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CH3 Br- -H + Nal product; Sy2 product of the reaction is : Acetone H- -CH3 CH2 – CH3 CH3 -H- CH3 CH3 CH3 H. (b) -CH3 it -CH3 I- -H- (а) H• -CH3 (c) -CH3 (d) H- CH H- CH2 – CH3 ČH2 – CH3 CH2 – CH3 CH2- CH3

CH3 Br- -H + Nal product; Sy2 product of the reaction is : Acetone H- -CH3 CH2 – CH3 CH3 -H- CH3 CH3 CH3 H. (b) -CH3 it -CH3 I- -H- (а) H• -CH3 (c) -CH3 (d) H- CH H- CH2 – CH3 ČH2 – CH3 CH2 – CH3 CH2- CH3

Chemistry
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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**Question 2:** Given the reaction: \[ \begin{align*} &\text{Reactant:} \\ &\quad \text{A molecular structure with bromine (Br), two hydrogens (H), two methyl groups (CH}_3\text{), and an ethyl group (CH}_2\text{-CH}_3\text{). Bromine is positioned on the left, with hydrogen opposite, methyl groups above and right, and the ethyl group below.} \\ &\text{Reaction:} \\ &\quad \text{+ NaI in Acetone} \quad \rightarrow \quad \text{product} \\ &\text{The S}_{\text{N}}^2 \text{ product of the reaction is:} \end{align*} \] **Multiple Choice Options for Product:** - (a) Iodine (I) at the left position, with hydrogen opposite, a methyl group above and right, and an ethyl group below. - (b) Hydrogen at the left position, with iodine opposite, a methyl group above and right, and an ethyl group below. - (c) Iodine at the left position, with a methyl group opposite, a methyl group above and right, and an ethyl group below. - (d) Iodine at the left position, with hydrogen opposite, a methyl group above, and hydrogen below. In this question, the reaction involves nucleophilic substitution (\(S_N2\)), where iodine from NaI displaces bromine in an inversion of configuration.
Transcribed Image Text:**Question 2:** Given the reaction: \[ \begin{align*} &\text{Reactant:} \\ &\quad \text{A molecular structure with bromine (Br), two hydrogens (H), two methyl groups (CH}_3\text{), and an ethyl group (CH}_2\text{-CH}_3\text{). Bromine is positioned on the left, with hydrogen opposite, methyl groups above and right, and the ethyl group below.} \\ &\text{Reaction:} \\ &\quad \text{+ NaI in Acetone} \quad \rightarrow \quad \text{product} \\ &\text{The S}_{\text{N}}^2 \text{ product of the reaction is:} \end{align*} \] **Multiple Choice Options for Product:** - (a) Iodine (I) at the left position, with hydrogen opposite, a methyl group above and right, and an ethyl group below. - (b) Hydrogen at the left position, with iodine opposite, a methyl group above and right, and an ethyl group below. - (c) Iodine at the left position, with a methyl group opposite, a methyl group above and right, and an ethyl group below. - (d) Iodine at the left position, with hydrogen opposite, a methyl group above, and hydrogen below. In this question, the reaction involves nucleophilic substitution (\(S_N2\)), where iodine from NaI displaces bromine in an inversion of configuration.
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