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A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?
- a)2-ethyl-1 -butanol
- b)4-methyl-1-pentanol
- c)3-methyl-3-penta
- d)3-methyl-1-pentanol
Correct answer is option 'D'. Can you explain this answer?
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A C6H14O chiral alcohol is converted to a bromide by treating with PBr...
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A C6H14O chiral alcohol is converted to a bromide by treating with PBr...
The Original Alcohol is 3-Methyl-1-Pentanol
To determine the original alcohol, we need to analyze the reactions step by step.
Reaction 1: Conversion to Bromide
The chiral alcohol C6H14O is treated with PBr3 to convert it into a bromide. This reaction involves the substitution of the hydroxyl group with a bromine atom. The reaction can be represented as follows:
C6H14O + PBr3 -> C6H13Br + HBr
The resulting product is a chiral bromide, as the carbon atom to which the bromine is attached is asymmetric.
Reaction 2: Reaction with Mg in Ether
The bromide obtained in the previous step is then treated with Mg in ether. This reaction is a Grignard reaction, which involves the formation of a carbon-carbon bond between the bromide and the Mg reagent. The reaction can be represented as follows:
C6H13Br + Mg -> C6H14MgBr
The resulting product is an organometallic compound known as a Grignard reagent.
Reaction 3: Quenching with HCl
The Grignard reagent obtained in the previous step is then quenched with 0.1 N HCl. This reaction involves the protonation of the carbon atom bonded to Mg and the formation of a new carbon-hydrogen bond. The reaction can be represented as follows:
C6H14MgBr + HCl -> C6H14 + MgClBr
The resulting product is an achiral hydrocarbon, as there is no longer an asymmetric carbon atom present.
Determining the Original Alcohol
To determine which of the given options is the original alcohol, we need to compare the structures of the original alcohol and the resulting hydrocarbon. By examining the structures of the options given, we can see that only 3-methyl-1-pentanol has a chiral center.
Therefore, the correct answer is option D, 3-methyl-1-pentanol.
To determine the original alcohol, we need to analyze the reactions step by step.
Reaction 1: Conversion to Bromide
The chiral alcohol C6H14O is treated with PBr3 to convert it into a bromide. This reaction involves the substitution of the hydroxyl group with a bromine atom. The reaction can be represented as follows:
C6H14O + PBr3 -> C6H13Br + HBr
The resulting product is a chiral bromide, as the carbon atom to which the bromine is attached is asymmetric.
Reaction 2: Reaction with Mg in Ether
The bromide obtained in the previous step is then treated with Mg in ether. This reaction is a Grignard reaction, which involves the formation of a carbon-carbon bond between the bromide and the Mg reagent. The reaction can be represented as follows:
C6H13Br + Mg -> C6H14MgBr
The resulting product is an organometallic compound known as a Grignard reagent.
Reaction 3: Quenching with HCl
The Grignard reagent obtained in the previous step is then quenched with 0.1 N HCl. This reaction involves the protonation of the carbon atom bonded to Mg and the formation of a new carbon-hydrogen bond. The reaction can be represented as follows:
C6H14MgBr + HCl -> C6H14 + MgClBr
The resulting product is an achiral hydrocarbon, as there is no longer an asymmetric carbon atom present.
Determining the Original Alcohol
To determine which of the given options is the original alcohol, we need to compare the structures of the original alcohol and the resulting hydrocarbon. By examining the structures of the options given, we can see that only 3-methyl-1-pentanol has a chiral center.
Therefore, the correct answer is option D, 3-methyl-1-pentanol.
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A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer?
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A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer?.
A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A C6H14O chiral alcohol is converted to a bromide by treating with PBr3. Reaction of this bromide, first with Mg in ether, followed by quenching in 0.1 N HCI produces an achiral C6H14 hydrocarbon. Which of the following is the original alcohol ?a)2-ethyl-1 -butanolb)4-methyl-1-pentanolc)3-methyl-3-pentad)3-methyl-1-pentanolCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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