Chapter 10, Problem 10.40AP

Interpretation Introduction

(a)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with concentrated aqueous $\text{HCl}$ is to be stated.

Concept introduction:

Alcohols undergo nucleophilic substitution reaction in the presence of halogen acids. The hydroxide group leaves after protonation and halogen group comes in via ${\text{S}}_{\text{N}}\text{2}$ pathway making protonated hydroxide group leave as water to give alkyl halide.

Answer to Problem 10.40AP

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and concentrated aqueous $\text{HCl}$ is $\text{2-chloro-2-methylpropane}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and concentrated aqueous $\text{HCl}$ is shown below obtained via nucleophilic substitution.

Figure 1

The $\text{2-methyl-2-propanol}$ undergo nucleophilic substitution reaction in the presence of $\text{HCl}$. The hydroxide group leaves after protonation and chloride group comes in via ${\text{S}}_{\text{N}}\text{2}$ pathway attacking from behind and making protonated hydroxide group leave as water to give $\text{2-chloro-2-methylpropane}$.

Conclusion

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and concentrated aqueous $\text{HCl}$ is $\text{2-chloro-2-methylpropane}$ which is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with ${\text{CrO}}_{\text{3}}$ in pyridine is to be stated.

Concept introduction:

Primary and secondary alcohols can be oxidized into aldehydes and ketones using $\text{Cr(VI)}$ compounds as they are very good oxidizing agents. Few forms of chromium used to oxidize are chromate $\left({\text{CrO}}_{\text{4}}^{2-}\right)$ and dichromate $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$ and chromium trioxide $\left({\text{CrO}}_{\text{3}}\right)$. Primary alcohols are oxidized into aldehydes and secondary alcohols are oxidized into ketones.

Answer to Problem 10.40AP

No product is obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{CrO}}_{\text{3}}$ in pyridine.

Explanation of Solution

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{CrO}}_{\text{3}}$ in pyridine is shown below.

Figure 2

No reaction takes place between $\text{2-methyl-2-propanol}$ and ${\text{CrO}}_{\text{3}}$ in pyridine. The ${\text{CrO}}_{\text{3}}$ in pyridine is oxidizing reagent which oxidizes alcohols to aldehydes and ketones. This reagent have no effect on the tertiary alcohols. Tertiary alcohols cannot be oxidized into aldehyde and ketones. $\text{2-methyl-2-propanol}$ is a tertiary alcohol.

Conclusion

No product is obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{CrO}}_{\text{3}}$ in pyridine because $\text{2-methyl-2-propanol}$ is a tertiary alcohol.

Interpretation Introduction

(c)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, heat is to be stated.

Concept introduction:

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the base and simultaneously leaving the group is eliminated. An alkene is obtained as a product of $\beta -\text{elimination}$ reaction.

Answer to Problem 10.40AP

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, heat is $\text{2-methylprop-1-ene}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, heat is obtained via $\text{E2}$ pathway is shown below.

Figure 3

The alcohol $\text{2-methyl-2-propanol}$ is first protonated by the acid ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. After this step, the conjugate base of acid takes up the $\beta \text{-}$ proton from the alcohol and simultaneously removes the protonated hydroxide group to give $\text{2-methylprop-1-ene}$ as a product. This reaction is also known as dehydration of alcohols to alkenes. This reaction is a temperature controlled reversible reaction.

Conclusion

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, heat is $\text{2-methylprop-1-ene}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (dark) is to be stated.

Concept introduction:

Free radical reaction are the reaction in which the bond between the molecule is broken homolytically due to the presence of light energy. These kind of reactions are generally obtained with alkenes when reacted with halogen in the presence of light or peroxides.

Answer to Problem 10.40AP

No product is obtained when $\text{2-methyl-2-propanol}$ reacts with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (dark).

Explanation of Solution

No reaction product is obtained on the reaction of $\text{2-methyl-2-propanol}$ and ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (dark). This is because the ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (dark) is a reagent which proceeds via free radical mechanism. The alcohols do not undergo any kind of free radical reactions.

Conclusion

No product is obtained when $\text{2-methyl-2-propanol}$ reacts with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ (dark) because alcohols do not undergo any kind of free radical reaction.

Interpretation Introduction

(e)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with potassium metal is to be stated.

Concept introduction:

Acid-base reaction are among the fastest reaction in the chemistry. Acids and bases reacts vigorously generating heat and water normally. Metals are basic in nature due to the presence of free electrons to donate. Alcohols are both acidic and basic in nature.

Answer to Problem 10.40AP

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and potassium metal is potassium $\text{2-methylpropan-2-olate}$ is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and potassium metal is shown below.

Figure 4

The alcohol $\text{2-methyl-2-propanol}$ acts as acid here and donates its proton to potassium metal to donates its electrons to it. The hydrogen ion is then converted into hydrogen radical and two hydrogen radicals combine to give hydrogen. Metals react with acids to liberate salt and hydrogen gas.

