Given that, when an unsymmetrical alkene such as propene is treated with NBS in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Whether the orientation is Markovnikov or non-Markovnikov is to be explained. Concept introduction: NBS can be used as a source of bromine, as it is stable, easily handled compound that slowly decompose in aqueous dimethyl sulfoxide to yield bromine at controlled rate. Bromine from NBS adds on to the double bond to give a cyclic bromonium ion. Water, being a nucleophile, attacks the bromonium ion from the unshielded side to give an intermediate which loses a proton to yield a bromohydrin as the product. The addition occurs with anti stereochemistry. According to Markovnikov rule, in the addition of HX to unsymmetrical alkenes the H (positive part) attaches itself to the carbon with fewer alkyl substituents and X (negative part) attaches itself to the carbon with more alkyl substituents. To explain: Whether the orientation of groups in the addition reaction between propene and NBS in aqueous dimethyl sulfoxide ,where in the major product has the bromine atom bonded to the less highly substituted carbon atom, is Markovnikov or non-Markovnikov.

Question

Wyerone, a compound found in broad beans, has antifungal properties. A Wyerone Identify a trans and a cis alkene in wyerone. Explain the distinction. Draw the product that would be formed by addition of Br2 to the double bond labelled A. Draw the product formed by addition of H2 to wyerone using Lindlar's catalyst. Draw the product formed by addition of excess H2 to wyerone using Pd metal as catalyst.

Answer 1

Question

Chapter 8.3, Problem 6P

Interpretation Introduction

Interpretation:

Given that, when an unsymmetrical alkene such as propene is treated with NBS in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Whether the orientation is Markovnikov or non-Markovnikov is to be explained.

Concept introduction:

NBS can be used as a source of bromine, as it is stable, easily handled compound that slowly decompose in aqueous dimethyl sulfoxide to yield bromine at controlled rate. Bromine from NBS adds on to the double bond to give a cyclic bromonium ion. Water, being a nucleophile, attacks the bromonium ion from the unshielded side to give an intermediate which loses a proton to yield a bromohydrin as the product. The addition occurs with anti stereochemistry.

According to Markovnikov rule, in the addition of HX to unsymmetrical alkenes the H (positive part) attaches itself to the carbon with fewer alkyl substituents and X (negative part) attaches itself to the carbon with more alkyl substituents.

To explain:

Whether the orientation of groups in the addition reaction between propene and NBS in aqueous dimethyl sulfoxide ,where in the major product has the bromine atom bonded to the less highly substituted carbon atom, is Markovnikov or non-Markovnikov.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

Wyerone, a compound found in broad beans, has antifungal properties. A Wyerone Identify a trans and a cis alkene in wyerone. Explain the distinction. Draw the product that would be formed by addition of Br2 to the double bond labelled A. Draw the product formed by addition of H2 to wyerone using Lindlar's catalyst. Draw the product formed by addition of excess H2 to wyerone using Pd metal as catalyst.

Both alkenes and alkynes under hydrogenation by the same reaction mechanism. Draw the line angle formula for chalcone. How many nonaromatic pi bonds does it have? Is it an alkene or an alkyne? Aromatic pi bonds are much more resist to chemical reactions than those of alkenes or alkynes, and will not react in ordinary alkene and alkyne reactions. Given that, draw the structure of the product you expect to be formed. Will the addition of hydrogen be syn or anti? Only second part of question draw structure of product

Ethylene oxide reacts readily with HO- because of the strain in the three-membered ring. Explain why cyclopropane, a compound with approximately the same amount of strain, does not react with HO-.