Chapter 10, Problem 10.39AP

Interpretation Introduction

(a)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with concentrated aqueous ${\text{HBr, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst, heat is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an nucleophile. The rate of reaction depends upon the nucleophilicity and concentration of the nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and concentrated aqueous ${\text{HBr, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst, heat is $\text{2-bromobutane}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and concentrated aqueous ${\text{HBr, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst, heat is shown below.

Figure 1

The alcohol $\text{1-butanol}$ reacts with concentrated aqueous ${\text{HBr, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst, heat via ${\text{S}}_{\text{N}}\text{2}$ reaction pathway. The hydroxide group is protonated with the use of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst in order to make it a good leaving group. After protonation the bromide ion attacks from behind and eliminates the protonated hydroxide group as water to give $\text{2-bromobutane}$ as product.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and concentrated aqueous ${\text{HBr, H}}_{\text{2}}{\text{SO}}_{\text{4}}$ catalyst, heat is $\text{2-bromobutane}$ shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with cold aqueous ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and cold aqueous ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{but-1-ene}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and cold aqueous ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Figure 2

The alcohol $\text{1-butanol}$ 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 $\text{but-1-ene}$ 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 $\text{1-butanol}$ and cold aqueous ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{but-1-ene}$ shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with pyridinium chlorochromate (PCC) in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$

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 10.39AP

The products obtained on the reaction of $\text{1-butanol}$ and pyridinium chlorochromate (PCC) in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is butyraldehyde which is shown below.

Explanation of Solution

The products obtained on the reaction of $\text{1-butanol}$ and pyridinium chlorochromate (PCC) in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is shown below.

Figure 3

Primary alcohols on oxidation give aldehyde. Aldehydes can further undergo oxidation to give carboxylic acids. The oxidizing reagent pyridinium chlorochromate (PCC) in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ stops the process of oxidation of primary alcohol at the stage of aldehyde only therefore, oxidation of $\text{1-butanol}$ produces butyraldehyde as a product.

Conclusion

The products obtained on the reaction of $\text{1-butanol}$ and pyridinium chlorochromate (PCC) in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is butyraldehyde shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with $\text{NaH}$ is to be stated.

Concept introduction:

Acid-base reaction are among the fastest reaction in the chemistry. Acids and bases react 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 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and $\text{NaH}$ is sodium $\text{butan-1-olate}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and $\text{NaH}$ is shown below.

Figure 4

The sodium hydride is a very strong base and it undergoes acid-base reaction with alcohol $\text{1-butanol}$. The alcohol is acting as an acid due to the presence of polar hydrogen available to donate. The product obtained after the acid-base reaction is sodium $\text{butan-1-olate}$.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and $\text{NaH}$ is sodium $\text{butan-1-olate}$ shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The product expected when the product of part (d) is reacted with ${\text{CH}}_{\text{3}}\text{I}$ in DMSO is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an nucleophile. The rate of reaction depends upon the nucleophilicity and concentration of the nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 10.39AP

The product obtained on the reaction of sodium $\text{butan-1-olate}$ and ${\text{CH}}_{\text{3}}\text{I}$ in DMSO is $\text{1-methoxybutane}$ which is shown below.

Explanation of Solution

The product of part (d) is sodium $\text{butan-1-olate}$.

The product obtained on the reaction of sodium $\text{butan-1-olate}$ and ${\text{CH}}_{\text{3}}\text{I}$ in DMSO is shown below.

Figure 5

The sodium $\text{butan-1-olate}$ is basic in nature and also acts as nucleophile. The reaction undergoes via ${\text{S}}_{\text{N}}\text{2}$ pathway. $\text{butan-1-olate}$ ion attacks from behind on the ${\text{CH}}_{\text{3}}\text{I}$ to substitute the iodide ion and make it leave to give rise to as a product.

Conclusion

The product obtained on the reaction of sodium $\text{butan-1-olate}$ and ${\text{CH}}_{\text{3}}\text{I}$ in DMSO is $\text{1-methoxybutane}$ shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with $p\text{-toluenesulfonyl}$ chloride in pyridine is to be stated.

Concept introduction:

The hydroxide group in alcohols is not a good leaving group in order to 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 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and $p\text{-toluenesulfonyl}$ chloride in pyridine is $\text{butyl-4-methylbenzenesulfonate}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and $p\text{-toluenesulfonyl}$ chloride in pyridine is shown below.

