Organic Chemistry

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8 of 10 12/23/2011 3:38 AM

Review of Concepts

Chapter 13 Alcohols

Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter I 3. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulal)'·

• When naming an alcohol, the parent is the longest chain containing the -----group.

• The conjugate base of an alcohol is called an ion. • Several factors determine the relative acidity of alcohols, including ____ _

______ ,and ___________ __

• The conjugate base of phenol is called a , or ion.

• When preparing an alcohol via a substitution reaction, primary substrates will require SN_ conditions, while tertiary substrates will require SN_ conditions.

• Alcohols can be fonned by treating a carbonyl group (C=O bond) with a agent.

• Grignard reagents are carbon nucleophiles that are capable of attacking a wide range of , including the carbonyl group of ketones or aldehydes, to produce an alcohol.

• groups, such as the trimethylsi lyl group, can be used to circumvent the problem of Grignard incompatibility and can be easily removed after the desired Grignard reaction has been performed.

• Tertiary alcohols will undergo an SN_ reaction when treated with a hydrogen halide.

• Primary and secondary alcohols will undergo an SN- process when treated with either HX, SOCh, PBr3, or when the hydroxyl group is converted into a tosylate group followed by nucleophilic attack.

• Tertiary alcohols undergo E I elimination when treated with ____ _ • Primary alcohols undergo oxidation twice to give a -----------• Secondary alcohols are oxidized only once to give a ____ _ • PCC is used to convert a primary alcohol into an -------• NADH is a biological reducing agent that functions as a delivery

agent (very much like NaBH4 or LAH). while NAD+ is an agent. • The are two key issues to consider when proposing a synthesis is whether there is:

l . a change in the---------2.achangeinthe ________ _



9 of 10 12/23/2011 3:38 AM

CHAPTER 13 275

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 13. The answers appear in the section entitled Skill Builder Review.

13.1 Naming an Alcohol

PROVIDE ASYSTEMATIC NAME FOR THE FOLLOWING COMPOUND

t) IDENTIFY THE PARENT

2) IDENTIFY AND NAME SUBSTfTUENTS

3) ASSIGN LOCANTS TO EACH SUBSTITUENT

4) ALPHABETIZE

5) ASSIGN CONFIGURA TtON

13.2 Comparing the Acidity of Alcohols

FOR EACH PAIR OF COMPOUNDS BELOW, CIRCLE THE COMPOUND THAT IS MORE ACIDIC:

c56 /"'--aH Cl)('aH Cl Cl

13.3 Identifying Oxidation and Reduction Reactions

IN THE FOLLOWING REACTION. DETERMINE WHETHER THE STARTING MATERIAL HAS BEEN OXIDIZED. REDUCED. OR NEITHER:

a _A_

RO OR )(_

13.4 Drawing a Mechanism, and Predicting the Products of Hydride Reductions

COMPLETE THE MECHANISM BELOW BY DRAWING ALL CURVED ARROWS, INTERMEDIATES AND PRODUCTS.

H IP-:a·· H- A - H ·a·

d' I H;. .:._H H

13.5 Preparing an Alcohol via a Grignard Reaction

IDENTIFY REAGENTS THAT CAN ACHIEVE EACH OF THE FOLLOWING TRANSFORMATIONS

a I)

PhA Et 2) ( ~ 0 t)( aH

Ph)lMe 2) ( ) PhiEt

a / Me)lEt

I)(

2) (



10 of 10 12/23/2011 3:38 AM

276 CHAPTER 13

13.6 Proposing Reagents for the Conversion or 110 Alcohol into an Alkyl Halide

IDENTIFY REAGENTS THAT CAN ACHIEVE EACH OF THE FOLlOWING TRANSFORMATJONS

1) ( )

2) c=::J

~ 0

13.7 P-redicting the Products of an O~idation Reaction

DRAW THE EXPECTED PRODUCT OF THE FOLLOWING REACTION

~OH Cr03 -Hao•

acetone

13.8 Com·erting Functional Groups

D IDENTIFY REAGENTS THAT CAN ACHiEVE EACH OF THE FOU.OWJNG FUNCTIONAL. GROUP TRANSFORMATJONS



1 of 10 12/23/2011 3:39 AM

CHAPTER 13 277

13.9 Proposing a Synthesis

AS A GUIDE FOR PROPOSING A SYNTHESIS, ASK THE FOLLOWING TWO OUESTIONS:

I) IS THERE A CHANGE IN THE ____ SKELETON?

2) IS THERE A CHANGE IN THE LOCATION OR iDENTITY OF THE

AFTER PROPOSING A SYNTHESIS. USE THE FOLLOWING TWO OUESTIONS TO ANAL 'IZE YOUR ANSWER:

I) IS THE _______ OUTCOME OF EACH STEP CORRECT?

2)/S THE OUTCOME OF EACH STEP CORRECT?

Review of Reactions Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook at the end of Chapter 13. The answers appear in the section entitled Review of Reactions.

Preparation of Alkoxides

ROH e Na

Preparation of Alcohols via Reduction

+ MeOH

Preparation of Alcohols via Grignard Reagents

0 OH

A- ~ R



2 of 10 12/23/2011 3:39 AM

278 CHAPTER 13

Protection and Deprotcction of Alcohols

R-OH

SNl Reactions with Alcohols

R R-1----0H

A

SN2 Reactions with Alcohols

OH

~

R- 0 -TMS

Cl ! ~

El and E2 Reactions with Alcohols

--j-oH

-j-oH -j-ors

Oxidation of Alcohols and Phenols



3 of 10 12/23/2011 3:39 AM

OH

6

Solutions

13.1. a) 5,5-dibromo-2-methylhexan-2-ol b) (2S,3R)-2,3,4-trimethylpentan-l.-ol c) 2,2,5,5-tetramethylcyclopentanol d) 2,6-diethylphenol e) (S)-2,2,4,4-tetramethylcyclohexanol

13.2.

&., a) Br

b)~ c) £.b--oH

CHAPTER 13 279

0

¢ 0

13.3. Nonyl mandelate has a longer alkyl chain than octyl mandelate and is therefore more effective at penetrating cell membranes, rendering it a more potent agent. Nonyl mandelate has a shorter alkyl chain than decyl mandelate and is therefore more watersoluble, enabling it to be transported through aqueous media and to reach its target destination more effectively.

MEvertson

Highlight

Solutions



4 of 10 12/23/2011 3:39 AM

280 CHAPTER J3

13.4. .. . .e c:B

a) a~H ~ a~: Na

~~H NaH .. e 0

b)

~~: Na

:OH .. e (£)

Li :o: Li

+ - + c)

:OH .. o e :o: Na

0 NaH 0 -d)

13.5. F F

a) ....... ~oH The electron-withdrawing effects of the fluorine atoms stabilize the conjugate base.

b)~OH

The conjugate base of a primary alcohol will be more easily solvated than the conjugate base of a tertiary alcoho l.

