Chapter 11, Problem 11.44AP

Interpretation Introduction

(a)

Interpretation:

The structure of the compound which is a nine-carbon ether and can not be prepared by the Williamson synthesis is to be predicted.

Concept introduction:

Williamson synthesis is a way of preparing ethers by the reaction between an alkyl halide and alkoxide. Williamson synthesis takes place through ${\text{S}}_{\text{N}}2$ mechanism. The carbon atom at which the substitution takes place undergoes inversion of configuration.

Answer to Problem 11.44AP

The structure of the compound which is a nine-carbon ether that can not be prepared by the Williamson synthesis is shown below.

Explanation of Solution

The number of carbon atoms in the ether is $9$.

The general formula of ether is $\text{R'}-\text{O}-\text{R}$.

Tertiary ether cannot be synthesis by Williamson synthesis process. Therefore, the expected structure of ether that cannot be synthesized by Williamson synthesis with nine carbon atoms is shown below.

Figure 1

Conclusion

The structure of the compound which is nine-carbon ether that could not be prepared by the Williamson synthesis is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The structure of the compound which is a nine-carbon ether and can be prepared by the Williamson synthesis is to be predicted.

Concept introduction:

Williamson synthesis is a way of preparing ethers by the reaction between an alkyl halide and alkoxide. Williamson synthesis takes place through ${\text{S}}_{\text{N}}2$ mechanism. The carbon atom at which the substitution takes place undergoes inversion of configuration.

Answer to Problem 11.44AP

The structure of the compound which is a nine-carbon ether and can be prepared by the Williamson synthesis is shown below.

Explanation of Solution

The number of carbon atoms in the ether is $9$.

The general formula of ether is $\text{R'}-\text{O}-\text{R}$.

Tertiary ether cannot be synthesis by Williamson synthesis process, however it can be synthesis primary ether easily. Therefore, the expected structures of ether that can be synthesized by Williamson synthesis with nine carbon atoms are shown below.

Figure 2

Conclusion

The structure of the compound which is nine-carbon ether that could be prepared by the Williamson synthesis is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The structure of the compound which is a four-carbon ether that would yield $1,4\text{-diiodobutane}$ after heating with an excess of $\text{HI}$ is to be predicted.

Concept introduction:

The ether reacts with a strong acid such as hydrogen iodide to give an alcohol and alkyl halide. A general reaction between hydrogen iodide and ether is represented as shown below.

$\text{ROR'}+\text{HI}\to \text{ROH}+\text{R'I}$

Answer to Problem 11.44AP

The structure of the compound which is a four-carbon ether that would yield $1,4\text{-diiodobutane}$ after heating with an excess of $\text{HI}$ is shown below.

Explanation of Solution

The number of carbon atoms in the ether is $4$.

The formation of $1,4\text{-diiodobutane}$ after heating with an excess of $\text{HI}$ indicates that the compound is cyclic ether. When tetrahydrofuran reacts with excess of $\text{HI}$, it produces $1,4\text{-diiodobutane}$. The corresponding chemical reaction is shown below.

Figure 3

Conclusion

The structure of the compound which is a four-carbon ether that would yield $1,4\text{-diiodobutane}$ after heating with an excess of $\text{HI}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The structure of an ether that would react with $\text{HBr}$ to give propyl bromide as the only alkyl halide is to be predicted.

Concept introduction:

The ether reacts with a strong acid such as hydrogen bromide to give an alcohol and alkyl halide. A general reaction between hydrogen bromide and ether is represented as shown below.

$\text{ROR'}+\text{HBr}\to \text{ROH}+\text{R'Br}$

Answer to Problem 11.44AP

The structure of an ether that would react with $\text{HBr}$ to give propyl bromide as the only alkyl halide is shown below.

Explanation of Solution

The number of carbon atoms in the ether is $4$.

The formation of propyl bromide after reaction $\text{HBr}$ indicates that the compound is symmetrical ether. When dipropylether reacts with excess of $\text{HBr}$, it produces propyl bromide. The corresponding chemical reaction is shown below.

Figure 4

Conclusion

The structure of an ether that would react with $\text{HBr}$ to give propyl bromide as the only alkyl halide is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The structure of the compound which is a four-carbon alkene that would give different glycols after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is to be predicted.

Concept introduction:

Potassium permanganate is a strong oxidizing agent. Potassium permanganate can oxidize an alkene into alcohol. It can also oxidize an alcohol into carbonyl compound and the carbonyl compound can be further oxidized to form a carboxylic acid.

Answer to Problem 11.44AP

The structure of the compound which is a four-carbon alkene that would give different glycols after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is shown below.

Explanation of Solution

The number of carbon atom in the alkene is four.

The formation of different glycols after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ indicates that the alkene has symmetry group attached to it. The attachement of symmetrical group around double bond gives cis and trans isomers.

The alkene $\text{but-2-ene}$ reacts with alkaline ${\text{KMnO}}_{\text{4}}$ to form cis and trans glycols. The corresponding chemical reaction is shown below.

