1.
Rank the following series by the trend requested.
a.(15) Rank the following compounds by rate of SN1 reactivity. Molecule that will react the
fastest in an SN1 reaction is 1 while the slowest is 5.
Cl
Br
3
2
Cl
Cl
I
5
1
4
b.(15) Rank the following compounds by radical stability. The most stable radical is 1 while the
least stable radical is 5.
•
•
•
•
•
4
5
3
2
1
c.(15) Rank the following compounds by anion stability. The most stable anion is 1 while the
least stable anion is 5.
CH3 O
CH3 NH
CH3 CH2
(CH3 )3C
CH3 CHCH3
1
2
3
5
4
2.
a.(7)
Provide answers for the following halogenation questions.
Supply the correct monobromination product.
Br2, hν
b.(7)
Br
Supply the correct monobromination product.
Br
Br2, hν
c.(7)
Supply the correct monobromination product.
Br
Br2, hν
d.(9)
Supply all monochlorination products.
Cl2, hν
Cl
Cl
Cl
Cl
e.(5) In part d would the rate be faster or slower if a monofluorination was attempted instead of
the monochlorination?
fluorination reaction is faster than a chlorination
3.(8 pts each) Supply the correct product for the following reactions. Only supply one structure
representing the preferred pathway. Indicate any regio or stereochemistry as required.
a.
H3CH2C
H3C
Br
H
NaOCH3
CH3
CH2CH 3
E2
b.
H3C
H3C
Br
H
H
NaOCH3
H3C
H3C
H
H
H
H
OCH3
SN2
c.
Br
NaOC(CH3 )3
CH3
CH3
E2
d.
Br
KI
E2
Br
e.
AgNO3
Br
CH3 OH
OCH3
SN1
f.
CH3 SNa
SN2
S
I
g.
NaOCH3
Br
any one of these E1 type
h.
NaI
Cl
SN2
I
i.
Br
CH3 NH2
NHCH3
excess
SN2
j.
Cl
CH3 CH2 OH
Δ
OCH2CH3
SN1
4.(15) Indicate how the following molecule can be synthesized starting with an alkyl halide.
Need to show what reagents will produce this product with the quickest rate.
O
O
I
5.(15) The starting material shown, 1-propanol, can be converted to 1-bromopropane with HBr.
When 1-propanol is reacted with NaBr, however, no reaction occurs.
HBr
OH
NaBr
Br
no reaction
a.
Draw the mechanism indicating the SN2 pathway that creates 1-bromopropane with HBr.
In order to do this use the appropriate arrow conventions which are used in organic chemistry to
show a mechanism, thus indicating what is the nucleophile, what is the electrophile and what is
the leaving group.
HBr
OH
OH2
Br
Br
the first step MUST protonate the hydroxy group in order to make a good leaving group (the
hydroxide will NEVER be a leaving group in an SN2 reaction) and then in the second step the
nucleophilic bromide displaces the water to give the product
b.
Why does the reaction not work when NaBr is used instead?
If no acid is used then the hydroxy cannot be protonated and hence will never leave as a leaving
group (strong bases are horrible leaving groups)
6.
When converting chlorocyclohexane to bromocyclohexane a student should consider
whether the reaction occurs by an SN1 or SN2 pathway. Compare the RATE of each of these
pathways if the following changes are made to the reaction. Therefore for each question (a-f)
answer by stating if that change will make either an SN1 or SN2 pathway faster, slower, or no
change compared to the reaction shown. Each question has two answers: what happens to the
SN1 and what happens to the SN2 pathways with the indicated change.
Cl
Br
NaBr
CH3 OH
a.(10) The chlorocyclohexane is changed to 1-chloro-1-methylcyclohexane.
SN2: the rate is slowed (SN2 never occurs at tertiary carbon)
SN1: the rate is faster (tertiary carbon reacts faster than secondary carbon)
b.(10) The Chlorine is replaced with an Iodine.
SN2: rate is faster, better leaving group
SN1: the rate is faster, better leaving group
c.(10) Sodium bromide is replaced with sodium iodide.
SN2: rate is faster, more polarizable nucleophile
SN1: no change
d.(10) The temperature the reaction is run at is raised.
SN2: rate is faster
SN1: rate is faster
e.(10) Change chlorocyclohexane to the starting material shown.
Cl
SN2: small change, slightly faster
SN1: rate is dramatically faster, formed carbocation is resonance stabilized
f.(10) The solvent is changed from methanol to hexanes.
SN2: small change
SN1: rate is dramatically slower, less polar solvent will not stabilize carbocation intermediate