CHEM217 INTERMEDIATE CHEMISTRY II ASSIGNMENT #3 2/9/00 DUE 2/16/00
Answer Key
We will discuss only those questions that are starred (*) next week in class but it would behoove you to do all
questions.
1.
Arrange the following in decreasing order (i.e., greatest first) for the indicated property.
a. Rate of SN1 reaction with H2O:
i) p-H2NC6H4CH2Cl
ii) p-NO2C6H4CH2Cl
iii) p-ClC6H4CH2Cl
Variable is carbocation stability. The more stable the carbocation, the faster it is formed - the Hammond
postulate.
Electron-donating groups (EDG's) on the ring will stabilize a carbocation while electronwithdrawing groups (EWG's) will destabilize a carbocation. H 2N- is a powerful EDG (by resonance) and
NO2- is a powerful EWG (by resonance and induction). Cl- falls in between. You should draw the resonance
structures to confirm this.
Order is i) > iii) > ii)
b.* Rate of SN1 reaction with H2O:
i) (CH3)2CHBr
ii) (CH3)2CHI
iii) (CH3)2CHOCH3
Variable is leaving group ability. Good leaving groups are the weak (i.e., stable) conjugate bases of strong
acids. HI is the strongest acid, followed by HBr and the CH 3OH which is the weakest (very weak by
comparison). You should be able to explain acidity and basicity orders. Can you here?
Order is ii) > i) > iii)
c. Rate of SN2 reaction with CH3CH2I:
i) EtNH2
ii) EtOH
iii) EtNH-
Variable is nucleophilicity. Since the nucleophilic centers in question are all from the second period,
nucleophilicity varies directly with basicity. Conjugate bases will therefore be stronger nucleophiles than
their corresponding acid, i.e., EtNH- is a stronger base/nucleophile than EtNH2.
Oxygen is more
electronegative than nitrogen so O holds onto its lone pairs more tightly thus reducing basicity.
Order is iii) > i) > ii)
d.* Rate of SN2 reaction with CH3I:
i) p-CH3C6H4O-
ii) p-NO2C6H4O-
iii) PhO-
Variable is nucleophilicity. All oxygen-centered nucleophiles so nucleophilicity varies with basicity. Groups
on the ring with either stabilize or destabilize the negative charge. EWG's (i.e., NO 2) will stabilize and EDG's
(i.e., CH3) will destabilize. Make sure you can draw the resonance structures that show this. The less stable
the base, the better it is as a nucleophile.
Order is i) > iii) > ii)
e.* Rate of SN1 reaction with H2O:
i) CH3CH2Cl
ii) (CH3)3CCl
iii) Ph3CCl
Variable is carbocation stability. 3o carbocations are more stable than 1 o carbocations and phenyl groups
are more powerful EDG's than methyl groups (why?).
Order is iii) > ii) > i)
f.* Rate of E1 elimination:
i) Ph2C(Br)CH3
ii) (CH3)3CBr
iii) CH3CHBrCH3
Variable is carbocation stability. See part e above.
Order is i) > ii) > iii)
2.* When menthyl chloride is treated with sodium ethoxide in ethanol, 2-menthene is the only alkene product
observed. When neomenthyl is subjected to the same conditions, the alkene products are mostly 3-menthene (78%)
along with some 2-menthene (22%). Menthyl chloride and neomenthyl chloride only differ in the stereochemistry
of the chlorine substituent, the alkyl substutuents being in their most stable configuration.
a.
Draw chair conformations of the two alkyl chlorides.
Cl
CH3
Cl
CH(CH3)2
CH3
CH(CH3)2
One of these is menthyl chloride with the other being neomenthyl chloride. Which is which will be decided by
answering part b below.
b.
Explain why different alkene products are formed from the different alkyl chlorides, and why 3-menthene is the
major product in the second reaction.
Reaction conditions are E2 (strong base - ethoxide).
