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2 -phenylethanol may be prepared by the reaction of phenylmagnesium bromide with (a) HCHO (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (c) C1CO1 (d) \(\mathrm{CH}_{3} \mathrm{CHO}\)

Short Answer

Expert verified
The correct reagent to prepare 2-phenylethanol is option (c), C1CO1 (epoxide).

Step by step solution

01

Understanding the Reaction

2-phenylethanol can be synthesized using a Grignard reaction, where phenylmagnesium bromide acts as the nucleophile. It will attack the electrophilic carbon of a carbonyl group.
02

Analyzing the Reagents

Let's analyze each given option to determine if it has the right carbonyl group for the formation of 2-phenylethanol.(a) HCHO (Formaldehyde)(b) \(\mathrm{CH}_{3}\mathrm{COCH}_{3}\) (Acetone)(c) Epoxide, represented as C1CO1 in SMILES(d) \(\mathrm{CH}_{3}\mathrm{CHO}\) (Acetaldehyde)
03

Reagent A: HCHO

Phenylmagnesium bromide reacts with formaldehyde to form an alcohol by adding a single carbon that will eventually lead to methanol but cannot form 2-phenylethanol.
04

Reagent B: CH₃COCH₃

Acetone, when attacked by phenylmagnesium bromide, would add a phenyl group to the carbonyl carbon, creating 2-phenylpropan-2-ol, which is not the desired product.
05

Reagent C: Epoxide (C1CO1)

Phenylmagnesium bromide can open the epoxide ring by attacking the less substituted carbon, leading to the formation of a 2-carbon chain ended with an alcohol group, forming 2-phenylethanol.
06

Reagent D: CH₃CHO

Acetaldehyde reaction with phenylmagnesium bromide would add an ethyl group to the phenyl group, forming 1-phenylethanol, not the desired 2-phenylethanol.
07

Determine the Correct Reagent

Among the options, only the epoxide (C1CO1) leads to the formation of 2-phenylethanol when reacted with phenylmagnesium bromide.

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Key Concepts

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

Phenylmagnesium Bromide
Phenylmagnesium bromide is a key reagent in the classic Grignard reaction. This compound, a member of the organomagnesium halides, is used to create new carbon-carbon bonds, leading to a wide variety of organic compounds. When dissolved in an ether solvent, phenylmagnesium bromide is formed from bromobenzene by the reaction with magnesium. It serves as a powerful nucleophile due to the high reactivity of the C-Mg bond. Grignard reagents like phenylmagnesium bromide attack electrophilic centers, commonly carbon atoms within carbonyl groups, to form alcohols. This makes them highly valuable in creating complex molecules and finding widespread applications in both academic and industrial settings.
2-Phenylethanol Synthesis
The synthesis of 2-phenylethanol involves creating a carbon-carbon bond using phenylmagnesium bromide. The desired product, 2-phenylethanol, is characterized by having a phenyl group attached to the ethanol moiety. One effective method to achieve this synthesis involves reacting phenylmagnesium bromide with an epoxide. In this specific reaction, the Grignard reagent opens the epoxide ring, enabling the addition of the phenyl group to form a secondary alcohol. This method is particularly advantageous because it leads directly to the desired product in one step. The reaction conditions are typically mild, and the process is generally straightforward, making it a practical choice for creating 2-phenylethanol.
Epoxide Ring Opening
Epoxides are highly reactive organic compounds due to the strained three-membered cyclic ethers. They serve as excellent intermediates in organic synthesis, particularly in the formation of alcohols via nucleophilic ring opening reactions. When phenylmagnesium bromide interacts with an epoxide, the Grignard reagent nucleophilically attacks the less substituted carbon, causing the ring to open. This reaction is a vital step towards forming alcohols, as it provides a straightforward way to increase the carbon chain length while introducing a hydroxyl group. The mild reaction conditions and high reactivity of epoxides make them excellent substrates for Grignard reactions. The ring opening not only elongates the carbon skeleton but also creates the necessary alcohol functionality in the desired carbon position, effectively achieving the synthesis of 2-phenylethanol in this case.

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Most popular questions from this chapter

Alkene \(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_{2}\) reacts with \(\mathrm{B}_{2} \mathrm{H}_{6}\) in the presence of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to give (a) [R]C(C)=O (b) [R]OC(C)O (c) \(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CHO}\) (d) \(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether due to the presence of (a) H-bonding in dimethyl ether (b) H-bonding in ethanol (c) \(\mathrm{CH}_{3}\) group in ethanol (d) \(\mathrm{CH}_{3}\) group in dimethyl ether

Phenol is heated with a solution of mixture of \(\mathrm{KBr}\) and \(\mathrm{KBrO}_{3} .\) The major product obtained in the above reaction is (a) 3-Bromophenol (b) 4-Bromophenol (c) \(2,4,6\)-Tribromophenol (d) 2-Bromophenol

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Which of the following is the most suitable method for removing the traces of water from ethanol? (a) Heating with sodium metal (b) Hassing dry HCl gas through it (c) Distilling it (d) Reacting with \(\mathrm{Mg}\)

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