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The reagent required to convert propene to 1 -propanol is (a) \(\mathrm{B}_{2} \mathrm{H}_{6}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\) (b) concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by hydrolysis with boiling water (c) HBr followed by hydrolysis with aqueous KOH (d) \(\mathrm{Hg}\left(\mathrm{OCOCH}_{3}\right)_{2}\) followed by reduction with \(\mathrm{NaBH}_{4}\) -

Short Answer

Expert verified
Option (a) \(\mathrm{B}_2\mathrm{H}_6\) followed by \(\mathrm{H}_2\mathrm{O}_2\) / \(\mathrm{NaOH}\) is correct.

Step by step solution

01

Identify the Reaction

The conversion of propene (an alkene) to 1-propanol (an alcohol) can typically be achieved via a hydration reaction. The goal is to add water across the double bond of propene to form an alcohol.
02

Evaluate Each Option for Correctness

Each reagent or sequence of reagents provided in the options corresponds to a known reaction mechanism. Let's evaluate them: (a) In hydroboration-oxidation (\(\mathrm{B}_2\mathrm{H}_6\) and \(\mathrm{H}_2\mathrm{O}_2\) / \(\mathrm{NaOH}\)), an alkene is converted into an alcohol by syn addition of water across the double bond, typically creating primary alcohols from terminal double bonds like in propene.(b) Using concentrated \(\mathrm{H}_2\mathrm{SO}_4\) followed by hydrolysis involves anti-Markovnikov addition and would not readily yield 1-propanol from propene.(c) Using HBr and base does not typically yield alcohols but rather bromoalkanes, not suitable here.(d) Oxymercuration-demercuration (\(\mathrm{Hg}\left(\mathrm{OCOCH}_3\right)_2\) followed by \(\mathrm{NaBH}_4\)) forms Markovnikov alcohols, not primarily targeting 1-propanol from propene.
03

Select the Right Reagent

Option (a), the hydroboration-oxidation method, is most suitable to convert propene into 1-propanol. This method allows the formation of a primary alcohol by employing anti-Markovnikov addition, suitable for converting terminal alkenes such as propene to the corresponding alcohol like 1-propanol.

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Key Concepts

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

Propene to Alcohol Conversion
Propene, a simple three-carbon alkene, is a common starting material in organic synthesis. It holds a reactive double bond which makes it easy to transform into other useful compounds. In the conversion of propene to alcohol, specifically 1-propanol, we focus on breaking this double bond and forming a new bond with an oxygen atom from water.
The process involves adding water across the double bond of propene, a reaction known as a hydration reaction. Several methods exist for achieving this transformation, but the hydroboration-oxidation method is often chosen for converting propene to a primary alcohol like 1-propanol. This method selectively adds water in a way that the hydroxyl group (OH) attaches to the less substituted carbon of the double bond. This method gives us our desired primary alcohol with high specificity and yield, making it the preferred choice for converting propene to 1-propanol.
Primary Alcohol Formation
In organic chemistry, primary alcohols are characterized by the presence of a hydroxyl group (OH) connected to a primary carbon atom. A primary carbon is only connected to one other carbon. When producing primary alcohols from alkenes, it's important to ensure that the hydroxyl group bonds correctly to achieve the desired product.
The hydroboration-oxidation method is particularly effective in forming primary alcohols from terminal alkenes. Here, borane (B_2H_6) initiates the reaction by forming a complex with the alkene. This is called syn addition because the components add to the same side of the alkene. The borane adds in a way that places the atom which will eventually become OH at the end carbon of the alkene. Following this, oxidation with hydrogen peroxide (H_2O_2) and a base such as sodium hydroxide (NaOH) finalizes the transformation by converting the boron-containing moiety into the alcohol group, thus resulting in a primary alcohol product such as 1-propanol.
Reaction Mechanism Analysis
Understanding the underlying reaction mechanisms is crucial for mastering organic chemistry reactions like the hydroboration-oxidation of propene. The initial part of the mechanism includes adding the trialkylborane (BHR_2) to propene via syn-addition. This reaction is not only regioselective, favoring less hindered sites, but also stereospecific because of the syn-addition leading to retention of configuration.
After the initial addition step, the oxidation phase takes place. Here, hydrogen peroxide acts as an oxidizing agent, converting the boron group into an alcohol. During this stage, coordination to the oxygen allows for a smooth transition where the boron site is replaced by a hydroxyl. Using NaOH ensures that there is a suitable environment in which the hydrogen peroxide can effectively oxidize the boron.
Paying attention to these details elucidates why hydroboration-oxidation is such an effective and reliable method for forming primary alcohols from alkenes like propene. Recognizing how each step contributes to the overall transformation enhances the ability to predict reaction outcomes in similar synthetic routes.

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Most popular questions from this chapter

The compound obtained by the reaction of ethene with diborane, followed by hydrolysis with alkaline hydrogen peroxide is (a) ethanal (b) ethanol (c) triethyl boride (d) propanol

Periodic acid oxidizes (a) 1,4 -diols (b) 1, 3-diols (c) 1,2 -diols (d) \(\beta\)-ketoaldehyde

When diethyl ether is treated with an excess of \(\mathrm{Cl}_{2}\) in the presence of sunlight, the product formed is (a) \(\mathrm{CH}_{3} \mathrm{CHCl}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHCl}-\mathrm{O}-\mathrm{CHClCH}_{3}\) (c) \(\mathrm{CCl}_{3} \mathrm{CCl}_{2}-\mathrm{O}-\mathrm{CCl}_{2} \mathrm{CCl}_{3}\) (d) \(\mathrm{CH}_{7} \mathrm{CCl}_{2}-\mathrm{O}-\mathrm{CHClCH}_{3}\)

Luca's test of alcohols involves following reaction: \(\mathrm{R}-\mathrm{OH}+\mathrm{HCl} \frac{\text { anhydrous }}{\mathrm{ZnCl}_{2}} \underset{\text { white turbidity }}{\mathrm{RCl}+\mathrm{H}_{2} \mathrm{O}}\) Select the correct statement(s) for the Luca's test. (a) Lesser the acidic character of alcohol, greater is its reactivity towards Luca's reagent. (b) ROH behaves as a base. (c) \(\mathrm{CH}_{3} \mathrm{OH}\) gives Luca's test most quickly. (d) Reactivity of \(1^{\circ}, 2^{\circ}, 3^{\circ}\) alcohol, lies in the following sequence (for Luca's reaction \(3^{\circ}>2^{\circ}>1^{\circ}\) ).

The reagent used for the preparation of higher ethers from halogenated ethers is (a) sodium alkoxide (b) Grignard reagent (c) concutrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) dry silver oxide

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