EBK GET READY FOR ORGANIC CHEMISTRY
EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 9780100576377
Author: KARTY
Publisher: YUZU
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Chapter 17, Problem 17.53P
Interpretation Introduction

(a)

Interpretation:

The complete mechanism that proceeds when the given species treated with triphenylphosphine followed by butyllithium is to be drawn.

Concept introduction:

The Wittig reagents can be generated from alkyl halides. The alkyl halide is first reacted with triphenylphosphine and the product of that reaction is treated with a strong base. In the first step reaction fovors SN2 reaction in which the nucleophile Ph3P attacks on the carbon attached to the halide anion. The organic product accommodates the positive charge on the phosphorus atom due to the large size and its electronegativity. Finally, deprotonation carried by butyllithium. It is a very strong base and which is necessary because deprotonation takes place at a carbon atom. However, deprotonation is facilitated because the negatively charged carbon is stabilized by the adjacent positive charge on the P atom.

Expert Solution
Check Mark

Answer to Problem 17.53P

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  1

Explanation of Solution

The given alkyl halide is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  2

The Wittig reagent is generated from the given alkyl halide and triphenylphosphine followed by butyllithium. Triphenylphosphine, Ph3P is a very good nucleophile, so it favors SN2 reaction. Firstly, nucleophile treated with the alkyl halide and Br- departs from the substrate in the same step. The product accommodates the positive charge on the P atom.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  3

Lastly, the product obtained from the first step is treated with butyllithium to form Wittig reagent. CH3CH2CH2CH2Li, strong base abstracts the proton from a Carbon atom. The carbon atom that is deprotonated possesses the negative charge and it is stabilized by the positive charge on P.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  4

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  5

Conclusion

The complete mechanism for the given species treated with triphenylphosphine followed by butyllithium is drawn.

Interpretation Introduction

(b)

Interpretation:

The complete mechanism that proceeds when the given species treated with triphenylphosphine followed by butyllithium is to be drawn.

Concept introduction:

The Wittig reagents can be generated from alkyl halides. The alkyl halide is first reacted with triphenylphosphine and the product of that reaction is treated with a strong base. In the first step reaction fovors SN2 reaction in which the nucleophile Ph3P attacks on the carbon attached to the halide anion. The organic product accommodates the positive charge on the phosphorus atom due to the large size and its electronegativity. Finally, deprotonation carried by butyllithium. It is a very strong base and which is necessary because deprotonation takes place at a carbon atom. However, deprotonation is facilitated because the negatively charged carbon is stabilized by the adjacent positive charge on the P atom.

Expert Solution
Check Mark

Answer to Problem 17.53P

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  6

Explanation of Solution

The given alkyl halide is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  7

The Wittig reagent is generated from the given alkyl halide and triphenylphosphine followed by butyllithium. Triphenylphosphine, Ph3P is a very good nucleophile, so it favors SN2 reaction. Firstly, nucleophile treated with alkyl halide and the leaving group, Br- departs from the substrate in the same step. The product accommodates the positive charge on the P atom.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  8

Lastly, the product obtained from the first step is treated with butyllithium to form Wittig reagent. CH3CH2CH2CH2Li, strong base abstracts the proton from a Carbon atom. The carbon atom that is deprotonated possesses the negative charge and it is stabilized by the positive charge on P.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  9

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  10

Conclusion

The complete mechanism for the given species treated with triphenylphosphine followed by butyllithium is drawn.

Interpretation Introduction

(c)

Interpretation:

The complete mechanism that proceeds when the given species treated with triphenylphosphine followed by butyllithium is to be drawn.

Concept introduction:

The Wittig reagents can be generated from alkyl halides. The alkyl halide is first reacted with triphenylphosphine and the product of that reaction is treated with a strong base. In the first step reaction fovors SN2 reaction in which the nucleophile Ph3P attacks on the carbon attached to the halide anion. The organic product accommodates the positive charge on the phosphorus atom due to the large size and its electronegativity. Finally, deprotonation carried by butyllithium. It is a very strong base and which is necessary because deprotonation takes place at a carbon atom. However, deprotonation is facilitated because the negatively charged carbon is stabilized by the adjacent positive charge on the P atom.

