Chapter 8, Problem 29P

Interpretation Introduction

Interpretation:

Structures of the major organic products formed in the reaction of each of the given reagents with $\text{2-methyl-2-butene}$ are to be drawn.

Concept introduction:

Since alkenes have the capacity to react with substances that add to them, they are unsaturated hydrocarbons. Therefore, alkenes show addition reactions.

According to Markovnikov’s rule. when an unsymmetrically substituted alkene reacts with a hydrogen halide, hydrogen adds to the carbon that has the greater number of hydrogens and halogen adds to the carbon that has fewer hydrogens.

In a hydroboration oxidation reaction, hydrogen atom gets bonded to the carbon that has fewer hydrogens and the hydroxyl to the carbon that has a greater number of hydrogens. This is a rule opposite of Markovnikov’s addition.

Answer to Problem 29P

Solution:

a)

b)

c)

d)

e)

f)

g)

h)

i)

Explanation of Solution

a) The reaction is as follows:

The given alkene $\text{2-methyl-2-butene}$ is an unsymmetrical alkene. When $\text{2-methyl-2-butene}$ is treated with hydrogen chloride, the major organic product formed is $\text{2-chloro-2-methylbutane}$.

Hydrogen chloride adds to the double bond of $\text{2-methyl-2-butene}$ in accordance with the Markovnikov’s rule. The hydrogen atom from hydrogen chloride gets added to the carbon $\text{(C2)}$ that has a greater number of hydrogens and chlorine gets added to the carbon $\text{(C3)}$ that has a fewer number of hydrogens.

b) The reaction is as follows:

This reaction is an acid catalyzed electrophilic addition reaction of alkenes. In this reaction, water molecule gets added to the double bond in $\text{2-methyl-2-butene}$.

In this reaction, the addition of water molecule to alkene follows Markovnikov’s rule. The hydrogen atom in water molecule adds to the carbon $\text{(C3)}$ that has a greater number of hydrogens and hydroxyl group adds to the carbon $\text{(C2)}$ that has a fewer number of hydrogens.

The addition mechanism for this reaction follows Markovnikov’s rule, and so the major organic product for the above acid-catalyzed electrophilic addition reaction is $\text{2-methyl-2-butenol}$.

c) The reaction is as follows:

This reaction is the hydroboration-oxidation reaction of alkenes. In this reaction, water molecule gets added to the double bond in $\text{2-methyl-2-butene}$. In hydroboration-oxidation reaction, the addition of H and OH is syn with regioselectivity and opposite to that of Markovnikov’s rule.

The hydrogen atom from water molecule adds to the carbon $\text{(C2)}$ that has a fewer number of hydrogens and the $\text{-OH}$ group adds to the carbon $\text{(C3)}$ that has a greater number of hydrogens. In the case of $\text{2-methyl-2-butene}$ the major product is $\text{3-methyl-2-butanol}$.

d) The reaction is as follows:

Bromine reacts rapidly with alkenes by electrophilic addition. The products are called vicinal dibromides, meaning that the bromine atoms are attached to adjacent double bonded carbon atoms. It is carried out in suitable solvents such as ${\text{CCl}}_{\text{4}}$ or ${\text{CHCl}}_{\text{3}}$.

Bromine adds across the double bond in $\text{2-methyl-2-butene}$ to form $\text{2,3-dibromo-2-methylbutane}$ as a major organic product. It is a vicinal dibromide.

e) The reaction is as follows:

In aqueous solution, chlorine and bromine react with alkenes to give corresponding vicinal halohydrins. Halohydrin compounds have a halogen and hydroxyl group on adjacent carbon atoms. In alkene, halogen atom bonds to that carbon atom which has a larger number of hydrogens and hydroxyl group bonds to that carbon atom which has a smaller number of hydrogens.

In the reaction of $\text{2-methyl-2-butene}$ with bromine water, $\text{3-bromo-2-methyl-2-butanol}$ (vicinal bromohydrin) forms as a major organic product. Bromine atom gets attached to the $\text{(C1)}$ atom while the hydroxyl group gets attached to the $\text{(C2)}$ atom in $\text{2-methyl-2-butene}$.

f) The reaction is as follows:

Peroxyacids transfers oxygen to the double bond of an alkene to yield epoxides. An epoxide is a three-membered oxygen-containing ring.

When $\text{2-methyl-2-butene}$ reacts with peroxyacetic acid, $\text{2-methyl-2,3-epoxybutane}$ is the major organic product, where the oxygen atom has formed a three-membered ring with both the carbon atoms in an alkene.

g) The reaction is as follows:

Ozone is a powerful electrophile and reacts with alkenes to cleave the double bond, forming an ozonide.

$\text{2-methyl-2-butene}$ when treated with ozone forms corresponding ozonide as shown in the above reaction. The three oxygen atoms in ozone form a five-membered ring with the two carbon atoms containing the double bond in $\text{2-methyl-2-butene}$.

h) The reaction is as follows:

Ozonides are formed as a result of the reaction of ozone with an alkene. Ozonides undergo hydrolysis in the water giving carbonyl compounds. According to the structure of starting alkene, various carbonyl products are formed such as formaldehyde, aldehydes, or ketones.

When the corresponding ozonide of $\text{2-methyl-2-butene}$ reacts further with zinc in the presence of water, two major organic products are formed, acetone and acetaldehyde.

i) The reaction is as follows:

Ozonides are formed as a result of the reaction of ozone with an alkene. Ozonides undergo hydrolysis in water, giving carbonyl compounds. According to the structure of starting alkene, various carbonyl products are formed such as formaldehyde, aldehydes, or ketones.

When corresponding ozonide of $\text{2-methyl-2-butene}$ reacts further with dimethyl sulfide, two products are formed as major products acetone and acetaldehyde.