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Chapter 6: Problem 6

Propene \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\) can be converted into 1 -propanol by oxidation. Indicate which sets of reagents amongst the following is ideal to effect the above conversion. (A) \(\mathrm{KMnO}_{4}\) (alkaline) (B) Osmium tetroxide \(\left(\mathrm{OsO}_{4} / \mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) (C) \(\mathrm{B}_{2} \mathrm{H}_{6}\) and alkaline \(\mathrm{H}_{2} \mathrm{O}_{2}\) (D) \(\mathrm{O}_{3} / \mathrm{Zn}\)

Short Answer

Expert verified
The correct set of reagents to convert propene (CH3CH=CH2) into 1-propanol is (C) B2H6 and alkaline H2O2.

Step by step solution

01

Identify the functional groups

Propene is an alkene with a double bond between the second and third carbons. So, the desired transformation is to convert this double bond into an alcohol functional group (OH).
02

Examine each set of reagents

Now, let's look at each of the given reagents: (A) KMnO4 (alkaline): Potassium permanganate (KMnO4) is a strong oxidizing agent. In alkaline conditions, it will cleave the double bond and convert it to a diol, which is not what we want. (B) Osmium tetroxide (OsO4/CH2Cl2): Osmium tetroxide is also a strong oxidizing agent, but instead of cleaving the double bond, it will add two OH groups across the double bond, forming a diol. Again, this is not the desired transformation. (C) B2H6 and alkaline H2O2: Borane (B2H6) reduces the alkene by hydroboration, which adds the boron and a hydrogen atom across the double bond. In the next step, alkaline hydrogen peroxide (H2O2) oxidizes the boron to form an alcohol (OH) group. This is the desired transformation. (D) O3/Zn: Ozone (O3) will react with the alkene to form ozonides, which break down to carbonyl compounds (ketones or aldehydes) under reduction conditions (Zn). This is not the desired transformation.
03

Conclusion:

The correct set of reagents to convert propene (CH3CH=CH2) into 1-propanol is (C) B2H6 and alkaline H2O2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkene Functional Group Transformation
Transforming an alkene functional group involves converting the carbon-carbon double bond into another type of functional group. This is a fundamental process in organic chemistry, especially in synthesizing complex molecules. In our case, propene features a double bond between the second and third carbon atoms that we aim to transform into an alcohol, characterized by an OH group.

The double bond in an alkene is a reactive site where transformations can happen quite easily with the right reagents. It provides opportunities for various chemical reactions, including addition reactions, where atoms or groups are added across the double bond. In converting propene to 1-propanol, we specifically convert the double bond into an alcohol group via a chemical reaction. This is achieved through a reaction that first involves the addition of substances across the double bond, leading to the introduction of an OH group.
Hydroboration-Oxidation
Hydroboration-oxidation is a two-step organic reaction quite useful in converting alkenes into alcohols. The process begins with hydroboration, where borane ( B_2H_6 ) or derivatives are added across the double bond of the alkene.

  • During hydroboration, boron attaches to the less substituted carbon, which is the carbon with fewer hydrogen atoms.
  • The hydrogen from the borane adds to the more substituted carbon, where more carbon atoms are attached.

Following hydroboration, the oxidation step uses alkaline hydrogen peroxide ( H_2O_2 ). This results in the replacement of the boron atom with an OH group, thereby forming an alcohol. This method is advantageous as it proceeds without rearrangement of the carbon skeleton and is highly regioselective, meaning it consistently yields the alcohol on the same carbon atom regardless of the starting material's structure.
Organic Chemistry Reagents
Understanding the various reagents is crucial in organic chemistry as they dictate how and what transformations can occur in molecular structures. In the case of propene conversion to 1-propanol, the reagents play a key role in defining the reaction pathway.

  • Borane ( B_2H_6 ): Used for hydroboration, borane is an electron-deficient compound that willingly adds to alkenes.
  • Alkaline Hydrogen Peroxide ( H_2O_2 ): This provides the necessary conditions for the oxidative step, which replaces the boron with an OH group.

Each reagent has a specific function and together they achieve the desired transformation with precision. While other reagents like KMnO_4 and OsO_4 also interact with alkenes, their resultant products do not lead to the specific type of alcohol formation desired in this case. This highlights the importance of selecting the proper reagents for the transformation of interest.

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Most popular questions from this chapter

$$ (\mathrm{P})+\mathrm{NaOH} \longrightarrow(\mathrm{Q}) \stackrel{\mathrm{CH}_{3}-\mathrm{I}}{\longrightarrow}(\mathrm{R}) $$ (P) has molecular formula ' \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}\) ' and ( \(\mathrm{R}\) ) is aromatic in nature. Then choose the correct option(s). (A) (R) has common name anisole (B) In the conversion (P) is used as phenoxide moiety (C) (P) is known as carbolic acid (D) Order of electron density in ring \(:(\mathrm{Q})>(\mathrm{R})>\)

The decreasing order of rate of reaction is $$ \mathrm{R}-\mathrm{OH}+\mathrm{HCl} \stackrel{\mathrm{ZnC}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{H}_{2} \mathrm{O} $$ \((A) 1^{\circ}>2^{\circ}>3^{\circ}\) (B) \(1^{\circ}<2^{\circ}>3^{\circ}\) (C) \(3^{\circ}>2^{\circ}>1^{\circ}<\mathrm{CH}_{3} \mathrm{OH}\) (D) \(3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_{4} \mathrm{OH}\)

Choose the incorrect options if \((\mathrm{X})\) is and (P) is the minor hydrocarbon product which is obtained when (X) reacts with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \Delta\). (A) \(\mathrm{X} \longrightarrow{\mathrm{PCl}_{5}}{\longrightarrow} \stackrel{\Theta_{\mathrm{OAc}}}{\longrightarrow} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{P}\) (Major product) (B) \(\mathrm{X} \stackrel{\mathrm{PCl}_{3}}{\longrightarrow} \frac{\mathrm{AgF} /}{\mathrm{DMF}} \stackrel{\text { Alc. KOH }}{\longrightarrow} \mathrm{P}\) (Major product) (C) \((\mathrm{X})\) on reaction with red P/HI gives a compound which is optically active. (D) \((\mathrm{X})\) and \((\mathrm{P})\) both can show stereoisomerism.

The reactions which does not correctly match with major product is/are (A) \(\mathrm{R}-\mathrm{OH}+\mathrm{NaBr}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{R}-\mathrm{Br}+\mathrm{NaH}_{2} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{R}-\mathrm{OH}+\mathrm{NaI}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{R}-\mathrm{I}+\mathrm{NaH}_{2} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (C) \(\mathrm{R}-\mathrm{OH}+\mathrm{NaBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{R}-\mathrm{Br}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{R}-\mathrm{OH}+\mathrm{NaI}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{R}-\mathrm{I}+\mathrm{NaHSO}_{4}+\mathrm{H}_{2} \mathrm{O}\)

Ethoxy ethane is obtained as a major product in the reaction: (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{443 \mathrm{~K}}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}-\frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{413 \mathrm{~K}}\) (C) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \frac{(\mathrm{i}) \mathrm{Na}}{(\mathrm{i}) \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl}}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OCH}_{2}-\mathrm{CH}_{3} \underset{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}\)
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