Cl

ClyYOH

CIY CI c) Cl The electron-withdrawing effects of the chlorine atoms stabilize the conjugate base.

d) YnOH The conjugate base is more hjghly stabilized by resonance, with the negative charge spread over two oxygen atoms, rather than just one oxygen atom.



5 of 10 12/23/2011 3:39 AM

CHAPTER 13 281

0r'OH

e) V The conjugate base is stabilized by resonance.

13.6. 2-nitrophenol is expected to be more acidic (lower pK.) because the conjugate ba e has a resonance structure in which the negative charge is spread onto an oxygen atom of the nitro group, shown below. In contrast, 3-nitrophenol does not have such a resonance structure:

13.7.

a) d Br

b) ~Br NaOH ~OH

d) ~ 1) BH3 • THF ~OH

e)~ 1) Hg(0Ac)2, H20

2) NaBH4

HOr-C



6 of 10 12/23/2011 3:39 AM

282 CHAPTER 13

13.8

~ 1) BH3 • THF ~OH

a) 2) H20 2, NaOH

OH

b)~ 1) Hg(OAc)2. H20

~ 2) NaBH4

c)~~ dilute H2S04 VOH ''-....,/' '-.../ ......__

13.9. a) (+2) ~ (+2). The starting material is neither oxidized nor reduced.

b) (+ I) ~ (+3). The starting material is oxidized.

c) (+3) ~ (-1 ). The starting material is reduced.

d) (+3) ~ (+3). The starting material is neither oxidized nor reduced.

e) (0) ~ (+2). The starting material is oxidized.

l) (+2) ~ (+3). The starting material is oxidized.

13.10. One carbon atom is reduced from an oxidation state of 0 to an oxidation state of - I, while the other carbon atom is oxidized from an oxidation state of 0 to an oxidation state of+ J. Overall, the starting material does not undergo a net change in oxidation state and is, therefore, neither reduced nor oxidized.

13.11. One carbon atom is reduced from an oxidation state of 0 to an oxidation state of -2, while the other carbon atom is oxidized from an oxidation state of 0 to an oxidation state of +2. Overall. the starting material does not undergo a net change in oxidation state and is. therefore, neither reduced nor oxidized.

13.12. a)

at~ -lA

H- AI-H I H

.. e : o:~ ..

\ ·O· H,....~,H

H



7 of 10 12/23/2011 3:39 AM

b)

OJLO ~ lA

c)

H-At-"'-H I H

~ ·lA

d )

H-1¥-H I H

ot·r H-\_H

c)

t)

lA H A~H

I H

---

:o·· o

~Mo 1A

H- 1¥-H I H

.. e

cr1: :o··

--- dtH---\_H

lA H- AI-'-H

~

CHAPTER 13 283

•• H :o'

crtH

H~ u OMe



8 of 10 12/23/2011 3:39 AM

284 CHAPTER 13

13.13.

H H

H t O: q:a·· ~AI2-H ~en H ,r n - I .Ct- ~0/'H"-A\2-H

H 8

13.14. a)

v . -

.. e o:

0 II 1) MeMgBr

H~~ OH

c)

~ 0~

)lH 2) H20

1) PrMgBr

2) H20

OH

~

e .. :0 H

~) J H @

~9.-H



9 of 10 12/23/2011 3:39 AM

d)

n 9 ~H

0

H~

e)

~ 0

~ 0

"11' 0

1) EtMgBr

1)~MgBr/ 2) H20

1) MeMgBr

~ 1) EtMgBr

1) MeMgBr

1 )~MgBr /

2) H20

CHAPTER 13 285

13.15 Each of the following two compounds can be prepared from the reaction between a Grignard reagent and an ester, because each of these compounds has two identical groups connected to the a position:

~H Me

The other four compounds from Problem 13.14 do not contain two identical groups connected to the a position, and cannot be prepared from the reaction between an ester and a Grignard reagent.



10 of 10 12/23/2011 3:39 AM

286 CHAPTER 13

13.16 Each of the following three compounds can be prepared from the reaction between a hydride reducing agent (NaBH4 or LAH) and a ketone or aldehyde, because each of these compounds has a hydrogen atom connected to the a position:

OH

~ OH

~

The other three compounds from Problem 13.14 do not contain a hydrogen atom connected to the a position and, therefore, cannot be prepared from the reaction be tween a hydride reducing agent (NaBIL or LAH) and a ketone or aldehyde.

13.17.

In this case, H30+ must be used as a proton source because water is not sufficiently acidic to proton ate a phenolate ion (see Section 13.2, Acidity of Alcohols and Phenols).

13.18. a)

~Br TMSO l!.._,yJ 1) Mg

OH I

TMSO~

! TBAF

OH I

HO~



1 of 10 12/23/2011 3:40 AM

13.19. a)

OH

if

b)

c)

o-oH

TMSOV Br

1) TsCI, py

/ 2) NaBr

" PBr3

HBr -1) TsCI, py

"' /

/ 2) NaCI ~

"-~_SO_C_I=-2 ------'/ py

TBAF -

~r

(('

1) Mg

0 2) II

,............_OMe

C HAPTER 13 287

(0.5 equivalents)

3) H~



2 of 10 12/23/2011 3:40 AM

288 CRAPTER 13

d) 1) TsCI, py

Q-oH

/ 2) NaBr \..

e) 1) TsCI. py

AH

/ 2) NaCI ~ __,lc1

" SOCI2 / py

t) 1) TsCI, py

/ 2) NaBr \..

~OH HBr ~Br

" PBr3 /

13.20.

f ::-

1) TsCI, py 1

Q-oH 0· .. ·•1 NaCI

Q-c1 2) Nal

13.21. a)

~ H2S04 ~ I heat + ~

major minor

b)

0--oH 1) TsCI, py 0 2) NaOEt



3 of 10 12/23/2011 3:40 AM

13.22. a)

HO\,. ... (

b)

c)

0

H

d-OH

H~OH 0

xsOOa HaO+

acetone

0

HO~OH 0

c)

;so e)

Q-oH PCC

C)=o CH2CI2

0

0-oH Na2Cr20 7

Q o H2S04 , H20

13.23. a)

0

~Br l ) NaOH

~H 2) PCC, CH2CI2

b)

1) BH3 ·THF 0

C1 2) H20 2, NaOH ~H 3) PCC, CH2CI2

CHAPTER 13 289



4 of 10 12/23/2011 3:40 AM

290 CHAPTER 13

c)

d )

13.24. a)

b)

