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1-phenylethanol can be prepared by the reaction of benzaldehyde with (a) methyl iodide and magnesium (b) methyl bromide and aluminium bromide (c) ethyl iodide and magnesium (d) methyl bromide

Short Answer

Expert verified
The preparation of 1-phenylethanol from benzaldehyde is best achieved with methyl iodide and magnesium.

Step by step solution

01

Identify the Reaction Type

To prepare 1-phenylethanol from benzaldehyde, one must identify a reaction that introduces a methyl group (–CH₃) to the compound, specifically at the carbonyl carbon of benzaldehyde. This is typically done using a Grignard reaction.
02

Choose the Suitable Reagents

The Grignard reaction requires the use of an organometallic reagent, typically formed from an alkyl halide and magnesium. Among the options, option (a) "methyl iodide and magnesium" fits this requirement, as methyl iodide (CH₃I) can react with magnesium to form the Grignard reagent CH₃MgI.
03

Outline the Reaction Mechanism

In the Grignard reaction, the Grignard reagent CH₃MgI acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde. This nucleophilic addition leads to the formation of an alkoxide intermediate.
04

Completion of the Reaction

The alkoxide intermediate generated in the previous step undergoes protonation, usually by the addition of a dilute acid, to yield 1-phenylethanol (C₆H₅CH(OH)CH₃).

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Key Concepts

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

Organometallic Reagent
In a Grignard reaction, the organometallic reagent plays a crucial role. It is typically formed by reacting an alkyl or aryl halide with magnesium in a non-reactive ether solution. This creates a compound such as CH₃MgI, where the carbon atom is bonded to the metal, magnesium. This bond gives the carbon a partial negative charge.

This negative charge equips the carbon with nucleophilic properties, allowing it to readily participate in reactions. The organometallic reagent acts as a powerful tool in organic synthesis, making it possible to form carbon-carbon bonds. This capability is essential for synthesizing various complex molecules, including alcohols, by attacking electrophilic centers in other compounds.
Nucleophilic Addition
Nucleophilic addition is a key step in many organic reactions, particularly those involving carbonyl groups. In the context of synthesizing 1-phenylethanol, the nucleophilic addition occurs when the negatively charged carbon atom of the Grignard reagent (CH₃MgI) attacks the electrophilic carbonyl carbon of benzaldehyde.

This attack forms a temporary bond, creating an alkoxide intermediate. The nucleophile, in this case, drives the reaction forward by breaking the C=O double bond and forming a new C-C single bond. This step is crucial as it sets the stage for the subsequent protonation, which will yield the desired alcohol.
Protonation
Protonation is the process of adding a proton (H⁺) to a molecule, often converting an intermediate into a stable product. After the nucleophilic addition of the Grignard reagent to benzaldehyde, an alkoxide ion intermediate is formed.

To transform this intermediate into the desired alcohol, a protonation step is necessary. Typically, this involves treating the intermediate with a dilute acid. The acid donates a proton to the alkoxide, transforming it into 1-phenylethanol. Protonation is essential to stabilize the molecule and complete the reaction, yielding a neutral alcohol.
1-Phenylethanol Synthesis
The synthesis of 1-phenylethanol from benzaldehyde via a Grignard reaction is a classic example of carbon-carbon bond formation. Starting with benzaldehyde, the process involves two main steps: the nucleophilic addition and subsequent protonation.

The choice of reagents is vital. Using methyl iodide and magnesium forms the necessary Grignard reagent, CH₃MgI. This reagent allows the introduction of a methyl group at the carbonyl carbon of benzaldehyde. After the build-up of the alkoxide intermediate through nucleophilic addition, protonation finalizes the structure.
  • The Grignard reagent introduces the methyl group.
  • Nucleophilic addition forms a carbon-carbon bond.
  • Protonation ensures the stability of 1-phenylethanol.
This method effectively transforms benzaldehyde into 1-phenylethanol, demonstrating the power and versatility of Grignard reactions in synthetic organic chemistry.

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Most popular questions from this chapter

Among the following, the most stable compound is (a) cis-1, 2-cyclohexanediol (b) trans- 1,2 -cyclohexanediol (c) cis-1, 3-cyclohexenediol (d) trans- 1,3 -cyclohexanediol

n-propyl alcohol and isopropyl alcohol can be chemically distinguished by (a) reduction (b) \(\mathrm{PCl}_{5}\) (c) ozonolysis (d) oxidation with potassium dichromate

Ethanol, on reacting with \(\mathrm{PCl}_{s}\) gives ' \(\mathrm{A}\) ', \(\mathrm{POCl}_{3}\) and HCl. 'A' reacts with silver nitrate to form 'B' (major product) and \(\mathrm{AgCl}\). 'A' and ' \(\mathrm{B}\) ' respectively are (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NO}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\)

A compound 'A' has a molecular formula \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{OH}\). It reduces Fehling's solution and on oxidation, gives a monocarboxylic acid 'B'. 'A' can be obtained by the action of chlorine on ethyl alcohol. 'A' is (a) chloroform (b) chloral (c) methyl chloride (d) monochloroacetic acid

Identify the correct order of boiling points of the following compounds: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (I) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (II) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (III) (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}\) (b) III > I > II (c) \(\mathrm{I}>\mathrm{III}>\mathrm{II}\) (d) \(\mathrm{III}>\mathrm{II}>\mathrm{I}\)

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