chapter 9 hw solutions a ethers - chemistry 21 · page 1 chapter 9 hw solutions: alcohols + ethers...
TRANSCRIPT
Page 1
CHAPTER 9 HW SOLUTIONS: ALCOHOLS + ETHERS
ALCOHOL + ETHER NOMENCLATURE
1. Give the IUPAC name for each compound. Include cis/trans or R/S stereochemistry if necessary.
Structure
Name 3-methyl-1-heptanol 1-t-butylcyclopentanol (2R,3S)-3-methyl-
2-pentanol
Structure
Name 4,5-dimethyl-3-decanol (1S,2S)-
2-chlorocycloheptanol
trans-4-(3-methylbutyl) cyclohexanol
(can’t use R,S because it’s achiral)
2. Give the IUPAC or common name for each compound. Include cis/trans or R/S stereochemistry if
needed.
Structure
Name hexyl isopropyl ether
or 1-isopropoxyhexane dicyclohexyl ether
cyclopentyl propyl ether or propoxycyclopentane
(no R,S- not chiral)
Structure
Name (R)-4-cyclopropoxyoctane (1S,4R)-
4-t-butoxycyclooctanol 1,6-diethoxy-2-methylheptane
HO OH OH
CH3
OH
Cl
OH HO
O OOCH2CH2CH3
OO OH
O
O
Page 2
REVIEW OF ALCOHOL SYNTHESES
3. Provide the starting alkyl halide and reagents needed in order to produce each alcohol through a substitution reaction.
WILLIAMSON ETHER SYNTHESIS
4. What is the purpose of the sodium hydride (NaH) in the following reaction?
H– (hydride) is a strong base and removes the H from the alcohol (acid-base). This converts CH3OH (poor nucleophile) into CH3O– (good nucleophile), so the reaction is faster.
5. Give the curved arrow mechanism for the following reactions.
a. c.CH3OH OHCH3XNaOH
SN2
NaOH
SN2X
b. d.OHOHX
XH2O
SN1
H2O
SN1
CH3OH CH3OCH3a. NaHb. CH3I
a.
H CH3 ICH3O H H3C Oacid-base SN2
H3C OCH3
b.OH O
a. NaHb. I
O OO IH H
c. Bra. NaHb. CH3OH +CH3OH
+ CH3OH
Br
HHCH3O H
H3C Oacid-base E2 because 2o
or 3o RX
CH3OH CH3OCH3a. NaHb. CH3I
Page 3
6. Give the major organic product for the following reactions.
7. Using the Williamson Ether Synthesis, show a synthetic route (complete with reagents) that efficiently produces each ether below.
8. Provide the reagents needed to complete each reaction.
a. OH a. NaHb. CH3Br
OCH3
d.a. NaHb.
OHI
O
b.OH a. NaH
b. CH3CH2CH2BrO
e.a. NaHb. BrOH OH
c.a. NaHb.CH3OH Cl
E2f.
OH a. NaHb. Cl
O
a. OCH3 OCH3OH
OCH3CH3OHBr
a. NaH
b. CH3Br
a. NaH
b.
Both methods work.
b. O
a. NaH
Br
OHO
b. CH3CH2Br
b.E2
2o RX.
a. NaH
CH3CH2OH CH3CH3O Br+
Below doesn't produce the ether well because E2 is the main pathway with a 2o RX.
c.OCH2CH2CH3 OH
I
OCH2CH2CH3
OHI
a. NaH
b.
a. NaH
b.This doesn't work well: E2 occurs instead.
Cl OCH2CH3a.
OH OCH2CH3
b.CH3CH2OH
SN1
a. NaH
Williamson Ether Synth.b. CH3CH2X
Page 4
INTRAMOLECULAR REACTIONS
9. Give the curved arrow mechanism for the following reaction.
10. Give the major organic product of each reaction.
DEHYDRATION REACTIONS
11. Give the curved arrow mechanism for each reaction. Include the Lewis structure of the acid in your mechanism.
Cl
OHNaH O
Cl
OO
H
Cl
OHacid-base Intramolecular SN2
a. ClNaH
OHb.
HOBr
NaHO O
15
1 5
3
3 12
3 41
23
4
a.con. H2SO4
heatOH
O
H
OH OS
OH
O OH
OS
OH
O O
H
H
E1
b.OH
con. H2SO4
heat
OH2
H
OH OS
OH
O OH
HSO4
c.OH con. H3PO4
heat
OHO
P
O
OHOH
H OH2
H OPO
OHOH
E2
1o dont form carbocations
Page 5
12. For the following reaction, a. Draw the curved arrow mechanism.
b. Use the mechanism to identify two reasons why “acid” is a catalyst in the dehydration reaction.