1) dilute H2S04

/2)Na2Cr20], H~04 , H20

1} Br2

2) xs NaNH2 3) H20

4) H2S04. H20 , HgSO•

1) Hg(0Ac)2, H~ 2) NaBH4

1) Br2 2) xs NaNH2

3) H20

1) H2, Lindlar's Catalyst

2) BH3 -THF

3) H20 2, NaOH

1) 9·BBN

2) H20 2, NaOH

~OH ~

\ 1) TsCI, py I 2) I·BuOK



5 of 10 12/23/2011 3:40 AM

CH APTER 13 291

d ) 1) BH3• THF

0

~ 2) H20 2, NaOH

~H 3) PCC. CH2CI2

e ) 1) H2so • • heat

I 2) H2. Pt \ /""-../ /""-../

OH \ 1) TsCI, py 2) NaOEt

3) H2, Pt

f) H2S04, heat

0 OH I \ '?H

1) LAH y y 2) H20 1) TsCI, py 2) I·BuOK

g) H2S04• heat

0 OH

~ 1) LAH

~ 2)H~ ~ \:) TsCI, py

2) NaOEt

13.25.

s H2so • • heat

~ 0 \]l TsCI, py I

2) NaOEt

13.26. H2S04, heat

X: ~ 1) BH3 ·THF

~OH 2) H20 2, NaOH \ 1) TsCI, py I

2) NaOEt



6 of 10 12/23/2011 3:40 AM

292 C IIAPTER 13

13.27. 1) E1MgBr

C)JH 2)H~

3) Na~r2C>7 .

H2S04 . H20 a)

1) MeMgBr

I o 2) H20

~H 3) Na;2Cr201 .

b)

13.28. a)

b)

#

c)

H~U. , H~

1) Br2

/ 2) xs NaNH2 ..

3) Hz0

1) NaNH2

2)~1

3) Na2Cr20 7 ,

H2S04 . H20

~

~

1)9·BBN 2) H20 2, NaOH

1) 9-BBN

2) H2~. NaOH

0 yv

1) MeMgBr

2) H20

3) Na2Cr207,

H2SO •• H20

0

~H 1) EtMgBr

2) H20

3) Na2Cr20 1 •

H2so • . H2o

0

~



7 of 10 12/23/2011 3:40 AM

CHAPTER 13

d ) OH r 1) TsCI, py

~ 1) BH3·THF

~OH 2) 1-BuOK 2) H20 2, NaOH

J PCC, CH,CI,

1) MeMgBr n/ 2) H20 n H

3) Na2Cr20 7 , 0

e)

#

0 1) 9-BBN

t)

OH

A

13.29. a)

0

_)lH

b)

~Br

2) H20 2, NaOH

1) Na2Cr20 7 • H2S04 , H20

2) MeMgBr

3) H20

1) MeMgBr

2) H20

e G> H-C:=C. Na

H2S04 , H20

O'[H

1) MeMgBr

<J( 2) H20

3) Na2Cr20 7 ,

H2S04 , H20

OH

~

OH

A

H2

dilute H2S04

.. ~ ""' I Lindlar's Catalyst

\ H2S04 , H2o., HgS0 4 ~

0

293



8 of 10 12/23/2011 3:40 AM

294 CHAPTER J3

c) H2

,...---....;;...-; .. ~ /~ I Lindlar"s

Catalyst

1) HBr, ROOR

2) NaOH ~ B 0 H- c::c. Na

/""'--sr

d)

/'er H-c: cf> Na(i) 1)9-BBN

13.30. a) 2-propyl-1-pentanol

! 1) NaNH2 2) EtBr

~~'--../"

b) (R)-4-methyl-2-pentanol c) 2-bromo-4-methylphenol d) (1 R,2R)-2-methylcyclohexanol

13.31.

a)

CCOH

OH

b) OH

~

c)

H2so •• H2o H9so.

0

~H 1) EtMgBr

2) H20

~OH

OH

~



9 of 10 12/23/2011 3:40 AM

d)

~ OH

HO _ __....__ e) -.......- ·oH

f)HO~

13.32. OH

HO~ ~ 1-butanol 2-butanol

13.33.

a) Increasing acidity

y OH

b) Increasing acidity

OH

c)

0 OH

Increasing acidity

~OH

v

2-methyl-2-propanol

Cl Cl CICI

CIY CI OH

y OH

CHAPTER 13 295

~OH 2-methyl-1-propanol



10 of 10 12/23/2011 3:40 AM

296 CHAPTER 13

13.34. a)

[(1-"1\"'8? ""': .. I _.. ,:;:;

b)

[ .. e . . l % .--- )\

c)

:o··

r

.. e

-~ - ~ 13.35. a) l-bromobutane

0

e)~H i)~OTMS

13.36.

13.37. a)

OH

~

. -.- ~ a 8'-.J

b) l -chlorobutane 0

t)~OH

j) ~OTs

0

~H

- ~] 8

c) 1-chlorobutane

.. 8 (£) ~O; Li

g) ..

./'-._ ./'-._ .. e CB ..-- '-./ ' O: Na

k) ..

d) trans-2-butene

/'-. /'-. .. e e _., '-./ -o: Na h) ..

~ .. ee I) ~: K



1 of 10 12/23/2011 3:41 AM

b) OH

~

c)

Na2Cr207

H2S04 , H20

0

~OH

OH

~

d)

OH

~

e)

OH

~

13.38. a)

b) 0

1) PCC, CH2CI2 2} EtMgBr 3) H20

4) Na2Cr20 7, H2S04 , H20

5} MeMgBr 6) H20

1) PCC, CH2CI2 2) PrMgBr

3) H20

4) Na2Cr20 7, H2S04 , H20

~ 1) MeMgBr

0

if 1) PrMgBr

~~ 2) H20

OH

~

0

~

OH

~

CHAPTER 13 297



2 of 10 12/23/2011 3:41 AM

298 CHAPTER 13

c)

~ 1) EtMgBr

~H~ OH

~ 0

H~

d)

0

~

0

~

0

~

13.39.

13.40. a)

1) MeMgBr

~ 1) EtMgBr

2) H20

~ 2) H20

~ b)u

1) EtMgBr

2) H20

3) Na2Cr20 7 ,

H2S04, H20

OH

~

0

c)~



3 of 10 12/23/2011 3:41 AM

b) 1) LAH

0

~H ~

~ MeOH

13.41.

~8 ··~f\ H :H

:Br .. o;

CHAPTER 13 299

.. e :Br:

13.42. The major product is 1-methylcyclohexanol (resulting from Markvonikov addition), which is a tertiary alcohol. Tertiary alcohols do not generally undergo oxidation. The minor product (2-methylcyclohexanol) is a secondary alcohol and can undergo oxidation to yield a ketone.

13.43. 0 cfH cY ()Br Mg ()MgBr 1) A H2S04

2) H20 heat

Compound A Compound B Compound C



4 of 10 12/23/2011 3:41 AM

300 C liAJ'TER 13

13.44.