Conclusion

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and potassium metal is potassium $\text{2-methylpropan-2-olate}$ is shown in Figure 4.

Interpretation Introduction

(f)

Interpretation:

The product expected when $\text{2-methyl-2-propanol}$ reacts with methanesulfonyl chloride in pyridine is to be stated.

Concept introduction:

The hydroxide group in alcohols is not a good leaving group that can perform a nucleophilic substitution reaction on alcohols to produce more compounds. Hydroxide group is made a good leaving group with the help of some compounds like methanesulfonyl chloride and p-toluenesulfonyl chloride.

Answer to Problem 10.40AP

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and methanesulfonyl chloride is $\text{tert-butylmethanesulfonate}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and methanesulfonyl chloride is shown below.

Figure 5

The reaction of an alcohol with sulfonate derivatives such as methanesulfonyl chloride and p-toluenesulfonyl chloride are done to make hydroxide group and good leaving group. Pyridine being more basic takes up the proton of the hydroxide group make it a good nucleophile at first. This nucleophiles then substitute the chloride group of methanesulfonyl chloride leading to the formation of an ester. Same happens here in this reaction $\text{2-methyl-2-propanol}$ reacts methanesulfonyl chloride to produce an ester $\text{tert-butylmethanesulfonate}$.

Conclusion

The product obtained on the reaction of $\text{2-methyl-2-propanol}$ and methanesulfonyl chloride is $\text{tert-butylmethanesulfonate}$ shown in Figure 5.

Interpretation Introduction

(g)

Interpretation:

The product expected when the product of part (f) is reacted with $\text{NaOH}$ in DMSO is to be stated.

Concept introduction:

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the base and simultaneously leaving the group is eliminated. An alkene is obtained as a product of $\beta -\text{elimination}$ reaction.

Answer to Problem 10.40AP

The product expected when the product of part (f) is reacted with $\text{NaOH}$ in DMSO is $\text{2-methylprop-1-ene}$ which is shown below.

Explanation of Solution

The product obtained in part (f) is $\text{tert-butylmethanesulfonate}$.

The product obtained on the reaction of $\text{tert-butylmethanesulfonate}$ and $\text{NaOH}$ in DMSO is shown below.

Figure 6

The $\text{tert-butylmethanesulfonate}$ undergoes $\beta -\text{elimination}$ reaction on reaction with $\text{NaOH}$ in DMSO. This reaction undergoes $\text{E2}$ reaction pathway because of steric hindrance of the tertiary group to not allow substitution reaction. The $\beta \text{-}$ proton from the $\text{tert-butylmethanesulfonate}$ is taken by the base and in turn, eliminates the methylsulfonate group to yield $\text{2-methylprop-1-ene}$ as a product.

Conclusion

The product expected when the product of part (f) is reacted with $\text{NaOH}$ in DMSO is $\text{2-methylprop-1-ene}$ shown in Figure 6.

Interpretation Introduction

(h)

Interpretation:

The product expected when the product of part (e) is reacted with the product of part (a) is to be stated.

Concept introduction:

An $\text{E2}$ reaction is a base-promoted $\beta -\text{elimination}$ reaction. This reaction follows a concerted reaction mechanism. In this reaction, a $\beta \text{-}$ proton is taken by the base and simultaneously leaving group is eliminated. An alkene is obtained as a product of $\beta -\text{elimination}$ reaction.

Answer to Problem 10.40AP

The products obtained on the reaction of $\text{2-chloro-2-methylpropane}$ and potassium $\text{2-methylpropan-2-olate}$ are $\text{2-methylprop-1-ene}$ and $\text{2-methyl-2-propanol}$ which is shown below.

Explanation of Solution

The product of part (a) is $\text{2-chloro-2-methylpropane}$ and the product of part (e) is potassium $\text{2-methylpropan-2-olate}$.

The products obtained on the reaction of $\text{2-chloro-2-methylpropane}$ and potassium $\text{2-methylpropan-2-olate}$ are shown below.

Figure 7

The reaction undergoes via $\text{E2}$ reaction pathway. The $\beta \text{-}$ proton alkyl halide $\text{2-chloro-2-methylpropane}$ is taken away by the base $\text{2-methylpropan-2-olate}$ and simultaneously removes the chloride group to give $\text{2-methylprop-1-ene}$ as a product. and $\text{2-methylpropan-2-olate}$ itself convert into the alcohol $\text{2-methyl-2-propanol}$.

Conclusion

The products obtained on the reaction of $\text{2-chloro-2-methylpropane}$ and potassium $\text{2-methylpropan-2-olate}$ are $\text{2-methylprop-1-ene}$ and $\text{2-methyl-2-propanol}$ shown in Figure 7.