Chapter 9, Problem 9.44AP

Interpretation Introduction

(a)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can exist as enantiomers are to be identified.

Concept introduction:

The compounds which have the same molecular formula but have different arrangements of atoms are known as isomers. The phenomenon is called isomerism. The isomers are generally classified as structural isomers and stereoisomers. Stereoisomers are further divided into two categories diastereomers and enantiomers.

Answer to Problem 9.44AP

The isomers of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ that can exist as enantiomers are $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The compounds that have chiral centers can exist as enantiomers. The compound $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$ have chiral centers in them. The chiral centers of $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$ are indicated with star mark as shown below.

Figure 2

Therefore, $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$ can exist as enantiomers.

Conclusion

The compounds $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$ can exist as enantiomers.

Interpretation Introduction

(b)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can exist as diastereomers are to be identified.

Concept introduction:

The compounds which have the same molecular formula but have different arrangements of atoms are known as isomers. The phenomenon is called isomerism. The isomers are generally classified as structural isomers and stereoisomers. Stereoisomers are further divided into two categories diastereomers and enantiomers.

Answer to Problem 9.44AP

The isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ that can exist as diastereomers is $\text{2-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The compounds that have two chiral centers can exist as enantiomers. The compound $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$ have chiral centers in them. The chiral centers of $\text{2-bromo-3-methylpentane}$ are indicated with star mark as shown below.

Figure 3

Therefore, $\text{2-bromo-3-methylpentane}$ can exist as diastereomers.

Conclusion

The compound $\text{2-bromo-3-methylpentane}$ can exist as diastereomers.

Interpretation Introduction

(c)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can give the fastest ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide are to be identified.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving the group by attacking on the electron-deficient carbon atom.

Answer to Problem 9.44AP

The fastest ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide is of $\text{1-bromohexane}$ among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction primary halide is greater than that of secondary halides. The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction secondary halide is greater than that of tertiary halides. The reason for this trend is the steric hindrance. The steric hindrance is very less than in the case of primary halide than secondary and tertiary halides.

The compound $\text{1-bromohexane}$ is primary halide with lowest steric hindrance. Therefore, $\text{1-bromohexane}$ undergoes the fastest ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide among the isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$.

Conclusion

The compound $\text{1-bromohexane}$ has fastest rate of ${\text{S}}_{\text{N}}\text{2}$ reaction.

Interpretation Introduction

(d)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can is least reactive to sodium methoxide in methanol are to be identified.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving the group by attacking on the electron-deficient carbon atom.

Answer to Problem 9.44AP

The isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ which has least reactivity towards sodium methoxide in methanol is $\text{1-bromo-2, 2-dimethy1butane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The compound $\text{1-bromo-2, 2-dimethy1butane}$ has preferred to undergo an elimination reaction to form alkene instead to undergo substitution reaction due to the bulky alkyl part. Therefore, the reactivity of $\text{1-bromo-2, 2-dimethy1butane}$ towards sodium methoxide in methanol is lowest among the reactivity of rest of the isomers of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$.

Conclusion

The compound, $\text{1-bromo-2, 2-dimethy1butane}$, has the least reactivity towards sodium methoxide in methanol.

Interpretation Introduction

(e)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can give only one alkene in the $\text{E2}$ reaction with ${\text{K}}^{+}{}^{-}\text{OtBu}$ are to be identified.

Concept introduction:

The elimination reaction of alkyl halide involves removal of the halogen atom and hydrogen atom from the adjacent carbon atoms, which leads to the formation of the alkene. A bulky base increases the chance of elimination reaction of substitution reaction.

Answer to Problem 9.44AP

The compound $\text{1-bromohexane}$ will give only one alkene after $\text{E2}$ reaction with ${\text{K}}^{+}{}^{-}\text{OtBu}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The compound $\text{1-bromohexane}$ will give only one alkene after $\text{E2}$ reaction with ${\text{K}}^{+}{}^{-}\text{OtBu}$ because the halide is attached to the terminal carbon atom. The possibilities of rearrangement in the $\text{E2}$ reaction are least for this compound.

Conclusion

The $\text{E2}$ reaction of $\text{1-bromohexane}$ with ${\text{K}}^{+}{}^{-}\text{OtBu}$ will produce only one alkene.

Interpretation Introduction

(f)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can that give an $\text{E2}$ but no ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide in methanol are to be identified.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving the group by attacking on the electron-deficient carbon atom.

Answer to Problem 9.44AP

The alkyl halide among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can that give an $\text{E2}$ but no ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide in methanol is $\text{3-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction varies primary halide greater than secondary halides. The rate of ${\text{S}}_{\text{N}}\text{2}$ reaction varies secondary halide greater than tertiary halides. The reason for this trend is the steric hindrance. The steric hindrance is very less than in case of primary halide than secondary and tertiary halides.

The steric hindrance in case of $\text{3-bromo-3-methylpentane}$ is maximum. Therefore, it will not undergo ${\text{S}}_{\text{N}}\text{2}$. The compound, $\text{3-bromo-3-methylpentane}$, can easily undergoes $\text{E2}$ reaction to form alkene.

Conclusion

The isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can that give an $\text{E2}$ but no ${\text{S}}_{\text{N}}\text{2}$ reaction with sodium methoxide in methanol is $\text{3-bromo-3-methylpentane}$.

Interpretation Introduction

(g)

Interpretation:

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can undergo an ${\text{S}}_{\text{N}}1$ reaction to give rearranged products are to be identified.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving the group by attacking on the electron-deficient carbon atom.

Answer to Problem 9.44AP

The alkyl halides among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can undergo an ${\text{S}}_{\text{N}}1$ reaction to give rearranged products are $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The stability of tertiary carbocation is more than the stability of secondary carbocation. Therefore, in case of $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$, the secondary carbocation formed will undergo rearrangement to form a more stable tertiary carbocation.

Conclusion

The isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can undergo an ${\text{S}}_{\text{N}}1$ reaction to give rearranged products are $\text{3-bromo-2-methylpentane}$ and $\text{2-bromo-3-methylpentane}$.

Interpretation Introduction

(h)

Interpretation:

The alkyl halide among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can gives the fastest ${\text{S}}_{\text{N}}1$ reaction is to be identified.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving the group by attacking on the electron-deficient carbon atom.

Answer to Problem 9.44AP

The alkyl halide among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can gives the fastest ${\text{S}}_{\text{N}}1$ reaction is $\text{3-bromo-3-methylpentane}$.

Explanation of Solution

The structures of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ isomers are shown below.

Figure 1

The rate of ${\text{S}}_{\text{N}}1$ reaction primary halide is lower than that of secondary halide. The rate of ${\text{S}}_{\text{N}}1$ reaction secondary halide is lower than that of tertiary halides. The reason for this trend is the formation of a stable carbocation. The compound $\text{3-bromo-3-methylpentane}$ a stable tertiary carbocation.

Therefore, $\text{3-bromo-3-methylpentane}$ undergoes the fastest ${\text{S}}_{\text{N}}1$ reaction among the isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$.

Conclusion

The isomer of ${\text{C}}_{\text{6}}{\text{H}}_{\text{13}}\text{Br}$ among $\text{1-bromohexane}$, $\text{3-bromo-3-methylpentane}$, $\text{1-bromo-2, 2-dimethy1butane}$, $\text{3-bromo-2-methylpentane}$, and $\text{2-bromo-3-methylpentane}$ that can gives the fastest ${\text{S}}_{\text{N}}1$ reaction is $\text{3-bromo-3-methylpentane}$.