0

ci'H OH

~ 0

I Na2Cr201 ~ H2S04, H20

1) PhMgBr c1 ll) TsCI, py

f 2) 1-BuOK

2) H20

1) BH3 ·THF "<::: dv 2) H20 2, NaOH dO ~ OH

c/o PCC l

CH2CI2

0

~ 13.45.

l)Mg

0 2) Jl

H H

3) H20

! PBr3

Br

()0

1) HBr, ROOR

2} Mg j 1) PrMgBr

2) Hp



5 of 10 12/23/2011 3:41 AM

CHAPTER 13 301

13.46.

a)

·o·· .. ) :CYH :o: (

l.. , :o·· ~ ~ H"' t ' H - H H

...._H

H- Atz-H

H

b)

:o·· 0 ..

"6 ~RH '6 \ :o·· H"' J ' H

I H

c)

~ e ..

"6 :0: H , :o··

6 H • ' Me -H H P H

I H

.. e py + so2 + :c1:

-



6 of 10 12/23/2011 3:41 AM

302 CHAPTER 13

b)

:sr", ,. r __,_ p.!..:s~:

:oH ·sr~ ~ __:_'· ·---

c)

:a·· .. e :a·· ~ l~o;'

~o:

~~~ -H .. 4 H ~H

I~ H-A~H

~

13.48.

OH

~ a)

Na2Cr207

H;>S04, H20

V OH PCC

c) CH2CI2

0 1) LAH

e) '(H 2) H20

13.49

a)

0

~

I~ H-A~H

I H

0

cY'H

OH

y

0 1)03

2) DMS _ ___;_ _____ HO~OH

3) Excess LAH

4)H~

:~r:

~ + PBr20H

OH PCC 0

b) y CH2CI2 "( OH 0

d)y Na2Cr207

'(oH H2S04, H20

0 1) LAH OH

f)~ ~ 2) H20



7 of 10 12/23/2011 3:41 AM

b)

0

c)

D

d)

0

1) 0 3

2) OMS

3) Excess LAH

4) H20

1) 0 3

2) OMS

3) Excess LAH

4) H20

1) EtMgBr

HO~OH

HO. _.,...__ _.,...__ "'-"' "'-"' ~OH

OH 2) H20

H~ 3) Na2Cr207 , H2S04 , H20 + 4) EtMgBr

5) H20

e) 1) LAH

2) H20

~0 3) TsCI • pyridine 0--LOTs H

f) 1) H3o+

0 2) Na2Cr20 7 , H2S04 , H20

13.50. a)

3) PhMgBr

4) H20

·o· ~0 C5 H-rf' MgB< .. e

HaC :o:---.__ · · X , ·o· u H'\ J' H

CHAPTER 13 303



8 of 10 12/23/2011 3:41 AM

304 CHAPTER 13

b)

13.51.

8 .·n cH -~(· 3 l5: -

H~H3 L CH3

OH

.. 1) 9·BBN

1) Br2

~:o/:~~M~r H -~H ___ .,_

.. e ~:

.. ?a~· CH3 ·o· ' .. H_...V' H CH

3

-oH

2) xs NaNH2

8 ..

=;z;~3

La?) .

J ..

H <;' •. p. H

3) H20 H2 , Lindlar's Catalyst

1) TsCI

~OH 2) NoOEt \

NaoH( )~ PBr3 2) H202, NaOH

HBr, ROOR

~Br t-BuOK



9 of 10 12/23/2011 3:41 AM

CHAPTE R 13 305

13.52.

a) 0 1) MeMgBr 0

ci'H 2) H20 d" 3) Na2Cr20 7 , H2S04 , H20

b) 0 1) LAH

ci'H 2) H20

3) TsCI, py 4) NaOEt

c) 0

1) LAH

ci'H ()0 2) H20

3) TsCI, py 4) NaOEt

5) 0 3 6) OMS

d) 0 1) MeMgBr

ci'H 2) H20 0 3) TsCI, py

4) 1-BuOK

e) 0

1) NaOH ~H ~CI

2) PCC, CH2CI2

f) 1) NaOH

0

~CI 2) PCC, CH2CI2 ~ 3) MeMgBr

4) H20

5) Na2Cr20 7 , H2S0 4 , H20



10 of 10 12/23/2011 3:41 AM

306 C IIAPTER J3

g)

0 1) dilute H2S04 0 0

2) Na2Cr20 7 , H2S04 , H20

h) 1) dilute H2S04

d OH 0 2) Na2Cr20 7 , H2S04 , H20

3) MeMgBr

4)H20

i) 1) dilute H2S04

0 2) Na2Cr207 , H2S04 , H20 (( 3) MeMgBr

4) cone. H2S04, heat

j )

1) HgS04, H2S04, H20 OH

¢;- + / 2) MeMgBr

3)H20

k) 1) dilute H2S04

~ 2) Na2Cr20 7, H2S04, H20

3) MeMgBr ~ 4) H20

I)

0 1) EtMgBr OH

~H 2)H20 ~

m) 0

1) LAH

~ 2) cone. H2S04 ry



1 of 10 12/23/2011 3:42 AM

n)

0

)y

o)

0

)l_,

p)

0

/y

q) 0

~

r ) 0

~H

s)

1) LAH

2) H20

3) TsCI, Et3N ~ 4) I·BuOK

1) LAH 2) H20

HOI 3) TsCI, Et:JN

4) I·BuOK

5) BH3 ·THF

6) H20 2, NaOH

1) MeMgBr

2) H20

3) cone. H2S04. heat ~ 1) EtMgBr

2) H~

1) LAH

OH

* 2) H20 ~Br

3) PBr3

OH

2) H20 2, NaOH ~ 3) PCC, CH2CI2

4) MeMgBr

5) H20

CHA PTER 13 307



2 of 10 12/23/2011 3:42 AM

308 CHAPTER 13

13.53.

HO~

~ 13.54. /'--../OH

13.55. OH

~

13.56. ~OH

v 13.57

-

:OH

~ HO:

e.. H :O

·c.~e :O t 0: .. ~ .. H •• 8

o:



3 of 10 12/23/2011 3:42 AM

13.59

0 1) Br2, hv

2) NaOMe

Pt

-H e .. I 8 <±l :~ CH3

H-~:~gBr r cH3

·o· 0 ··· r.··. ~t\ 0 .

2) H20 2, NaOH

.. e o:

3) Na2Cr20 7, H2S04, H20

CHAPTER 13 309

J

1) MeMgBr

2) H20



4 of 10 12/23/2011 3:42 AM

310 CHAPTER 13

13.60.

:o·· Sl_ '::o:( .. o( \_H

lA H- At=:-H

I H

13.61.

-

0 ,--....... (>, II · ·. . · · H'-·0- S- 0-H HO .. OH II K o

:J-t

. . H,(±) HO: :o H

K

:o··

OptH-IA

H- At=:-H I H

· HP

.. 0 :o:)

,f' i• ·H .o-:-t .. H

3 .• j . cH o:e

13.62. One carbon atom is oxidized from an oxidation state of+ l to an oxidation state of +2, while the other carbon atom is reduced from an oxidation state of+ I to an oxidation state of 0. Overall, the starting material does not undergo a net change in oxidation state and is, therefore, neither reduced nor oxidized.



5 of 10 12/23/2011 3:42 AM

Review of Concepts

Chapter 14 Ethers and Epoxides;

Thiols and Sulfides

Fi II in the blanks below. To verify that your answers are correct, look in your textbook at the end o f Chapter 14. Each of the sentences below appears verbatim in the sectio n entitled Review of Concepts and Voc:ahulary.

• Ethers are often used as for organic reactions. • Cyclic polyethers, or ethers, are capable of solvating metal ions in

organic (nonpolar) solvents. • Ethers can be readily prepared from the reaction between am alkoxide ion and an ----------• a process called a Williamson ether synthesis. This process works best for or alkyl haUdes. ______ alkyl halides are significantly less efficient, and -------alkyl halides cannot be used.

• When treated with a strong acid, an ether will undergo acidic------ in which it is converted into two alkyl haUdes.

• When a phenyl ether is cleaved under acidic conditions, the products are ______ and an alkyl halide.

• Ethers undergo autooxidation Ln the presence of atmospheric oxygen to form

• Substituted oxiranes are also called --------• can be converted into epoxides by treatme nt with peroxy acids

or via halohydrin formation and subsequent e poxidation. • catalysts can be used to achieve the enantioselective epoxidation

of aLiylic alcohols. • Epoxides will undergo ring-opening reactions in: I) conditions involving a

strong nucleophile, or under 2) __ -catalyzed conditions. When a strong nucleophile is used, the nucleophile attacks al the ___ -substituted position.

• Sulfur analogs of alcohols contain an SH group rather than an OH group, and are called ____ _