Note: “acid” is any strongly acidic source, such as H2SO4, H3O+, or ROH2+.
• Acid accelerates the reaction by turning a bad leaving group (OH–) into a good leaving group (OH2). This makes the catalytic pathway have a lower activation barrier.
• The acid is not consumed in the reaction. It is used in step 1, but is regenerated in step 3.
c. Draw the energy diagram.
13. Draw all probable dehydration products for these reactions, including stereoisomers. Then decide which should be the major product and briefly explain your answer.
OH H2O
HStep 1
OS
OH
O OH
Step 2
HSO4
Step 3
a.con. H2SO4
heatOHmost substituted (di) and trans has fewer repulsions, so lowest energy.
b.OH con. H3PO4
heatTrisubsituted is lower energy than disub.
c.con. H2SO4
OH
ΔNo other possibility.
d.con. H3PO4
heatOH
Trisub lower E than disub. Major put bulkiest groups opposite. (In truth all 4 trisub are similar E.)
OH con. H2SO4
heat
E
OH
H2O
Alkene is higher energy than alcohol reactant.
Step 1 Acid-base reaction is favorable (check pKas)
Page 6
HYDRIDE AND ALKYL SHIFTS
14. Draw the intermediate formed after each mechanistic step.
15. What is a possible driving force (or reason) for the rearrangement in:
a. Problem 14a?
A secondary carbocation is converted into a tertiary carbocation through the rearrangement, which is more stabilized by hyperconjugation. The motivation for the shift may be to lower the energy of the carbocation.
b. Problem 14c?
A four-membered ring is converted into a five-membered ring through the rearrangement, which has less ring strain. The motivation for the shift may be to relieve the ring strain (lower the energy of the system).
16. Give the curved arrow mechanism for each reaction. Include the Lewis structure of the acid.
Reaction
a.
b.
CC
H
CC
H
H
H
H
H H
a.H CH
HH
CC
CC
H
H
H
H
H H
CH
HH
H H
b.
H3C CH3
CH3
CH3
c.
H3C
1
2
34
56
12
65
H3C 4 3
d.1
23 1
2
34
45
5
6
6
77 88
con. H2SO4
heatOH
OHO
SOH
O OH
OH2
H
HHH2O
H-shift
or HSO4
OH2H
HHH
H-shift H-shiftanother
H
con. H2SO4
Δ
OH OS
OH
O OH
HSO4
Page 7
c.
d.
17. Identify which of the reactions W-Z would synthesize 3,3-dimethylcyclopentene most efficiently (with
the fewest competing products). Then explain why the other routes are less efficient.
Reaction Y would most efficiently produce 3,3-dimethylcyclopentene as there are no significant alternative products. There is only one type of beta-hydrogen, so the E2 reaction can make only one possible product. (Also SN2 does not compete much when using such a bulky base.)
In Reaction W, two different beta-hydrogens are present, leading to a likely mixture of the products shown. This would split the yield, lessening the quantity of 3,3-dimethylcyclopentene
obtained. Also this mechanism proceeds through a carbocation, so rearrangements could further divert the yield.
In Reaction X, a bulky base would more likely remove a beta-hydrogen from position “a,” creating the wrong alkene isomer as the major product. Removal of a beta-hydrogen from position “a” is still possible, but the intended alkene would be a minor product.
In Reaction Z, the carbocation generated in this mechanism is neighboring a quaternary center, so rearrangements involving a methyl shift are very likely, producing the wrong product.
con. H3PO4
heatOH
OH2
CH3
CH3H
methylshift
OPO
OHOHH
H2PO4
OHcon. H2SO4
OH2 =
Halkyl
shift
(ring expansion)
Δ
OS
OH
O OH
HSO4
3,3-dimethylcyclopentene
OHcon. H2SO4
heatW
Br KOC(CH3)3X
Br
KOC(CH3)3Y
OHcon. H2SO4
heatZ
+
majorminor
+
b
a
BEST
+
likely
Page 8
ALCOHOL REACTION WITH HX
18. Give the curved arrow mechanism for each reaction.
Reaction
a.
b.
c.
