The mechanism for the given reaction is to be proposed. Concept introduction: The dienes are the compounds containing two double bonds. These two double bonds may or may not be conjugated. The non-conjugated dienes undergo electrophilic addition reactions in the presence of a strong acid. The first step is the protonation of the double bond, which would form the most stable carbocation intermediate. This generates the conjugate base of a strong acid, which can serve as a nucleophile. The addition of proton across the double bond follows Markovnikov’s rule. In this, the proton gets attached to the carbon atom in the double bond, which is least substituted or has greater number of hydrogens attached. For long chain isolated dienes, an intramolecular reaction may take place in which the other double bond serves as a nucleophile and attacks the carbocation intermediate within the same compound. The driving force for this step is the formation of stable five or six membered ring. In the third step, the conjugate base produced in the first step serves as a nucleophile and attacks the carbocation to yield the final product.

Question

Want to see the full answer?

Answer 1

Question

Chapter 11, Problem 11.57P

Interpretation Introduction

Interpretation:

The mechanism for the given reaction is to be proposed.

Concept introduction:

The dienes are the compounds containing two double bonds. These two double bonds may or may not be conjugated. The non-conjugated dienes undergo electrophilic addition reactions in the presence of a strong acid. The first step is the protonation of the double bond, which would form the most stable carbocation intermediate. This generates the conjugate base of a strong acid, which can serve as a nucleophile. The addition of proton across the double bond follows Markovnikov’s rule. In this, the proton gets attached to the carbon atom in the double bond, which is least substituted or has greater number of hydrogens attached. For long chain isolated dienes, an intramolecular reaction may take place in which the other double bond serves as a nucleophile and attacks the carbocation intermediate within the same compound. The driving force for this step is the formation of stable five or six membered ring. In the third step, the conjugate base produced in the first step serves as a nucleophile and attacks the carbocation to yield the final product.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Part V. Label ad match the carbons in compounds Jane and Diane w/ the corresponding peak no. in the Spectra (Note: use the given peak no. To label the carbons, other peak no are intentionally omitted) 7 4 2 -0.13 -0.12 -0.11 -0.10 -0.08 8 CI Jane 1 -0.09 5 210 200 190 180 170 160 150 140 130 120 110 100 -8 90 f1 (ppm) 11 8 172.4 172.0 f1 (ppr HO CI NH Diane 7 3 11 80 80 -80 -R 70 60 60 2 5 -8 50 40 8. 170 160 150 140 130 120 110 100 90 -0 80 70 20 f1 (ppm) 15 30 -20 20 -60 60 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -0.01 10 -0.17 16 15 56 16 -0.16 -0.15 -0.14 -0.13 -0.12 -0.11 -0.10 -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 17.8 17.6 17.4 17.2 17.0 f1 (ppm) -0.03 -0.02 550 106 40 30 20 20 -0.01 -0.00 F-0.01 10 0

Consider the reaction of 2-methylpropane with a halogen. With which halogen will the product be almost exclusively 2-halo-2-methylpropane? 1. F2 2. Cl2 3. Br2 4. I2

None