• Thiols can be prepared via an SN2 reaction between sodium hydrosulfide (NaSH) and a suitable-----------

• The sulfur analogs of ethers (thioethers) are called _____ _ • Sulfides can be prepared from thiols in a process that is essentially the sulfur

analog of the Williamson ether synthesis, involving a ion, rather than an alkoxide.



6 of 10 12/23/2011 3:42 AM

312 C HAPTER 14

Review of Skills Fill in the blanks and empty boxes below. To veri fy that your answers are correct, look in your textbook at the end of Chapter l4. The answers appear in the section entitled SkillBuilder Review.

14.1 Naming an Ether

PROVIDE A SYSTEMATIC NAME FOR THE FOLLOWING COMPOUND

('yO'--'"' ct-t-'

Cl

I) IOENTIPY r HE P~RENT

2/ IDENTIFY A NO NAME SUSSTtTUEN TS

3' ASSIGN I.OGANTS TO EAC H SUBSTJTUENT

4J ALPHABETIZE

5} ASSIGN CONFIGURATION

14.2 Preparing an Ether via a Williamson Ether Synthesis

IDENTIFY REAGENTS THAT WILL. ACHIEVE THE FOLLOWIN(J TRANSFORMATION:

~OH

v 1)

2)

14.3 Prer1a ring E poxides

IDENTIFY REAGENTS THAT WII.L ACHIEIIE THE FOLLOWING TRANSFORMATION.

/ "" ~:~ 0

) M~''1Et

"" 1)

;f Me

2) )

14.4 Drawing the Mecha nism a nd Predicting the Product of the Reaction between a Strong Nucleophile and an Epoxide

COMPLETE THE MECHANISM BELOW BY DRAWING ALL CURVED ARROWS. INTERMEDIATES AND PRODUCTS.

NaCN



7 of 10 12/23/2011 3:42 AM

CHAPTER 14 313

14.5 Drawing the Mechanism and Predicting the Product of Acid-Catalyzed Ring-Opening

COMPLETE THE MECHANISM BELOW BY DRAWING ALL CURVED ARROWS, INTERMEDIATES AND PRODUCTS .

:o·· H- OS03H DE6H .. o" D Et"'n ''Et ~ ~

Me H

14.6 Ins ta lling Two Adjac.ent Functional Groups

IDENTIFY WHETHER EACH RING·OPENING REACTION BELOW REQUIRES ACIDIC CONDITIONS OR BASIC CONDITIONS:

) CONDITIONS HO

R~ R X R

)= R

MCPBA 0

R--P R

) CONDITIONS

R OH R--j-1

X

14.7 Choosing the Appropriate Grignard Reaction

IOENT/FY REAGENTS THAT WILL ACHIEVE EACH OF THE FOtLOWING TRANSFORMA T/ONS:

R- Br

1) [ ) OH

RA_ R- Br

1) [ )

~~o· 3) [ )

Review of Reactions Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook at the end of Chapter 14 . The answers appear in the section entitled Review of Reactions.

Preparation of Ethers

Williamson ether synthesis

R- OH R- 0 - R

A lkoxyme rcuration-deme rcu ration

R H

>=< R R



8 of 10 12/23/2011 3:42 AM

3 14 CHAPTER14

Reactiot1S of E thers

Acidic cleavage

R- 0 - R R- X +

Q-o-R Autooxidation

Preparation of Epoxides

H HI / " >==< R R

"' cis /

Enantioselectivc Epoxidation

R~OH

Ring-Opening Reactio ns of Epoxides

Strong Nucleophile

r OH

RO;-\-OH

Hsrf-OH

R;-f-

Q-oH +

OOH

/"oA__

0

H1'M''1

H R R

cis

0

R- X

P·~ R '-OH

Acid-Catalyzed

OH

H;-f--HO~

X

H0'---0 OH

HO'--t-

OR



9 of 10 12/23/2011 3:42 AM

Tbiols and utndcs

Thiol\ Br

ASH + ASH

Su/fidt'l A- SH

Solutions

14.1. a) 2-ethoxypropane b) (S)-2-chloro-1-ethoxypropane c) 2.4-dichloro-1-etho>.} benzene

/

d) (I R.2R)-2-ethoxycyclohexanol e) 1-etho>. ycyclohe\ene

14.2.

a )

~EI

b )

v oY

CHAPTER 14 315

A-S-S-A

--~/

A- S- A

e + X

0 II

A- S- A to

0

MEvertson

Highlight

Solutions



10 of 10 12/23/2011 3:42 AM

316 CH APTER 14

14.3.

Common names in parentheses:

/0~

1· methoxybutane (butyl methyl ether)

1-elhoxypropane (ethyl propyl ether)

1-methoxy-2-methylpropane

(isobutyl methyl ether)

14.4.

Br d KF benzene

a) 18-Crown-6

Br d UF benzene

c) 12-Crown-4

14.5.

/oj-: ------ Chirality center

2·methoxybutane (sec-butyl methyl ether)

2-atMxyplopaM (ethyl isopropyl ether)

2-methoxy-2-methylpropane

(tart-butyl methyl ether)

F Br NaF

d b)

~ benzene

15-Crown-5

F

d KMn04 a- benzene

d) 18-Crown-6

F

d cPOH

a) A Williamson ether synthesis will be more efficient with a less sterically hindered substrate. since the process involves an SN2 reaction. Therefore, in this case, it is berter to stan with a secondary alcohol and a primary alkyl halide, rather than a primary alcohol and a secondary alkyl halide:

1) Na 00~, 2) I~



1 of 10 12/23/2011 3:43 AM

CHA PTER 1.4 3 17

b) In this case, it is better to start with a secondary alcohol and a primary alkyl halide, rather than a primary alcohol and a secondary alkyl halide:

uo~ HO~ 1) Na

2)UI c) In this case, it is better to start with a tertiary alcohol and a methyl halide, rather than methanol and a tertiary alkyl halide:

d OH 1)Na d OMe 2) Mel

14.6. Cli:t NaH

14.7. No. The Williamson ether synthesis employs an SN2 process, which cannot occur readily at tertiary or vinylic positions. Making this ether would require at least one of these two processes, neither of which can be used.

14.8. a)

~ 1) Hg(OAc)2, EtOH ~Et 2)NaBH4

b) OH

~--( ~ 1) Hg(0Ac)2 , A 2)NaBH4

c)

0 1) Hg(OAcb MeOH Q--OMe 2) NaBH4

d) HO

() 1) Hg(OAcb b d ob 2) NaBH4



2 of 10 12/23/2011 3:43 AM

318

14.9.

c 0

14.10.

~

14.11. a)

CHAPTER 14

1) Hg(0Ac)2, HO-o

2) Na8H4

1) BH3· THF

2) H20 2. NaOH

1) Hg(0Acl2, ~OH 2) NaBH4

~0~ HBr - 2 ~Br

HI - 1~1

d )

co ..6

HBr -Hl - r(YOH

~I

EtBr



3 of 10 12/23/2011 3:43 AM

.14.12.

HI -HBr -

~I (racemic mixture)

EtBr +

a) 2-methyl- 1,2-epoxypropane or 1,1-dlmethyloxirane b) I , 1-diphenyl- 1 ,2-epoxyethane or I , 1-diphenylox irane c) I ,2-epoxycyclohexane

14.13.

CHAPTER 14 319

a) (S)-2-phenyl- 1 ,2-epoxypropane or (S)-1-methyl-1-phenyloxirane b) (3R,4R)-3,4-epoxyheptane or (I R,2R)-I-ethyl-2-propyloxirane c) (2R,3S)-4-methyl-2,3-epoxypentane or (I S,2R)- I -i opropyl-2-methyloxirane

14.14. a)

vW b)

MCPBA

0

~ MCPBA O····A ··'H

c)

~ MCPBA

d)

D MCPBA

H Me

0 /''"J--\ ,,H 0J Me

Qjo

"H H



4 of 10 12/23/2011 3:43 AM

320 CHAPTER 14

