353 MT Winter 1999
Here is an post-mortem anaylsis / "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
Anions are more nucleophilic than neutral species (high e density) so
ii>i
and nucleophilicity increases as you go down a group becuase the larger
atoms are more polarisable and this facilitates the bond formation so
iii>ii
Qu2: B
Good LG are the conjugate bases of strong acids so by considering the
acids
involved HBr, H2O and NH4+ we can see that i > iii > ii
Qu3: D
AgNO3 / aq EtOH indicates an SN1 reaction, so our first concern should
be the stability of the carbocations formed. ii is the very favourable
benzyl cation (resonance stabilised), iii is secondary and i is the
very
poor phenyl cation (no resonance due to perpendicular nature of the
orbitals)
so rate should be ii > iii > i
Qu4: B
Two things we can use to get the answer to this. First the trends from
the periodic table. Basicity increases R to L so C- > N- > O-.
But
that is only based on comparing atoms in similar states. In this
question
the C is only sp hybridised and this makes it more stable and therefore
less basic (inc. s character moves e closer to the +ve nucleus). SO how
do we deal with that ? Need to recall how we prepare the acetylide
anion.....
CH3CCH + NaNH2 ---> CH3CC- Na+ and NH3
This tells us that -NH2 (pKa=33) is a stronger base than CH3CC-
(pKa=25).
If an alkoxide were strong enough then it would be the base of choice
for
the reaction... but it isn't. (pKa=16) So i > iii > ii
Qu5: AB
Radical stability follows the same order as for carbocations 3 > 2
>
1. SO in the question we have i = 2, ii = 3 but iii is 2 and benzylic
so
iii > ii > i
Qu6: D
Reaction is of cyanide (good nucleophile) in a polar aprotic solvent
(DMSO)
so SN2. Therefore we need to look at degree of substitution and the LG.
They are all Cl, so we have i = 3, ii = Me and iii = 1 therefore rate
is
ii > iii > i since the nore hindered the system the slower the
reaction.
Qu7: D
Reaction conditions tells us that we are dealing with an elimination of
water under acidic condtions so probably E1 and therefore controlled by
the stability of the carbocation. i = phenyl, ii = 2o and
benzylic
and iii = secondary. Therefore ii > iii > i. Infact the phenol
(i)
will not dehydrate at all under these types of conditions.
Qu8: AB
Nucleophilicity. Due to polarisability (see above) S systems are better
Nu than O therefore iii > ii. Since steric bulk tends to inhibit Nu
properties due to the difficulties in getting at the reactive center,
ii
> i. So iii > ii > i
Qu9: A
Slow reaction with AgNO3 / aq EtOH (SN1) and fast with NaI/acetone
(SN2)
indicates an primary alkyl halide
Qu10: E
Colourless test with Br2 indicates an alkene or alkyne. The alcohol
that
gives the ketone BD on oxidation is 2-butanol which is AD. Dehydration
of 2-butanol gives mainly trans-2-butene, E.
Qu11: AB
Colourless test with Br2 indicates an alkene or alkyne. The Lucas test
tells us that the alcohol is tertiary so t-butanol, AE, which when
dehydrated
gives alkene AB.
Qu12: BD
Precipitate with 2,4-DNP indicates either an aldehyde or a ketone. The
Lucas test tells us that the alcohol is secondary, so it has to be
2-butanol,
AD, which when oxidised gives BD
Qu13: C
Slow reaction with NaI/acetone (SN2) but rapidly with AgNO3 / aq EtOH
tells
us that we are after an tertiary alkyl halide. The halide obtained from
the Lucas test of t-butanol AE must be t-butyl chloride, C.
Qu14: BE
AC is a primary alcohol so it can be oxidised to either the aldehyde BC
or the acid BE. The clear red colour with 2,4-DNP is not a positive
test
(a precipitate is required) so we must have the acid BE.
Qu15: C
The alcohol would react thionyl chloride (SN2) to give the alkyl
chloride,
which in turn reacts with the Mg to generate the Grignard reagent.
Grignard
reagents are good bases so any acidic H will just protonate them to
generate
the alkane (C).
Qu16: A
The alcohol reacts with HBr to give the bromide (SN2). React with Li
gives
the alkyl lithium which on treatment with Cu generates the lithium
dialkyl
cuprate (CH3CH2CH2)2CuLi.
Unlike
Grignards or alkyl lithiums, cuprates react with alkyl bromides to give
what looks like an SN2 react and "couples" the alkyl groups,
so here we get butane CH3CH2CH2CH3
Qu17: A
Radical bromination of pentane should give 2-bromopentane as the major
product (2o 4H). Substitution with water gives 2-pentanol
which
is oxidised with the chromate (see Cr, think [O] ! ) to give the
ketone,
2-pentanone, A.
Qu18: C
A Gabriel synthesis of primary amines. Phthalimide is treated with a
base
to remove the NH proton generating an N-ve nucleophile. This reacts in
an SN2 reaction with the n-propyl bromide which is then reacted with
hydrazine
(NH2NH2) to release the primary amine, C.
