ALKYL HALIDE SOLUTIONS
1.
Do not forget to study the generic functional groups and the common alkyl groups
For SN2 reactions, less hindered substrates react faster (all else being equal)
a)
Br
b)
or
1° benzylic is
less hindered
than vinylic
substrate
CH3CH2CH2Br
CH2
I
1° substrate
less hindered
or
CH2
CH
I
2.
a) (CH3)2CH-Br or
(CH3)2CH-Cl
in an SN1 reaction
Br is a better leaving group
b) CH3CH2-Cl in methanol
or
CH3CH2-Cl in acetone
in an SN2 reaction
Acetone is a polar aprotic solvent
c) (CH3)3Br
or
CH3-Br
with NO3-
The 3° substrate forms a C+ intermediate that can react with the very weak Nu
via SN1 or E1. Methyl bromide cannot ionize. Methyl C +’s are too unstable.
d) C6H5-Br
or
CH2=CH-CH2-Br
with I-
This 1° allylic substrate is unhindered in the S N2 reaction with I-. The phenyl
substrate is badly hindered and cannot react via SN2 (or SN1) reactions.
3.
a) 3° substrates can react only by SN1 or E1. Rate increases, 3 since for SN1 and E1
reactions, Rate = k[RX] (unimolecular). Concentration of Nu does not affect the rate.
b) A 1° alkyl halide will react with a v.gd. Nu by SN2. Rate is unchanged, (i.e., 1). For SN2
reactions, like this one, Rate = k[RX][Nu]. The doubled concentration of substrate
doubles the rate, but the halved concentration of Nu halves the rate. The net effect is no
change in rate.
c) Rate increases, 9. For this SN2 reaction, tripling the substrate concentration triples the
rate and tripling the Nu triples the rate again. The net result is a 9-fold rate increase.
PAGE 1
ALKYL HALIDE SOLUTIONS
4.
Nu’s are electron donors.
a)
AlH3
or
b)
NH2-
or
AlH3 has an incomplete octet and an empty 2pz orbital. It is an E+.
NH3 ‘s non bonded electron pair makes it nucleophilic .
NH3
N and P are both in Group 5A. Nucleophilic strength increases
down each vertical group in the periodic table because atomic size
and polarizability are increasing
PH2-
5. Substrates that form the most stable C+ will react fastest in SN1 reactions, i.e., 1 = fastest, 4 = slowest
a)
1° allylic
CH2
CH
CH2
2° benzylic
phenyl
Br
CH3
CH
methyl
CH3Br
Br
Br
2
3
4
1
F is electron withdrawing. It destabilizes the C+ formed in SN1 reactions
b)
(CH2F)3CBr
(CH3)3CBr
(CF3)3CBr
2
1
3
6. For SN2 reactivity, the least hindered (blocked) substrate is fastest (1= fastest, 4 = slowest).
1°
Br
1°
3°
2°
Br
Br
Br
n-butyl bromide
isobutyl bromide
sec-butyl bromide
1
2
3
tert-butyl bromide
4
7. Protic solvents, i.e., solvents with polar H’s, favor SN1 reactions over SN2 reactions. Examples include
H2O, CH3OH, CH3CH2OH (and other alcohols), CH3COOH (and other organic liquid carboxylic acids)
and even liquid NH3. Protic solvents stabilize the C+ intermediates that form in SN1 reactions. They
also solvate and weaken the nucleophile (thus inhibiting SN2 reactions). Recall that the nucleophile is
not a factor in SN1 reactions.
8. Solvents that favor SN2 reactions over SN1 reactions are polar aprotic. They are polar but have no
polar H’s (aprotic). They solvate the cationic counterion of nucleophiles, e.g., they solvate the Na + in
NaCN, thus leaving the CN- bare and more reactive. Examples include acetonitrile (CH3CN),
dimethylformamide (DMF) [(CH3)2NCHO], dimethyl sulfoxide, DMSO [(CH3)2SO], and
dimethylacetamide (DMA)
O
H
C
N
O
O
CH3
H3C
S
CH3
H3C
C
N
CH3
CH3
DMF
CH3
DMSO
DMA
PAGE 2
ALKYL HALIDE SOLUTIONS
9. Iodide is a non basic, very good Nu that reacts via SN2 with 2° substrates.
sec-butyl bromide
Br
K+I-
+
2°
sec-butyl iodide
SN 2
KBr
+
I
10.
