Chapter 16 homework solutions PDF

Preview

Summary

Chapter 16 homework solutions...

Description

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Chapter 16: Aldehydes and Ketones: 16-2 Structure and Nomenclature Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning

16.14. Name each compound, showing stereochemistry where relevant. a.

b.

c.

d.

e.

f.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

g.

h.

i.

16.15. Draw a structural formulas for each compound. a. 1-Chloro-2-propanone

b. 3-Hydroxybutanal

c. 4-Hydroxy-4-methyl-2-pentanone

d. 3-Methyl-3-phenylbutanal

e. 1,3-Cyclohexanedione

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

f. 3-Methyl-3-buten-2-one

g. 5-Oxohexanal

h. 2,2-Dimethylcyclohexanecarbaldehyde

i. 3-Oxobutanoic acid

16.16. The infrared spectrum of Compound A, peak at

. From this information and its

, shows a strong, sharp spectrum, deduce

the structure of compound A. Compound A:

Compound A has an aldehyde or ketone function judging from the peak at

in the IR spectrum. The rest of the structure can be

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

deduced from the

The

, with assignments as listed below.

hydrogens with signals at

and

above are not

equivalent because they are adjacent to a tetrahedral chiral center in the molecule. 16.17. Following are

spectra for compounds B

, and C

. On warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds B and C. Compound B:

Compound C:

From the molecular formulas it is clear that compound B undergoes acid-catalyzed dehydration to create compound C. Thus, compound B must have an

group. Furthermore, because its index of hydrogen

deficiency is one, compound B must have one ring or

bond. However,

compound C has an index of hydrogen deficiency of two, so it must

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

have two

bonds or rings. The

spectrum of compound B

shows all singlets. Especially helpful are the methyl group resonances; the singlet integrating to

at

and the singlet integrating to

at

. This latter signal is assigned as a methyl ketone by comparison to the previous problem. The other two methyl groups are equivalent. There is the

hydrogen at

and a

resonance at

. The only structure consistent with these signals is 4-hydroxy4-methyl-2-pentanone.

Upon dehydration, Compound B would be converted to 4-methyl3-pentene-2-one.

The structure of 4-methyl-3-pentene-2-one is entirely consistent with the spectrum, especially the presence of the signal at and the methyl singlets at and

(integrating to

(integrating to

)

).

Chapter 16: Aldehydes and Ketones: 16-2 Structure and Nomenclature Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning © 2021 Cengage Learning Inc. All rights reserved. No part of this work may by reproduced or used in any form or by any means graphic, electronic, or mechanical, or in any other manner - without the written permission of the copyright holder.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Chapter 16: Aldehydes and Ketones: 16-3 Addition of Carbon Nucleophiles Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning

16.18. Draw structural formulas for the product formed by treating each compound with propylmagnesium bromide followed by aqueous

.

The products after acid hydrolysis are given in bold. a.

b.

c.

d.

16.19. Suggest a synthesis for the following alcohols starting from an aldehyde or a ketone and an appropriate Grignard reagent. Below each target molecule is the number of combinations of Grignard reagent and aldehyde or ketone that might be used.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

a.

b.

c.

Note that each product is actually a racemic mixture. 16.20. Show how to synthesize the following alcohol using 1-bromopropane, propanal, and ethylene oxide as the only sources of carbon atoms.

This synthesis is divided into two stages. In the first stage, 1-bromopropane is treated with magnesium to form a Grignard reagent and then with propanal followed by hydrolysis in aqueous acid to give 3-hexanol.

In the second stage, 3-hexanol is treated with thionyl chloride followed by magnesium in ether to form a Grignard reagent. Treatment of this Grignard reagent with ethylene oxide followed by hydrolysis in aqueous acid gives 3-ethyl-1-hexanol. A chiral center is created in this synthesis.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

The product is a racemic mixture. Note that the alcohol could have also been treated with

followed by

to make a bromo Grignard.

16.21. 1-Phenyl-2-butanol is used in perfumery. Show how to synthesize this alcohol from bromobenzene, 1-butene, and any necessary inorganic reagents.

a. Bromobenzene is treated with magnesium in diethyl ether to form phenylmagnesium bromide, in preparation for part (c).

b. Treatment of 1-butene with a peroxycarboxylic acid gives 1,2epoxybutane.

c. Treatment of phenylmagnesium bromide with 1,2-epoxybutane followed by hydrolysis in aqueous acid gives 1-phenyl-2-butanol.

A chiral center is created in this synthesis. The product is a racemic mixture. A Gilman reagent could also have been used in this step. 16.22. With organolithium and organomagnesium compounds, approach to the carbonyl carbon from the less hindered direction is generally preferred. Assuming this is the case, predict the structure of the major product formed by reaction of methylmagnesium bromide with 4-tert-butylcyclohexanone. The bulky tert-butyl group lies in an equatorial position. Approach to the carbonyl carbon may be by way of a pseudo-axial direction or a pseudoequatorial direction. The less hindered approach is from the pseudo-

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

equatorial direction which then places the incoming group equatorial and the

axial.

