CHAPTER 9 HW SOLUTIONS A ETHERS -...

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

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

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

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

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)

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

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

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

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

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

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+

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

CHAPTER 9 HW SOLUTIONS A ETHERS -...

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