Chapter 11, Problem 11.52P

Interpretation Introduction

(a)

Interpretation:

The major product of the given electrophilic addition reaction is to be predicted.

Concept introduction:

Strong Bronsted acids add to the double bond in a two-step electrophilic addition reaction to replace the $\text{π}$ bond with two $\text{σ}$ bonds. The reaction generally follows Markovnikov’s rule. The first step is electrophilic addition of ${\text{H}}^{+}$ to generate a carbocation. In case the double bond is not symmetric, the electrophile is added to the less substituted carbon so that the most stable carbocation is formed. The negatively charged conjugate base of the acid adds to the carbocation in the second step to give the product. If the conjugate base of the acid is neutral, a final, deprotonation step is necessary for the formation of a neutral product.

Answer to Problem 11.52P

The major product of the given reaction is

Explanation of Solution

The given reaction is $\text{4-Chlorobut-1-ene + HBr }\to \text{ ?}$

The substrate is an alkene, and the other reactant is the strong Bronsted acid HBr. The overall reaction is the addition of the acid to the double bond.

In the first step, the $\text{π}$ bond pair from the double bond breaks heterolytically and deprotonates the acid. This forms a carbocation. Out of the two possible carbocations that can form, the secondary carbocation on C2 is more stable than the primary carbocation on C1 and therefore forms preferentially. The proton (electrophile) is added to the primary carbon.

In the second step, the conjugate base ${\text{Br}}^{-}$ forms a bond with the carbocation to give the product $\text{3-bromo-1-chlorobutane}$.

Thus, for the given reaction, the major product formed is

Conclusion

Strong Bronsted acids add to a double bond, replacing the $\text{π}$ bond with two $\text{σ}$ bonds.

Interpretation Introduction

(b)

Interpretation:

The major product of the given electrophilic addition reaction is to be predicted.

Concept introduction:

In a two-step electrophilic addition reaction, the strong Bronsted acids add to the double bond to replace the $\text{π}$ bond with two $\text{σ}$ bonds. The reaction generally follows Markovnikov’s rule. The first step is electrophilic addition of ${\text{H}}^{+}$ to generate a carbocation. In case the double bond is not symmetric, the electrophile is added to the less substituted carbon so that the most stable carbocation is formed. The negatively charged conjugate base of the acid adds to the carbocation in the second step to give the product. If the conjugate base of the acid is neutral, a final, deprotonation step is necessary for the formation of a neutral product.

Answer to Problem 11.52P

For the given reaction, the major product formed is,

Explanation of Solution

The given electrophilic addition reaction is $\text{1-Chlorobut-1-ene + HBr }\to \text{ ?}$

The first step of the reaction is electrophilic addition of the proton from the acid. Two carbocations are possible when the $\text{π}$ bond pair breaks off to form a bond with the proton. One of these will have a positive charge on the terminal carbon, the one bonded to chlorine. The other carbocation will have a positive charge on the secondary carbon, the carbon next to the one that is bonded to chlorine. The first one is an unstable primary carbocation, which would be further destabilized by the electron-withdrawing effect of chlorine. The secondary carbocation will, therefore, form preferentially.

The conjugate base ${\text{Br}}^{-}$ forms a bond with the carbocation in the second step to give the final product.

Thus, for this reaction, the major product formed is,

Conclusion

Strong Bronsted acids add to a double bond, replacing the $\text{π}$ bond with two $\text{σ}$ bonds.

Interpretation Introduction

(c)

Interpretation:

The major product of the given electrophilic addition reaction is to be predicted.

Concept introduction:

In a two-step electrophilic addition reaction, the strong Bronsted acids add to the double bond to replace the $\text{π}$ bond with two $\text{σ}$ bonds. The reaction generally follows Markovnikov’s rule. The first step is electrophilic addition of ${\text{H}}^{+}$ to generate a carbocation. In case the double bond is not symmetric, the electrophile is added to the less substituted carbon so that the most stable carbocation is formed. The negatively charged conjugate base of the acid adds to the carbocation in the second step to give the product. If the conjugate base of the acid is neutral, a final, deprotonation step is necessary for the formation of a neutral product.

