To prove that a tetrahedral intermediate is formed in the given reaction. Concept introduction: The nucleophilic acyl substitution reaction in esters is an S N 1 reaction. It is a two-step reaction. In the first step, a tetrahedral intermediate is formed which then collapses in the second step to give the product. For the formation of the tetrahedral intermediate, the nucleophilic acyl substitution reaction takes place on the carbonyl carbon in the first step of the reaction. For the collapse of the tetrahedral intermediate, the group attached to the substituent attached to the acyl group leaves leading to the formation of carbonyl compound and a weaker base.

Question

Want to see the full answer?

Answer 1

Question

Chapter 15, Problem 62P

Interpretation Introduction

Interpretation:

To prove that a tetrahedral intermediate is formed in the given reaction.

Concept introduction:

The nucleophilic acyl substitution reaction in esters is an SN1 reaction. It is a two-step reaction. In the first step, a tetrahedral intermediate is formed which then collapses in the second step to give the product.

For the formation of the tetrahedral intermediate, the nucleophilic acyl substitution reaction takes place on the carbonyl carbon in the first step of the reaction.

For the collapse of the tetrahedral intermediate, the group attached to the substituent attached to the acyl group leaves leading to the formation of carbonyl compound and a weaker base.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

The following reaction involves two sequential Heck reactions. Draw structural formu- las for each organopalladium intermediate formed in the sequence and show how the final product is formed. Note from the molecular formula given under each structural formula that this conversion corresponds to a loss of H and I from the starting material. Acetonitrile, CH,CN, is the solvent. 1% mol Pd(OAc), 4% mol Ph,P CH,CN C4H171 C4H16

Step 4: In an acid-catalyzed epoxide ring opening, the epoxide is first protonated and can then be attacked by weak nucleophiles. H HO: :Nuc Nuc Weak nucleophiles such as water and alcohols can react with the protonated epoxide. One of the main differences between base-catalyzed and acid-catalyzed epoxide ring openings is the regiochemical product expected. In an acid-catalyzed reaction, if one of the epoxide carbons is tertiary and the other primary, the nucleophile will attack at the more substituted carbon. If the epoxide is monosubstituted, the nucleophile will attack preferentially at the primary carbon vs. the secondary carbon. Which reaction will not give the indicated product? N 1. CH,CH,O 2. H₂O OH CH₂CH₂OH, H* N CH₂CH₂OH, H 1 CHICH,O 2. H₂O HO HO H- H

When n-butyl bromide is treated with sodium iodide (Nal) in ethanol (EtOH), the concentration of iodide quickly decreases, but then slowly returns to its original concentration. Provide an explanation for these observations, and identify the final product of the reaction. Olodide displaces the bromide leaving group, and then bromide displaces the iodide leaving group. Olodide displaces the bromide leaving group, and the resulting n-butyl iodide undergoes an E2 elimination. Olodide displaces the bromide leaving group, and the resulting n-butyl iodide undergoes an E2 elimination. O lodide displaces the bromide leaving group, and the resulting n-butyl iodide undergoes a substitution with EtOH. Draw Your Solution Save for Later Urin maltinale atheneint Attempts: 0 of 3 used Submit Answer