ORGANIC CHEMISTRY W/BIOLOGICAL TOPICS
ORGANIC CHEMISTRY W/BIOLOGICAL TOPICS
6th Edition
ISBN: 9781260325294
Author: SMITH
Publisher: RENT MCG
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Chapter 21, Problem 36P
Interpretation Introduction

(a)

Interpretation: The major enolate formed by the treatment of given compound with LDA is to be identified.

Concept introduction: Lithium diisopropylamide (LDA), molecular formula [(CH3)2CH]2NLi is a strong base and has been widely accepted due to its good solubility in non-polar organic solvent. It is a bulky base which usually extracts hydrogen from less hindered carbon.

Interpretation Introduction

(b)

Interpretation: The major enolate formed by the treatment of given compound with LDA is to be identified.

Concept introduction: Lithium diisopropylamide (LDA), molecular formula [(CH3)2CH]2NLi is a strong base and has been widely accepted due to its good solubility in non-polar organic solvent. It is a bulky base which usually extracts hydrogen from less hindered carbon.

Interpretation Introduction

(c)

Interpretation: The major enolate formed by the treatment of given compound with LDA is to be identified.

Concept introduction: Lithium diisopropylamide (LDA), molecular formula [(CH3)2CH]2NLi is a strong base and has been widely accepted due to its good solubility in non-polar organic solvent. It is a bulky base which usually extracts hydrogen from less hindered carbon.

Interpretation Introduction

(d)

Interpretation: The major enolate formed by the treatment of given compound with LDA is to be identified.

Concept introduction: Lithium diisopropylamide (LDA), molecular formula [(CH3)2CH]2NLi is a strong base and has been widely accepted due to its good solubility in non-polar organic solvent. It is a bulky base which usually extracts hydrogen from less hindered carbon.

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9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00
Assign all the carbons
C 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00

Chapter 21 Solutions

ORGANIC CHEMISTRY W/BIOLOGICAL TOPICS

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