7.1
Classify the alkyl halide as 1°, 2°, or 3° by counting the number of carbons bonded directly
to the carbon bonded to the halogen.
C bonded to 1 C
1° alkyl halide
a.
J
CH3CH 2 CH 2 CH 2 CH 2 -Br
b.(k
C bonded to 2 C's
2° alkyl halide
?H3
l
c. CH3-C-CHCH3
I
C bonded to 3 C's
3° alkyl halide
I
CH 3 Cl
\ }
d. ~
C bonded to 3 C's
3° alkyl halide
7.2 Use the directions from Answer 7 .1.
3°, neither
2°, neither
2°, neither
!
J
a, b.
~'::a'
Cl
3o, njther
telfairine
Br~ vinyl
!
Cl
Br - 1 °, neither
,,,Cl-3°, allylic
Br
l
2°, neither
halomon
Cl -vinyl
7.3 Draw a compound of molecular formula C6 H13 Br to fit each description.
Br
a.
Br~
b.
1° alkyl halide
one stereogenic center
C.
2° alkyl halide
two stereogenic centers
3° alkyl halide
no stereogenic centers
7.4 To name a compound with the IUPAC system:
[I] Name the parent chain by finding the longest carbon chain.
[2] Number the chain so the first substituent gets the lower number. Then name and number
all substituents, giving like substituents a prefix (di, tri, etc.). To name the halogen
substituent, change the -ine ending to -o.
[3] Combine all parts, alphabetizing substituents, and ignoring all prefixes except iso.
a.
(CH3hCHCH(Cl)CH2CH 3
l re-draw
[1] ~
2-methyl
[2)~
f
[3] 3-chloro-2-methylpentane
2 Cl-3-chloro
Cl
5 carbon alkane = pentane
b.
[1]
1tz>,
e>
(I)
C:
w
Reaction coordinate
7,23 To draw the products of SN2 reactions, replace the leaving group by the nucleophile, and
then draw the stereochemistry with inversion at the stereogenic center.
CH3CH2
q
D CH CH
':
a.
C-Br + -OCH 2CH 3
I
:_/
-
CH 3CH 20-C
\
H
2
3
H
7.24 Increasing the number of R groups increases crowding of the transition state and decreases the
rate of an SN2 reaction.
Cl
a.
or
2° alkyl halide
~Cl
I
1° alkyl halide
faster reaction
b
. Q-sr or
2° alkyl halide
faster reaction
3° alkyl halide
7.25
loss of
---
a proton
IV
\__/'Br
,,,Q
O
N
H
CH3
nicotine
7.26 ln a first-order reaction, the rate changes with any change in [RX]. The rate is independent
of any change in [:Nu-].
a. [RX] is tripled, and [:Nu-] stays the same: rate triples.
b. Both [RX] and [:Nu-] are tripled: rate triples.
c. [RX] is halved, and [:Nu-] stays the same: rate halved.
d. [RX] is halved, and [:Nu-] is doubled: rate halved.
7 27 [n Sr,1 reactions,_ racemization_ always occurs at a stereogenic center. Draw two products,
· with the two possible configurations at the stereogenic center.
leaving group
?H3
l
nucleophile
l
/ C,,.,Br
a. (CH3)2CH \H2CH3
CH 3
H20
I
+
(CH3hCH ,......-C\ 'OH
+
HBr
CH 2CH 3
enantiomers
nucleophile
.
l
CH3Coo-
H,° ' C H
, 3
b
CH 3CH{
,,,I
dlastereomers
leaving group
?.28 Carbocations are classified by the number of R groups bonded to the carbon:
oR groups= methyl, 1 R group= 1°, 2 R groups= 2°, and 3 R groups= 3°.
+
a.
+
b. (CH3l3CCH2
2 R groups
2° carbocation
1 R group
1° carbocation
c.(y
d.(X
3 R groups
3° carbocation
2 R groups
2° carbocation
7.29 For carbocations: Increasing number of R groups= Increasing stability.
+
CH 3-C-CH3
I
1° carbocation
least stable
CH 3
2° carbocation
intermediate
stability
3° carbocation
most stable
7.30 For carbocations: Increasing number of R groups= Increasing stability.
+
1° carbocation
least stable
2° carbocation
intermediate
stability
3° carbocation
most stable
7.31 The rate of an SNl reaction increases with increasing alkyl substitution.
o-
CH3Br
a.
(CH 3lJCBr
or
(CH 3)JCCH 2Br
3° alkyl halide
1° alkyl halide
faster SN1 reaction slower SN1 reaction
b.
or
3° alkyl halide
faster SN1 reaction
2° alkyl halide
slower SN1 reaction
7·32 • For methyl and 1° alkyl halides, only SN2 will occur.
• For 2° alkyl halides, SNI and SN2 will occur.
• For 3° alkyl halides, only SNI will occur.
CH 3 H
I
I
a. CH 3-C-C - Br
I
/ ' y Br
b. ~
I
CH 3 CH 3
c. 0 - Br
Br
/ \
3° alkyl halide
SN1
2° alkyl halide
SN1 and SN2
1° alkyl halide
SN2
2° alkyl halide
SN1 and SN2
d.
7.33 • Draw the product of nucleophilic substitution for each reaction.
• For methyl and 1° alkyl halides, only SN2 will occur.
• For 2° alkyl halides, SNI and SN2 will occur and other factors determine which mechanism
operates.
• For 3° alkyl halides, only SNI will occur.
Strong nucleophile
favorsi5N2.
\ /
a. ~
c.
CH 30H
Cl
-
~
OCH3
U ' CH,CH,0- U OCH,CH:
I
+ HCI
3° alkyl halide
only SN1
2° alkyl halide
Both SN 1 and SN2
are possible.
Weak nucleophile
favors SN1.
( " rsr
b.v
( " rsH+
V
j
Br-
CH30 H
d.~
Br
1° alkyl halide
only SN2
I
+
HBr
OCH 3
2° alkyl halide
Both SN1 and SN2
are possible.
7.34 First decide whether the reaction will proceed via an SNI or SN2 mechanism. Then draw the
products with stereochemistry.
a.
