Can you explain the reasons and concepts for why these answers are true? **Question 14:** Why is the nitro group a *meta* director?A. Because it is sterically very large. B. Because it adds electron density to the *meta* position, thus activating it. C. Because it stabilizes the intermediate cation. D. **Because it removes more electron density from the *ortho* and *para* positions than the *meta* position, thus deactivating the *meta* position less.** The correct answer is highlighted in yellow: Option D. This option explains that the nitro group directs other substituents to the *meta* position during electrophilic aromatic substitution because it withdraws electron density most from the *ortho* and *para* positions, leaving the *meta* position relatively less deactivated. **Question 13:**Which of the following halides will *not* work as an electrophile in a Friedel-Crafts alkylation reaction?**Options:**- **I.** (Bromobenzene)- **II.** (Benzyl bromide)- **III.** (Isopropyl bromide)- **IV.** (2-Bromopropane)**Correct Answer:**- **A. I** (Bromobenzene)**Explanation:**In a Friedel-Crafts alkylation reaction, an alkyl halide reacts with an aromatic compound in the presence of a Lewis acid to form an alkylated aromatic compound. The electrophile in this reaction is usually a carbocation formed from the alkyl halide. Bromobenzene (I) cannot form the necessary carbocation, as the carbon-bromine bond in aryl halides (such as bromobenzene) is too strong and stable, making it ineffective as an electrophile in this reaction.

Since you have posted two different questions. According to guidelines I am solving 1st question.If…

14. Why is the nitro group a meta director? A. Because it is sterically very large. B. Because it adds electron density to the meta position, thus activating it. C. Because it stabilizes the intermediate cation. D. Because it removes more electron density from the ortho and para positions than the meta position, thus deactivating the meta position less.

14. Why is the nitro group a meta director? A. Because it is sterically very large. B. Because it adds electron density to the meta position, thus activating it. C. Because it stabilizes the intermediate cation. D. Because it removes more electron density from the ortho and para positions than the meta position, thus deactivating the meta position less.

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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