14.15. This process for epoxide formation involves deprotonation of the hydroxyl group, followed by an intramolecular SN2 attack. The SN2 step requires back-side attack, which can only be achieved when both the hydroxyl group and the bromine occupy axial positions. Due to the steric bulk of a tert-butyl group, Compound A spends most of time in a chair conformation that has the tert-butyl group in an equatorial position. In this conformation, the OH and Br are indeed in axial positions, so the reaction can occur quite rapidly. In contrast, Compound B spends most of its time in a chair conformation in which the OH and Br occupy equatorial positions. The SN2 process cannot occur from this conformation.

14.16. a)

U OH

b)

~OH

c)

={_OH

d)

14.17.

-j-o- o - H

Ti[OCH{CH3)2)4 Cfi~OH

{+)-DET

- ifOH

-j-o - o - H

Ti[OCH(CH3bl4 {-) -DET

OH

- y

OH

X

-j-o - o - H

Ti[OCH(CH3)2]4 (+)-DET

-j-o - o - H

Ti[OCH(CH3)2)4 (- )-DET



5 of 10 12/23/2011 3:43 AM

CHAPTER 14 321

b )

e ""o: N:=c:~ -

Me

c)

.. e ~9: H~:~ '\ -

~OH HS i .. Me Me

d)

e)

e.(" :o: H ,, .. , \_ O~Me

Et :~H

t)

q.~ .. .o: !i "-... ·o ·

0 L....;;.Me "' , ... , - \ H \f H Et H

.------... :o·· .r.e i·f'"'v'H :o: :OH

A A

HO: H

0 \_:: Me

""1 \ Et :SH

HO: H \'''£ --\_ O~Me

Et H

OH

- ut no

chirality centers

14.18. The reaction yields a meso compound, regardless of which electrophil ic position is attacked by hydroxide.

NaOH HO Me

>-< Me' OH

HO OH )-......(

Me'" /Me

meso



6 of 10 12/23/2011 3:43 AM

322 CHAPTER 14

14.19. 0 NaOH

Me~Me H20

HO Me

Mr-<OH +

Ill OH

Jy-OH

enantiomers

14 .. 20. a)

· . ,r--... H-C.I' ·o· . D ..

c)

H :(f1?. .. e L..X' + :CI :

"--__)'

H e' · .. e ~o· ·s. ~ + ·. !· ~

Me>---tH

HO Me

Ill OH

+ OH

EIOH



7 of 10 12/23/2011 3:43 AM

d)

e)

0

:o··~ ~·

0 _u_ H-'-Br :

'\''J \ ''JH •• Et Me

H

(±)~t. .. ~ . ·o· Me'''A \ ·•H H' ' H

Et Me

H <Dot .

r • ~

0'"1t\'~'~H + : ~·r ~ Et~·

CHAPTER 14 323



8 of 10 12/23/2011 3:43 AM

324 CHAPTER 14

14.21.

14.22. a)

CJ'sr

b)

(J'Br

c)

~r ~

14.23. a)

0 SH

b)

Br

OH

NaSH

NaSH

NaSH

1) NaOH

2) ,..............Br

............... SNa

H ~6·.

Et1''~ '''H Me

C/"sH

()-SH

SH



9 of 10 12/23/2011 3:43 AM

CHAPTER 14 325

c)

14.U . a)

~ 1) MCPBA

H N HO 2) NaCN

3) H20

b ) 1) MCPBA

/ 2)NaOH OH

0 1) Br2. hv 0 3) H20 •• >OH

2) NaOEI + En

1) MCPBA

2) H3o·

c) OH

~ 1) MCPBA ~SH 2) NaSH

3) H20

d )

~ 1) MCPBA

2) (H2S04l, H2S ~ OH

c)

1) LAH OH 0

~ ~ Na2Cr~7 ~ 2)H~ H~4. H20



10 of 10 12/23/2011 3:43 AM

326 CHAPTER 14

14.25. I·BuOK

~I -:::Y

14.26. a)

1) Mg

0Br 2) 6

3)H20

b)

c)

d) 1) Mg

2):K 0

Br

3) H20

e) 1) Mg

!TYCI

v 2) 6

1) MCPBA

2) H3o•

U OH

QI\OH

OH

dY

~OH

v

HO~OH

Na2Cr20 7

H2S0 4, H20

SOCI2

py

0

dY

~CI

v



1 of 10 12/23/2011 3:44 AM

f)

0('CI

v

g)

h)

O,..CI

I

i)

14.27.

1) Mg

2)>-<

1) MCPBA

2) PrMgBr

3) H20

Undlar's Catalyst

~OH

SOCI2

py

~OH VA

OH

~

I ~

\ 1) MCPBA 2) H3Q+ ..-

CHAPTER 14 327

Cl

~



2 of 10 12/23/2011 3:44 AM

328 CHAPTER 14

14.28.

4-H2

Lindlar's Catalyst

14.29.

?> 1) MeMgBr

2) H20

14.30

OH v PCC 0

~H

1) Na (2 eq)

1) EtMgBr

2 } H20

3} NazCr207, H2S04 , H20

1) NaH

2) Etl

a) (IS. 2S)-J-ethoxy-2-methylcyclohexane b) (R)-2-ethoxybutane c) (S)-3-hexanethiol d) ethyl propyl sul foxide e) (£)-2-ethoxy-3-methyl-2-penteue t) I ,2-dimethoxybenzene g) ethyl propyl sulfide

14.31 a)

()0" HBr ()Br

b) ao" I HBr V OH

.;;::;-

c)

~0~ HBr 2 )-sr

d)

1 0 HBr Br~Br

~

+ CH3Br + H20

+ CH3Br + H20

+ H20

+ H20

0

~

j 1) EtMgBr

2) H20



3 of 10 12/23/2011 3:44 AM

14.32

/0~

1·methoxypropane (methyl propyl ether)

2-methoxypropane (Isopropyl methyl ether)

14.33 a)

0 b)

0 c)

0 14.34 a)

1) Hg(0Ac)2, EtOH

2) NaBH4

1) MCPBA

2) MeOH, [H2S04]

HOX 1) Hg(OAcb,

2) NaBH4

Comp ound A

0 0 '-/'

o:OH

'OMe

()o X

CHAPTE R 14 329

ethoxyethane (dlethyl ether)

b) Two moles of Compound A are produced for every one mole of I ,4-dioxane.



4 of 10 12/23/2011 3:44 AM

330 C HAPTER 14

c)

:o··~ r •.. c ) H---;1,:

·.0:

H 0 'o

C~)/j.C?J + ·.I.·-

·.0:

'I- e .. .. f .· ·····.-~··~ _r•: . . " ,-.... "

(

+ :? ---:--:- • I 15 • H ,_,. , .

H

"~'' j

14.35

(

''"" /"-- _QH H~r "-/ "-/ ·.

14.36

14.37

.1'1 " " :'I- H -~·I~ ,-.1: .. ·~ '-- _) ..

·.0:

:o·· C)

H03SO~H~H .. ..

a) Neither alkyl group (on either side of the oxygen atom) can be installed via a Williamson ether synthesis. Installation of the tert-butyl group would require a tertiary alkyl halide, which is too sterically hindered to serve as an electrophile for an SN2 process. Installation of the phenyl group would require an SN2 reaction taking p lace at an s/ hybridized center, which does not readi ly occur.



5 of 10 12/23/2011 3:44 AM

CHAPTER14 33 1

b) Oxymercuration-demercuration can be used to prepare tert-butyl phenyl ether:

I 1) Hg(0Ac)2,

~ 2)NaBH4

o-'\; OH

14.38 Ethylene oxide has a high degree of ring strain, and readily functions as an electrophile in an SN2 reaction. The reaction opens the ring and alleviates the ring strain. Oxetane has less ring strain and is, therefore, less reactive as an electrophile towards SN2. The reaction can still occur, albeit at a slower rate, to alleviate the ring strain associated with the four membered ring. THF has almost no ring strain (very little) and does not function as an electrophile in an SN2 reaction.

14.39

#' 1} NaNH2 / 2) Mel #'

J 1) "'""' 2) Mel

H2 / Na r F\ #' Lindlar's / NH3 I MCPBA

Catalyst

l MCPBA

0 0

~ ,,.·~

14.40

HO~Br NaH -b



6 of 10 12/23/2011 3:44 AM

332 CHAPTER 14

14.41 a)

# 1) NaNH2

~ 2)~Br

b) 1) NaNH2

2)~Br

c) 1) NaNH2 1) NaNH2

-¢:- -¢:-2)Etl 2) Mel ~

d) 1) NaNH2 1) NaNH2

-¢:-2) Ell -:P 2) Ell

~

1) NaNH2

2) Mel

1) NaNH2

2) Ell

/ -¢:-~

/'--.... -¢:-~

11

) ~~d lar's cat.

2) MCPBA

0

