**Task:** Draw the major product of this reaction. Ignore inorganic byproducts.**Chemical Structure Provided:**- A hexagonal ring with a ketone group (C=O) at the top point and a methyl group attached to the carbon adjacent to the ketone.**Reaction Conditions:**1. LDA (Lithium Diisopropylamide), at -78°C2. PhCH₂Br (Benzyl bromide), 1 equivalent**Instructions:**The diagram shows the starting material undergoing treatment with LDA, a strong base, at low temperature to form an enolate. This intermediate then reacts with benzyl bromide to form the major organic product. Draw the resulting compound in the designated area below.The image represents a typical enolate alkylation reaction, where the ketone forms an enolate ion that is then alkylated by the benzyl bromide.

Answered: Image /qna-images/answer/762bb6b8-3ce4-4dc5-8731-9f93f3f52226.jpg

Answered: Draw the major product of this…

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Chemistry

Draw the major product of this reaction. Ignore inorganic byproducts. 1. LDA, -78 °C 2. PhCH2Br (1 equiv) Drawing O

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Alkanes and Cycloalkanes

Alkanes are saturated organic compounds which are made up of C and H atoms. Alkanes have the general formula of CnH2n+2. Alkane contains sp3 hybridized carbon atoms with four sigmas (σ) bonds & every hydrogen atom is connected with one of the carbon atoms. The size of the alkane could be determined by the number of carbon atoms located in it.

Question

**Task:** Draw the major product of this reaction. Ignore inorganic byproducts.

**Chemical Structure Provided:**
- A hexagonal ring with a ketone group (C=O) at the top point and a methyl group attached to the carbon adjacent to the ketone.

**Reaction Conditions:**
1. LDA (Lithium Diisopropylamide), at -78°C
2. PhCH₂Br (Benzyl bromide), 1 equivalent

**Instructions:**
The diagram shows the starting material undergoing treatment with LDA, a strong base, at low temperature to form an enolate. This intermediate then reacts with benzyl bromide to form the major organic product. Draw the resulting compound in the designated area below.

The image represents a typical enolate alkylation reaction, where the ketone forms an enolate ion that is then alkylated by the benzyl bromide.

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