Chemistry 14D Winter 2016
Midterm Exam 1 Solutions
Page 1
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1.
(a) This reaction is fastest when X (the leaving group) is iodide or a sulfonate (triflate in particular).
(b) The reaction is fastest when the nucleophile is HS- or S2- (followed by protonation).
(c) The reaction is fastest when the solvent is acetone (best answer) or DMF. Methanol and ethanol are
moderately polar but protic, and therefore reduce nucleophilicity due to hydrogen bonding.
Cl δ-
H
2.
δ− Cl
C
H3 C
H
3.
I δ−
CH2CH3
and
H
C
I δ-
CH2CH3
Models make this task much, much easier.
CH3
I
Models make this task easier.
4.
Mostly (S) with some (R). SN2 always causes inversion of configuration. (R)-2-chlorobutane, 80% of the
starting material, becomes (S)-2-iodobutane, 80% of the product.
5.
(a) Slower. Triflate ion is a poorer nucleophile than iodide ion due to resonance and inductive effects.
(b) Faster. Triflate ion is better leaving group than chloride ion due to resonance and inductive effects, as well
as oxygen's high electronegativity.
(c) Faster. In an aprotic solvent, fluoride ion is a better nucleophile than iodide ion due to fluorine's small
atomic radius and compact electron density.
(d) Slower. This change increases the steric hindrance at the carbon bearing the leaving group.
(e) Slower. Methanol is more polar than acetone. Methanol is protic whereas acetone is aprotic.
6.
ε = 20. Compared to ethanol, propanol has the same number of polar bonds but more nonpolar bonds, so
propanol is less polar than ethanol.
7.
Protic (it has an OH group) and moderately polar or nonpolar (ε = 20; right at the moderately/nonpolar cutoff).
8.
(a) Decreased. The nucleophile has a negative formal charge, and is therefore its energy decreases significantly
as solvent polarity increases.
(b) Decreased. The transition state has two δ charges, and is therefore its energy decreases a bit as solvent
polarity increases.
−
(c) Reactants. The nucleophile has a full negative formal charge, whereas the transition state has partial
charges. Therefore the energy of the reactants is more strongly influenced by solvent choice than the
energy of the transition state.
Chemistry 14D Winter 2016
(d)
Midterm Exam 1 Solutions
Page 2
Small stabilization
Large stabilization
Energy
ΔG‡ acetone
ΔG‡ acetonitrile
Acetone
Acetonitrile
Reaction coordinate
(e) Acetone. ΔG‡ (acetone) < ΔG‡ (acetonitrile). Note that the problem statement indicates low solubility is not
an issue in this case.
Ph
9.
Ph
CH3
CH3
OH
OH
Ph
Ph
H
CH3
CH3
Among possible 1,2-shifts, the most likely shift leads to the most stable carbocation. 1,3-shifts are forbidden
regardless of how stable the resultant carbocation is.
10. Capture a nucleophile, and be deprotonated to form a pi bond.
H3C
H3C
I
11.
CH3OH
Cl
OCH3
Cl
Displacement of the best leaving group, formation of
the more stable carbocation, and inversion of
configuration.
H
12.
H3C
Cl
H3C HOCH3
I
Cl
H3C
Cl
HOCH3
H3C
OCH3
OCH3
Cl
Same mechanism applies for both product enantiomers.
13. (a) Slower. CH3CH2CH2OH is less polar than CH3OH.
(b) Faster. The carbocation intermediate is more stable.
(c) Slower. Chloride ion is a poorer leaving group than iodide ion.
14. (a) This reaction does not occur by the SN2 mechanism because the carbon bearing the leaving group is tertiary
(i.e., there is too much steric hindrance for the nucleophile to approach).
(b) This reaction does not occur by the SN2 mechanism because methanol is a poor nucleophile.
Chemistry 14D Winter 2016
H3C
Midterm Exam 1 Solutions
Page 3
OCH3
15.
Other answers are possible.
Cl
Cl
Cl
From SN1
From E1
From E1
The reaction involves a carbocation, so any carbocation fate can occur. The question does not limit these
products to SN1.
16. C is not the most likely ionization product because hydroxide ion is not a leaving group in SN1 or E1.
D is the mostly likely ionization product.
E is not the most likely ionization because carbocation E is less stable than carbocation D.
17. Z. The highest priority groups (Br and CH2CH3) are on the same side of the alkene. Cis and trans do not apply
because both alkene carbons do not have a carbon group.
H3C
18.
CH2CH3
C
H3C
Internal, tetrasubstituted alkene. Steric strain is not too severe.
C
CH3
H
19.
OH2
HO
H3C
H3C
OH2
H
H3C
OH2
Also acceptable:
H3C
CH3
CH3
CH3
OH2
CH3
CH3
H
All other alkene products are less stable, and therefore probably formed in lesser amounts.
20. Reason #1: There is no strong base. E2 requires a strong base. Like methanol and ethanol, CH3CH2CH2OH is a
poor base.
Reason #2: Fluoride ion is a poor leaving group. It can be made to leave in an E2 reaction, but only under
extreme conditions.
Reason #3: The molecule cannot easily achieve a periplanar (anti or syn) H–C–C–LG arrangement. Verify this
with a model.