~~~H +En

j 1) Na, NH3

2) MCPBA

0

ry-f\~'Et +En

1) H2 0 - H1'A '"H Lindlar's Catalyst Et Me 2) MCPBA +En

1) Na, NH3 0

2) MCPBA H' 'A '''Et Et H

+En



7 of 10 12/23/2011 3:44 AM

CliAPTER 14 333

1-'.42 a)

~ 1) AC03H

2) MeMgBr ~ 3) H20 OH

b)

~ 1) Hg(0Ac)2 , MeOH

2) NasH. ~ OMe

(racemiC) c)

~ 1) MCPBA

2) NaSH ~SH OH

(racemic) d)

o-oH 1) Na

o-oet 2) EtCI

e)

o-oH 1) Na Q"0~0H 0 2) L\ 3) H20

f)

o-cl 1) Mg, dtethyl ether ey--oH

2) 13. 3) H20

g)

o-oH 1) Na Q"0

Y OH 2) 0

-F 3) H~

h)

o-cl 1) Mg, diethyl ether ~OH 2) 0

3) H20~



8 of 10 12/23/2011 3:44 AM

334 CHAPTER 14

14.43 a)

c)

. . 8 -1\./--: c=c- H

d)

e)

e r--.....___ .. . •·• " ;o.:._

- ~ H V H

{) ~.He Cl~q:-H - ·- --

f)

-j-.)OH MeS



9 of 10 12/23/2011 3:44 AM

14.44

6 14.45

0

b

14.47 a)

0 b)

