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Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 7 - Structure and Synthesis of Alkenes; EliminationProblem 49c
Chapter 7, Problem 49c

Predict the products of E1 elimination of the following compounds. Label the major products.
(c)

Verified step by step guidance
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Step 1: Understand the E1 elimination mechanism. E1 elimination involves two steps: (1) the leaving group departs, forming a carbocation intermediate, and (2) a proton is removed from a β-carbon by a base, resulting in the formation of a double bond. The reaction typically favors the formation of the most stable alkene (Zaitsev's rule).
Step 2: Identify the structure of the given compound, (CH3)3C-CH(CH3)Br. The molecule consists of a tertiary carbon bonded to a bromine atom (the leaving group) and a secondary carbon with a methyl group attached.
Step 3: Determine the carbocation intermediate. When the bromine leaves, a tertiary carbocation forms at the central carbon atom (C1), as tertiary carbocations are more stable due to hyperconjugation and inductive effects.
Step 4: Analyze the β-hydrogens available for elimination. The β-carbons are adjacent to the carbocation (C2 and C3). C2 has one hydrogen atom, and C3 has three hydrogen atoms. Removing a β-hydrogen from either carbon will result in the formation of a double bond.
Step 5: Apply Zaitsev's rule to predict the major product. Zaitsev's rule states that the most substituted alkene is the major product. Removing a β-hydrogen from C2 will form a double bond between C1 and C2, resulting in a more substituted alkene compared to removing a β-hydrogen from C3. Label the alkene formed by elimination at C2 as the major product.

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Key Concepts

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

E1 Elimination Mechanism

The E1 elimination mechanism involves two main steps: the formation of a carbocation intermediate after the leaving group departs, followed by the removal of a proton to form a double bond. This mechanism is favored in tertiary substrates due to their ability to stabilize the carbocation, leading to the formation of alkenes as products.
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Drawing the E1 Mechanism.

Carbocation Stability

Carbocation stability is crucial in determining the pathway of elimination reactions. Tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects from surrounding alkyl groups. In the given compound, the formation of a tertiary carbocation will significantly influence the major product of the E1 reaction.
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Determining Carbocation Stability

Zaitsev's Rule

Zaitsev's Rule states that in elimination reactions, the more substituted alkene is typically the major product. This is because more substituted alkenes are generally more stable due to hyperconjugation and steric factors. When predicting the products of the E1 elimination for the given compound, identifying the more substituted alkene will help in labeling the major product.
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Defining Zaitsev’s Rule
Related Practice
Textbook Question

A double bond in a six-membered ring is usually more stable in an endocyclic position than in an exocyclic position. Hydrogenation data on two pairs of compounds follow. One pair suggests that the energy difference between endocyclic and exocyclic double bonds is about 9 kJ/mol. The other pair suggests an energy difference of about 5 kJ/mol. Which number do you trust as being more representative of the actual energy difference? Explain your answer.

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Textbook Question

Predict the products of E1 elimination of the following compounds. Label the major products.

(a)

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Textbook Question

Predict the products formed by sodium hydroxide-promoted dehydrohalogenation of the following compounds. In each case, predict which will be the major product.

d. cis-1-bromo-2-methylcyclohexane

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Textbook Question

Predict the products formed by sodium hydroxide-promoted dehydrohalogenation of the following compounds. In each case, predict which will be the major product.

a. 1-bromobutane

b. 2-chlorobutane

c. 3-bromopentane

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Textbook Question

The energy difference between cis- and trans-but-2-ene is about 4 kJ/mol; however, the trans isomer of 4,4-dimethylpent-2-ene is nearly 16 kJ/mol more stable than the cis isomer. Explain this large difference.

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Textbook Question

Predict the products formed by sodium hydroxide-promoted dehydrohalogenation of the following compounds. In each case, predict which will be the major product.

e. trans-1-bromo-2-methylcyclohexane