Br
a)
v. strong base
good Nu
2°
CH3CHCH2CH3
Br
b)
Na
+
2°
CH3CHCH2CH3
Zaitsev product
E2
CH3CH=CHCH3
CH2CH3
v. good Nu
moderate base
+ Na+CN-
CH3CHCH2CH3
SN1
2°
Br
CH3CHCH2CH3
+
non basic
weak Nu
H2O
CH3CH3
+
NaBr
SN 2
C
c)
+
NaBr
+
N
CH3CHCH2CH3
+
HBr
+
H3O+
OH
heat
E1
CH3CH=CHCH3
+
Br-
Zaitsev product
d)
Br
water is weakly acidic and reacts
with the strongly basic Grignard
CH3CH2CH2CH3 + MgOHBr
MgBr
CH3CHCH2CH3
Mg, ether
+
CH3CHCH2CH3
H2O
Grignard
e)
Br
2°
CH3CHCH2CH3 +
v. good Nu
moderate base
Na+CH3S-
SN2
CH3CHCH2CH3 +
S
NaBr
CH3
11. 1-chloro-1-propylcyclopentane is a 2° substrate. NaOH is a strong base, good Nu. The reaction
mechanism will be E2.
H
Na+OH-
CH3CH2CH
CH3CH2CH
Cl
Zaitsev product
+
H2O
+
NaCl
E2
propylidenecyclopentane
PAGE 3
ALKYL HALIDE SOLUTIONS
12.
2°
H3C
CH
a)
fair Nu
wk base
O
Na+ -O
C
+
Br
mainly E2
CH3
CH3CH=CH2
CH3COOH
+
+
NaBr
H3C
O
non basic
O
v. wk Nu
2°
H3C
b)
CH
+
Br
HO
C
t-butoxide
a strong bulky base
good Nu
2°
CH
CH
HBr
+
CH3CH=CH2
+
CH3COOH
+
HBr
CH3CH=CH2
+
(CH3)3COH
+
NaBr
E2
(CH3)3CO- Na+
+
Br
O
CH3
E1
H3C
C
CH3
H3C
c)
H3C
SN1
CH3
H3C
d)
v. strong base
good Nu
2°
H3C
CH
E2
C- Na+
CH3C
+
Br
CH3CH=CH2
+
CH3C CH
+
NaBr
H3C
strong base
good Nu
1°
e)
CH2CH2CH2Br
CH3CH2O- Na+
+
SN2
CH2CH2CH2OCH2CH3
NaBr
+
13. Trans-1-bromo-2-methylcyclopentane is a 2° substrate and reacts with the following:
a)
CH3
H
The H on the #2 Carbon is not antiperiplanar to the Br on the #1 Carbon.
H
+
H
Br
H
b)
strong base
good Nu
E2
+
E1
CH3CH2OH
SN1
Zaitsev product
H
CH3CH2OH2+
+
H2O
Br-
+
CH3CH2O
CH3
CH3
+
+
H
+
HBr
H
OCH2CH3
After a hydride H
shift.
H
CH3
H
I
CH3
+-
+
H
H
KBr
H
CH3
OCH2CH3
H
+
H
3-methylcyclopentene
CH3
H
CH3
1-methylcyclopentene
non basic
weak Nu
c)
non Zaitsev product
K+OH-
Br
KI
non basic
v. good Nu
SN2
H
+
KBr
H
cis-1-iodo-2-methylcyclopentane
PAGE 4
ALKYL HALIDE SOLUTIONS
14.
a)
3°
H3C
CH3
C
..
O
..
CH3
H+Br-
+
H
C
H3C
CH3
H
+
..
O
CH3
SN1
H3C
H
C+
CH3
BrH3C
CH3
CH3
C
Br
+
H2O
CH3
t-butyl bromide
A Hydrogen that is antiperiplanar to iodine is removed
H
b)
I
+
H
Zaitsev product
strong base
good Nu
+
K OH
2°
+
E2
-
KI
+
H2O
cyclopentene
H
15.
a)
O
CH3
strong base
good Nu
O- Na+
SN2
+
CH3
CH3
methyl substrate
CH3Br
This is the only possible set of reagents.