Figure 6

The reaction of an alcohol with sulfonate derivatives such as methanesulfonyl chloride and $p\text{-toluenesulfonyl}$ chloride are done to make hydroxide group and good leaving group. Pyridine being more basic takes up the proton of the hydroxide group and make it a good nucleophile. This nucleophile then substitute the chloride group of methanesulfonyl chloride leading to the formation of an ester. Same happens here in this reaction $\text{1-butanol}$ reacts with $p\text{-toluenesulfonyl}$ chloride to produce an ester $\text{butyl-4-methylbenzenesulfonate}$.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and $p\text{-toluenesulfonyl}$ chloride in pyridine is $\text{butyl-4-methylbenzenesulfonate}$ shown in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with $\text{PrMgBr}$ in anhydrous ether is to be stated.

Concept introduction:

The reaction of an alkyl halide with a metal like magnesium in the presence of dry ether leads to the formation of $\text{RMgX}$ (organometallic compounds) also known as a Grignard reagent. These compounds are very sensitive to moisture or polar hydrogens and react immediately leading to the formation of alkane of the alkyl group.

Answer to Problem 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and $\text{PrMgBr}$ in anhydrous ether is propane which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and $\text{PrMgBr}$ in anhydrous ether is shown below.

Figure 7

The $\text{PrMgBr}$ is a grignard reagent and reacts very violently with polar hydrogen containing compounds. Alcohols contains polar hydrogen. Grignard reagent reacts with alcohols to give alkane. The alcohol $\text{1-butanol}$ reacts with Grignard reagent $\text{PrMgBr}$ to give propane as a product.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and $\text{PrMgBr}$ in anhydrous ether is propane shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with ${\text{SOCl}}_{\text{2}}$ in pyridine is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an nucleophile. The rate of reaction depends upon the nucleophilicity and concentration of the nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{SOCl}}_{\text{2}}$ in pyridine is $\text{1-chlorobutane}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{SOCl}}_{\text{2}}$ in pyridine is shown below.

Figure 8

The alcohol $\text{1-butanol}$ undergo ${\text{S}}_{\text{N}}\text{2}$ type reaction with ${\text{SOCl}}_{\text{2}}$ in pyridine. The reagent ${\text{SOCl}}_{\text{2}}$ in pyridine on reaction with alcohols give alkyl chloride as a product. Therefore, the product of reaction between $\text{1-butanol}$ and ${\text{SOCl}}_{\text{2}}$ in pyridine is $\text{1-chlorobutane}$.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{SOCl}}_{\text{2}}$ in pyridine is $\text{1-chlorobutane}$ shown above in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The product expected when $\text{1-butanol}$ reacts with ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF is to be stated.

Concept introduction:

The nucleophilic substitution reactions are the reactions in which a group is substituted by an nucleophile. The rate of reaction depends upon the nucleophilicity and concentration of the nucleophile.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution bimolecular reaction in which the addition of nucleophile and elimination of a leaving group takes place simultaneously. The rate of reaction depends upon both the substrate and reactant in the rate law equation.

Answer to Problem 10.39AP

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF is $\text{1-bromobutane}$ which is shown below.

Explanation of Solution

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF is shown below.

Figure 9

The alcohol $\text{1-butanol}$ undergo ${\text{S}}_{\text{N}}\text{2}$ type reaction with ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF. The reagent ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF on reaction with alcohols give alkyl bromide as a product. Therefore, the product of reaction between $\text{1-butanol}$ and ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF is $\text{1-bromobutane}$.

Conclusion

The product obtained on the reaction of $\text{1-butanol}$ and ${\text{Ph}}_{\text{3}}{\text{PBr}}_{\text{2}}$ in DMF is $\text{1-bromobutane}$ shown above in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The product expected when product of part (a) is reacted with $\text{Mg}$ in dry ether is to be stated.

Concept introduction:

The reaction of an alkyl halide with a metal like magnesium in the presence of dry ether leads to the formation of $\text{RMgX}$ (organometallic compounds) also known as a Grignard reagent. These compounds are very sensitive to moisture or polar hydrogens and react immediately leading to the formation of alkane of the alkyl group.

Answer to Problem 10.39AP

The product obtained on reaction of $\text{2-bromobutane}$ and $\text{Mg}$ in dry ether is $\text{sec-}$ butylmagnesium bromide which is shown below.

