Chapter 11, Problem 11.46AP

Interpretation Introduction

(a)

Interpretation:

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring-opening reactions which are acid-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are acidic, then the reaction will occur at the more substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown below.

Explanation of Solution

The compound $\text{2-ethyl-2-methyloxirane}$ undergoes ring opening reaction in the presence of acid. The water molecule acts a nucleophile and attacks on the more substituted carbon atom of the epoxide ring to form diol. The corresponding chemical reaction is shown below.

Figure 1

Conclusion

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and $\text{NaOH}$ followed by heat supply is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring opening reactions which are base-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are basic, then the reaction will occur at the less substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and $\text{NaOH}$ followed by heat supply is shown below.

Explanation of Solution

The compound $\text{2-ethyl-2-methyloxirane}$ undergoes ring-opening reaction in the presence of base. The water molecule acts a nucleophile and attacks on the less substituted carbon atom of the epoxide ring to from diol. The corresponding chemical reaction is shown below.

Figure 2

Conclusion

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with water and $\text{NaOH}$ followed by heat supply is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ in ${\text{CH}}_{\text{3}}\text{OH}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring-opening reactions which are base-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are basic, then the reaction will occur at the less substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ in ${\text{CH}}_{\text{3}}\text{OH}$ is shown below.

Explanation of Solution

The compound $\text{2-ethyl-2-methyloxirane}$ undergoes ring-opening reaction in the presence of base ${\text{CH}}_{\text{3}}{\text{O}}^{-}$. The methanol molecule acts a nucleophile and attacks on the less substituted carbon atom of the epoxide ring to from the product. The corresponding chemical reaction is shown below.

Figure 3

Conclusion

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{Na}}^{+}{\text{CH}}_{\text{3}}{\text{O}}^{-}$ in ${\text{CH}}_{\text{3}}\text{OH}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{CH}}_{\text{3}}\text{OH}$ and a catalytic amount of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring-opening reactions which are acid-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are acidic, then the reaction will occur at the more substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{CH}}_{\text{3}}\text{OH}$ and a catalytic amount of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

Explanation of Solution

The compound $\text{2-ethyl-2-methyloxirane}$ undergoes ring opening reaction in the presence of acid. The methanol molecule acts a nucleophile and attacks on the more substituted carbon atom of the epoxide ring to from diol. The corresponding chemical reaction is shown below.

Figure 4

Conclusion

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with ${\text{CH}}_{\text{3}}\text{OH}$ and a catalytic amount of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with dilithium dimethylcyanocuprate, followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring opening reactions which are acid-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are acidic, then the reaction will occur at the more substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with dilithium dimethylcyanocuprate, followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown below.

Explanation of Solution

The compound $\text{2-ethyl-2-methyloxirane}$ undergoes ring-opening reaction in the presence of acid. The dilithium dimethylcyanocuprate molecule generates a nucleophile $\left({}^{-}{\text{CH}}_{\text{3}}\right)$. This nucleophile will attack the more substituted carbon atom of the epoxide ring to from diol. The corresponding chemical reaction is shown below.

Figure 5

Conclusion

The product of the reaction of $\text{2-ethyl-2-methyloxirane}$ with dilithium dimethylcyanocuprate, followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The product of the reaction of the product of part (c) with $\text{HBr}$ at $25°\text{C}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring-opening reactions which are acid-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are acidic, then the reaction will occur at the more substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of the product of part (c) with $\text{HBr}$ at $25°\text{C}$ is shown below.

Explanation of Solution

In the presence of $\text{HBr}$, the methoxy-group and hydroxyl groups of $\text{1-methoxy-2-methybutan-2-ol}$ forms a protonated epoxy ring. The ring breaks after the addition of nucleophile ${}^{-}\text{Br}$ at more substituted carbon atom of the ring. The corresponding chemical reaction is shown below.

Figure 6

Conclusion

The product of the reaction of the given compound with $\text{HBr}$ at $25°\text{C}$ is shown in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The product of the reaction of the product of part (d) with $\text{HBr}$ at $25°\text{C}$ is to be predicted.

Concept introduction:

Epoxides undergo nucleophilic ring-opening reactions which are acid-catalyzed. If the epoxide is unsymmetrical, then the anionic nucleophile will attack the less-hindered carbon atom of the ring. If the reaction conditions are acidic, then the reaction will occur at the more substituted carbon atom.

Answer to Problem 11.46AP

The product of the reaction of the product of part (d) with $\text{HBr}$ at $25°\text{C}$ is shown below.