+
~
H Br
2° alkyl halide
SN1 and SN2
H20
l
+ HBr
-
Weak nucleophile
favors SN1.
HOH
HOH
enantiomers
HC=C)("'
b.
~Cl
H D
1° alkyl halide
SN2 only
SN1 = racemization at the
stereogenic C
D H
+
CI-
SN2 = inversion at the stereogenic C
. , . , , ..... , ,~vu
7.3 5 c ompounds wi
a. cH
th
3CH 2CH2CI
l 't U \, l t;U t,JIIII IV
V\..l"-,1 -.J u , ...... ,-
I
hi larz~;t~:or
ei
Q)
CH~
fH 3
C=C
I
H
\
H
cis-2-butene
C:
w
+
H2
--
CH3CH2CH,CH3
111-f'
= -120 kJ/mol
---
______ ii~~er in energy
butane
smaller 2 H' for cis-2-butene
lower in energy, more stable
8.35
a.
CH CH=CHCH2CH2CH (CH3}i +
3
(loss of P1 H)
monosubslituted
(loss of P2 H)
major product
disubstituted
Oo~
DBU
b.
c.
only product
+
CH 3CH 2C(CH 3)=C(CH 3)CH2CH2CH3
cH CH 2CH(CH 3)C(CH3)=CHCH2CH3
3
(loss of P2 H)
(loss of P1 H)
major product
telrasubstituted
+
lriwbst'"I•~
(loss of P3 H)
disubstituted
p
d.
Cl
-
OC(CH3)J
only product
P2
e.
-OH
~
I
CH3CH2CH2CH2CH=CHCH3
P1
f.
+
(loss of P2 H)
monosubstituted
(loss of P1H)
major product
disubstituted
+t"''
- OH
1
JJ fl
+
(loss of P1 H)
disubstituted
(loss of P2 H)
major product
trisubstituted
8.36 To give only one alkene as the product of elimination, the alkyl halide must have either:
• only one pcarbon with a hydrogen atom
• all identical pcarbons, so the resulting elimination products are identical
CH 3
9H3
a. CH3-9-9-H
H
b.
Cl
0---{I
H
'
I
c=cI
CH3
\
H
- - Q-cH=CH 2
9H3~
CH3 - C-C - H
I
Cl
I
H
o - - rCI
Alkyl Halides and Elimination Reactions 8-17
c
cl -
00 --
~
Cl
s.37 Draw the products of the E2 reaction and compare the number of C's bonded to the C=C.
~(!)
1
Br
~@]
P1
~2
I½
+
(CH3)iCHCH=CHCH 3
a
major product
trisubstituted
disubstituted
a
(CH3)iCHCH=CHCH 3
P1
major product
disubstituted
+
monosubstituted
A yields a t~isubst_ituted alkene as the ma)or product and a disubstituted alkene as minor product.
8 yields a d1subst1tuted alkene as the maJor product and a monosubstituted alkene as minor product.
Since the major and minor products formed from A have more alkyl groups on the C=C
(making them more stable) than those formed from B, A reacts faster in an elimination reaction.
8.38
a. Mechanism:
~
.
I-..!
0
by-products
..
81/-=\?C(CH3)J
HJ
(CH3)JCOH
b. Rate= k[R-Br]LOC(CH3)3]
[ l] Solvent changed to DMF (polar aprotic) = rate increases
[2] tOC(CH3)J] decreased = rate decreases
[3] Base changed to -oH = rate decreases (weaker base)
[4] Halide changed to 2° = rate increases (More substituted RX reacts faster.)
[5] Leaving group changed to i- = rate increases (better leaving group)
8.39
[>-Q
2° halide Cl
CH3CH2CH=CH 2
-OC(CH 3)J
~I
Cf
+
(cis and trans)
E1
1° halide
d.
CH 3CH=CHCH 3
weak base
2° halide
CH3CH2CH=CH2
+
(cis and trans)
E2
2° halide
C.
CH 3CH=CHCH 3
strong base
Br
strong base
E2
OH
strong base
E2
CrCHCH2CH3 +
CH 3
q-CH2CH 2CH3
CH 3
>-0
>-0
+
>--0
+
q-
CH2CH 2CH 3
CH 3
g,4J The order of reactivity is the same for both E2 and EI : I < 20 < 3o.
O
CH3
b--c1
Q s,
QcH3
CH3
Cl
2° halide
3° halide
3° halide+
better leaving group
Increasing reactivity in E1 and E2
s.44
Cl
('r(CH3
a.
v
H2O
b.
, CI
3° halide-faster reaction
-OH
C.
(CH3)JCCI
I
c=rBr
-OH
l
s,,oog base-E2
DMSO
C)
cis-cyclodecene
bromocyclodecane
H2O
CH3 Cl
/ 3° halide-faster reaction
8.45
(CH3)JCCI
1
polar aprolic solvent
faster reaction
In a ten-membered ring, the cis isomer is more
stable and, therefore, the preferred elimination
product. The trans isomer is less stable because
strain is introduced when two ends of the double
bond are connected in a trans arrangement in this
medium-sized ring.
8.46 With the strong base -0CH2CH3, the mechanism is E2, whereas with dilute base, the
mechanism is E1. E2 elimination proceeds with anti periplanar arrangement of H and X. In
the EI mechanism there is no requirement for elimination to proceed with anti periplanar
geometry. In this case the major product is always the most stable, more substituted alkene.
Thus, C is the major product under El conditions. (In Chapter 9, we will learn that additional
elimination products may form in the El reaction due to carbocation rearrangement.)
OCH2CH3
A
strong base
E2
rO
B
.. .
Since this is an E2 mechanism,
dehydrohalogenalion needs an anti
periplanar H to form the double bond.
There is only one H trans to Cl, so the
disubstituted alkene B must form .