Answer to Problem 11.52P

For this reaction, the major product formed is,

Explanation of Solution

The given reaction is ${\text{4, 4-dimethylcyclopentene + H}}_{\text{2}}{\text{O, H}}^{\text{+}}\to ?$

The overall reaction is one of hydration of the double bond, i.e., the addition of water. Water is a weak acid, but in the presence of a strong acid (denoted by the ${\text{H}}^{\text{+}}$), it is protonated to the strong acid ${\text{H}}_3{\text{O}}^{\text{+}}$. This transfers a proton to the alkene in an electrophilic addition step to produce a carbocation. The double bond is symmetrically substituted, so only one carbocation can form.

In the next step, the conjugate base ${\text{H}}_{\text{2}}\text{O}$ adds to the carbocation through one of the lone pairs on the oxygen atom. This produces a protonated alcohol. A deprotonation step then gives the final, neutral product.

Thus, for this reaction, the major product formed is,

Conclusion

Strong Bronsted acids add to a double bond, replacing the $\text{π}$ bond with two $\text{σ}$ bonds.

Interpretation Introduction

(d)

Interpretation:

The major product of the given electrophilic addition reaction is to be predicted.

Concept introduction:

In a two-step electrophilic addition reaction, the strong Bronsted acids add to the double bond to replace the $\text{π}$ bond with two $\text{σ}$ bonds. Since there are two $\text{π}$ bonds in an alkyne, the reaction occurs twice, provided the acid is present in excess. The reaction generally follows Markovnikov’s rule. The first step is electrophilic addition of ${\text{H}}^{+}$ to generate a carbocation. In case the double bond is not symmetric, the electrophile is added to the less substituted carbon so that the most stable carbocation is formed. The negatively charged conjugate base of the acid adds to the carbocation in the second step to give the product. If the conjugate base of the acid is neutral, a final, deprotonation step is necessary for the formation of a neutral product.

Answer to Problem 11.52P

For the given reaction, the major product formed is,

Explanation of Solution

The given reaction is $\text{propyne}+2\text{HCl}\to ?$

There are two $\text{π}$ bonds in propyne, so two moles of HCl will react with it in successive electrophilic addition reactions.

In the first step, an alkenyl carbocation is formed by addition of proton the triple bond. A more stable carbocation, with the positive charge on the secondary carbon, is produced.

In the second step, the conjugate base ${\text{Cl}}^{-}$ adds to the carbocation to form $\text{2-chloropropene}$.

Since the product has a $\text{π}$ bond and another mole of HCl is present, a second addition reaction will occur to produce the final product.

Thus, for this reaction, the major product formed is,

Conclusion

Strong Bronsted acids add twice to a triple bond, replacing each $\text{π}$ bond with two $\text{σ}$ bonds.

Interpretation Introduction

(e)

Interpretation:

The major product of the given electrophilic addition reaction is to be predicted.

Concept introduction:

In a two-step electrophilic addition reaction, the strong Bronsted acids add to the double bond to replace the $\text{π}$ bond with two $\text{σ}$ bonds. The reaction generally follows Markovnikov’s rule. The first step is electrophilic addition of ${\text{H}}^{+}$ to generate a carbocation. In case the double bond is not symmetric, the electrophile is added to the less substituted carbon so that the most stable carbocation is formed. The negatively charged conjugate base of the acid adds to the carbocation in the second step to give the product. If the conjugate base of the acid is neutral, a final, deprotonation step is necessary for the formation of a neutral product.

Answer to Problem 11.52P

For the given reaction, the major product formed is,

Explanation of Solution

The given electrophilic addition reaction is

The acid in this case is ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$. The $\text{π}$ bond pair from cyclopentylethene will deprotonate ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ to form a carbocation. Since the double bond is a terminal double bond, two carbocations are possible, one of them a primary carbocation and the other a secondary carbocation. The secondary carbocation will form preferentially as it is more stable.

This carbocation is prone to rearrangement. A $\text{1, 2-hydride}$ shift from the carbon at the point of attachment to the ring will produce the most stable tertiary carbocation.

The conjugate base ${\text{H}}_{\text{2}}\text{O}$ will then add to this carbocation to form a protonated tertiary alcohol. This is followed by its deprotonation to form the final product.

Thus, for the given reaction, the major product formed is,

Conclusion

Strong Bronsted acids add to a double bond, replacing the $\text{π}$ bond with two $\text{σ}$ bonds.