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.
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
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.
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
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....
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.
Qu42: BD
Qu43: B
Qu44: CE
Qu45: AC
Qu46: DE
Qu47: CE
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.