Chapter 16: Aldehydes and Ketones: 16-3 Addition of Carbon Nucleophiles Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning © 2021 Cengage Learning Inc. All rights reserved. No part of this work may by reproduced or used in any form or by any means - graphic, electronic, or mechanical, or in any other manner - without the written permission of the copyright holder.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Chapter 16: Aldehydes and Ketones: 16-4 Wittig Reaction Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning

16.23. Draw structural formulas for (1) the alkyltriphenylphosphonium salt formed by treatment of each haloalkane with triphenylphosphine, (2) the phosphonium ylide formed by treatment of each phosphonium salt with butyllithium, and (3) the alkene formed by treatment of each phosphonium ylide with acetone. In the following, the triphenylphosphonium salts are listed as (1), the phosphonium ylides as (2), and the alkene formed upon reaction with acetone as (3). a.

b.

c.

d.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

e.

f.

16.24. Show how to bring about the following conversions using a Wittig reaction. a.

Start with 1-halopentane and, by treatment with triphenylphosphine followed by butyllithium, convert it to a Wittig ylide. Treatment of this ylide with acetone gives the desired alkene.

b.

Treatment of acetophenone with the Wittig ylide derived from methyl iodide gives the desired alkene.

c.

Treatment of cyclopentanone with the Wittig ylide derived from 3,4dimethoxybenzyl bromide gives the desired alkene.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

16.25. The Wittig reaction can be used for the synthesis of conjugated dienes, as for example, 1-phenyl-1,3-pentadiene.

Propose two sets of reagents that might be combined in a Wittig reaction to give this conjugated diene. Allylic halides can be used as starting materials for preparation of Wittig reagents. Below are combinations of allylic halides and aldehydes that can be used to prepare the desired diene.

These other sets of reagents may also be used as shown below.

16.26. Wittig reactions with the following

-chloroethers can be used for the

synthesis of aldehydes and ketones:

a. Draw the structure of the triphenylphosphonium salt and Wittig reagent formed from each chloroether.

b. Draw the structural formula of the product formed by treatment of each

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Wittig reagent with cyclopentanone. Note that the functional group is an enol ether, or, alternatively, a vinyl ether.

c. Draw the structural formula of the product formed on acid-catalyzed hydrolysis of each enol ether from part (b). The acid catalyzed hydrolysis leads to an aldehyde and methyl ketone for (A) and (B), respectively.

16.27. It is possible to generate sulfur ylides in a manner similar to that used to produce phosphonium ylides. For example, treating a sulfonium salt with a strong base gives the sulfur ylide.

Sulfur ylides react with ketones to give epoxides. Suggest a mechanism for this reaction.

Step 1: Make a bond between a nucleophile and an electrophile.

Step 2: Make a bond between a nucleophile and an electrophile and simultaneously break a bond to give stable molecules or ions.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

16.28. Propose a structural formula for Compound D and for the product, , formed in this reaction sequence.

The product of this transformation is an epoxide with both a threemembered and a six-membered ring bonded through so-called “spiro” attachments.

Chapter 16: Aldehydes and Ketones: 16-4 Wittig Reaction Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning © 2021 Cengage Learning Inc. All rights reserved. No part of this work may by reproduced or used in any form or by any means - graphic, electronic, or mechanical, or in any other manner - without the written permission of the copyright holder.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Chapter 16: Aldehydes and Ketones: 16-5 Addition of Oxygen Nucleophiles Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected]) © 2018 Cengage Learning, Cengage Learning

16.29. 5-Hydroxyhexanal forms a six-membered cyclic hemiacetal, which predominates at equilibrium in aqueous solution.

a. Draw a structural formula for this cyclic hemiacetal. 5-Hydroxyhexanal forms a six-membered cyclic hemiacetal.

b. How many stereoisomers are possible for 5-hydroxyhexanal? Two stereoisomers are possible for 5-hydroxyhexanal; a pair of enantiomers. c. How many stereoisomers are possible for this cyclic hemiacetal? Four stereoisomers are possible for the cyclic hemiacetal; two pairs of enantiomers. Following are planar hexagon formulas for each pair of enantiomers of the cyclic hemiacetal.

d. Draw alternative chair conformations for each stereoisomer and label groups axial or equatorial. Also predict which of the alternative chair conformations for each stereoisomer is the more stable. Alternative chair conformations are drawn for (A), one of the cis

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

enantiomers, and for (C), one of the trans enantiomers. Methyl and hydroxyl groups are

kJ/mol and

kJ/mol more stable in the

equatorial position, respectively. For (A), the diequatorial chair is the more stable by

kJ/mol. For (C), the chair with

the methyl group is equatorial (structure on the left) is kJ/mol more stable.

