EBK ORGANIC CHEMISTRY STUDY GUIDE AND S
EBK ORGANIC CHEMISTRY STUDY GUIDE AND S
6th Edition
ISBN: 9781319385415
Author: PARISE
Publisher: VST
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Chapter 14, Problem 14.35AP
Interpretation Introduction

(a)

Interpretation:

The possible enol structures for the given structure are to be drawn.

Concept introduction:

The carbonyl compound contains a C=O group. They are of two types, aldehyde (CHO) and ketone (RC=OR). Ketone also exist in two forms which are results of keto enol tautomerism. Tautomers are isomers which differ only in the position of the protons and electrons of double bond of the electronegative atom in the compound. There is no change in the carbon skeleton of the compound. This phenomenon which involves simple proton transfer in an intramolecular fashion is known as tautomerism.

The very common example of tautomerism is keto-enol tautomerism. It can be acid or base catalysed.

Interpretation Introduction

(b)

Interpretation:

Whether the alkyne hydration is a good method for the preparation of given compound or not is to be stated.

Concept introduction:

There are two classes of hydrocarbon compounds, Saturated and unsaturated hydrocarbons. Unsaturated hydrocarbon are of two types, alkenes and alkynes. The alkene contains a double bnd between two carbon atoms. The alkynes contains a triple bond between two carbon atoms and follow general formula CnH2n-2. The name of the alkyne compounds ends with suffix yne.

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1. Refer to the compounds below to answer the following questions: CO₂Et 0 C. H O O₂N-CH2-C-CH3 0 OEt || 111 A. Indicate all the acidic hydrogens in Compounds I through IV. IV B. Indicate which hydrogens in Compound II are the most acidic. Explain your answer C. Choose the most acidic compound from Compounds I - IV. Explain your choice.
Show how you would accomplish the following transformations. More than one step may be required. ow all reagents and all intermediate structures [one ONLY] A. H Br H CH3 NHz CH3 CH3 B. CH3CH2C-Br CH3CH2C-CN CH3 CH3.
Show how you would accomplish the following transformations. More than one step may be required. now all reagents and all intermediate structures [one ONLY] A. H Br H CH3 NHz CH3 CH3 B. CH3CH2C-Br CH3 CH3CH2C-CN CH3
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