ORG.CHEM W/TEXT+SOLU.MANUAL
ORG.CHEM W/TEXT+SOLU.MANUAL
15th Edition
ISBN: 9780393252125
Author: KARTY
Publisher: W.W.NORTON+CO.
Question
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Chapter 18, Problem 18.6P
Interpretation Introduction

(a)

Interpretation:

The complete and detailed mechanism of the given reaction is to be drawn, and the product is to be predicted.

Concept introduction:

There are two locations in an α,β-unsaturated carbonyl that are subject to a nucleophilic attack. One is the carbonyl carbon, leading to a 1, 2- or direct addition product. The second is the β-carbon, leading to a 1, 4- or conjugate addition product. Uncharged nucleophiles heavily favor conjugate addition over direct addition. An uncharged nucleophile favors conjugate addition because the first step of the mechanism, nucleophilic addition, is reversible and occurs under thermodynamic control. Conjugate addition involves a resonance-stabilized enolate anion. The addition of the uncharged nucleophile to α,β-unsaturated carbonyl compound does not require acidic condition. Although HCN (CN) is a charged nucleophile, it adds in a reversible manner.

Expert Solution
Check Mark

Answer to Problem 18.6P

The product of given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  1

The mechanism is drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  2

Explanation of Solution

Even though NC- is a charged nucleophile, it adds to an α,β-unsaturated carbonyl reversibly. Catalytic amount of KOH increases the concentration of NC- as a result of the acid-base reaction with HCN. NC- then adds to the β-carbon, shifting the π electron pair toward the carbonyl carbon and the carbonyl π electron pair to the oxygen atom.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  3

The negatively charged oxygen extracts a proton from the solvent molecule (H2O) to form a neutral enol form of the product. This proton transfer step involves a solvent molecule rather than HCN because of the basic conditions.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  4

The enol form being relatively unstable rapidly tautomerizes to the more stable keto form of the product.

Thus, the final product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  5

And the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  6

Conclusion

The mechanism and the product of the given reaction were determined on the basis of reversible addition of cyanide ion to the α,β-unsaturated carbonyl.

Interpretation Introduction

(b)

Interpretation:

The complete and detailed mechanism of the given reaction is to be drawn, and the product is to be predicted.

Concept introduction:

There are two locations in an α,β-unsaturated carbonyl that are subject to a nucleophilic attack. One is the carbonyl carbon, leading to a 1, 2- or direct addition product. The second is the β-carbon, leading to a 1, 4- or conjugate addition product. Uncharged nucleophiles heavily favor conjugate addition over direct addition. An uncharged nucleophile favors conjugate addition because the first step of the mechanism, nucleophilic addition, is reversible and occurs under thermodynamic control. Conjugate addition involves a resonance-stabilized enolate anion. The addition of the uncharged nucleophile to α,β-unsaturated carbonyl compound does not require acidic condition.

Expert Solution
Check Mark

Answer to Problem 18.6P

The product of the given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  7

The mechanism is drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  8

Explanation of Solution

PhCH2SH is an uncharged nucleophile that favors conjugate addition to an α,β-unsaturated carbonyl compound. The uncharged nucleophile adds reversibly at the β- carbon.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  9

Two proton transfer steps follow and produce an uncharged enol. The enol form is relatively unstable and rapidly tautomerizes to the more stable keto form.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  10

Thus, the final product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  11

And the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  12

Conclusion

The mechanism and the product of the given reaction were determined on the basis of reversible addition of the uncharged nucleophile to the α,β-unsaturated carbonyl.

Interpretation Introduction

(c)

Interpretation:

The complete and detailed mechanism of the given reaction is to be drawn and the product is to be predicted.

Concept introduction:

There are two locations in an α,β-unsaturated carbonyl that are subject to nucleophilic attack. One is the carbonyl carbon, leading to a 1, 2- or direct addition product. The second is the β-carbon, leading to a 1, 4- or conjugate addition product. Uncharged nucleophiles heavily favor conjugate addition over direct addition. An uncharged nucleophile favors conjugate addition because the first step of the mechanism, nucleophilic addition, is reversible and occurs under thermodynamic control. Conjugate addition involves a resonance-stabilized enolate anion. The addition of uncharged nucleophile to α,β-unsaturated carbonyl compound does not require acidic condition.