Qu19: C
Treatment of an alcohol with conc. sulphuric acid should cause a
dehydration
reaction (elimination of water). Protonation of the -OH to make a
better
leaving group, then loss of water gives a 2o carbocation. A
1,2 methyl shift can then occur to form the 2,3-dimethylbutane skeleton
and the more stable 3o carbocation. Loss of a proton gives
the
tetra-substituted alkene, 2,3-dimethyl-2-butene, C.
Qu20: B
Reaction of a terminal acetylene (alkyne) with the NaNH2
generates
the anion (acid/base reaction). This good carbon nucleophile then
undergoes
an SN2 reaction with methyl iodide to produce 2-pentyne, B.
Qu21: C
The only molecule there that can react with NaOH other than to give a
simple
deprotonation is C. The hydroxide function as a Nu, attacks the less
substituted
end in SN2 fashion, opening the epoxide, which on protonation gives the
1,2-diol.
Qu22: B
If we are reacting with Mg, we must be preparing a Grignard reagent, so
we must be starting from an alkyl halide, so it has to be B.
Qu23: C
A tricky one. TsCl is used to prepare tosylates from alcohols, but all
choices are alcohols, R-OH. Once we have our R-OTs group we are
substituting
using an acetate, CH3COO-, to give an ester, CH3COO-R
which is hydrolysed with aq. acid to give back the alcohol R-OH. Note
how
we have not changed the system by addding any atoms (C, H or O).
However
this final step gives an inversion of stereochemistry because it is an
SN2 type process. Therefore we must have started form the enantiomer of
the product, so C.
Qu24: D
PBr3 reacts with alcohols to give bromides, so we can narrow
it to C or D. We know we must have a halide becasue the next step is
the
preparation of a Grignard reagent which is then reacted with H2C=O
(formaldehyde) followed by an acidic work-up to give a primary alcohol,
which on oxidation with Cr gives the carboxylic acid. Therefore, the
starting
material only contains 6C (the 7th in the product is from the
aldehyde),
so the SM = D.
Qu25: D
Product we are after is a cyclic ether. Final reagent is Na metal,
which
could be used to prepare an alkoxide from an alcohol as part of a
Williamson
ether synthesis, in this case an intramolecular example. Therefore we
need
to break open the cyclic ether to a 5-halo-pentan-1-ol (count C !)
Working
back from this NaI /acetone is SN2 method and would have put in a -I
(to
create a better LG for the Williamson step. The aq NaOH has probably
been
used to introduce the -OH group. Given the trend, the only reasonable
conclusion
is D. It has the right number of C, and means we can change the -Br to
-OH, then the -Cl to the -I then use the Na to prepare the nuclophile
for
the ether formation.
Qu26: D
A Wittig reaction, given away by the PPh3 and the alkene in
the product. By counting C, D has to be the answer. The sequence is the
PPh3 reacts with ethyl bromide via SN2 to give a phophonium
bromide. Deprotonation with a strong bsae (BuLi) gives the ylid, or
Wittig
reagent (Ph3P=CHCH3) which will react with the
ketone
to give the exocyclic, tri-substituted alkene.
Qu27: D
Reaction of the tolyl chloride (p-chlorotoluene) with Li generates the
aryl lithium. This reacts with the epoxide, followed by an acid work-up
to give 2-(p-tolyl)ethanol. H-nmr of this should be characterised by 3H
singlet for the Ar-CH3 around 2.3ppm, 4 ArH (7-8ppm), 2
triplets
for the two coupled -CH2- units (one deshielded due to -O-)
and a singlet for the -OH. This is spectrum D.
Qu28: E
2-(p-tolyl)ethanol is just an alcohol that will be dehydrated by H2SO4
to give the alkene, p-tolylethene. The H-nmr of this will show a 3H
singlet
for the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm) and a 3H
mulitplet
pattern between 5-7ppm, characteristic of alkene H. This is spectrum E.
Qu29: A
Catalytic hydrogenation of the p-tolylethene will reduce the alkene to
the alkane, p-tolylethane. The H-nmr of this will show a 3H singlet for
the Ar-CH3 (@ 2.3ppm), 4 ArH (7-8ppm), then very
characteristic
2H quartet and 3H triplet for the ethyl group. This is spectrum A.
Qu30: B
Radical chlorination of p-tolylethane will give
1-chloro-1-(p-tolyl)-ethane
as the major product due to the stability of the secondary, benzylic
radical.
The H-nmr will be characterised by 3H singlet for the Ar-CH3
(@ 2.3ppm), 4 ArH (7-8ppm), a deshielded 1H triplet (above 4ppm due to
Cl and Ar) and a 3H doublet, as shown by spectrum B.
Qu31: AE
Water will substitute the benzylic chloride to give 1-(p-tolyl)-ethanol
via an SN1 reaction. The H-nmr will show 3H singlet for the Ar-CH3
(@ 2.3ppm), 4 ArH (7-8ppm), a deshielded 1H quartet (above 4ppm due to
O and Ar), a 3H doublet (-CH3) and a 1H for the -OH. See AE.