Organic Chemistry: Principles And Mechanisms
Organic Chemistry: Principles And Mechanisms
2nd Edition
ISBN: 9780393663549
Author: KARTY, Joel
Publisher: W. W. Norton and Company
Question
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Chapter 19, Problem 19.73P
Interpretation Introduction

(a)

Interpretation:

It is to be shown how to carry out the given synthesis by using any reagents necessary.

Concept introduction:

Alkylation of carbonyl compound can take place at the α-carbon via nucleophilic substitution reaction. In this reaction acidic proton of the carbonyl compound is abstracted by the strong base an enolate anion is formed. The attack of this nucleophile takes place at the electrophilic carbon of the alkyl bromide and bromine atom is displaced through the SN2 mechanism. The reduction of the carbonyl group of a ketone to a methylene (CH2) group and an aldehyde to methyl (CH3) group by the treatment of hydrazine followed by heating under basic conditions is called Wolff-Kishner reduction. This method is useful for acid-sensitive compound (aldehyde or ketone). LDA is the strong non-nucleophilic base, due to bulky isopropyl groups. Therefore, it abstracts proton of less sterically hindered α-carbon of the carbonyl compound.

Expert Solution
Check Mark

Answer to Problem 19.73P

The given synthesis is carried out as:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  1

Explanation of Solution

The given synthesis:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  2

So the given synthesis is carried out below:

In the first step, the nucleophilic addition of the enolate anion to the ethyl bromide takes place shown below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  3

Note than LDA is sterically hindered strong base, therefore, abstract less sterically hindered proton selectively.

Finally, the carbonyl group is reduced to a corresponding methylene group (CH2) by Wolff-Kishner reduction to form the final required product.

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  4

Conclusion

It is shown how to carry out the given synthesis by using any reagents necessary.

Interpretation Introduction

(b)

Interpretation:

It is to be shown how to carry out the given synthesis by using any reagents necessary.

Concept introduction:

Alkylation of carbonyl compound can take place at the α carbon via nucleophilic substitution reaction. In this reaction acidic proton of the carbonyl compound is abstracted by the strong base an enolate anion is formed. The attack of this nucleophile takes place at the electrophilic carbon of the alkyl bromide and bromine atom is displaced through SN2 mechanism. The reduction of the carbonyl group of a ketone to a methylene (CH2) group and an aldehyde to methyl (CH3) group by the treatment of hydrazine followed by heating under basic conditions is called a Wolff-Kishner reduction. This method is useful for acid-sensitive compound (aldehyde or ketone).

Expert Solution
Check Mark

Answer to Problem 19.73P

The given synthesis is carried out as:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  5

Explanation of Solution

The given synthesis is

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  6

So the given synthesis is carried out below:

In the first step, the nucleophilic addition of the enolate anion to the ethyl bromide takes place shown below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  7

The require nucleophile is generated by the treatment of the strong base NaOH.

Finally, the carbonyl group is reduced to a corresponding methylene group (CH2) by Wolff-Kishner reduction to form the final required product.

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  8

Conclusion

It is shown how to carry out the given synthesis by using any reagents necessary.

Interpretation Introduction

(c)

Interpretation:

It is to be shown how to carry out the given synthesis by using any reagents necessary.

Concept introduction:

Lithium dialkylcuprate, (R)2CuLi, is called a Gilman reagent which possesses a weak nucleophile R-. Since R- is weak nucleophile adds to the β- carbon of the α,β- unsaturated carbonyl compound selectively (rather than the direct addition) and this addition is called conjugate or 1, 4-addtion. Alkylation of carbonyl compound can take place at the α carbon via nucleophilic substitution reaction. In this reaction acidic proton of the carbonyl compound is abstracted by the strong base an enolate anion is formed. The attack of this nucleophile takes place at the electrophilic carbon of the alkyl bromide and bromine atom is displaced through SN2 mechanism. The reduction of the carbonyl group of a ketone to a methylene (CH2) group and an aldehyde to methyl (CH3) group by the treatment of hydrazine followed by heating under basic conditions is called a Wolff-Kishner reduction. This method is useful for acid-sensitive compound (aldehyde or ketone).

