The approach of the Br - on the (CH 3 ) 3 C + carbocation from within the plane of the central C atom in the coordination step is to be discussed. Concept introduction: In the coordination step, a substrate undergoes substitution, in which the nucleophile is attached to the electrophile. The conjugate bases of strong acids act as nucleophiles. The negative charged nucleophile is attracted to an atom bearing a positive charge on the electrophile. However, for the bond formation between a nucleophile and an electrophile, only a single covalent bond is formed. Nucleophile is the electron rich species, and it will contribute the HOMO interaction, and electron poor C atom will contribute the LUMO interaction. The highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) must have axial overlap or parallel overlap. The axial overlap between HOMO and LUMO has substantial net overlap, so this reaction is allowed. If the HOMO and LUMO do not have a substantial overlap, this overlap is forbidden.

Question

Want to see the full answer?

Answer 1

Question

Chapter D, Problem D.6P

Interpretation Introduction

Interpretation:

The approach of the Br- on the (CH3)3C+ carbocation from within the plane of the central C atom in the coordination step is to be discussed.

Concept introduction:

In the coordination step, a substrate undergoes substitution, in which the nucleophile is attached to the electrophile. The conjugate bases of strong acids act as nucleophiles. The negative charged nucleophile is attracted to an atom bearing a positive charge on the electrophile. However, for the bond formation between a nucleophile and an electrophile, only a single covalent bond is formed. Nucleophile is the electron rich species, and it will contribute the HOMO interaction, and electron poor C atom will contribute the LUMO interaction. The highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) must have axial overlap or parallel overlap. The axial overlap between HOMO and LUMO has substantial net overlap, so this reaction is allowed. If the HOMO and LUMO do not have a substantial overlap, this overlap is forbidden.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

scratch paper, and the integrated rate table provided in class. our scratch work for this test. Content attribution 3/40 FEEDBACK QUESTION 3 - 4 POINTS Complete the equation that relates the rate of consumption of H+ and the rate of formation of Br2 for the given reaction. 5Br (aq) + BrO3 (aq) + 6H (aq) →3Br2(aq) + 3H2O(l) • Your answers should be whole numbers or fractions without any decimal places. Provide your answer below: Search 尚 5 fn 40 * 00 99+ 2 9 144 a [

(a) Write down the structure of EDTA molecule and show the complex structure with Pb2+ . (b) When do you need to perform back titration? (c) Ni2+ can be analyzed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange indicator. A solution containing 25.00 mL of Ni2+ in dilute HCl is treated with 25.00 mL of 0.05283 M Na2EDTA. The solution is neutralized with NaOH, and the pH is adjusted to 5.5 with acetate buffer. The solution turns yellow when a few drops of indicator are added. Titration with 0.02299 M Zn2+ requires 17.61 mL to reach the red end point. What is the molarity of Ni2+ in the unknown?

A compound has the molecular formula CH40, and shows a strong IR absorption at 2850-3150 cm. The following signals appear in the 'H NMR spectrum: 1.4 ppm (triplet, 6H), 4.0 ppm (quartet, 4H), 6.8 ppm (broad singlet, 4H). Which of the following structures is consistent with these data? Select the single best answer. OCH CH₂ x OCH2CH3 CH₂OCH3 OH CH₂OCH OH CH, OCH₁ CH₂OCH, CH₂OCH HO OH ° CH₂OCH3