Me
Me
Me
i
Me
Cl
-
Cl
E2
Cl
Pr
H
menthyl chloride
EtO
most stable conformation
-
i
EtOH
Pr
i
Pr
i
least stable conformation
but only one that puts the
Cl and the H anti-coplanar
Pr
2-menthene
only possible product
Cl
E2
Me
i
Me
i
Pr
Pr
+
Me
H
i
H
Me
neomenthyl chloride
minor product
less-substituted
alkene
Me
major product
more-substituted
alkene
i
Pr
2-menthene
i
Pr
3-menthene
Pr
c. When the reaction with menthyl chloride is carried out in 80% aqueous ethanol with no added base, the
following products from the elimination reaction are obtained. How do you explain these results?
CH3
CH3
CH3
20:80 H 2O/EtOH
+
Cl
CH(CH3)2
CH(CH3)2
CH(CH3)2
menthyl chloride
2-menthene
3-menthene
Conditions are now E1 (weak base - EtOH and more polar solvent - H2O/EtOH). No stereochemical
requirements for E1 but Saytzeff's rule still applies.
H
Me
Cl
E1
Me
Cl+
i
iPr
menthyl chloride
o
2 carbocation
Pr
H
H
Me
Me
i
Pr
+
Me
Me
i
i
Pr
Pr
i
2-menthene
3-menthene
Pr
3. The pKa of HCN is 9.40 and that of HN3 is 4.72. Suppose that CH3I is dissolved in aqueous acetone with one
equivalent each of KN3 and KCN. What product would you expect to isolate in greatest yield from the reaction that
occurs and why?
CH3I + Nu-  CH3Nu + IE2 mechanism
The two competing nucleophiles are N3- (azide ion) and CN- (cyanide ion). The major product will result
from attack by the strongest nucleophile. Assuming that nucleophilicity follows basicity (basic centers are
both second period), then CN- is a stronger base/nucleophile than N3- so CH3CN will be the major product.
CN- is the stronger base because its conjugate acid (HCN) is the weaker of the two acids (has more positive
pKa).
4.* Consider the following experiments with trityl chloride, Ph3CCl, a very reactive 3o alkyl halide.
In all that follows, the mechanism is SN1 because the substrate is a 3o alkyl chloride. Therefore, the reaction
rate will be dependent only on the concentration of the alkyl chloride (i.e., first order) and independent on the
concentration of the nucleophile. However, the identity of the product will depend on the relative
nucleophilicities of any nucleophiles present, and the stronger the nucleophile the more product containing
that nucleophile that will be formed.
Ph3CCl  Ph3C+ Cl-
slow (rate determing) step
Ph3C+ + Nu  Ph3CNu (fast step)
a)
In aqueous acetone, the reaction of trityl chloride follows a first order (in trityl chloride) rate law, and the
product is trityl alcohol, Ph3COH.
Only nucleophile present is water, so only product is the alcohol.
Ph3C+ + H2O  Ph3COH2+  Ph3COH + H+
b) In another reaction, when a small amount of NaN3 (sodium azide) is added to a solution that is otherwise
identical to that used in a) above, the reaction rate is virtually the same as in a); however, the product isolated in
good yield is Ph3CN3.
Two nucleophiles are now present: H2O and N3-. N3- must be the better nucleophile since the only product
isolated is Ph3CN3
Ph3C+ + N3- + H2O  Ph3CN3 + H2O
c) In a reaction mixture in which both NaN3 and NaOH are present in equal concentrations, Ph3COH is the product
and, once again, the reaction rate remains the same.
Three nucleophiles are now present: H2O, HO-, and N3-. HO- is the strongest nucleophile (base) of the three
and is the one that ends up in the product.
Ph3C+ + HO- + H2O + N3-  Ph3COH + H2O + N3Explain why the reaction rate is the same but the products are different in these three experiments.
5.
An unknown liquid A (C7H7Br, 1H NMR: 7.28, s, 5H and 4.44, s, 2H) is treated first with Mg in ether then
ethylene oxide followed by an acid work-up to give compound B. Treatment of B with hot sulfuric acid gives
compound C (C9H10). C is a mixture of cis and trans isomers. Ozonolysis of C followed by a reductive workup gives benzaldehyde (PhCHO) and ethanal (acetaldehyde, CH3CHO). Compound D is produced by reacting
C with bromine. D exists as 2 pairs of enantiomers. Treatment of D with 2 equivalents of the strong base,
NaNH2, gives compound E (C9H8). The IR of E shows a sharp, weak peak at 2220cm-1. From the above
information deduce the structures of compounds A through E. Show your reasoning.