Expert Solution
Check Mark

Answer to Problem 19.73P

The given synthesis is carried out as:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  9

Explanation of Solution

The given synthesis is

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  10

It is noticed that one ethyl group is added to the 3-position of the original compound, therefore, the conjugate addition must be needed.

So the given synthesis is carried out below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  11

The first conjugate addition is carried out by the treatment of lithium diethylcuprate, a Gilman reagent with the given starting material. Since the conjugate addition is irreversible to the carbonyl group and regenerated during the workup step.

The above product then treated with ethyl bromide in the presence of a strong base NaNH2. The alkylation takes place at α- carbon via nucleophilic substitution reaction.

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  12

Finally, the carbonyl group is reduced to a corresponding methyl group (CH3) by Wolff-Kishner reduction to form the final required product.

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  13

Conclusion

It is shown how to carry out the given synthesis by using any reagents necessary.

Interpretation Introduction

(d)

Interpretation:

It is to be shown how to carry out the given synthesis by using any reagents necessary.

Concept introduction:

LDA is the strong non-nucleophilic base, due to bulky isopropyl groups. Therefore, it abstract proton of less sterically hindered α-carbon of the carbonyl compound. The reduction of the carbonyl group of a ketone to a methylene (CH2) group and an aldehyde to methyl (CH3) group by the treatment of hydrazine followed by heating under basic conditions is called a Wolff-Kishner reduction. This method is useful for acid-sensitive compound (aldehyde or ketone). The protected alcohol in the form of sillylated ether (e.g. OTBDMS) are deprotected by using F- ion. Commonly for that purpose (Bu4N+F-) or HI are used.

Expert Solution
Check Mark

Answer to Problem 19.73P

The given synthesis is carried out as:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  14

Explanation of Solution

The given synthesis is

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  15

It is noticed that the phenyl group is added to the carbonyl carbon of the given starting material. So Grignard reaction carried out first as below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  16

Since acidic condition required in workup step of Grignard reaction, the protected carbonyl group deprotected automatically because in acidic condition acetal protection also deprotect and the carbonyl group regenerated (right to the above product).

The alkylation of right carbon of the regenerated carbonyl compound is needed and it is done by using the sterically hindered base, LDA followed by methyl bromide. The bromine atom is replaced by a nucleophile (an enolate anion) via SN2 mechanism. To carry out this operation alcoholic group must be protected because it will react with LDA.

The substrate is first treated with TBDMS-Cl to protect the OH group shown below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  17

Now the required alkylation is carried out as below:

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  18

Finally, the carbonyl group is reduced to a corresponding methylene group (CH2) by Wolff-Kishner reduction and the deprotection of the OH group is done to form the final required product.

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  19

Organic Chemistry: Principles And Mechanisms, Chapter 19, Problem 19.73P , additional homework tip  20

The protected OH in the form of OTBDMS is deprotected by the use of the source of F- ion, for that purpose (Bu4N+F-) is used.

Conclusion

It is shown how to carry out the given synthesis by using any reagents necessary.

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Chapter 19 Solutions

Organic Chemistry: Principles And Mechanisms

Ch. 19 - Prob. 19.11PCh. 19 - Prob. 19.12PCh. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Prob. 19.21PCh. 19 - Prob. 19.22PCh. 19 - Prob. 19.23PCh. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - Prob. 19.29PCh. 19 - Prob. 19.30PCh. 19 - Prob. 19.31PCh. 19 - Prob. 19.32PCh. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Prob. 19.37PCh. 19 - Prob. 19.38PCh. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - Prob. 19.70PCh. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.1YTCh. 19 - Prob. 19.2YTCh. 19 - Prob. 19.3YTCh. 19 - Prob. 19.4YTCh. 19 - Prob. 19.5YTCh. 19 - Prob. 19.6YTCh. 19 - Prob. 19.7YTCh. 19 - Prob. 19.8YTCh. 19 - Prob. 19.9YTCh. 19 - Prob. 19.10YT
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