(Y

HBr ~Br H20 + Br

DBr HBr

Br

1) Hg{OAch , MeOH

2) NaBH4

1) Hg(OAc)2 , MeOH

2) NaBH4

Et

G~ ~~ .e

o: ,-0' -

:fl:

Et ; Me

CtOH

Et

~OMe

d oMe

C HAPTE R 14 335

Et

~~. ~··

(_

J

H~ I I e (±} H-c-c : MgBr

I I H H



10 of 10 12/23/2011 3:44 AM

336 CHAPTER 14

c)

Q-oH

0 1) Hg(OAcb •

0-o-Q 2) NaBH4

d )

Q-oH

(r 1) Hg(0Ac)2 ,

d-o -Q 2) NaBH4

14.48

a) 1) TMSCI, Et3N

# 1) NaNH2 ~OH 2) NaNH2 ~OH

2) 'C> # 3) 'C>

# Ho/'..../

3) H20 4) H20 j H, 5) TBAF Lindlar's Catalyst

HO~OH

b) 1) TMSCI, Et3N

# 1) NaNH2 ~OH 2) NaNH2 ~OH

2) 'C> #

3) 'C> # HO/'..../

3) H20 4) H20

j H2

5) TBAF Pt

0

H~H PCC

HO~OH CH2CI2

0



1 of 10 12/23/2011 3:45 AM

CHAPTER 14 337

14.49

OH

d'oEt

~ J

MCPBA ~tMgBr /2) H20

1) NaOEt

MeOH cfaH

1) NaH

2) Ell

cFsH OH

d'c N

OH

a-

14.50

0 1) Hg(OAch , EtOH

2) NaBH4

) "c~' /

O O 1)NaSH

2) H20

HI -heat

.. o:OH

'C, + En " C

o:OH

+ En 'SH

HBr o:OH

+ En 'Br

'H



2 of 10 12/23/2011 3:45 AM

338 C IIAPTER14

14.51 a)

MCPBA 1) MeMgBr ~ ~ 2) H20 ~

OH

b) MCPBA 1) MeMgBr N~Cr207

~ ~ 2) H20 ~ H~04. H20 --Tf' OH 0

c) l)Mg soc~

~CI 2) ?.> ~OH py ~CI

3) H20

d)

1)Mg OH Cl

~ SOCiz ~ ~CI 2) ll____ py

3) H20

e)

X oH

1) NaH

X o /'.... 2) Eli

f)

0 0H 1) NaH 00~0H 2) <). 3)Hz0

g)

~ 1) BH3·THF 1) NaH

2) H20 2, NaOH ~OH 2) Ell ~o/'....



3 of 10 12/23/2011 3:45 AM

h)

~

i)

~ OH

1} NaH

2) Eli

CHAPTER 14 339

+ En

/ ¢;-

H2 MCPBA 0

~ ~ + En Lindlar's Catalyst

j) / H2 MCPBA 0 ¢;- F\ / -\. / Lindlar's

Catalyst

k)

#/ Na MCPBA 0 ~ +En /~ NH3 r ·· ... ,

I) H2 MCPBA 1) EtMgBr

~ - ~ ~ ~ Llndlar's 2) H20 OH

Catalyst

m)

0 / OH 0~

0 HI 6 1) NaOH 6 2) Ell

n)

()oD HBr ()Br 1)Mg U OH

2) <> 3) H20



4 of 10 12/23/2011 3:45 AM

340 CHAf>TER J4

1) NaH o)

~OH

2) <> 3) HzO

~O~OH

p)

~CI

q)

~CI

r)

()OH

s)

()OH

t)

()

u)

()

1) Mg

2) <c> ~OH

3) H20

1) Mg

2) <c> ~OH

3) H20

[H2S04]

-F

1) NaH

2) <[>( 3) H20

1) Br2. hv MCPBA

2) NaOEt

1) Br2, hv

2) NaOEt (:6~ +En

3) MCPBA

0

~H

o:· oH +En

OMe



5 of 10 12/23/2011 3:45 AM

14.52.

14.53. ~0~

0

14.54. 0

14.55. ~0~

14.56 0

M:::.tl"'H U Me

14.57

1) LiAI04

HO Me Me~j--\''H U D

.. e H :o; ~ u :X:

CHAPTER 14 341

14.58 When methyloxirane is treated with HBr, the regiochemical outcome is determined by a competition between steric and electronic factors, with steric factors prevailing- the Br is positioned at the less substituted position. However, when phenyloxirane is treated with HBr, electronic factors prevail in controlling the regiochemical outcome. Specifically, the position next to the phenyl group is a benzylic position and can stabilize a large partial positive charge. ln such a case, electronic factors are more powerful than steric factors, and the Br is positioned at the more substituted position.



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342 C HAPTER J4

14.59.

14.60

~Br

0

1) MCPBA

2) NaOEt 2) PrMgBr

3) H20

OH

~ + Enantiomer

1) Mg ~OH

0

~ ~ t1 (J ; ~ ~ J MCPBA

+ Enantiomer

~ 0

v j 1) MCPBA

2) PhMgBr

3) H20

14.61 Since the Grignard reagent is both a strong base and a strong nucleophile, substitution and elimination can both occur. Indeed, they compete with each other. As we discussed in Chapter 8, elimination will be favored when the substrate is secondary. Both electrophilic positions in this epoxide are secondary, and so, elimination predominates:



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Chapter 15 Infrared Spectroscopy and Mass Spectrometry

Review of Concepts FUJ in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 15. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.

• Spectroscopy is the study of the interaction between and ___ _ • The difference in energy (t.E) between vibrational energy levels is determined

by the nature of the bond. If a photon of light possesses exactJy th.is amount of energy. the bond can absorb the photon to promote a --------excitation.

• JR spectroscopy can be used to identify which are present in a compound.

• The location of each signal in an fR spectrum is reported in terms of a frequency-related unit called --------

• The wavenumber of each signal is determined primarily by bond-----and the of the atoms sharing the bond.

• The intensity of a signal is dependent on the of the bond giving rise to the signal.

• C=C bonds do not produce signals. • Primary amines exhibit two signals resulting from _____ stretching

and stretching. • Mass spectrometry is used to determine the and -----------of a compound.

• Electron impact ionization (EI) involves bombarding the compound witJ1 high energy , generating a radical cation that is symbolized by (M ) .... and is called the molecular ion, or the ion.

• Only t11e molecular ion and the cationic fragments are deflected, and they are then separated by t11eir (m/.::).

• The tallest peak in a mass spectrum is assigned a relative value of I 00% and is called the peak.

• The relative he ights of the (M t ' peak and the (M+ It' peak indicates the number of ________ _

• A signal at M- I 5 indicates the loss of a group: a signal at M-29 indicates the loss of an group.

• alkanes have a molecular formula of the form CnH2n+2·

• Each double bond and each ring represents one degree of-------

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