I call these types of problems "Sherlock Holmes" questions (for obvious reasons, I think!). These types of
questions are designed as "capstone" problems in that they require you to bring in information from many
different chapters. You know your organic chemistry if you can do these kinds of questions well!
Identity of A
C7H7Br has an IHD of 4, which probably means we are dealing with a benzene ring. The 1H NMR should
confirm this. The singlet peak at 7.28 integrating for 5 protons can only be due to a mono-substituted
benzene ring, C6H5. The singlet peak at 4.44 integrating for 2 protons must therefore be a highly deshielded
CH2 group. Highly deshielded because it's connected to both a benzene ring and a bromine atom. Putting
this all together leads to the identity of A as being benzyl bromide; PhCH2Br.
CH2Br
benzyl bromide
(1-bromo-1-phenylmethane)
A
Now we use our chemical knowledge to figure out the other structures.
Identity of B
Alkyl bromides react with magnesium to give Grignard reagents, which can then react with a number of
electrophiles (ethylene oxide in this case).
O
PhCH2Br + Mg
Et 2O
1)
PhCH2MgBr
Grignard
reagent
A
PhCH2CH2CH2OH
+
2) H
epoxide
ring-opening
B
Identity of C
B is an alcohol (3-phenyl-1-propanol) which can be dehydrated (lose water - elimination reaction) with hot
acid. The initial product formed (3-phenyl-1-propene) can not exist as cis and trans isomers, so, we need to
move the double bond to give the more stable alkene (1-phenyl-1-propene) which does exist as cis and trans
isomers.
CH3 Ph
Ph
PhCH2CH2CH2OH
H2SO4/
PhCH2CH=CH2
H2SO4/
not isolated
B
+
H
H
H
Ozonolysis cleaves carbon-carbon double bonds and replaces them with =O bonds. Reductive conditions
mean that any aldehyde formed remains as an aldehyde and is not oxidized to a carboxylic acid. The identity
of the products also confirms the identities of A through C.
PhCH=CHCH3
cis- and transisomers
C
2) reductive
workup
PhCH=O
benzaldehyde
CH 3
cis
trans
C
more stable alkene - conjugated
with benzene ring and nore
substituted
What's the mechanism
for this transformation?
1) O3
H
+
O=CHCH3
acetaldehyde
Identity of D
Bromine adds to alkenes to give 1,2-dibromoalkanes. Bromine addition is anti (i.e., the two bromines add
from opposite sides of the molecule - cyclic bromonium ion)
Ph
CH 3
Ph
CH 3
H
H
Br2
Br
H
anti addition
H
Ph
Br2
H
H
H
SS/RR pair of enantiomers
D
CH 3
H
Br
Ph
CH 3
H
Ph
Br
H
anti addition
Br
SR/RS pair of enantiomers
D
H
NH2-
NH2Ph
C
C
H
Br
CH 3
all 4 diastereomers of
D
PhCH
C(Br)CH 3
not isolated
Ph
C
C
CH 3
1-phenyl-1-propyne
E
Br
H
CH 3
Identity of E
Alkyl bromides undergo E2 elimination reactions with strong bases. A double elimination to produce an
alkyne takes place here because we have a dibromide and are using 2 equivalents of base.
Br
Br
CH 3
Br
+
CH 3
trans-isomer of
C
Ph
+
Ph
cis-isomer of
C
H
H
CH 3
H
H
Ph
H
Br
6.* Propose a mechanism for the following reaction:
OH
H2O
CH3
CH3
C
CH3
Br
CH3
rearrangement
SN1
OH2
+
CH 3
OH
CH 3
+
CH 3
-H+
Br
Br-
CH 3
CH 3
CH 3
o
o
3 carbocation
2 carbocation
ring strain
no ring strain
CH 3
CH 3