The curved arrow notation and products for each elementary step described by the given sequence are to be drawn. Concept introduction: The E2 is the bimolecular elimination step. The step is bimolecular because it involves two reactants – a strong base and the substrate. The strong base reacts with the substrate having the leaving group, and the hydrogen atoms are on adjacent carbons like H − C − C − L or H − C = C − L . The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene ( C=C ) or alkyne ( C ≡ C ). An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

Question

Want to see the full answer?

Answer 1

Question

Chapter 7, Problem 7.60P

Interpretation Introduction

Interpretation:

The curved arrow notation and products for each elementary step described by the given sequence are to be drawn.

Concept introduction:

The E2 is the bimolecular elimination step. The step is bimolecular because it involves two reactants – a strong base and the substrate. The strong base reacts with the substrate having the leaving group, and the hydrogen atoms are on adjacent carbons like H−C−C−L or H−C=C−L. The first curved arrow is drawn from the negatively charged atom of a strong base to the hydrogen adjacent to the leaving group in the substrate. The second curved arrow is drawn from the C-H bond to the region between the adjacent carbon atoms bearing the hydrogen and the leaving group. The third curved arrow is drawn representing the detachment of the leaving group to avoid exceeding octet of the carbon atom. The main product formed in such a step is alkene (C=C) or alkyne (C≡C).

An elementary step in which a proton is transferred from an electron-poor site to an electron-rich site and one bond is broken and another is formed simultaneously is called the proton transfer step.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Predict the major products of the following organic reaction.

1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…

What steps might you take to produce the following product from the given starting material? CI Br Он до NH2 NH2