Organic Chemistry
Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 16, Problem 42P
Interpretation Introduction

Interpretation:

The stereoisomers formed by the reduction of ketone group by sodium borohydride need to be predicted and the relationship between the stereoisomers has to be identified.

Concept introduction:

The compounds that have the same molecular formula of atoms but the structure of the compounds are different for the positions of atoms are called isomers.

Stereoisomers are those isomers that differ in their stereochemistry around any one chiral carbon. Those isomers that are mirror images of each other are called enantiomers and the isomers in which one of the chiral centers has the same stereochemistry but the other has a different are termed as diastereomers.

The objects or molecules that are superimposable with their mirror images are achiral objects or molecules and these objects have a centre of symmetry or plane of symmetry.

The achiral compounds in which plane of symmetry is present internally and consists of chiral centres are known as meso compounds, but they are optically inactive.

Enantiomers are designated as R and S. The designation is based on the rotation, while moving from higher priority group (one thathas higher molecular weight marked as 1) to the lower one (marked as 2,3, and 4). If the rotation is clockwise, the designation is R and if it is anti-clockwise, the designation is S.

Further, if both stereoisomers are formed in equal proportion, then the mixture is said to be racemic mixture. Ketone group bears a carbonyl bond, which is planar in nature.

Reagents, such as NaBH4/Methanol (MeOH), LiAlH4/H2O and H2/Pt, are reducing agents and convert aldehydes to alcohols.

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Chapter 16 Solutions

Organic Chemistry

Ch. 16 - Prob. 11PPCh. 16 - Practice Problem 16.12 What product would be...Ch. 16 - Prob. 13PPCh. 16 - Practice Problem 16.14 Dihydropyran reacts readily...Ch. 16 - Practice Problem 16.15 Show how you might use...Ch. 16 - Practice Problem 16.16 (a) Show how you might...Ch. 16 - Practice Problem 16.17 In addition to...Ch. 16 - Practice Problem 16.18 Triphenylphosphine can be...Ch. 16 - Prob. 19PPCh. 16 - PRACTICE PROBLEM 16.20 Give the structure of the...Ch. 16 - PRACTICE PROBLEM 16.21 What would be the major...Ch. 16 - Prob. 22PCh. 16 - 16.23 Write structural formulas for the products...Ch. 16 - Give structural formulas for the products formed...Ch. 16 - 16.25 What products would be obtained when...Ch. 16 - Predict the major organic product from each of the...Ch. 16 - 16.27 Predict the major product from each of the...Ch. 16 - 16.28 Predict the major product from each of the...Ch. 16 - Prob. 29PCh. 16 - 16.30 Write detailed mechanisms for each of the...Ch. 16 - Prob. 31PCh. 16 - Prob. 32PCh. 16 - Show how you would convert benzaldehyde into each...Ch. 16 - 16.34 Show how ethyl phenyl ketone could be...Ch. 16 - Show how benzaldehyde could be synthesized from...Ch. 16 - Give structures for compounds AE. Cyclohexanol...Ch. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - Prob. 42PCh. 16 - 16.43 The structure of the sex pheromone...Ch. 16 - Provide reagents that would accomplish each of the...Ch. 16 - Write a detailed mechanism for the following...Ch. 16 - Prob. 46PCh. 16 - Dutch elm disease is caused by a fungus...Ch. 16 - Prob. 48PCh. 16 - Compounds W and X are isomers; they have the...Ch. 16 - Compounds Y and Z are isomers with the molecular...Ch. 16 - Compound A (C9H18O) forms a phenylhydrazone, but...Ch. 16 - Compound B (C8H12O2) shows a strong carbonyl...Ch. 16 - Prob. 53PCh. 16 - Prob. 54PCh. 16 - Prob. 55PCh. 16 - (a) What would be the frequencies of the two...Ch. 16 - Prob. 57PCh. 16 - Prob. LGP
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