CH 3OH
+
weak base
A
E1
B
disubstituted alkene
C
trisubslituted alkene
more stable
Chapter B-20
8.47
+ CH 3CH=CHCHi
.,p'v
2-butene
1-butene
from loss of
from loss of
P2H
P, H
Na' -ocH2CH3
19%
81 %
K'-OC(CH 3lJ
33%
67%
The H's on the Cl-1 2 group of the ~2carbon are more sterically hindered than the ~~son the
C~ i group of the~, carbon . Since K•-oqCHi)i is a much bulkier base than Na OCH2CH 3,
1115
easier to remove the more accessible H on ~1, giving it a higher percentage of !-butene.
8 48
· Han_d Br must be anti during the E2 elimination . Rotate if necessary to make them anti ; then
eliminate.
CH 3
a.
C5H5~
C6H5
H
-
f y cH,cH,
E2
CH3
CH 3
CH,CH3
CH 3
Br
b.
",Q)'"'
C5H5
$
CH3
rotate
CH3
CH 2CH 3
C, Hs
CH 2CH 3
CH,
CH3
CH, CH 3
C, Hs
CH2CH 3
H
C6H5
~
'"'~"
Br
CH,
H
E2
CH 3
C6H5
C,
c , Hs
rotate
~",
C
H
E2
CH,CH 3
CH 3
CH 3
CH 3
8.49
a.
('y
Cl
two chair
H
H axial
_C',Ql)f9_1'!1)~~9~~-ma
V. •,,CH
H
3
:
CH(CH 3),
CIH
(CH 3>,CH
CH
(CH 3),CH A3H
H_
B
Choose this conformation.
axial Cl
H
(CH,l,CH
H
0
one axial H
H
(CH3),CH ~
CH
3
H
0..
:
only product
'CH 3
CH(CH 3)2
b.
ex
Cl two chair
H
_c_~~f_or~~tio~~ H
_
CH,
~
C
CH(CH,l:,
(CH3),CH A H
(CH, )2CCHH
rY~i,
H
Choose this conformation .
-
axial Cl
H
H
B
Pi
H
(CH3)2CH ~
CH3~
H
I!)
(CH,l:,CH ~ H axial
CH
P1
H
n
H
(CH3),CH ~ i : : \
CH 3~
(loss of P, H)
two axial H's
I re-draw
I re-draw
Q
:
CH 3
CH(CH3)2
U
CH3
CH(CH 3l,
(loss of P2 H)
D
60
D~
H H
D
l
This conformation reacts.
axial Cl
H
(loss of P2 H)
major product
trisubstituted
enantiomers
D
o--fi-o
H
= Q
D
(loss of P1 H)
(loss of P, D)
H
p~l
o~Pi
I
IL
o~
HffZH
...,___----1\
-~O.,_,_
H_ _
_ __ _ _
This conformation reacts .
axial Cl
H- - f i - H
D
(loss of P1 D)
enantiomers
Q
=0
Chapter 8-22
8.50
enantiomers
a.
enantiomers
9H3
•.H
6
CJH
3- ~~(CH
C
.•c-c•:--CH3
3
CH 3CH{ I
CH3
CJ
\9J
B
l
A
2-chloro-3-methylpentane
ll
-HCI
H and Cl are arranged anti in
each stereoisomer, for anti
periplanar elimination .
CH 3CH~
H
I
1
\
CH2CH3
\
I
C=C
CH3 CH3
c=c
CH3
-HCI
CH3
----,
identical
b. Two different alkenes are fom,ed as products.
. .d • I products A and B
.
.
(A and B)
c. Tl1e pro ducts are diastereomers:
Two enant1omers
. give .I ent1ca
lk
Thus·
are diastereomers of C and D. Each pair of enantiomers gives a sing 1e a ene.
'
diastereomers give diastereomeric products.
8.51
NaNH2
a.
(2 equiv)
b.
c.
9H3
CH3CH2-9-9HcH2Br
CH 3 Br
Cl
CH3
I
CH3-9-CH2CH 3
Cl
d.
(2 equiv)
9H3
CH 3CH 2-9-C=CH
(excess)
i\.ti-o
Cl Cl
;\__C=C-0
(2 equiv)
\_F
8.52
H H
Br
I
I
CH3 - 9-CH2CH3
or
Br
CH3
b. CH3-C - C=CH
I
CH3
CH3 Br
I
I
CH 3-9-9-CH3
CH3 Br
or
I
CH3-9 - 9-CH3
Br Br
9H3~r
CH3-9-cH-CH2Br
or
9H3
CH3-9-cH2CHBr2
CH3
CH 3
0-cH-cH-0
Br Br
or
I
I
\
/;
IJ.53
H
I
H
I
CH3 - 9 - 9 - CH 3
CH, - C =' C - CH 3
Br Br
11
spsp
2.3-dibromobutane
CH 3 - CH =C=CH 2
CH 2=C H - C H =CH 2
1
C
sp
B
A
S.54 Use the "S ummary chart on the four mechanisms: SN I, SN2, EI , or E2" on p. 8- 2 to answer the
quest10ns.
a. Both SN I and EI involve carbocation intermediates.
b. Both SN I and E 1 have two steps.
c. SN I, SN2, E 1, and E2 a ll have increased reaction rates with better leav ing groups.
d. Both SN2 and E2 have increased rates when changing from CH 30H (a protic solvent) to
(CH 3 )zSO (an aprotic solvent).
e. Jn SN I and EI reactions, the rate depends on on ly the alkyl halide concentration .
f Both SN2 and E2 are concerted reactions.
g. cH 3 CH 2 B~ and NaO~ react by an SN2 mechanism.
h. Racemizat1on occurs in SN 1 reactions.
_
i. In SNI, E I , and E2 mechanisms, 3° alkyl halides react faster than 1° or 2° hahdes.
· E2 and SN2 reactions follow second-order rate equations.
J·
8.55
a.
- OC(CHah
Br
sterically
11° halide hindered base
SN2 or E2
b.
~
- OCH2CH3
I
E2
~
OCH2CH3
strong
nucleophile
1 ° halide
SN2 or E2
Cl
I
c.
