ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
2nd Edition
ISBN: 9780393664034
Author: KARTY
Publisher: NORTON
Question
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Chapter 19, Problem 19.79P
Interpretation Introduction

(a)

Interpretation:

It is to be shown how to carry out the given synthesis.

Concept introduction:

The protection of the alcoholic OH group is done by the treatment with silyl chloride (either TMS-Cl or TBDMS-Cl) the protected form of alcohol is silyl ether (R-O-Si-). For deprotection of silyl ether, Bu4N+F- or concentrated HF acid is used. Alkyl halide on treatment with solid magnesium metal corresponding Grignard reagent is formed. Grignard reagent on treatment with an aldehyde corresponding secondary alcohol is formed as the product during the workup step while ketone produces tertiary alcohol.

Expert Solution
Check Mark

Answer to Problem 19.79P

The given synthesis is carried out as:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  1

Explanation of Solution

The given synthesis is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  2

It is noticed that from the given starting material, a tertiary alcohol is synthesized. So the required step-up reaction carried out by using a Grignard reagent and corresponding ketone. This synthesis is given below:

The given starting material is converted to corresponding Grignard reagent by the treatment with solid Mg metal in the ether as a solvent. But to avoid unwanted reactions of OH group with Mg metal this OH group protected first.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  3

Then the Grignard reaction is carried out with the corresponding ketone to form the required tertiary alcohol.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  4

Finally, the deprotection of the OH group is done to produce the final required compound.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  5

Conclusion

It is shown how to carry out the given synthesis.

Interpretation Introduction

(b)

Interpretation:

It is to be shown how to carry out the given synthesis.

Concept introduction:

The protection of the alcoholic OH group is done by the treatment with silyl chloride (either TMS-Cl or TBDMS-Cl) the protected form of alcohol is silyl ether (R-O-Si-). For deprotection of silyl ether, Bu4N+F- or concentrated HF acid is used. Alkyl halide on treatment with solid magnesium metal corresponding Grignard reagent is formed. Grignard reagent on treatment with an aldehyde corresponding secondary alcohol (note that formaldehyde produces primary alcohol) is formed as the product during the workup step while ketone produces tertiary alcohol.

Expert Solution
Check Mark

Answer to Problem 19.79P

The given synthesis is carried out as:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  6

Explanation of Solution

The given synthesis is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  7

It is noticed that only the bromine atom of the starting material is replaced by the CH2OH group in the product. So overall step-up reaction is carried out via Grignard reaction, the given starting compound is converted to corresponding Grignard reagent at bromine atom. For that purpose OH group of the starting material is protected first.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  8

Then the Br atom of the starting material subjected to Mg metal to form corresponding Grignard reagent.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  9

Above Grignard reagent is then treated with formaldehyde to carry out required step-up reaction via Grignard reaction. The acid workup step is needed for the Grignard reaction.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  10

Finally, the deprotection of the protected OH group in the first step is carried out to form the final product.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  11

Conclusion

It is shown how to carry out the given synthesis.

Interpretation Introduction

(c)

Interpretation:

It is to be shown how to carry out the given synthesis.

Concept introduction:

Alkyl halide on treatment with solid magnesium metal corresponding Grignard reagent is formed. Grignard reagent on treatment with an aldehyde corresponding secondary alcohol is formed as the product during the workup step while ketone produces tertiary alcohol. The specialty of the Grignard reagent is when it treated with carbon dioxide (CO2) produces corresponding carboxylic acid on the workup step. The protection of the carbonyl group of the ketone and aldehyde is carried out by the treatment with ethylene glycol in the presence of acid as the catalyst. The protected form of the carbonyl group is called acetal. For deprotection, simply acidic conditions required.

Expert Solution
Check Mark

Answer to Problem 19.79P

The given synthesis is carried out as:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  12

Explanation of Solution

The given synthesis is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  13

It is noticed that only a bromine atom of the starting material is replaced by a carboxylic acid group in the product. It is remembered that Grignard reagent when treated with carbon dioxide (CO2) forms corresponding carboxylic acid on the workup step. So the given synthesis is carried out below:

To carry out the required Grignard reaction the carbonyl group of the starting material is protected first to avoid an unwanted reaction.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  14

For the protection of ketone, it is treated with ethylene glycol in the presence of an acid catalyst. The protected form of the ketone is called ketal (a cyclic acetal).

Above protected ketal then treated with Mg metal in the ether as a solvent to form the corresponding Grignard reagent.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  15

The above Grignard reagent then treated with carbon dioxide to form the required product. Since acid workup step required in the Grignard reaction, the protected carbonyl group of the ketone is deprotected, too.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  16

Conclusion

It is shown how to carry out the given synthesis.

Interpretation Introduction

(d)

Interpretation:

It is to be shown how to carry out the given synthesis.

Concept introduction:

The catalytic hydrogenation of the alkene produces the corresponding alkane. ). 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. Enolate anion on treatment with an alkyl halide, the alkylation at α-carbon of the enolate anion takes place via an SN2 mechanism. A Wittig reaction is useful for the conversion of C=O bond of a ketone or aldehyde into a C=C bond.

Expert Solution
Check Mark

Answer to Problem 19.79P

The given synthesis is carried out as:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  17

Explanation of Solution

The given synthesis is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  18

To carry out the given synthesis following steps are carried out:

In the first step, the catalytic hydrogenation of the alkene group of the starting material is done as below:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  19

To reduce the only C=C bond of the given starting material selectively the catalytic hydrogenation with H2/ Pd is used. The methylation of the less sterically hindered α-carbon of the above product is done by using a sterically hindered base, LDA followed by the addition of CH3I. An enolate anion formed is the first step by an abstraction of less sterically hindered proton of the ketone. Which then adds to the electrophilic carbon of methyl iodide via SN2 mechanism.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  20

Finally, a Wittig reaction is carried out to produce the required product.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  21

In the above Wittig reaction, the oxygen atom of the ketone is replaced by CH2 group of the Wittig reagent.

Conclusion

It is shown how to carry out the given synthesis.

Interpretation Introduction

(e)

Interpretation:

It is to be shown how to carry out the given synthesis.

Concept introduction:

The protection of the carbonyl group of the ketone and aldehyde is carried out by the treatment with ethylene glycol in the presence of acid as the catalyst. The protected form of the carbonyl group is called acetal. For deprotection, simply acidic conditions are required. Catalytic hydrogenation of an alkyne with Lindlar catalyst always produces corresponding cis alkene.

Expert Solution
Check Mark

Answer to Problem 19.79P

The given synthesis is carried out as:

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  22

Explanation of Solution

The given synthesis is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  23

To carry out the given synthesis following steps are carried out:

In the first step, the protection of the carbonyl group of the given starting material is done by the treatment with ethylene glycol in the presence of an acid (e.g. HCl). The protected form of the corresponding aldehyde is called acetal. Otherwise unwanted nucleophilic addition will take place at electrophilic carbon of aldehyde carbon.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  24

Now the required nucleophilic addition is done to insert required moiety in the product.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  25

The triple bond of an above alkyne then subjected to catalytic hydrogenation with Lindlar catalyst to form the required cis alkene. Finally, a deprotection step carried out to produce the required product.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 19, Problem 19.79P , additional homework tip  26

Conclusion

It is shown how to carry out the given synthesis.

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Chapter 19 Solutions

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5

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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