ORGANIC CHEMISTRY 1 TERM ACCESS
ORGANIC CHEMISTRY 1 TERM ACCESS
3rd Edition
ISBN: 9781119661511
Author: Klein
Publisher: WILEY
Question
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Chapter 9, Problem 64IP

(a)

Interpretation Introduction

Interpretation:

The target molecules should be drawn for the given staring molecule by using its structure.

Concept Introduction:

Lindlar reduction: The alkenes or alkynes can be reduced to alkanes with H2 in the presence of metal catalyst (Pd) . This heterogeneous catalyst that consists of palladium deposited on calcium carbonate which is then poisoned with various forms of lead or Sulphur. The new C-H σ bonds are formed simultaneously from H atoms absorbed into the metal surface.

Soda amide ( NaNH2 ): The strong base of NaNH2 will deprotonate alkynes, alcohols and other organic functional groups with acidic protons such as asters and ketones. It is also a very strong nucleophile. It is a strong base and excellent nucleophile. It’s used deprotonated of weak acids and also for elimination reaction.

Birch Reduction: The conjugated alkynes and benzynes in the presence of sodium metal in liquid ammonia and alkyne produced a non-conjugated diene system. The alkyne involves sodium (Na)/NH3 . This end up reducing to alkyne to give the trans (E) alkene.

Meta-chloroperoxybenzoic acid (m-CPBA): This reagent is extremely useful reagent most frequently encountered in the synthesis of epoxides when added to alkenes or alkynes.

(b)

Interpretation Introduction

Interpretation:

The target molecules should be drawn for the given staring molecule by using its structure.

Concept Introduction:

Lindlar reduction: The alkenes or alkynes can be reduced to alkanes with H2 in the presence of metal catalyst (Pd) . This heterogeneous catalyst that consists of palladium deposited on calcium carbonate which is then poisoned with various forms of lead or Sulphur. The new C-H σ bonds are formed simultaneously from H atoms absorbed into the metal surface.

Soda amide ( NaNH2 ): The strong base of NaNH2 will deprotonate alkynes, alcohols and other organic functional groups with acidic protons such as asters and ketones. It is also a very strong nucleophile. It is a strong base and excellent nucleophile. It’s used deprotonated of weak acids and also for elimination reaction.

Birch Reduction: The conjugated alkynes and benzynes in the presence of sodium metal in liquid ammonia and alkyne produced a non-conjugated diene system. The alkyne involves sodium (Na)/NH3 . This end up reducing to alkyne to give the trans (E) alkene.

Meta-chloroperoxybenzoic acid (m-CPBA): This reagent is extremely useful reagent most frequently encountered in the synthesis of epoxides when added to alkenes or alkynes.

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Students have asked these similar questions
1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.
3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗ
2. Draw the arrows required to make the following reactions occur. Please ensure your arrows point from exactly where you want to exactly where you want. If it is unclear from where arrows start or where they end, only partial credit will be given. Note: You may need to draw in lone pairs before drawing the arrows. A. B. H-Br 人 C Θ CI H Cl Θ + Br O

Chapter 9 Solutions

ORGANIC CHEMISTRY 1 TERM ACCESS

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