CH3-?-CH3
Cl
dihalide
d. ( ) Br
1° halide
SN2 or E2
HC=C-CH3
equiv)
strong base
(2
DBU
sterically
hindered
base
er
e. ( XCH2CH 3 - OC(CH3}J
Br
2° halide
SN1, SN2, E1, E2
sterically
hindered
base
E2
( YCH2CH3
V
+
major product
E2
Chapter 8-2 4
f. r"r(Br
V., CH2CH3
weak base
3• halide
no SN2
g,
E1 products
2 NaNH2
(CH3)iCH-9HcH 2Br
V
('YCHCH3
H
OCH2C \
CH3CH 20H
(CH )iCH - C= CH
3
dihalide Br
h.
KOC(CH 3lJ
9H3
CH 3- 9-c = CH
(2 equiv)
CH 3
DMSO
dihalide
I
OCH 2CH 3
i.
CH3CH 20H
2° halide
SN1,SN2,E1,E2
weak base
-
H20
j.
3• halide
no SN2
weak base
+
SN1 product
%
+
E1 product
+
CH 3CH = CHCH3
(cis and trans)
E1 product
CH 3CH 2C(CH 3)= CHCH3
+
(cis and trans)
E1 product
SN1 product
E1 product
8.56 [ 1] NaOCOCH3 is a good nucleophile and weak base, and substitution is favored. [3] KOC(CH3)J
is a strong, bulky base that reacts by E2 elimination when there is a~ hydrogen in the alkyl
halide.
a.
CH3CI
[1] Na0C0CH 3
[2] Na0CH 3
CH30C0CH 3
C.
[1] Na0C0CH3
O-ococH 3
O-c1
[2] Na0CH3
CH30CH3
0 - 0CH3
SN2
[3] K0C(CH3)J
b.
CH30C(CH3)J
~ C l [1] Na0C0CH 3
~OCOCH3
[3] KOC(CH3)J
d. d c i [1] Na0C0CH 3
·O
0
dococH3
[2] Na0CH3
~ O C H3
[2] Na0CH 3
(J
[3] K0C(CH3)J
O=
[3] K0C(CH3)J
(J
E2
8.s7
two enantiomers:
a.
(CH3lJC
-o
(CH3lJC " "
A
Q
'
B
b. The bulky tert-butyl group anchors the cyclohexane ring and occupies the more roomy
equatorial po~ition. The cis isomer has the Br atom axial, while the trans isomer has the Br
atom equatonal. For dehydrohalogenation to occur on a halo cyclohexane, the halogen must
be axial to afford trans diax.ial elimination of Hand X. The cis isomer readily reacts since
the Br atom is axial. The only way for the trans isomer to react is for the six-membered ring
to flip into a highly unstable conformation having both (CH 3)3C and Br axial. Thus, the trans
isomer reacts much more slowly.
~ ~ t r a n s diaxial
(CH3'3C~
/
HH/
(CH 3l J C ~ B r
cis-1-bromo-4-tert-butylcyclohexane
lrans-1-bromo-4-lert-butylcyclohexane
c. two products:
OCH3
D= (CH 3) J C ~
d. cis-l-Bromo-4-tert-butylcyclohexane reacts faster. With the strong nucleophile -0CH3,
backside attack occurs by an SN2 reaction, and with the cis isomer, the nucleophile can
approach from the equatorial direction, avoiding 1,3-diaxial interactions.
1,3-diaxial interactions
Br
Hi
(CH 3l J C ~ ~ C H 3
~QCH3
(CH 3)Jc--MBr
equatorial approach preferred
cis-1-bromo-4-lert-butylcyclohexa ne
axial approach
lrans-1 -bromo-4-lert-butylcyclohexane
e. The bulky base -0C(CH3)J favors elimination by an E2 mechanism, affording a mixture of
two enantiomers, A and B. The strong nucleophile -oCH 3 favors nucleophilic substitution
by an SN2 mechanism. Inversion of configuration results from backside attack of the
nucleophile.
8.58
Cl
H
a. ~
2' halide
SN1, SN2, E1, E2
Cl
b.
OH
strong base
SN2 and E2
~+
+
+
SN2 product
major E2 product
inversion at
stereogenic center
minor E2 product
H
~
H~
weak base
2' halide
SN1, SN2, E1, E2 SN1 and E1
+
~
+
SN 1 products
+
major E1 product
+
minor E1 product
;=v
minor E1 product
minor E2 product
CH 3
CH3
c tCH 3
Cl
···csHs
C.
c tCH3
OCH3
C6Hs
CH30H
weak base
SN1 and E1
3° halide
no SN2
e.
CH 3COO
~Br
f.
~
···o
b.
C(c1
strong base
SN2 and E2
CH30H
weak base
SN1 and E1
major E2 product
• halide
3
minor E2 product
>;
u., D u
+
SN2 product
inversion at
stereogenic center
(trans diaxial elimination of D, :·,·;
E2 product
cc +C(+C(+C(
ex ex + cc
'OCH3
E1
SN1
--X-
E1
+
8.60
a.
C5H5
achiral
SN1 product
aOCH;
•,
SN1
KOH
strong base
E2
+ -OC(CH3h
.#
CH3
·00CCH3
x , ,OH
8.59
C(c1
0
KOH
2• halide
SN1,SN2,E1 , E2
a.
( :t°"'
E1 product
OH
SN2 product
weak base
good nucleophile
SN1
3° halide
no SN2
ct/
+
C6Hs
0CH3
~+
Cl
strong base
•
halide
SN2 and E2
2
SN1,SN2,E1,E2
0CH3
'Br
+
SN 1 products
NaOH
d.
CH3
CHC
(J
CH3
OC(CH3h
strong
bulky
base _ _ _ _ _ _ __
E2
[QJ
U
CH3 + c rCH2
major product
more substituted alkene
No substitution occurs with a strong bulky base and a 3° RX. The C with the leaving
group is too crowded for an SN2 substitution to occur. Elimination occurs instead by an
E2 mechanism.
E1
Alkyl Halides and Elimination Reactions 8-27
~
b.
+ - OCH 3
Br
1• halide
---)(-
strong nucleophile
I
SN2
- - - - - - - - [ ~OCH 3
All elimination reactions are slow with 1• halides
-----~
The strong nucleophile reacts by an SN2 mecha~ism instead.
a-
ct?
c.