Expert Solution
Check Mark

Answer to Problem 17.53P

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  11

Explanation of Solution

The given alkyl halide is

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  12

The Wittig reagent is generated from the given alkyl halide and triphenylphosphine followed by butyllithium. Triphenylphosphine, Ph3P is a very good nucleophile, so it favors SN2 reaction. Firstly, nucleophile treated with the alkyl halide and the leaving group, Cl- departs from the substrate in the same step. The product accommodates the positive charge on the P atom.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  13

After that, the product obtained from the first step is treated with butyllithium to form Wittig reagent. CH3CH2CH2CH2Li, strong base abstracts the proton from a Carbon atom. The carbon atom that is deprotonated possesses the negative charge and it is stabilized by the positive charge on P.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  14

The complete mechanism of the given alkyl halide is treated with triphenylphosphine followed by butyllithium as shown below.

EBK GET READY FOR ORGANIC CHEMISTRY, Chapter 17, Problem 17.53P , additional homework tip  15

Conclusion

The complete mechanism for the given species treated with triphenylphosphine followed by butyllithium is drawn.

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Chapter 17 Solutions

EBK GET READY FOR ORGANIC CHEMISTRY

Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - Prob. 17.26PCh. 17 - Prob. 17.27PCh. 17 - Prob. 17.28PCh. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - Prob. 17.34PCh. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - Prob. 17.38PCh. 17 - Prob. 17.39PCh. 17 - Prob. 17.40PCh. 17 - Prob. 17.41PCh. 17 - Prob. 17.42PCh. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Prob. 17.46PCh. 17 - Prob. 17.47PCh. 17 - Prob. 17.48PCh. 17 - Prob. 17.49PCh. 17 - Prob. 17.50PCh. 17 - Prob. 17.51PCh. 17 - Prob. 17.52PCh. 17 - Prob. 17.53PCh. 17 - Prob. 17.54PCh. 17 - Prob. 17.55PCh. 17 - Prob. 17.56PCh. 17 - Prob. 17.57PCh. 17 - Prob. 17.58PCh. 17 - Prob. 17.59PCh. 17 - Prob. 17.60PCh. 17 - Prob. 17.61PCh. 17 - Prob. 17.62PCh. 17 - Prob. 17.63PCh. 17 - Prob. 17.64PCh. 17 - Prob. 17.65PCh. 17 - Prob. 17.66PCh. 17 - Prob. 17.67PCh. 17 - Prob. 17.68PCh. 17 - Prob. 17.69PCh. 17 - Prob. 17.70PCh. 17 - Prob. 17.71PCh. 17 - Prob. 17.72PCh. 17 - Prob. 17.73PCh. 17 - Prob. 17.74PCh. 17 - Prob. 17.75PCh. 17 - Prob. 17.76PCh. 17 - Prob. 17.77PCh. 17 - Prob. 17.78PCh. 17 - Prob. 17.79PCh. 17 - Prob. 17.80PCh. 17 - Prob. 17.81PCh. 17 - Prob. 17.82PCh. 17 - Prob. 17.83PCh. 17 - Prob. 17.84PCh. 17 - Prob. 17.1YTCh. 17 - Prob. 17.2YTCh. 17 - Prob. 17.3YTCh. 17 - Prob. 17.4YTCh. 17 - Prob. 17.5YTCh. 17 - Prob. 17.6YTCh. 17 - Prob. 17.7YTCh. 17 - Prob. 17.8YTCh. 17 - Prob. 17.9YTCh. 17 - Prob. 17.10YTCh. 17 - Prob. 17.11YTCh. 17 - Prob. 17.12YTCh. 17 - Prob. 17.13YTCh. 17 - Prob. 17.14YT
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