CH3
HBr
3°
b)
H3C
H3C
CH3
C Br
C
OH
CH3
SN1
Several possible sets of reagents will work
but only via SN1 since the substrate is 3°.
CH3
CH3
KBr
H3C
C
Cl
CH3
Br
c)
H3C
H3C
C
E2
CH2
H3C
Br
H3C
2-methylpropene
or
isobutylene
d)
CH3C
4-methyl-2-pentyne
C
+
3°
CH3
+
non basic
weak Nu
H2O
CH3
SN2
- NaBr
Although this second set of reagents will
produce the product shown, it is not the
'correct' answer. A 'good' yield will not
be obtained. More SN1 product will
be formed.
an alkynide
CH3
methyl
CH3Br
CCH
CH3
CH3
CH3
E1
CH3
C
strong base
Other strong bases will also force an
good Nu
E2 reaction and produce the same product
KOH
+
v. strong base
good Nu
Na+ -C
CCH
CH3
This is the only possible
set of reagents. Of course
the cation can also be K+,
Li+, etc.
sodium 3-methylbutynide
PAGE 5
ALKYL HALIDE SOLUTIONS
16. Recall that terminal alkynes are weakly acidic (pKa = 25) and can be deprotonated to alkynides
(pKb = -11) using a very strong base such as NaNH2 (sodium amide, pKb = -21). The alkynide causes
SN2 reactions with methyl or 1° alkyl halides producing longer, internal alkynes.
H
C
..
Na+NH2-
C
acid/base
reaction
cyclopentylethyne
or
ethynylcyclopentane
+-
Na C
CH3CH2
Cl
CH3CH2
C
C
C
+
NaCl
SN2
1-cyclopentyl-1-butyne
sodium cyclopentylethynide
17. Substitution will be either SN1 or SN2. Choose reagents that go by SN2 so that only 1 product forms.
Your decision is based on the degree of substitution of the substrate and the basicity and nucleophilic
strength of the Nu.
a)
v. strong base
good Nu
SN2
(CH3)3C
b)
C
C
(CH3)3C
CH3
2°
Cl
CN
+
SN2
..-
C Na+
+
methyl
CH3I
v. good Nu
moderate base
NaCN
chlorocyclopentane
or
cyclopentyl chloride
strong base
good Nu
O- Na+
c)
O
C
CH2CH3
1°
SN2
This is the only possible
set of reagents.
CH3CH2Br
+
18. Gilman’s reagent undergoes SN2 substitution with all types of alkyl halides (except 3°), so make some
Gilman’s reagent out of half or the reagent and use the other half to react with the Gilman reagent .
Br 2°
excess
finely dispersed
Li metal in a
hydrocarbon solvent
CH3 CH3
CH3CH CHCH3
CH3CHCH3
-
LiBr
-
LiBr
Br
CH3CHCH3
Li
CuI
[(CH3)2CH]2CuLi
CH3CHCH3
2
-
LiI
PAGE 6
ALKYL HALIDE SOLUTIONS
19. Chlorocyclohexane is a 2° alkyl halide. Methanol is a non basic, weak Nu. A reaction can only occur
if the substrate ionizes and it will on heating, in which case the C+ intermediate is so reactive that it
reacts with even weak Nu’s and generally both SN1 and E1 occur together.
..
O
..
CH3
Cl
H
H
H

: OCH3
H
H
..
OCH3
..
H
+
SN1
2°
..
O
..
CH3
H
+
-
H
H
Cl-
H
H
H
H
H
H
and
+
CH3OH2+
+
Cl-
H
CH3
..
O
..