('Y strong base
V
OCH
minor prodru_
ct_o_nl.cy___ _ _ _ _ __
2
E ---------
3° halide
\
1
alkene is favored.
1
d.
3
More substituted
minor product only
good nucleophile,
weak base - - - - - - - - SN2 favored
I
major product
Cl
2° halide
I
The 2° halide can react by an E2 or SN2 reaction with a negatively charged nucleophile or base .
Since 1- is a weak base, substitution by an SN2 mechanism is favored.
8.61
3• halide, weak base:
SN1 and E1
CH 3CH 2OH
a.
\Cl
The steps:
l
overall ~
reaction
+
~
+ HCI
SN1
CH 3CH29H
or
E1H ~
)
:ci":
E1
+HCI
+
or
~
+HCI
+ ~
+HCI
Any base (such as CH 3CH 2OH or cI-) can
be used to remove a proton to form an
alkene. If cI- is used, HCI is formed as a
reaction by-product. If CH 3CH 2OH is used,
(CH 3CH 2OH 2)' is formed instead.
Chapter
8-28
o -CH
Cl
b.
3
_
OH
overall
reaction
3° halide
strong base
E2
Each product:
U
CH3
a CH, 'H,O'
+
ceJ 1
:
cf "'
_.. one step
L---9.H
+ H29 +
or
y
c.CI
Cl
ct
.CH - i ; ----:qH
a
Cl
-one step
8·62 Draw the products of each reaction with the I alkyl hal ide.
O
DBU
a.
~
NaOCH 2CH 3
Cl
b.
~
c.
strong
nucleophile
SN2
H ·.
KCN
Cl
l·.--.
Cl
H ·.
sterically
hindered base
E2
~
strong
nucleophile
SN2
H ·.
,~
CN
H ·.
8.63
ci-H\
H
H
-
H20:
.
+
Hi):
above
-
:
H
·Br· ·.. ·
H
'
:
: re-draw
'
+
-
::B,~
:
H
H 0:
2
below
..
:OH
ctf'"~. _ Cv•• ,
H
~
H
+
HBr
+
HBr
:
H
H
+p-H\
.· \l
H
\
:Br:~ ~
~··
H
HBr
:5:iH
.-
H.64
good nucleophile
0
II
CH;(c ' o -
CH3- HCH 3
7
o , c , CH 3
II
CH3COQ- is a good nucleophile and a weak base
and so it favors substitution by SN2 ·
0
(only)
strong base
CH3 ~ 7HCH3 + CH 3CH = CH The strong base gives both SN2 and
2
OCH,CH 3
E2 products. but since the 2• RX is
20%
8.65
80%
somewhat hindered to substitution ,
the E2 product is favored.
c5
6
3• halide
weak base
SN1 and E1
HCI
+
HCI
HCI
or
+
HCI
AJJ:,vvc;- 1 .:,
9.1
Lv,
,v._,,,._,,,-->
0
• Alcohols are classified as I , 2°, or 3°, depending on the number of carbon atoms bonded to
the carbon with the OH group.
• Symmetrical ethers have two identical R groups, and unsymmetrical ethers have R groups
that are different.
OH
.._____,,,o'-./
CHa , O ~
2' alcohol
symmetrical ether
unsymmetrical ether
~OH
CHa, ~
O
unsymmetrical ether
~OH
1" alcohol
XH
3' alcohol
1' alcohol
Chapter 9-6
9.2 Use the definitions in Answer 9.1.
CH 20 H -
10
.oH-30
·•·CH3
0
9.3 To name an alcohol:
bstituent. Name the molecule as a
[I] Find the longest chain that has the OH gro~p a:h: ~; ending of the alkane to the suffix ·of.
derivative of that number of carbons by changing the tower number. When the OH
[2] Number the carbon chain to give the OH group. ·ng with the OH group, and the "I''
group is bonded to a ring, the ring is numbered beginni
is usually omitted.
h name
[3] Apply the other rules of nomenclature to complete t e
·
1
a. [1]
IZ X S JOH
[2]
~
t t
5 carbons = pentanol
b. [1]
[3] 3,3-dimethyl-1-pentanol
OH
~ C H3
[ 2]
MoH
[RCH 3-2-methyl
[3] cis-2-methylcyclohexanol
~OH
6 carbon ring = cyclohexanol
1
-6-methyl
, . l1] ~
0H
121 ~
:
[3] 5-ethyl-6-methyl-3-nonanol
t
9 carbons = nonanol
5-ethyl
9.4 To work backwards from a name to a structure:
[I] Find the parent name and draw its structure.
[2] Add the substituents to the long chain.
OH
7
, . 7,Mmothy1·4-octaool
q+
3~
o 2"erl·bo\y1·3•methylcyoloh"'ool
4
b. 5-methyl-4-propyl-3-heptanol
2
1OH
/ 5
d. trans-1,2-cyclohexanediol
cr
1
OH
'OH
( \ ' ,,OH
or
~OH
To name simple ethers:
9,5 [ 1] Name both alkyl groups bonded to the oxygen.
[2] Arrange these names alphabetically and add the word ether. For symmetrical ethers, name
the alkyl group and add the prefix di.
To name ethers using the IUPAC system:
[I] Find the two alkyl groups bonded to the ether oxygen. The smaller chain becomes the
substituent, named as an alkoxy group.
[2] Number the chain to give the lower number to the first substituent.
a. common name:
IUPAC name:
CH 3-0-CH2CH2CH2CH 3
I
~ C H2CH 2CH 2CH 3 -larger group- 4 C's
I
methyl
methoxy
butyl methyl ether
1-methoxybutane
IUPAC name:
b. common name:
y
(YOCH3
(YOCH3 --substituent -
V
m1thyl
methoxy
t
larger group - 6 C's
cyclohexane
cyclohexyl
cyclohexyl methyl ether
methoxycyclohexane
IUPAC name:
c. common name:
ICH3CH2CH2-ofCH2CH 2CH3
I
1
propoxy
propane
CH 3CH 2CH 2-0-CH2CH 2CH 3
I
propyl
butane
Isubstituent:
butyl
I
propyl
dipropyl ether
1-propoxypropane
9.6 Name the ether using the rules from Answer 9.5 .
1 CH 3
I
CH 3 C
I
2-methyl
CH3
0-CH 3
I
)
2-methoxy
propane
2-methoxy-2-methylpropane
9.7 Three ways to name epoxides:
[ l] Epoxides are named as derivatives of oxirane, the simplest epoxide.
