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Which of the following reaction conditions results in Markovnikov addition to an unsymmetric alkene? A) HBr B) HBr with peroxides C) Br₂ D) H₂ with Pd-C E) KMnO with NaOH

Which of the following reaction conditions results in Markovnikov addition to an unsymmetric alkene? A) HBr B) HBr with peroxides C) Br₂ D) H₂ with Pd-C E) KMnO with NaOH

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
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### Question 8 of 30 **Which of the following reaction conditions results in Markovnikov addition to an unsymmetric alkene?** - A) HBr - B) HBr with peroxides - C) Br₂ - D) H₂ with Pd-C - E) KMnO₄ with NaOH #### Explanation: In organic chemistry, the Markovnikov's rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom from HX will attach to the carbon with the greater number of hydrogen atoms already present, and the halide (X) will attach to the carbon with fewer hydrogen atoms. This rule helps predict the outcome of addition reactions involving asymmetrical alkenes. - **Option A (HBr)**: This condition follows Markovnikov's rule. HBr adds to the double bond of an unsymmetrical alkene in such a way that the hydrogen atom attaches to the carbon with more hydrogen atoms. - **Option B (HBr with peroxides)**: This condition leads to anti-Markovnikov addition due to the formation of free radicals. - **Option C (Br₂)**: Br₂ addition typically leads to vicinal dibromide formation, not specifically following Markovnikov's rule. - **Option D (H₂ with Pd-C)**: This condition is for catalytic hydrogenation, adding hydrogen atoms to both carbons of the double bond. - **Option E (KMnO₄ with NaOH)**: This condition results in dihydroxylation, not a Markovnikov or anti-Markovnikov addition. ### Correct Answer: **A) HBr**
Transcribed Image Text:### Question 8 of 30 **Which of the following reaction conditions results in Markovnikov addition to an unsymmetric alkene?** - A) HBr - B) HBr with peroxides - C) Br₂ - D) H₂ with Pd-C - E) KMnO₄ with NaOH #### Explanation: In organic chemistry, the Markovnikov's rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom from HX will attach to the carbon with the greater number of hydrogen atoms already present, and the halide (X) will attach to the carbon with fewer hydrogen atoms. This rule helps predict the outcome of addition reactions involving asymmetrical alkenes. - **Option A (HBr)**: This condition follows Markovnikov's rule. HBr adds to the double bond of an unsymmetrical alkene in such a way that the hydrogen atom attaches to the carbon with more hydrogen atoms. - **Option B (HBr with peroxides)**: This condition leads to anti-Markovnikov addition due to the formation of free radicals. - **Option C (Br₂)**: Br₂ addition typically leads to vicinal dibromide formation, not specifically following Markovnikov's rule. - **Option D (H₂ with Pd-C)**: This condition is for catalytic hydrogenation, adding hydrogen atoms to both carbons of the double bond. - **Option E (KMnO₄ with NaOH)**: This condition results in dihydroxylation, not a Markovnikov or anti-Markovnikov addition. ### Correct Answer: **A) HBr**
### Question 9 of 30 **What is the classification of the carbocation shown here?** On the left side of the screen, there is a diagram of a carbocation. The chemical structure of the carbocation indicates a central carbon atom bonded to three other carbon atoms, none of which are hydrogen atoms. There is also a plus sign (indicating a positive charge) on the central carbon atom. On the right side of the screen, there are four multiple-choice options provided: A) 1° B) 2° C) 3° D) 4° **Answering the Question:** In order to classify the carbocation, we need to determine the degree of the carbocation. This is done by counting the number of carbon atoms directly bonded to the positively charged carbon atom. - A 1° (primary) carbocation has the positively charged carbon bonded to only one other carbon. - A 2° (secondary) carbocation has the positively charged carbon bonded to two other carbons. - A 3° (tertiary) carbocation has the positively charged carbon bonded to three other carbons. - A 4° (quaternary) carbocation does not exist as carbocations are positively charged and cannot have four carbon bonds since it would carry a neutral or negative charge otherwise. Based on the diagram, the central carbon with the positive charge is connected to three other carbon atoms. Therefore, this carbocation is classified as a tertiary (3°) carbocation. **Correct Answer:** C) 3°
Transcribed Image Text:### Question 9 of 30 **What is the classification of the carbocation shown here?** On the left side of the screen, there is a diagram of a carbocation. The chemical structure of the carbocation indicates a central carbon atom bonded to three other carbon atoms, none of which are hydrogen atoms. There is also a plus sign (indicating a positive charge) on the central carbon atom. On the right side of the screen, there are four multiple-choice options provided: A) 1° B) 2° C) 3° D) 4° **Answering the Question:** In order to classify the carbocation, we need to determine the degree of the carbocation. This is done by counting the number of carbon atoms directly bonded to the positively charged carbon atom. - A 1° (primary) carbocation has the positively charged carbon bonded to only one other carbon. - A 2° (secondary) carbocation has the positively charged carbon bonded to two other carbons. - A 3° (tertiary) carbocation has the positively charged carbon bonded to three other carbons. - A 4° (quaternary) carbocation does not exist as carbocations are positively charged and cannot have four carbon bonds since it would carry a neutral or negative charge otherwise. Based on the diagram, the central carbon with the positive charge is connected to three other carbon atoms. Therefore, this carbocation is classified as a tertiary (3°) carbocation. **Correct Answer:** C) 3°
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