PBr3, SOCl2, AND TOSYLATE REACTIONS
19. Fill in the boxes with the organic product from each reaction.
HBrOH Br
OH2SN2
OH H BrBr
Br
HICH3
ICH3
OH
CH3
OH H I
CH3
OH2
CH3H
hydride
shiftCH3
I
SN1
CH3
I
OH
HClCl
OH2
ethylshift Cl Cl
(same as above)
OH HNaH CH3I
O OCH3
TsCl
py.KN3
OH H OTs N3N
NN
SOCl2 KN3OH H Cl N3
Page 9
COMBINED ALCOHOL REACTIONS
20. Give the major organic product for each reaction. Consider plausible rearrangements.
1˚ don’t form carbocations, so no rearrangement occurs
Double SN2
Single SN2
1˚ alcohol goes through SN2, so inversion occurs
a. HClOH Cl
i.Cl SO
OCH3 OH CH3 OTspy.
b.HI
OH IOH
a. SOCl2
b. KOCH2CH3j. OCH2CH3
c. HBrHO Br OH a. TsCl, py.
b. NaN3k.
N3
d. PBr3OH Br OH a. PBr3
b. NaCNl.
Br CN
e. SOCl2OH Cl
OH a. TsCl, py.
b. OONa
m. O
O
f. CH3OHTsClpy.
CH3OTs
OHHBr
CH3
n.
CH3
Br
g.OH
NaHO
D
OHHClo.
D
Cl
h.HO PBr3
Br
Page 10
21. The synthesis of the product shown in reaction L (call it product Q) is best achieved by this method. Explain why methods M-O will not effectively synthesize product Q.
Best synthesis of Q: Double SN2 gives the right stereochemistry of the SH group (PBr3 inverts the center, SH inverts the center again).
Reaction (M) doesn’t work: OH– is not a good leaving group for SN2 reactions.
Reaction (N) is not efficient: the reaction goes through a flat carbocation intermediate, so attack of H2S gives both isomers (with SH out and SH back). Carbocations also often rearrange, and a significant product might be with the SH on the same carbon as the methyl group. E1 may also compete.
Reaction (O) doesn’t product Q: single SN2 produces the wrong isomer.
EPOXIDE REACTIONS
22. Give the curved arrow mechanism for the following reactions.
a.NaOHO
OH
OH
H2O
O
OH
OH
OHOH
O H OH
b.H+O
OCH2CH3
OH
CH3CH2OH
H+O O
H
CH3CH2OHO
OH
HCH3CH2OH CH2CH3OCH2CH3
OH
c. KOCH2CH3O
ClCH3CH2O
O
DMF
O
Cl
CH3CH2OO
CH3CH2O
O
Cl
CH3CH2O
OH
CH3a. PBr3
b. NaSHSH
CH3
(L)
Q
OH
CH3
(M) NaSH
OH
CH3
(N) H2SO4H2S
OH
CH3
(O)a. TsCl, py.
b. NaSH
Page 11
23. Concerning the following two reactions:
a. Explain the regioselectivity of the reactions.
The first reaction involves a negatively charged nucleophile (good Nu) so reacts SN2-style at the least hindered side of the epoxide, at site “a.”
The second reaction involves protonation of the epoxide before addition of the nucleophile (Cl–). Since the LG is better now, the LG has already partially left (SN1-like). There are partial positive charges on the two carbons of the epoxide, but the more substituted side (3˚ vs. 2˚) stabilizes a partial carbocation better. Thus there is a greater d+ on the more hindered side, which is where the nucleophile attacks.
b. Explain the stereoselectivity of the reactions.
Both reactions occur with inversion of the reacting carbon, SN2-style, with backside attack.
Even though the protonated epoxide in the second reaction has a partial carbocation, it is not a full carbocation. The reaction mechanism is really in between SN1 (strong d+ on the C) and SN2 (backside attack).
24. Give the major organic product for each reaction. Indicate if a racemic mixture is formed.
OCH3
NaSHH2O
OH
SH
CH3
a
bHCl
OCH3
CH3
OH
Cl
a. CH3CH2OKO
OH
OCH3CH2OH
e.
O
CH3
H+
CH3OH
OH
CH3 OCH3
b.H+
(CH3)2CHOH
OOH
O
racemic
f.
O
CH3
KOHH2O
OHCH3
OH
c. HBrO Br
OHg.
OHI
OH
I
d.NaCNH2O
O OH
CNh.
O HC C
b. H2O
a. OH
CCH
OCH3
δ+ H
δ+
δ+
Page 12
COMBINED ALCOHOL + EPOXIDE REACTIONS
25. Give the major organic product for these reactions. Consider plausible rearrangements and indicate if a racemic mixture is formed.
a.a. NaH
OH
CH2CH3 b. CH3CH2I
OCH2CH3
CH2CH3
e. HBrO
HO
Br
b.NaOHH2O
O
OH
OH
racemic
f. HI
OH I
c.con. H2SO4
OH
CH3
heatCH3
g. NaIOTs
H3C
H
H3C I
H3C
OTs
H3C
Ior or
d. PBr3
OH
CH3O
Br
CH3O
h.KCNH2OO
CH3OH
CH3
CN