[2] Epoxides can be named by considering the oxygen as a substituent called an epoxy group,
bonded to a hydrocarbon chain or ring. Use two numbers to designate which two atoms the
oxygen is bonded to.
[3] Epoxides can be named as alkene oxides by mentally replacing the epoxide oxygen by a
double bond. Name the alkene (Chapter I0) and add the word oxide.
Three possibilities:
[1] methyloxlrane
[2] 1,2-epoxypropane
[3] propene oxide
p
0
a.
Three possibilities:
[ ) cis-2-methyl-3-propyloxiran
1
[2) cis-2,3-epoxyhe~ane
e
[ ) cis-2-hexene oxide
3
CH 3
1
(;o-
\ CH3- 1 •methyl
b.
epoxy group
2
Two possibilities:
ne
[1] 6 carbons = cyclohh~~~clohexane
1,2-epoxy-1-met y oxide
[2] 1-methylcycloheXene
9.8 Two rules for boiling point:
[ 1] The stronger the forces the higher the bp,
d'
[2] Bp increases as the extent of the hydrogen b_on
same number of carbon atoms: hydrogen bondtng
1° ROH .
a.
0
CH
t
vow
lowest bp
3
a
CH3
t
0
OH
vow
intermediate bp
hydrogen
bonding
highest bp
DD
b. ~
OH
~
Oii
t
t
3' ROH
lowest bp
DD
yv
OH
t
t
vow
increases. For alcohols with the
bp's increase: 3° ROH < 2° ROH
HO
(1R,2SJ-2-isobutylcyclopentanol
d. constitutional isomer
i'()-oH
TsCI
(1S,3S)-3-isobutylcyclopentanol
[6]~
HO
pyridine
e. constitutional isomer with an ether
butoxycyclopentane
9.37
HO
H
HBr
a.
Br
H
+
s
HO
R
H
H
= racemization
SN2 = inversion.
R
HO
2° alcohol
SN1
PBr3 follows
b.
C.
H Br
HCI
Cl
H
2° alcohol
+
s
'
R
SN1
= racemization
Alcohols, Ethers, and Epoxldes 9-17
HO
H
~
~ \ .Cl
pyridine ,,,,.
d. ~
soc12 follows
SN2 =Inversion.
R
9 38 Draw the structure of each alcohol using the d. ti . .
,
OH
a.
[>-I
'
A
10
.
c m111ons m Answer 9. 1.
OH
~
2'
OH
b.
H
3'
enol
9.39 Use the directions from Answer 9.3.
(CH 3),CHCH2CH2CH2OH
a.
[1]
H
[2]
[ CH, -?-CH2CH2CH20 B]
1
[3]
@:Ha- y- CH, CH, CH, oBJ
CH3
s carbons = pentanol
4-methyl-1-pentanol
CH, -4-methyl
b.
(CH 3),CHCH,CH(CH2CH3)CH(OH)CH CH
2 3
[1]
H
?H
[2]
[GH3- 9 CH2-y CH CH2C':fil
H
H O.!:!.-- 3
9 CH, 9 -c:H-CH CH2J
2
- ; ;cl:H
::-,~-±
c-:c
H-, c,-H,------''--.::J
f 3
@,
cH,
CH2CH,
7 carbons = heptanol
6-methyl
[3)
4-ethyl-6-methyl-3-heptanof
4-ethyl
c.
[1]
[2] ~
:thyl
[3]
s carbons =octanol
4-ethyl-5-methyl-3-octanol
OH 4-ethyl
M
3
d.
[1]
4
2
[ ] HO-bLOH
HO-OOH
cyclohexanediol
[1]
[2]
[2]
OH
1
[2]
00
5 carbons
=cyclopentanof
[3]
OH
5-methyl-2,3,4-heptanetriol
OH
5-methyl
g,
HO° °'CH(CH3)2
(2R,3R)-2,3-butanediol
2R,3R
7 carbons =heptanetriol
[1]
[3]
HOH
=butanediol
¼
OH
cis-1 ,4-cyclohexanediol
HOH
HOH
f.
[3]
cis
et1J
4 carbons
1
.G
[3]
HO' ' ~CH(CH3)2
I
3-isopropyl
trans-3-lsopropylcyclopentanol
Chapter g_ 18
97
9 .4o Use the rules from Answers 9.5 and · ·
a.
c.
x .2-rnethylhexane
1.2~b~tJ1-2-rnethylox_irane
or 2 thylhexene oxide
or 2-rne
cH 3 GH3
I
I
c - o - c - GH3
cH, - ,
,
d.
cH, GH3
O o-0
dlcycloheJ1yl ether
~
4-fk
3
I I
~
CH3
h. 1-ethoxy-3-ethylheptane
. R )_ 3.jsopropyl-2-hexanol
I. (2 ,35
2
d. 6 -sec-butyl-7.7-diethyl-4-decanol OH
j. (ZS)-z-ethoxy- 1,1-dirnethylcyclopen~ane
e. 3-chloro-1 ,2-propanedlol
1
3
HO~
CI
OH
Alcohols , Ethers, and Epoxldes 9-19
9,42
Eight constitutional isomers of
mo 1ecular formula C H
s 120 containing an OH group:
OH 1-pentanol
~
1° alcohol
®
OH
2-pentanol
2° alcohol
[zspJ
OH
2
3-pentanol
~
1 OH J-me th Y1·1-butanol
1° alcohol
1° alcohol
H
2,2-dimethyl-1-propanol
1° alcohol
0H
20 alcohol
OH 2-methyl-1-butanol
~
OH
3-methyl-2-butanol
2° alcohol
2-methyi-2-butanol
2
3° alcohol
9.43 Use the boiling point rules from Answer 9_8_
a.
