Name:
CHEM 331
FOURTH EXAMINATION
All answers should be written on the exam in the spaces provided. Clearly indicate your answers
in the spaces provided; if I have to guess as to what or where your answer is, it is wrong. Where
applicable, outline the logic or mystical principle you used to arrive at your answer, as partial credit may
be awarded for correct approaches. Work pages are provided at the end of the exam. Make sure your
name is on each page of the examination; the management is not responsible for lost pages.
You may assume that standard work-up conditions ( i.e. those required to obtain a stable, neutral
product, like aqueous acid as example) follow each reaction on the exam. Clearly indicate
stereochemistry where appropriate. Be careful not to show stereochemistry where none exists.
Please take this test wisely for the time provided. You are strongly advised to read it through
completely before you begin.
(1)..20 pts....................._____
(2)..24 pts....................._____
(3)..18 pts....................._____
(4)..12 pts....................._____
(5)..12 pts....................._____
(6).. 6 pts....................._____
(7).. 8 pts....................._____
Bonus…4 pts……...…_____
_______________________
TOTAL (100 pts)…....….._____
Percentage …………….–––––
1. [20 pts] Answer the following questions about radical reactions:
a. [6 pts] Label each of the following steps as Initiation, propagation or termination
+
O
O
O
O
termination
CH4
Cl2
+
Cl
2 Cl
CH3
+
H-Cl
propagation
initiation
b. [6 pts] There are three constitutional isomers of molecular formula C6H14 shown below. Assign the
structures to their identities (A – C) based on the information below:
When treated with Cl2 and light, isomer A gives a mixture of five monochlorinated products. Under the
same conditions, isomer B gives a mixture of three monochlorinated products. Isomer C produces two
different monochlorinated products when reacted under these conditions. Assign structural formulas to
isomers A, B and C.
Isomer A:
Isomer B:
Isomer C:
c. [4 pts] Draw the most stable radical that could form upon hydrogen atom abstraction from the
following molecule.
d. [4 pts] Explain why bromine is regioselective for radical bromination, as shown in the reaction below:
Br
Br2
light
the most stable radical forms on the benzylic
position
2. [24 pts] Predict whether substitution and/or elimination will predominate in the following reactions by:
a.
Identify what type of alkyl halide is present (1°, 2°, 3°, allylic or benzylic if so)
b.
Identify whether you have a strong or weak base, strong or weak nucleophile, or what the solvent
indicates for the reaction.
c.
Identify the most likely mechanism(s) (SN1, SN2, E1, E2)
d.
Draw the structures of the resulting product(s) and do not forget stereochemistry in your product if
necessary.
(i)
Br
OH
2º benzylic halide
weak nuc/weak base
SN1 (E1 not possible)
H2O
(ii)
NaOH
Br
OH
1º halide
strong nuc/strong base
SN2 only on 1º systems
(E2 only with SNNB)
(iii)
OMs
NaOCH3
OCH3
2º halide
strong nuc/strong base
SN2 and E2
(iv)
Br
KOtBu
tBuOH
3º halide
Strong, Non-Nucleophilic Base
Presence of tBuOH solvent Hofmann Product
3. [18 pts] Answer the following general questions about Substitutions and Eliminations (Circle the
correct answer):
a. Which of the following isomers reacts fastest in an SN2-type reaction?
i.
ii.
iii.
2-chloro-4,4-dimethylpentane
1-chloro-3,3-dimethylpentane
3-chloro-2,3-dimethylpentane
What is the effect on the rate of the E2 reaction of tert-butyl bromide, (CH3)3CBr, and NaOH if the
concentration of the base is halved?
b.
iv.
v.
vi.
Faster
Slower
No Change
c. Which would be the best alkyl halide to react with sodium methoxide, NaOCH3, to form the following
ether:
OCH3
i.
ii.
iii.
1-fluoro-4-methylpentane
1-iodo-4-methylhexane
1-bromo-4-methylpentane
d. What is the effect on the SN2 reaction of 1-bromobutane if the solvent is switched from
dimethylsulfoxide (DMSO) to methanol (CH3OH)?
i.
ii.
iii.
Faster
Slower
No Change
e. Which produces the most selective reaction and highest yielding product formation of 2-pentene from
2-bromopentane?
1.
NaOCH3 in DMSO
ii.
KOtBu (KOC(CH3)3) in DMSO
iii.
KOtBu (KOC(CH3)3) in tBuOH
f. What is the effect on the SN1 reaction of 3-bromo-3-methylpentane with water, H2O, if the
concentration of the nucleophile is doubled?
i.
ii.
iii.
Faster
Slower
No Change
4. [12 pts] For each of the following reactions, draw the starting reagent or product(s) or provide the
reagent necessary for each one-step conversion to occur. Assume standard conditions to produce stable,
neutral product formations.
a.
HCl
Cl
b.
OH
Br
HBr
c.
OH
SOCl2, pyridine
Cl
d.
Br
(CH3)2CuLi
CH3
5. [12 pts] Synthesis: Devise a short sequence (2 steps) of reactions for the construction of the following
product. Draw the intermediate product that may form in your sequence to receive full credit.
a.
HBr
Br
(CH3CH2)2CuLi
b.
Br
Mg
MgBr
H2O
6. [6 pts] Name the following compound according to IUPAC rules:
Br
5-bromo-7-cyclopropyl-3-ethyl-6-isobutyl-1-decene
7. [16 pts] Short Answer Questions: Answer 2 of the following 3 questions:
a. Explain why 1-bromo-3,4-dimethyloctane does not undergo SN1 type reactions.
Primary alkyl halides cannot form a carbocation intermediate and this compound is primary.
b. Explain why SN2 reactions proceed slower in a protic solvents like methanol, CH3OH.
Protic solvents surround the nucleophilic preventing it from being able to react with the electrophilic
carbon of the alkyl halide.
c. Front side attack is disfavored for an SN2-type substitution reaction because the leaving group and the
nucleophile both carry negative charges and thus repel each other. What is the other reason front side
attack is disfavored?
Sterics – not enough physical space for a nucleophile to attack while the leaving group is leaving.
Bonus: [4 pts] Explain why trans-1-bromo-2-methylcyclohexane yields the Hofmann product 3methylcyclohexene on treatment with any strong base.
H
CH3
Br
For an E2 reaction on a cyclohexane ring, anitperiplanar means that the
leaving halide must be trans diaxial to a H atom. The H atom that would
lead to the Zaitsev product is cis, not trans to the bromide. Zaitsev cannot
form. Hofmann is the only elimination product able to form.