Explanation of Solution

The product of part (a) is $\text{2-bromobutane}$.

The product obtained on reaction of $\text{2-bromobutane}$ and $\text{Mg}$ in dry ether is shown below.

Figure 10

The alkyl halide reacts with magnesium metal in dry ether to produce the Grignard reagent. The $\text{2-bromobutane}$ reacts with the $\text{Mg}$ in dry ether to produce $\text{sec-}$ butylmagnesium bromide.

Conclusion

The product obtained on reaction of $\text{2-bromobutane}$ and $\text{Mg}$ in dry ether is $\text{sec-}$ butylmagnesium bromide shown in Figure 10.

Interpretation Introduction

(k)

Interpretation:

The product expected when the product of part (f) is reacted with ${\text{K}}^{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{–}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COH}$ 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 10.39AP

The product obtained on the reaction of $\text{butyl-4-methylbenzenesulfonate}$ and ${\text{K}}^{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{–}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COH}$ is $\text{but-1-ene}$ which is shown below.

Explanation of Solution

The product of part (f) is $\text{butyl-4-methylbenzenesulfonate}$.

The product obtained on the reaction of $\text{butyl-4-methylbenzenesulfonate}$ and ${\text{K}}^{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{–}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COH}$ is shown below.

Figure 11

The ester $\text{butyl-4-methylbenzenesulfonate}$ undergoes $\beta -\text{elimination}$ reaction with the strong base ${\text{K}}^{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{–}$. The base takes up the $\beta \text{-}$ proton of the alkyl substitutent which in turn eliminates the p-toluenesulfonate group or $\left(-\text{OTs}\right)$ group to give alkene $\text{but-1-ene}$ as product.

Conclusion

The product obtained on the reaction of $\text{butyl-4-methylbenzenesulfonate}$ and ${\text{K}}^{+}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-{\text{O}}^{–}$ in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COH}$ is $\text{but-1-ene}$ shown in Figure 11.

Interpretation Introduction

(l)

Interpretation:

The product expected when $\text{1-butanol}$ is reacted with triflic anhydride and pyridine is to be stated.

Concept introduction:

The hydroxide group in alcohols is not a good leaving group. In order to 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 $p\text{-toluenesulfonyl}$ chloride.

Answer to Problem 10.39AP

The product obtained when $\text{1-butanol}$ is reacted with triflic anhydride and pyridine is butyl trifluoromethanesulfonate which is shown below.

Explanation of Solution

The product obtained when $\text{1-butanol}$ is reacted with triflic anhydride and pyridine is shown below.

Figure 12

The triflic anhydride is also one of the compounds used to make hydroxide group a good leaving group. This compound is used much more reactive ester as compared to given by tosylates and mesylates. The yield of the product obtained by this ester of substitution is very high.

The $\text{1-butanol}$ reacts with triflic anhydride to give an ester (butyl trifluoromethanesulfonate) as a product.

Conclusion

The product obtained when $\text{1-butanol}$ is reacted with triflic anhydride and pyridine is butyl trifluoromethanesulfonate shown above in Figure 12.

Interpretation Introduction

(m)

Interpretation:

The product expected when the product of (l) is reacted with anhydrous ${\text{K}}^{+}{\text{F}}^{–}$ in acetonitrile is to be stated.

Concept introduction:

The hydroxide group in alcohols is not a good leaving group in order to 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 $p\text{-toluenesulfonyl}$ chloride.

Answer to Problem 10.39AP

The product obtained when butyl trifluoromethanesulfonate is reacted with anhydrous ${\text{K}}^{+}{\text{F}}^{–}$ in acetonitrile is $\text{1-Fluorobutane}$ which is shown below.

Explanation of Solution

The product obtained when butyl trifluoromethanesulfonate is reacted with anhydrous ${\text{K}}^{+}{\text{F}}^{–}$ in acetonitrile is shown below.

Figure 13

The compound butyl trifluoromethanesulfonate undergoes nucleophilic substitution reaction with ${\text{K}}^{+}{\text{ F}}^{–}$ to give $\text{1-Fluorobutane}$ as a product. Flouride ion acts as a nucleophile to substitute trifluoromethanesulfonate leaving group.

Conclusion

The product obtained when butyl trifluoromethanesulfonate is reacted with anhydrous ${\text{K}}^{+}{\text{ F}}^{–}$ in acetonitrile is $\text{1-Fluorobutane}$ shown in Figure 13.