CH 3CH20CH3
(CH3hCHOH
I
ether
no hydrogen bonding
lowest bp
I
2° alcohol
hydrogen bonding
intermediate bp
b. CH3CH2CH2CH2CH2CH3
no OH group
lowest water solubility
CH3CH2CH 20H
I
1° alcohol
hydrogen bonding
highest bp
CH 3CH 2CH 2CH CH cH 0H
2 2 2
HOCH2CH2CH2CH2CH 2CH 20H
one OH group
two OH groups
intermediate water solubility
highest water solubility
9.44 Melt_ing po_ints depend on intennolecular forces and symmetry. (CHihCHCHiOH has a lower
melting pomt than CH1CH2CH2CH20H because branching decreases surface area and makes
(CH1)2CHCH20H less symmetrical so it packs less well. Although (CH 3) 3COH has the most
branch in~ ~nd _least s~rface area,_it is the most symmetrical, so it packs best in a crystalline
lattice, g1vmg 11 the highest melting point.
~OH
-108 °C
lowest melting point
-90°C
intermediate melting point
26 ' C
highest melting point
9.45 Stronger intermolecular forces increase boiling point. All of the compounds can hydrogen
bond, but both dials have more opportunity for hydrogen bonding since they have two OH
groups, making their bp 's higher than the bp of 1-butanol. 1,2-Propanediol can also
intramolecularly hydrogen bond. lntramolecular hydrogen bonding decreases the amount of
intermolecular hydrogen bonding, so the bp of 1,2-propanediol is somewhat lower.
Increasing boiling point
HOT
1-butanol
118 °C
OH
1,2-propanediol
187°C
HO~OH
1,3-propanediol
215 °C
9.46
NaH
HCI
ZnCl 2
HBr
8•
CH3CH2CH 20H
f. CH3CH 2CH 20H
g. CH3CH 2CH 20H
SOCl2
pyridine
PBr3
CH 3CH2CH2Br
TsCI
CH 3CH 2CHz0TS
pyridine
[1] NaH
[1] TsCI
(Y,
9.47
a. CfoH
b. CfoH
c. CfoH
d. CfoH
NaH
C fo H ~
NaCl
N.R.
HBr
C fs,
HCI
e. C foH
( fo - Na '
+ H2
Cfc1
g.
(foe
__I'!""'--
U
+H20
a,N.R.
1,1ce,oa,a, Cfo-ce,ce,
h. ( fo H ~ U
pynd1ne
•
u
9.48 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major
product.
a.
Jx
OH
/
~/T Jy
tetrasubstituted
major product
TsOH
b.
c.
U
disubstituted
CH 2C:
3
CX ·CX
0
trisubstituted disubstituted
major product
CHCH
3
d.
CH 3CH2CH2CH20H
OH
-~, ,..,,..,, '- ll l t:,1..:, 1 C I I U
-
~~UAIU ~t;
'd-
TsOH
-
TsOH
e.
tetrasubslituted
disubstituted
major product
Two products formed
by carbocation rearrangement
9.49 The most stable alkene is the major product.
OH
+
trans and disubstituted
monosubstituted
major product
cis and disubstituted
9,50 OTs is a_go?d leaving group_an? will easily be replaced by a nucleophile. Draw the products
by substttutrng the nucleophlle in the reagent for OTs in the starting material.
NaOH
SN2
K+ -oc(CH3l3
E2
9.51
a.
~ -
HO H
b.
~
OH
H D
_!::!fL.
ZnCl2
l'-0
pyridine
HO H
d.
/ ·,_
H Br
Cl ~ -
H Cl
pyridine
HO H
Configuration is maintained.
C-O bond is not broken.
2° Alcohol will undergo SN 1.
racemization
1° Alcohol will undergo SN2·
inversion
SOCl2
C.
+ ~ -
Br H
Ts0 H
SOCl 2 always implies SN2.
inversion
_!SL_
SN2
inversion
H I
'
Chapter 9-22
-oH
--
TsO H
c HP
9.52
A= ~
NaH
(b)
HO H
TsCI
pyridine
.
"'-
(c~
(2R) -2-hexanol
C = ,,~
E= ~
H Br
CH3I
B=
cHP
o=
---
F=
.
. entical products,
Routes (a) and (c) given rd
.
or
9.53 Acid-catalyzed dehydration follow s an El mechan1sm
.
make a good leaving group. The three steps ~re.
p
[ l] Protonate the oxygen to make a good leaving grou ·
[2] Break the C-0 bond to form a carbocation.
[3) Remove a phydrogen to form the re bond.
H OCH 3
labeled Band F.
,
2° and 3° ROH with an added step t
o
:~
A / CH3 H.L6so3H
V
overall
The steps~ I
reaction
~
U
cH 3 +
I
(_:0.-H
CH3
H-KY
+
CH3
--
-
+ HS04
H
H
~_)
.. _
+ H20
..
HS0 4
2° carbocation
and
&~H3
2° carbocation
1,2Hsh<
cr·~1Cr°"'\.
3° carbocation
and
~H
+ HS04
OCH3_
9.54 With POCh (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a
p hydrogen, only one product is formed. With H2S04, the mechanism of elimination is E1.
A 2° carbocation rearranges to a 3° carbocation, which has three pathways for elimination.
0
0
,,
:oH
cb
0
~(
r - - - , 11/l
Cl , IP' CI
Cl
N""
\_ H
t±>
cb
+
I/
POCl2
Cl
+
~ .o
+N
+ -0PHOCl2
0
+N
H
()
~ H- OS03H
:oH
cb
I/
+
COH2
cb
+ HS04-
cb
+ H20
3° carbocation
2° carbocation
Cb
3° carbocation
Hso,-
3° carbocation
j
j
j
m
co
m
9.55 To
(I]
[2]
(3]
[4]
Cb
1,2-CH3 shift
y
draw the mechanism :
Protonate the oxygen to make a good leaving group.
Break the C-O bond to form a carbocation.
Look for possible rearrangements to make a more stable carbocation.
