Major product of the sequence of reactions shown is to be determined. Concept introduction: The OH group in an alcohol is a poor leaving group as the hydroxide ion is a strong base. For substitution reactions, therefore, it is necessary to convert it to a good leaving group. One way to do this is to sulfonate the alcohol using a reagent such as tosyl chloride. This results in the replacement of the hydrogen in the alcoholic OH by the tosyl group. The bulky tosylate is an excellent leaving group. Further, as the reaction does not occur at the carbon atom, the configuration at the carbon is preserved if it is an asymmetric carbon. The tosylate can then be treated with a bulky, non-nucleophilic base to induce an E2 elimination. E2 elimintation involves elimination of the leaving group and a proton anticoplanar to it. Only the E isomer of the corresponding alkene is formed as a result.

Question

Want to see the full answer?

Answer 1

Question

Chapter 21, Problem 21.11P

Interpretation Introduction

Interpretation:

Major product of the sequence of reactions shown is to be determined.

Concept introduction:

The OH group in an alcohol is a poor leaving group as the hydroxide ion is a strong base. For substitution reactions, therefore, it is necessary to convert it to a good leaving group. One way to do this is to sulfonate the alcohol using a reagent such as tosyl chloride. This results in the replacement of the hydrogen in the alcoholic OH by the tosyl group. The bulky tosylate is an excellent leaving group. Further, as the reaction does not occur at the carbon atom, the configuration at the carbon is preserved if it is an asymmetric carbon. The tosylate can then be treated with a bulky, non-nucleophilic base to induce an E2 elimination. E2 elimintation involves elimination of the leaving group and a proton anticoplanar to it. Only the E isomer of the corresponding alkene is formed as a result.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

2. Specify the solvent and reagent(s) required to carry out each of the following FGI. If two reagent sets must be used for the FGI, specify the solvent and reagent(s) for each reagent set. If a reaction cannot be carried out with reagents (sets) class, write NP (not possible) in the solvent box for reagent set #1. Use the letter abbreviation for each solvent; use a number abbreviation for reagent(s). Solvents: CH2Cl2 (A); H₂O (B); Reagents: HBr (1); R₂BH (6); H2SO4 (2); CH3OH (C); Br₂ (3); CH3CO₂H (D) NaHCO3 (4); Hg(OAc)2 (5); H₂O2/HO (7); NaBH4 (8) Reagent Set #1 Reagent Set #2 FGI + enant OH Solvent Reagent(s) Solvent Reagent(s)

Germanium (Ge) is a semiconductor with a bandgap of 2.2 eV. How could you dope Ge to make it a p-type semiconductor with a larger bandgap? Group of answer choices It is impossible to dope Ge and have this result in a larger bandgap. Dope the Ge with silicon (Si) Dope the Ge with gallium (Ga) Dope the Ge with phosphorus (P)

Which of the following semiconductors would you choose to have photons with the longest possible wavelengths be able to promote electrons to the semiconductor's conduction band? Group of answer choices Si Ge InSb CdS