16.30. Draw structural formulas for the hemiacetal and then the acetal formed from each pair of reactants in the presence of an acid catalyst. a.

b.

c.

16.31. Draw structural formulas for the products of hydrolysis of the following acetals in aqueous

.

a.

b.

c.

Note that stereochemistry of this product will be that of the starting

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

material, because the configuration of the chiral center is not altered during the reaction. 16.32. Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If the carbonyl oxygen of 4-hydroxypentanal is enriched with oxygen-18, do you predict that the oxygen label appears in the cyclic acetal or in the water?

Step 1: Propose the initial formation of a hemiacetal, which is actually several steps.

Step 2: Add a proton. The hemiacetal is protonated on the

group.

Step 3: Break a bond to give stable molecules or ions. Water is lost to form a resonance-stabilized cation. If the carbonyl group of 4-hydroxypentanal is enriched with oxygen-18, the oxygen-18 label appears in the water that is released in this step.

Step 4: Make a bond between a nucleophile and an electrophile.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Step 5: Take a proton away.

16.33. Propose a mechanism for this acid-catalyzed hydrolysis.

Step 1: Add a proton. A reasonable mechanism for this reaction involves protonation of the

bond of the carbon-carbon double bond to give a

resonance stabilized cation intermediate.

Step 2: Make a bond between a nucleophile and an electrophile. Water reacts with the cation to form a hemiacetal after loss of a proton.

Step 3: Take a proton away.

Step 4: Add a proton.

Step 5: Break a bond to give stable molecules or ions. Methanol departs.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

Step 6: Take a proton away.

16.34. In Section 11.5 we saw that ethers, such as diethyl ether and tetrahydrofuran, are quite resistant to the action of dilute acids and require hot concentrated

or

for cleavage. However, acetals in which two ether

groups are linked to the same carbon undergo hydrolysis readily, even in dilute aqueous acid. How do you account for this marked difference in chemical reactivity toward dilute aqueous acid between ethers and acetals? The first step of the cleavage reactions in acid for both ethers and acetals is protonation of an oxygen to form an oxonium ion. For acetals, the following step is cleavage of a carbon-oxygen bond to form a resonance-stabilized cation. For ethers, similar cleavage occurs, but the cation formed has no comparable resonance stabilization. Therefore, it is the resonance-stabilization of the cation intermediate formed during the cleavage of acetals that lowers the activation energy for their hydrolysis much below that for the hydrolysis of ethers. 16.35. Show how to bring about the following conversion:

The most convenient way to convert an alkene to a glycol is to oxidize the alkene with osmium tetroxide in the presence of hydrogen peroxide. These conditions, however, will also oxidize an aldehyde to a carboxylic acid. Therefore, it is necessary to first protect the aldehyde by transformation to an acetal. In the following answer, ethylene glycol is the protecting agent.

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

16.36. A primary or secondary alcohol can be protected by conversion to its tetrahydropyranyl ether. Why is formation of THP ethers by this reaction limited to primary and secondary alcohols? It will help to refer to the mechanism given in Example 16.7. The concerns here are acid-catalyzed dehydration of the alcohol as a competing reaction and relative rates of nucleophilic reaction. Primary and secondary alcohols do not dehydrate nearly as readily as tertiary alcohols (due to relative carbocation stabilities). In addition, primary and secondary alcohols are relatively unhindered nucleophiles, while tertiary alcohols are so sterically hindered that they are sluggish nucleophiles. The net result is that tertiary alcohols dehydrate too readily, and react too slowly as nucleophiles, to be converted efficiently into tetrahydropyranyl ethers in acidic conditions. Primary and secondary alcohols, on the other hand, react fast enough as nucleophiles and dehydrate slowly enough to produce tetrahydropyranyl ethers in high yields with acid catalysis. 16.37. Which of these molecules will cyclize to give the insect pheromone frontalin?

The answer is B. Using models may help with this answer.

Chapter 16: Aldehydes and Ketones: 16-5 Addition of Oxygen Nucleophiles Book Title: Student Study Guide and Solutions Manual for Brown/Iverson/Anslyn/Foote’s Organic Chemistry Printed By: Clinton King ([email protected])

Print Preview

https://ng.cengage.com/static/nbreader/ui/apps/nbreader/print_preview/p

© 2018 Cengage Learning, Cengage Learning © 2021 Cengage Learning Inc. All rights reserved. No part of this work may by reproduced or used in any form or by any means - graphic, electronic, or mechanical, or in any other manner - without the written permission of the copyright holder.

Print Preview
...