HCl
+
H
+
E1
H
H
H
H
CH3OH2+
+
+
Cl-
20. For Nu’s consider the basicity and nucleophilic strength. For substrates, consider their degree of
substitution. Consider possible acid /base reactions
a)
phenyl chloride
or
chlorobenzene
(C6H5)2CuLi
+
Cl
b)
Substitution occurs by an unknown mechanism
with all alkyl halides
Gilman's reagent
Gilman's reagent
(C6H5)2CuLi
C
C6H5Cu
+
C6H5Cu
+
LiCl
biphenyl
ether
H3C
+
H3C
C
CH2
CH2
+
LiBr
ether
Br
2-phenylpropene
2-bromopropene
CH3CH2MgBr
Br
c)
+
ether
No reaction. Grignards are generally
not reactive enough to react with
alkyl halides.
Gilman reagent replaces halogens in all types of alkyl halides (except 3°).
d)
1° benzylic
CuLi
2
+
CH2I
ether
CH2
Gilman
(cyclopentylmethyl)benzene
CH3
e)
CH3
NaH
1°
CH3CHCH2Br
isobutyl bromide
or
1-bromo-2-methylpropane
SN2
CH3COOH
CH3CHCH3
isobutane
+
NaBr
Sodium hydride is a
strong base (pKb = -21)
and is a very powerful Nu.
No Reaction. Acetic acid is a very weak, non basic Nu.
A substitution or elimination reaction will not occur unless
a C+ forms-and it cannot as the substrate is primary.
PAGE 7
H2O
ALKYL HALIDE SOLUTIONS
21. Work backwards as well as forwards.
a)
O
OH
CH3CH2CH2
CH3CHCH3
C
O
O
H2O
H2SO4
SN2
dehydration
HBr
CH2
CH3CH
fair Nu
wk base
Na+ - O C
CH3CH2CH2Br
H2O2
1°
antiMarkovnikov hydrohalogenation
C
CH2CH3
b)
CHCH3
SN2
NBS
(PhCO2)2
N
2°
KCN
v. good Nu
moderate base
CHCH3
Br
Recall, NBS and peroxide
brominates an aromatic ring
at the benzylic Carbon.
This produces a 2° benzylic
halide that can undergo an SN2
reaction with CN-
1-iodopropane
or
CH3CH2CH2I n-propyl iodide
2-chloropropane
Cl 2°
or
isopropyl chloride CH3CHCH3
c)
HCl
dehydrohalogenation
of a 2° alkyl halide with a
strong base
SN2
KOH
CH3CH
CH2
HBr
CH3CH2CH2Br
H2O2
Only HBr works with H2O2 works in this fashion
d)

SN1
CH3CH2OH
OCH2CH3
non basic
weak Nu
+
non basic
v. good Nu
1°
antiMarkovnikov
addition product
Br
2°
NaI
HBr
This is a very poor method since
E1 will also occur. Later, we will
study better ways to produce ethers.
You cannot use CH3CH2O-Na+ as
this will cause E2.
PAGE 8
ALKYL HALIDE SOLUTIONS
e)
CH2
NH2
CH2
KOH
acid/base
methylidenecyclopentane
SN2
HBr
NaNH2
CH2Br
antiMarkovnikov
addition
f)
1°
NH3
fair Nu
moderate base
Do this without dehydrating the 1° alcohol
1°
CH3CH2CH2CH2OH
SN2
n-butyl bromide
or
1-bromobutane
NH3+Br-
CH2
SN 2
H2O2
HBr
CH3CH2CH2CH3

Water is acidic enough
to be deprotonated
completely by strongly
basic Grignard reagents
or organosodium reagents.
H2O
Mg
ether
CH3CH2CH2CH2MgBr
CH3CH2CH2CH2Br
or
excess
dispersed Na
CH3CH2CH2CH2Na
g)
FC
alkylation
CH3Cl
AlCl3
1°
Other similar routes are possible
using a Gilman synthesis.
CH2Br
CH3
CH2Br
ether
NBS
(PhCO2)2
CH2 CuLi
Li
CuI
2
Gilman
C
h)
2 NaNH2
AlCl3
CH3CH2Cl
Cl
Br2
FeBr3
CH2CH3
Br
NBS, (PhCO2)2
CHCH3
Br
CH
C
CH
2 HBr
Br
ether 0°C
1 2 Li ether 0°C
2 CuI ether 0°C
CuLi
CH
CH2
Br
2
Br2
KOH
CH
CH2
HBr
PAGE 9