Figure 5

The alkene $\text{but-2-ene}$ reacts with with meta-chloroperoxybenzoic acid followed by dilute aqueous acid to two glycols. The corresponding chemical reaction is shown below.

Figure 6

Conclusion

The structure of the compound which is a four-carbon alkene that would give different glycols after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The structure of the compound which a four-carbon alkene that would give the same glycol after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is to be predicted.

Concept introduction:

Potassium permanganate is a strong oxidizing agent. Potassium permanganate can oxidize an alkene into alcohol. It can also oxidize an alcohol into carbonyl compound and the carbonyl compound can be further oxidized to form a carboxylic acid.

Answer to Problem 11.44AP

The structure of the compound which a four-carbon alkene that would give the same glycol after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is shown below.

Explanation of Solution

The number of carbon atom in the alkene is four.

The formation of same glycols after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ and meta-chloroperoxybenzoic indicates that the alkene part is present at the termical carbon atom.

The alkene $\text{but-1-ene}$ reacts with alkaline ${\text{KMnO}}_{\text{4}}$ or meta-chloroperoxybenzoicacid followed by dilute aqueous acid to form same glycol. The corresponding chemical reactions are shown below.

Figure 7

Conclusion

The structure of the compound which a four-carbon alkene that would give the same glycol after treatment with alkaline ${\text{KMnO}}_{\text{4}}$ or treatment with meta-chloroperoxybenzoic acid followed by dilute aqueous acid is shown in Figure 7.

Interpretation Introduction

(g)

Interpretation:

The structure of a diene ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}$ that can form only one mono-epoxide and two di-epoxides is to be predicted.

Concept introduction:

Degree of unsaturation in a compound gives the number of double bonds and ring present in a compound. Degree of unsaturation is used to predict the structure of the compound. Degree of unsaturation is calculated by the formula as shown below.

$\text{Degree}\text{of}\text{unsaturation}=\frac{2N_{\text{C}}+2+N_{\text{N}}-N_{\text{X}}-N_{\text{H}}}{2}$

Answer to Problem 11.44AP

The structure of a diene ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}$ that can form only one mono-epoxide and two di-epoxides is shown below.

Explanation of Solution

The number of carbon atom in the diene is six.

The molecular fomula of dieneis ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}$.

Degree of unsaturation is calculated by the formula as shown below.

$\text{Degree}\text{of}\text{unsaturation}=\frac{2N_{\text{C}}+2+N_{\text{N}}-N_{\text{X}}-N_{\text{H}}}{2}$

Where,

• $N_{\text{C}}$ is the number of carbon atoms.
• $N_{\text{N}}$ is the number of nitrogen atoms.
• $N_{\text{X}}$ is the number of halogen atoms.
• $N_{\text{H}}$ is the number of hydrogen atoms.

Substitute the values of $N_{\text{C}}$, $N_{\text{N}}$, $N_{\text{X}}$ and $N_{\text{H}}$ in the above expression.

$\begin{array}{rll}\text{Degree}\text{of}\text{unsaturation}& =& \frac{2\times 6+2+0-\text{0}-8}{2}\\ & =& 3\end{array}$

The value of degree of unsaturation indicates that the diene also has a ring in it.

Therefore, the structure of a diene with molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}$ is shown below.

Figure 8

Conclusion

The structure of a diene ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}$ that can form only one mono-epoxide and two di-epoxides is shown in Figure 8.

Interpretation Introduction

(h)

Interpretation:

The structure of an alkene ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ that would give the same glycol either from treatment with a peroxycarboxylic acid, followed by acid catalyzed hydrolysis, or from glycol formation with ${\text{OsO}}_{\text{4}}$ is to be predicted.

Concept introduction:

Addition of two hydroxyl groups on double bond to form $1,2\text{-diol}$ is known as dihydroxylation. In presence of ${\text{OsO}}_{\text{4}}$, syndihydroxylation takes place. In this reaction two hydroxyl groups are added on the same side of the double bond.The ${\text{OsO}}_{\text{4}}$ is a toxic and expensive reagent.

Answer to Problem 11.44AP

The structure of the compound which an alkene ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ that would give the same glycol either from treatment with a peroxycarboxylic acid, followed by acid catalyzed hydrolysis, or from glycol formation with ${\text{OsO}}_{\text{4}}$ is shown below.

Explanation of Solution

The number of carbon atom in the alkene is six.

The molecular formula of alkene is ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$.

The formation of same glycols after oxidation with peroxycarboxylic acid, acid catalyzed hydrolysis or reaction with ${\text{OsO}}_{\text{4}}$ the alkene part is present at the termical carbon atom.

Therefore, the expected sturture of alkene with molecular formula ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ is shown below.

Fgure 9

Conclusion

The structure of the compound which an alkene ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ that would give the same glycol either from treatment with a peroxycarboxylic acid, followed by acid catalyzed hydrolysis, or from glycol formation with ${\text{OsO}}_{\text{4}}$ is shown in Figure 9.