give a tetra-chloride if there
were excess chlorine or a vinyldihalide. B lacks a methyl
group. E has the C=C in the wrong location to give the required
product. So is it C or D. To sort them out, you need to pay
attention to the stereochemistry of the reaction. You should redraw the
product Fischer
diagram first as a wedge-hash diagram then so that the two Cl are
drawn anti as this is the way they are added in the reaction itself
since the addition of Cl is an anti addition process. Once
drawn this shows that that the original alkene
is requires that the methyl and t-butyl groups are trans, i.e. D. The alkene C where the two methyl groups are
trans would give products that were
stereoisomers of the required answer.
Qu23: BD
Dissolving metal reduction of an alkyne with Na in ammonia gives the
trans alkene. Generation of a carbene (here loss of HCl gives :CCl2)
which then adds to the C=C to give a cyclopropane in a syn fashion so
the
stereochemistry of the C=C is preserved. B and D
are
enantiomers so both will be produced in equal amounts. A
has
the wrong alkene stereochemistry, C has the wrong carbene. E
has the wrong reaction ! It's an epoxide not a cyclopropane system.
Qu24: C
Catalytic hydrogenation of the alkyne using Lindlar's catalyst gives
the cis alkene. Permanganate oxidation of an alkene gives the syn
1,2-diol, i.e. HO-C-C-OH. Overall anti addition to a trans
alkene
means a meso product. A, D and E
are not
a diols. B has the wrong stereochemistry. Overall syn addition
to
a cis alkene means a meso product and B is not a meso compound
(it's
the S,S enantiomer).
Qu25: E
Reaction is the hydroboration-oxidation of an alkene. This gives the
anti-Markovnikov alcohol via a syn addition due to the concerted
addition
of the B and H across the C=C. Note that the -OH is formed with
retention
of stereochemistry when the B atom is replaced. A and B
have
the wrong regiochemistry. C and D are from anti
additions
- consider the location of the -OH group - if the -OH and the methyl
group
are cis then the -OH and the -H that added must be trans (i.e. from an
incorrect anti addition).
Qu26: C
Dissolving metal reduction of an alkyne with Na in ammonia gives the
trans alkene. Bromination of an alkene gives a 1,2-dibromide i.e.
Br-C-C-Br with anti stereochemistry - thus just by looking at the
added
groups, it must be either C or D or E. Overall
anti
addition to a trans alkene means a meso product. Of the possible viable
solutions then, C is the (R,S) or meso isomer, D and E
are
the (S,S) and (R,R) enantiomeric pair, and so have the wrong
stereochemistry.
Qu27: AE or CD
Need to have the same configurations.
Qu28: AB or BE
Need to have the opposite configurations and all centers.
Qu29: (A or B or E)
with (C or D)
If they aren't enantiomers, they are diastereomers.... they need to
have the same configuration at at least one chirality center.
Qu30: CD
Meso isomers have an internal mirror plane so are (R,S) configuration.
Qu31: AE
Fisher diagrams are defined based on an eclipsed conformation (so A).
Eclipsed means zero degrees as a torsional angle, so E.
Qu32: E
The specific rotation [a]D =
a /c.l where a
= obs. rotation, 5.1o, c = conc. g/ml and l = pathlength in
dm. So [a]D = +5.1o /
(1.5/10)(1) = +34o
Qu33: C
Since A is (R,R) and B is (S,S), and there is more A
than B, and the sample rotation has a +ve rotation, then
the
[a]D for (R,R) must be +ve. If we
calculate the enantiomeric excess of A to B = (1.2 -
0.3)/(1.2
+ 0.3) = 0.9/1.5 = 60% then based on optical purity, 60% [a]D
for (R,R) = +34o therefore [a]D
for (R,R) = 56.67o
Qu34: D
See answer to Qu 33.