Explanation of Solution

In the presence of $\text{HBr}$, the methoxy-group and hydroxyl groups of $\text{2-methoxy-2-methybutan-1-ol}$ forms a protonated epoxy ring. The ring gets break after the addition of nucleophile ${}^{-}\text{Br}$ at the more substituted carbon atom of the ring. The corresponding chemical reaction is shown below.

Figure 7

Conclusion

The product of the reaction of the given compound with $\text{HBr}$ at $25°\text{C}$ is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The product of the reaction of the product of part (c) with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}\text{I}$ is to be predicted.

Concept introduction:

The metal hydride reagents are good reducing agents such as $\text{NaH}$. Sodium hydride is used in many organic syntheses as a reducing agent. Sodium hydride can also be used as a base in the organic syntheses reactions. It abstracts the acidic proton from the organic compounds.

Answer to Problem 11.46AP

The product of the reaction of the product of part (c) with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}\text{I}$ is shown below.

Explanation of Solution

The base $\text{NaH}$ abstract the proton of the hydroxyl group of $\text{1-methoxy-2-methybutan-2-ol}$ to form an anion. The anion acts as nucleophile and attacks on the carbon atom of ${\text{CH}}_{\text{3}}\text{I}$. The corresponding chemical reaction is shown below.

Figure 8

Conclusion

The product of the reaction of the given compound with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}\text{I}$ is shown in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The product of the reaction of the product of part (d) with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is to be predicted.

Concept introduction:

The metal hydride reagents are good reducing agents such as $\text{NaH}$. Sodium hydride is used in many organic syntheses as a reducing agent. Sodium hydride can also be used as a base in the organic syntheses reactions. It abstracts the acidic proton from the organic compounds.

Answer to Problem 11.46AP

The product of the reaction of the product of part (d) with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown below.

Explanation of Solution

The base $\text{NaH}$ abstract the proton of the hydroxyl group of $\text{2-methoxy-2-methybutan-1-ol}$ to form an anion. The anion acts as nucleophile and attacks on the carbon atom of ${\text{CH}}_{\text{3}}{\text{CH}}_2\text{I}$. The corresponding chemical reaction is shown below.

Figure 9

Conclusion

The product of the reaction of the given compound with $\text{NaH}$ followed by addition of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{I}$ is shown in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The product of the reaction of the product of part (a) with periodic acid is to be predicted.

Concept introduction:

The periodic acid acts as a strong oxidizing agent. The periodic acid reacts with a vicinal diol to form two aldehyde after breaking the carbon-carbon bond between the vicinal diol. The gem diol does not react with periodic acid as vicinal diol reacts.

Answer to Problem 11.46AP

The products of the reaction of the product of part (a) with periodic acid are shown below.

Explanation of Solution

The given compound is vicinal diol. It reacts with periodic acid to form two aldehydes. The carbon-carbon bond between the carbon atoms attached to two adjacent hydroxyl groups gets breaks. The corresponding chemical reaction is shown below.

Figure 10

Conclusion

The products of the reaction of the given compound with periodic acid are shown in Figure 10.

Interpretation Introduction

(k)

Interpretation:

The product of the reaction of the product of part (f) with $\text{Mg}$ in dry ether is to be predicted.

Concept introduction:

Grignard reagents are organometallic compounds which are prepared using alkyl halides in the presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases. Grignard reagents react with carbonyl compounds to form alcohol.

Answer to Problem 11.46AP

The product of the reaction of the product of part (f) with $\text{Mg}$ in dry ether is shown below.

Explanation of Solution

The compound $\text{2-bromo-1-methoxy-2-methylbutane}$ has a bromo group and methoxy group in it. The compound $\text{2-bromo-1-methoxy-2-methylbutane}$ reacts with magnesium metal to form a Grignard reagent. The corresponding chemical reaction is shown below.

Figure 11

Conclusion

The product of the reaction of the given compound with $\text{Mg}$ in dry ether is shown in Figure 11.

Interpretation Introduction

(l)

Interpretation:

The product of the reaction of the product of part (k) with ethylene oxide followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in dry ether is to be predicted.

Concept introduction:

Grignard reagents are organometallic compounds which are prepared using alkyl halides in the presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases. Grignard reagents react with carbonyl compounds to form alcohol.

Answer to Problem 11.46AP

The product of the reaction of the product of part (k) with ethylene oxide followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in dry ether is shown below.

Explanation of Solution

Grignard reagent can act as a nucleophile. In the presence of an acid, it can attack the more substituted carbon atom of the epoxy ring. The Grignard reagent reacts with the ethylene oxide followed by protonolysis to form alcohol. The corresponding chemical reaction is shown below.

Figure 12

Conclusion

The product of the reaction of the given compound with ethylene oxide followed by addition of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ in dry ether is shown in Figure 12.