Homework problems –Chapters 6 and 7
1. Give the curved-arrow formalism for the following reaction:
: :
H3C
O: - + H
H
C
C
H2
CH3 OH + H2C
Br
CH + Br-
CH3
CH3
2. In each of the following sets, arrange the compounds in order of
decreasing pKa and explain your reasoning.
As pKa decreases, acidity increases. Hence,
(a)
Relative acidity: CH3CH2OH < ClCH2CH2OH < ClCH2CH2SH
(b)
Relative acidity: CH3—COOH
CH3OCH2—CH(OCH3)—COOH
(c)
Relative acidity:
(CH3)3N+—OH
Relative acidity:
(d)
CH3CH2OH
<
<
CH3O—CH2—COOH
<
(CH3)2N—CH2CH2OH
<
O
CH3 CH2CH2
O
C
<
OH
Cl—CH2 CH2CH2
C
O
H3CHC CH2
C
OH
<
O
OH
<
CH3CH2
Cl
HC
C
OH
Cl
In each case, the order of acidity is determined by inductive effects.
3. Rank the following compounds in order of increasing basicity:
O
H3N +
O
O
O-
< CH3
C
NHC
O< CH3
The donor atoms are from the same period (row) of the periodic
table, hence electronegativity wins, and the less electronegative
atom is more basic. For bases with identical donating atoms,
positive charge on the molecule decreases basicity.
CH2
C
1
4. In each pair, choose the species that reacts more rapidly with
methyl bromide in ethanol solvent, and explain your choices.
(a) –OC2H5 or C2H5OH
When the attacking atom is the same, nucleophilicity roughly
parallels Bronsted basicity. Since ethoxide (–OC2H5) is a much
stronger base than ethanol, it will react more rapidly with methyl
bromide via SN2 mechanism.
(b) F- or IIn protic solvents, nucleophilicity increases with increasing
atomic number within a group of the periodic table. Thus, I- is
the better nucleophile and will react more rapidly with methyl
bromide.
5. In each pair, choose the combination of nucleophile and solvent
that would give the faster SN2 reaction with ethyl iodide
(a) Na+ –OAc in formamide, or Na+ –OAc in DMF
Formamide [HC(O)NH2] is a protic polar solvent while DMF
[H(CO)NMe2] is an aprotic polar solvent
.A protic solvent
consists of molecules than can act as hydrogen-bond donors.
Polar protic solvents slow down SN2 reaction by stabilizing the
nucleophile via hydrogen bonds to the nucleophile and hence
decreasing its reactivity towards electrophiles. Since the
nucleophilicity of an anion is much greater in polar aprotic
solvents, Na+ –OAc in DMF would give the faster SN2 reaction
with ethyl iodide.
(b) Na+ –OAc in acetic acid, or Na+ –OAc in DMF.
Acetic acid is a protic solvent. Thus, Na+ –OAc in DMF would give
the faster SN2 reaction with ethyl iodide for the same reason as
above.
6. Choose the alkyl halide(s) from the following list of C6H13Br
isomers that meet each criterion below:
(1) 1-bromohexane
(2) 3-bromo-3-methylpentane
(3) 1-bromo-2,2-dimethylbutane (4) 3-bromo-2-methylpentane
(5) 2-bromo-3methylpentane
2
(1) Br
(2)
Br
(3) Br
Br
(4)
(5)
Br
(a)
the compound that gives the fastest SN2 reaction with
sodium methoxide
1-Bromohexane gives the fastest SN2 reaction with sodium
methoxide because primary alkyl halides undergo SN2
substitution with good nucleophiles (such as methoxide),
whether they are good or poor bases. 1-Bromohexane
reacts faster than 1-bromo-2,2-dimethylbutane (which is
also a primary alkyl halide) due to decreased steric
crowding of the transition state.
(b)
the compound that is least reactive to sodium methoxide
in methanol
As a primary alkyl halide that contains no C-H bonds
adjacent to the electrophilic carbon center, 1-bromo-2,2dimethylbutane can only undergo SN2 substitution and
cannot undergo E2 elimination. Similarly, it cannot
undergo SN1 or E1 reactions since primary carbocations
are too high in energy whereas secondary and tertiary
alkyl halides readily dissociate in protic solvents due to
stabilization of the resulting carbocation by solvation (via
hydrogen bonding). Since SN2 reactions are slowed down
in protic solvents due to solvation of the nucleophile and
the transition state for SN2 reaction is more crowded for 1bromo-2,2-dimethylbutane than for 1-bromobutane, it is
least reactive to sodium methoxide in methanol.
(c)
the compound(s) that give only one alkene in the E2
reaction
Anti periplanar geometry, meaning that all four reacting
atoms- the hydrogen, the two carbons, and the leaving
group- lie in the same plane, is required for E2 elimination.
