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Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions
Bruice - Organic Chemistry 8th Edition
Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook
All textbooksBruice 8th EditionCh. 6 - The Reactions of Alkenes • The Stereochemistry of Addition ReactionsProblem 75
Chapter 7, Problem 75

A student was about to turn in the products he had obtained from the reaction of HI with 3,3,3-trifluoropropene when he realized that the labels had fallen off his flasks and he did not know which label belonged to which flask. His friend reminded him of the rule that says the electrophile adds to the sp2 carbon bonded to the most hydrogens. In other words, he should label the flask containing the most product 1,1,1-trifluoro-2-iodopropane and label the flask containing the least amount of product 1,1,1-trifluoro-3-iodopropane. Should he follow his friend’s advice?

Verified step by step guidance
1
Step 1: Understand the reaction mechanism. The reaction of HI with 3,3,3-trifluoropropene is an electrophilic addition reaction. In this reaction, the hydrogen ion (H⁺) acts as the electrophile, and the iodide ion (I⁻) acts as the nucleophile.
Step 2: Apply Markovnikov's rule. Markovnikov's rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom (H⁺) will add to the sp² carbon atom that is bonded to the greater number of hydrogen atoms. This is because the more stable carbocation intermediate is formed during the reaction.
Step 3: Analyze the structure of 3,3,3-trifluoropropene. The molecule has a double bond between the first and second carbons. The first carbon (C1) is bonded to three fluorine atoms and no hydrogens, while the second carbon (C2) is bonded to two hydrogens. According to Markovnikov's rule, the H⁺ will add to C2, forming a carbocation on C1.
Step 4: Determine the major and minor products. After the carbocation is formed on C1, the iodide ion (I⁻) will attack the positively charged carbon. This results in the formation of 1,1,1-trifluoro-2-iodopropane as the major product. The minor product, 1,1,1-trifluoro-3-iodopropane, would result from the addition of H⁺ to C1 and I⁻ to C2, which is less favorable due to the instability of the carbocation formed on C2.
Step 5: Conclude based on the friend's advice. The friend’s advice is correct. The flask containing the most product should be labeled as 1,1,1-trifluoro-2-iodopropane, and the flask containing the least product should be labeled as 1,1,1-trifluoro-3-iodopropane, as this aligns with Markovnikov's rule and the stability of the intermediates.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrophilic Addition

Electrophilic addition is a fundamental reaction mechanism in organic chemistry where an electrophile reacts with a nucleophile, typically involving alkenes or alkynes. In this context, the electrophile (HI) adds to the sp2 carbon of the alkene (3,3,3-trifluoropropene), which is more substituted and has more hydrogen atoms, leading to the formation of more stable carbocation intermediates.
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Markovnikov's Rule

Markovnikov's Rule states that in the addition of HX to an alkene, the hydrogen atom from HX will attach to the carbon with the greater number of hydrogen atoms already attached. This rule helps predict the major product of the reaction, guiding the student to label the flask with 1,1,1-trifluoro-2-iodopropane as the major product due to the more favorable formation of a more stable carbocation.
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Product Stability and Regioselectivity

Product stability and regioselectivity refer to the preference for the formation of certain products over others based on their stability. In this reaction, the product with the more substituted carbon (1,1,1-trifluoro-2-iodopropane) is favored due to its lower energy and greater stability compared to the less substituted product (1,1,1-trifluoro-3-iodopropane), which aligns with the student's friend's advice.
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Related Practice
Textbook Question

Draw the products of the following reactions. If the products can exist as stereoisomers, show which stereoisomers are formed.

c.

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Textbook Question

b. Which step is the rate-determining step?

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Textbook Question

a. Propose a mechanism for the following reaction (show all curved arrows):

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Textbook Question

The second-order rate constant (in units of M-1s-1) for acid-catalyzed hydration at 25 °C is given for each of the following alkenes:

a. Calculate the relative rates of hydration of the alkenes. (Hint: Divide each rate constant by the smallest rate constant of the series: 3.51 × 10-8.)

b. Why does (Z)-2-butene react faster than (E)-2-butene?

c. Why does 2-methyl-2-butene react faster than (Z)-2-butene?

d. Why does 2,3-dimethyl-2-butene react faster than 2-methyl-2-butene?

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Textbook Question

c. What is the electrophile in the first step?

d. What is the nucleophile in the first step?

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Textbook Question

Draw the products of the following reactions. If the products can exist as stereoisomers, show which stereoisomers are formed.

b.

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