Expert Solution
Check Mark

Answer to Problem 18.6P

The product of given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  13

The mechanism of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  14

Explanation of Solution

Pyrrolidine is uncharged nucleophile and favors conjugate addition to an α,β-unsaturated carbonyl compound. The uncharged nucleophile adds reversibly to the β- carbon.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  15

After the nucleophilic addition, two proton transfer steps occur to produce an uncharged enol. The enol form is relatively unstable and rapidly tautomerizes to the more stable keto form.ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  16

Thus, the final product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  17

And the complete mechanism of the reaction can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 18, Problem 18.6P , additional homework tip  18

Conclusion

The mechanism and the product of the given reaction were determined on the basis of reversible addition of the uncharged nucleophile to the α,β-unsaturated carbonyl.

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Chapter 18 Solutions

ORG.CHEM W/TEXT+SOLU.MANUAL

Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - Prob. 18.23PCh. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - Prob. 18.26PCh. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31PCh. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - Prob. 18.35PCh. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - Prob. 18.39PCh. 18 - Prob. 18.40PCh. 18 - Prob. 18.41PCh. 18 - Prob. 18.42PCh. 18 - Prob. 18.43PCh. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - Prob. 18.46PCh. 18 - Prob. 18.47PCh. 18 - Prob. 18.48PCh. 18 - Prob. 18.49PCh. 18 - Prob. 18.50PCh. 18 - Prob. 18.51PCh. 18 - Prob. 18.52PCh. 18 - Prob. 18.53PCh. 18 - Prob. 18.54PCh. 18 - Prob. 18.55PCh. 18 - Prob. 18.56PCh. 18 - Prob. 18.57PCh. 18 - Prob. 18.58PCh. 18 - Prob. 18.59PCh. 18 - Prob. 18.60PCh. 18 - Prob. 18.61PCh. 18 - Prob. 18.62PCh. 18 - Prob. 18.63PCh. 18 - Prob. 18.64PCh. 18 - Prob. 18.65PCh. 18 - Prob. 18.66PCh. 18 - Prob. 18.67PCh. 18 - Prob. 18.68PCh. 18 - Prob. 18.69PCh. 18 - Prob. 18.70PCh. 18 - Prob. 18.71PCh. 18 - Prob. 18.72PCh. 18 - Prob. 18.73PCh. 18 - Prob. 18.74PCh. 18 - Prob. 18.75PCh. 18 - Prob. 18.76PCh. 18 - Prob. 18.77PCh. 18 - Prob. 18.78PCh. 18 - Prob. 18.79PCh. 18 - Prob. 18.80PCh. 18 - Prob. 18.81PCh. 18 - Prob. 18.82PCh. 18 - Prob. 18.83PCh. 18 - Prob. 18.84PCh. 18 - Prob. 18.85PCh. 18 - Prob. 18.86PCh. 18 - Prob. 18.87PCh. 18 - Prob. 18.88PCh. 18 - Prob. 18.89PCh. 18 - Prob. 18.90PCh. 18 - Prob. 18.91PCh. 18 - Prob. 18.92PCh. 18 - Prob. 18.93PCh. 18 - Prob. 18.94PCh. 18 - Prob. 18.95PCh. 18 - Prob. 18.96PCh. 18 - Prob. 18.97PCh. 18 - Prob. 18.98PCh. 18 - Prob. 18.99PCh. 18 - Prob. 18.100PCh. 18 - Prob. 18.101PCh. 18 - Prob. 18.102PCh. 18 - Prob. 18.103PCh. 18 - Prob. 18.1YTCh. 18 - Prob. 18.2YTCh. 18 - Prob. 18.3YTCh. 18 - Prob. 18.4YTCh. 18 - Prob. 18.5YTCh. 18 - Prob. 18.6YTCh. 18 - Prob. 18.7YTCh. 18 - Prob. 18.8YTCh. 18 - Prob. 18.9YTCh. 18 - Prob. 18.10YTCh. 18 - Prob. 18.11YTCh. 18 - Prob. 18.12YTCh. 18 - Prob. 18.13YTCh. 18 - Prob. 18.14YTCh. 18 - Prob. 18.15YT
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