Only 1-bromobutane will give only one alkene because it
contains only primary C-H bonds adjacent to the
electrophilic carbon.
The remaining alkenes either have
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no C-H bonds adjacent to the electrophilic carbon or more
than one type of C-H bond (primary, secondary, or
tertiary).
(d)
the compound(s) that give an E2 but no SN2 reaction with
sodium methoxide in methanol
Tertiary alkyl halides are so hindered that they never
undergo SN2 substitution. E2 elimination occurs when a
good base, such as methoxide, is used. Thus, only 3bromo-3-methylpentane undergoes no SN2 reaction.
Remember, secondary alkyl halides undergo both SN2 and
E2 elimination, and a mixture of products often results.
(e)
the compound(s) that undergo an SN1 reaction to give
rearranged products
When
3-bromo-2-methylpentane
and
2-bromo-3methylpentane dissociate bromide, they form secondary
carbocations. The latter can undergo a hydride shift to give
a more stable tertiary carbocation, which then reacts with a
nucleophile to produce a rearranged product.
CH3
H 3C
CH
CH
H2
C
-BrCH3
H 3C
H
CH3
C +
C
H
H2
C
hydride
shift
CH3
CH3
H2
+C
C
C
CH3
H 3C
H2
Br
Nu-
Nu-
CH3
H
H 3C
C
CH
H2
C
CH3
H2
Nu
+
C
C
C
CH3
CH3
H 3C
H2
Nu
(f)
the compound that gives the fastest SN1 reaction
3-bromo-3-methylpentane gives the fastest SN1 reaction
because it forms the most stable carbocation, and
carbocation formation is rate determining in SN1 reactions.
7. Give the products expected when 1-bromo-3-methylbutane
reacts with the following reagents:
(a) KI in aqueous acetone
4
KI
Br
I
H2O/acetone
Under the above conditions, which are essentially neutral,
primary alkyl halides undergo SN2 substitution with good
nucleophiles.
(b)
KOH in aqueous ethanol
KOH
Br
HO
H2O/ethanol
Under basic conditions, primary alkyl halides undergo SN2
substitution with good nucleophiles, whether they are good or
poor bases.
K+ t-Bu—O- on t-BuOH
KOC(CH3)3
(c)
Br
t-butanol
Under basic conditions, primary alkyl halides undergo E2
elimination with good bases/poor nucleophiles.
8. Give the products expected when 2-bromo-2-methylhexane
reacts with the following reagents:
(a) KI in aqueous acetone
I
A
Br
KI
H2O/acetone
OH
+
B
This is a tertiary halide in a polar protic solvent mixture
(water/acetone) containing a good nucleophile that is a weak
base (I-). The polar protic solvent mixture promotes carbocation
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formation, and hence SN1 and E1 reactions are observed. The
SN1 products are the alkyl iodide (A) and the alcohol (B). A will
predominate, the more iodide is present since it is a much better
nucleophile than water. Some alkenes will also be formed since
E1 reaction always accompanies the SN1 reaction of an alkyl
halide with β-hydrogens. No rearrangement products are
predicted because the carbocation intermediate is tertiary.
(b) Sodium ethoxide in ethanol
This tertiary alkyl halide is subjected to a strong base in a protic
solvent. Under these conditions, tertiary alkyl centers are so
hindered that they never undergo SN2 substitution but the E2
elimination can occur. While protic solvents promote SN1 and E1
reactions, high concentration of base will favor E2 elimination
since rates of SN1 and E1 reactions are independent of the
concentration of base. The alkene products are shown below.
Note that the more substituted alkene D will predominate
according to Zaitsev’s rule.
C
Br
NaOC2H5
ethanol
+
D
If the concentration of ethoxide is low, then SN1 and E1
reactions will compete more effectively with E2 elimination. E1
elimination will give the same alkene products as above while
SN1 result in the ether product.
(c)
1:1 ethanol-water
Neither SN2 nor E2 reactions can occur for tertiary alkyl halides
in polar protic solvents in the presence of a poor nucleophile and
a weak base. However, polar protic solvents promote SN1 and
E1 reactions, and hence these reactions are observed. The
substitution products would predominate because of high
concentration of nucleophile, and the alcohol will be the
predominant substitution product since water is the better
nucleophile. The more substituted alkene will predominate
according to Zaitsev’s rule.
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OH
Br
major
+
ethanol/H2O (1:1)
major
+
OC2H5
minor
minor
9. Rank the following compounds in order of increasing SN2
reaction rate with KI in acetone.
