Chem 331, PS #6
due Wed, Oct 23, 2001
Name
KEY
1) Provide a mechanism
OH
Br H
Br–
HBr
Br
H2O
H
H
O+
Br–
+
+
resonance forms
2) Which reaction is faster (A or B)? Why?
A
Br H
B
HO
HBr
HBr
Br
OH
Br
Br–
Br–
H
H
O+
SLOW
+
+
Severe distortions are not required in order to put a
cyclopentane in a planar arrangement awith 120 ° angles.
Remember: carbocation formation is rate determining!
The ideal geometry for a carbocation is planar with 120° angles.
Thus, the carbocation formation is slow because it is severly distorted by
the 60 ° angle of the cyclopropane ring.
This is the faster reaction
3) Which reaction is faster (A or B)? Why?
The arrow pushing is the same as for question 2. The rate differences again relate to carbocation stability.
Et
B
A
HBr
HO
Et
HO
Br
Et
Br
Et
+
Such resonance structures do not apply
for the cation below
Et
Et
Et
the π orbital of the carbocation
cannot overlap with the
π system of the benzene ring.
No resonance stabilization
side view
Et
Et
Et
+
+
The carbocation
intermediate is resonance
stabilized (with the benzene
ring— see below). This is
the faster reaction
Et
HBr
Et
+
+
4) Provide a detailed explanation
Br
Me
X
but
Me
Br
O
Me
Me
H
H(eq)
Br
Me (eq)
H E2 elimination can
Me occur via an antiperiplanar conformation
i.e. both the hydrogen
and the bromine are axial
H
NaOtBu
Br
NaOtBu
H
Br(ax)
H( ax)
the molecule cannot adopt an conformation where Br and H are
anti-periplanar. At least one group must be equatorial.
–
5) Draw the product of the following reaction. Explain your assignment and provide a mechanism.
NaOtBu
tBu
OAc
13
C12H20O2
21.4 (q)
22.6 (t)
28.9 (t)
29.2 (q, 3 carbons)
32.9 (s)
45.9 (d)
70.5 (d)
128.1 (d)
133.0 (d)
170.9 (s)
OAc
The most stable conformer has
t-Bu in the axial position. Two antiperiplanar E2 eliminations
are possible to give structures A and B.
–
H(ax)
H NMR:
0.90 (s, 9H)
1.40 (m, 1H)
1.85 (m, 2H)
2.10 (s, 3H)
2.15 (m, 2H)
5.25 (m, 1H)
5.60 (m, 1H)
5.80 (m, 1H)
would expect singlet
this is not observed
O
O–
H( ax)
tBu
OAc (eq)
tBu
1
C NMR:
H( ax)
O
CH3
OAc (ax)
O
A
tBu
H(ax)
OAc
tBu
O
CH3
OAc (ax)
B
O
We need to use the spectral data to pick the right product. We can tell right away by looking at the 13 C NMR that A is not correct—
we would expect one of the alkene carbons to be a singlet. The complete assignments are below:
1.85 (m, 2H)
22.6 (t) 28.9 (t)
45.9 (d)
2.15 (m, 2H)
5.80 (m, 1H)
1.40 (m, 1H)
70.5 (d)
170.9 (s)
O
29.2
(q, 3 carbons)
O
0.90 (s, 9H)
CH3
O
32.9 (s)
128.1 (d)
133.0 (d)
CH3
21.4 (q)
O
5.25 (m, 1H)
5.60 (m, 1H)
2.10 (s, 3H)