Remove a p hydrogen to form the 7t bond.
Dark and light circles are meant to show where the carbons in the starting material appear in the product.
~-H
1+
H
+
Hso.-
- : - a -
2° carbocation
+ HS04 -
J
3° carbocation
C(
+ H2S04
Chapter 9-24
The 2' alcohol reacts b
an SN1 mechanism 1/
form a carbocation
that rearranges.
er
9.56
-
OH
l-methyl-2-butanol
2-malhyl-1-propanol
~ ~ J-.
{1}HBr
, carbocalion
3
[2]-H 20
"/'-y
HCl
2' carbocat1on
HBr
\-
H
,q7T ---:li'r.
The 1° alcohol reacts with HBr
by an SN2 mechanism.
no carbocatlon Intermediate =
no rearrangement possible
er~
• H,Q.
·5N2
no carbocation
\H-Br
"
9.57
+ :d: -
9.58
CH,CH,CH,CH,OH
H,so,, NaBr
overall reaction
two resonance structures
tor the carbocation
CH CH CH CH 2Br + CH 3CH 2CH=CH 2
3
2
2
+ CH CH CH 2CH 20CH 2CH 2CH 2CH 3
3
2
• ormat,on of a good leaving group
•
step 111for all product5 • F
r--.HQ,S0 H
CH,CH 2CH 2CH,-OH
3
CH,CH2CH 2CH,-?;-H
Formation of CH3CH2CH=CH2:
CH,CH CH-CH ~0-H _ _ _ _ _ _ r - -- - 2
2 H
Hso,-rH, J
E2cH,cH,cH=cH,I
,+
+
HS04
H
I
• H,9.
+
H,so,
Ether formation (from the protonated alcohol):
CH3CH2CH2CH2~?,.- CH2 CH2CH2 CH3 + H 2Q
rH
HSO,-
j
1,2-shift
l,) ..)
9
+ Hso,-
+Hso,-
+
-CY
9
+ H2SO,
9,60
~QisoaH
H
QH
~
,:;H2 _
~QH
+ Hso,-
9.61
a.
/t+
Br~
2° halide
b.
I
/'-Br + --0~
1° halide
less hindered RX
preferred path
2 2 3
/ a j H CH CH
a
l o- + BrCH2CH2CH3 cY''
.
V
1' halide
less hindered RX
preferred path
2' halide
_JH,CH,CH,
Chapter 9-26
C.
TH,CH,CH,
CH 3CH.f)CH2CH 2CH~
1
CH3CH20- + BrCH2CH2CH3
CH 3CH 2Br + --OCH2CH2CH3
1• halide
1° halide
Neither path preferred.
9.62 A tertiary halide is loo hindered and an aryl halide too unreactive to undergo a Williamson
ether synthesis.
. sets of start·m9 materials:
Two possible
9H3
O- C- CH
I
cr
3
CH 3
Br
+
yH3
-o-c-CH3
yH3
Br-C-CH3 +
er
~I
o-
I
I
CH3
CH3
3° alkyl halide
too sterically
hindered for SN2
aryl halide
unreactive in SN2
9.63
B.
b.
(CH3l3COCH 2CH 2CH 3
Q-o--Q ( ~:~iv)
2
9.64
(CH 3lJCBr + BrCH 2CH 2CH3
(2 equiv)
+
H,O
C.
0 - - 0CH3
__!!L 0-sr +
(2 equiv)
CH, 3:
+ H,o
2 Q -sr + H,o
-~\ :3
overall reaction
/ \ CH3
a. (_____)'(
l
H-1
The steps:
/ \ CH3 ..
\o~CH3 + :_v
·1
H
:i j ~,b. Cl ~
~H
0.p.
+ H2 + NaCl
q,66
0
0
HBr
I \
a.
,,c - C" •H
H/
'H
H
0
I \
c.
e.
0
H
I \
CH3CH20CH2CH 20H
[2] H20
f. HJ c-c,.H
H
H
9,67
0
a.
a
b
CH30H
CH3CH20H
CH{ A · H
CH3
H
I
I
I
I
c.
CH 3-C-C-H
H2S04
~H3
CH3CH20
(1) CH3CH20- Na+
a
H
CH3
OH
d.
(2] H20
OCH 2CH 3
9.68
a.
H
Cl
CH3~ .
/
c-c..
(I I
\'''CH 3H- 0 :
H
\~'
HOCH2CH20H
[2] H20
(1] CH3CH20 -
I \
.c-c ... H
H'/
'H
HC c: C- CH 2- CH20H
[2] H20
(1) OH
I \
HOCH 2CH 20H
H2S0 4
0
H
0
H20
' H
H
[1]HC = C-
d.
H
,c - C·,, H
b H/
I \
BrCH 2CH 20H
ct
ct
H
Cl.,_
CH3-.:,.
/ ,../
c-c,,
\ ''CH 3:q:
H
[1] CH3S-
CH 3SCH2CH20H
[2] H20
~
HBr
r
OH
OH
(1)NaCN
C rCN
(2) H20
The 2 CH 3 groups are anti in the
starting material, making them
trans in the product.
/_J
Na'H:
CH3
,CH 3 The 2 CH 3 groups are gauche in the
H.::·c-c·::.H
starting material, making them
b.
'}f:
cis in the product.
C4H80
Cl
c.
\
OH
I
,C-C,,
CH{ '
\'''CH3
H
H
rotate
CI
/C~3
C \c-c·· ,'
,•'JI
CH:i H
\
'-- :V
backside
attack
CH3
,H
H.::•c-c·::.cH 3
\;
:o :
C4H 80
Chapter 9-28
.
t rial and both products. Because
I starting Jl1a e
d
n the epoxide oxygen, an one must
t coille 1rOI
9,69 First, use the names to draw the structures oft ic '
the product has two OH groups, one OH mus
come from the nuclcophile, either -oH or H20 .
OH
With -oH, the nucleophile
attacks at the less
o __.substituted end of the _
[2] H20
epoxide to form the R
isomer of the product.
(2R)-2-ethyl-2-methyloxirane
~c@J
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