(a) methyl bromide
(b) sec-butyl bromide
(c)
3-(bromomethyl)-3-methylpentane
(d) 1-bromopentane
(e) 1-bromo-2-methylpentane
(a) H
H
(d)
Br
Br
H
(b)
(c)
Br
(e)
Br
Br
(b) < (c) < (e) < (d) < (a)
SN2 reactions can occur only at relatively unhindered sites.
Thus, methyl halides are by far the most reactive in SN2
reactions, followed by primary alkyl halides. Branching one
carbon removed from the leaving group greatly slows
nucleophilic displacement, with steric hindrance increasing as
branching increases. Thus, (c) reacts more slowly than (e).
Alkyl branching next to the leaving slows the reaction even
more, and hence only a few simple secondary alkyl halides are
useful substrates in this reaction while tertiary alkyl halides do
not undergo SN2 reactions.
10. Give all of the product(s) expected, with pertinent
stereochemistry, when each of the following compounds reacts
with sodium ethoxide in ethanol.
(a)
(R)-2-bromopentane
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This secondary alkyl halide is subjected to a strong base/strong
nucleophile in a protic solvent. Under these conditions, the SN2
reaction is retarded by branching at the electrophilic carbon
center and by solvation of the nucleophile but the E2 elimination
can occur.
Although protic solvents promote SN1 and E1
reactions, high concentration of base will favor E2 elimination
since rates of SN1 and E1 reactions are independent of the
concentration of base. Thus, the products are the following two
alkenes:
Br
E and Z , with E predominanting
major
H 3C
H
H
H
minor
The ether product resulting from the minor SN2 reaction pathway
will show complete inversion of configuration at the electrophilic
carbon stereocenter.
(b)
Under basic conditions, primary alkyl centers undergo SN2
substitution with good nucleophiles whether they are good or
poor bases. Thus, ethoxide will be substituted for bromide with
complete inversion of configuration at the electrophilic carbon
stereocenter.
H
CH3 CH2CH2
OC2H5
CH3 CH2CH2
H
OC2H5
Br
ethanol
D
S-configuration
-
D
R- configuration
11.Which of the following alkyl halides can give only one alkene,
and which can give a mixture of alkene in the E2 reaction?
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CH3CH2
Br
CHCH2Br
(a)
(b)
CH3
Br
H2
C
(c)
C
CH3
CH3
Br
B:(a)
Br
(b)
H
=
CH3CH2
C
CH3
CH2
H
-
B:
Br
H
Both compounds (a) and (b) can only one alkene product in the
E2 reaction: (a) because there is only one C-H bond adjacent
to the C-Br bond; (b) because abstraction of either axial
hydrogen by base will lead to the same product. In contrast,
(c) contains primary and secondary C-H bonds adjacent to the
C-Br bond and can thus give more than one product.
(c)
H
Br
C
C
H
CH3
CH3
H
Br
H
C
C
C
H
CH3
H
H
OR
B:-
B:-
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12. Which alkyl halide and what conditions should be used to
prepare the following alkene in good yield by an E2 elimination?
CH2
10
13.What products are formed, and by what mechanisms, in each of
the following cases?
(a) methyl iodide and NaCN in ethanol
Because a methyl halide has no β-hydrogens (C-H bonds
adjacent to the electrophilic center), it cannot undergo
elimination reaction. The only possible reaction is SN2 although
polar protic solvents are not as effective for SN2 reaction as polar
aprotic ones. Thus, the product is CH3—CN because –CN is an
excellent nucleophile.
(b)
2-bromo-3-methyl butane in ethanol
This is a secondary halide under conditions involving no
nucleophile or base other than solvent, and a polar aprotic
solvent. Because ethanol is a poor nucleophile and a weak base,
neither SN2 nor E2 reactions can occur. Because polar protic
solvents promote SN1 and E1 reactions, these will be the only
reactions observed:
Br
OC2H5
ethanol
+
SN1 product
+
E1 products
+
OC2H5
+
carbocation rearrangement products
Notice the rearrangement products! Rearrangements should be
considered anytime the SN1 or E1 reaction is expected,
especially if the initially formed carbocation is secondary.
(c)
(d)
(e)
2-bromo-3-methylbutane in anhydrous acetone
2-bromo-3-methylbutane in ethanol containing
excess of sodium ethoxide
2-bromo-2-methylbutane in ethanol containing
excess of sodium iodide
an
an
11
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(f)
neopentyl bromide in ethanol containing an excess of
sodium ethoxide.
Br
Neopentyl
bromide
contains
no
βhydrogens hence no elimination is possible. Although it
is a primary halide, it is does not readily undergo SN2
reaction because of great steric hindrance in the
transition state of the reaction. And since primary alkyl
halides do not form carbocations, neither SN1 nor E1
reaction is possible. Thus, this alkyl halide is essentially
inert and no reaction is predicted. (If the reaction
mixture is strongly heated, some SN2 reaction might
occur after several days)
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