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3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed is
- a)a pure enantiomer is formed
- b)racemic mixture is formed
- c)optically inactive product is formed
- d)meso-product is formed
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-meth...
The correct answer is Option A.
3−Methyl−pent−2−ene in reaction with HBr forms an additional product.
A molecule of HBr is added to C=C double bond.
The addition follows anti-Markovnikov's rule. Br is added to less substituted C atoms.
The number of possible stereoisomers for the products is four. The product has 2 chiral C atoms and is unsymmetrical.
An enantiomer will rotate the plane of polarization the equal angle but opposite direction. A racemic mixture will not rotate the plane of polarization.
The addition follows anti-Markovnikov's rule. Br is added to less substituted C atoms.
The number of possible stereoisomers for the products is four. The product has 2 chiral C atoms and is unsymmetrical.
An enantiomer will rotate the plane of polarization the equal angle but opposite direction. A racemic mixture will not rotate the plane of polarization.
Most Upvoted Answer
3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-meth...
When a pure enantiomer of 3-methyl-1-pentene is treated with HBr, a racemic mixture is formed. This means that both enantiomers of the product are formed in equal amounts. Let's break down the reaction and explain why this occurs.
1. Reaction Mechanism:
The reaction between 3-methyl-1-pentene and HBr proceeds via an electrophilic addition mechanism. The first step involves the formation of a carbocation intermediate. The HBr molecule acts as an electrophile, donating a proton to the alkene to form a secondary carbocation.
2. Formation of the Carbocation:
The carbon atom in 3-methyl-1-pentene that becomes chiral after the reaction is the one that initially forms the carbocation. The carbocation can be formed on either of the two sp² hybridized carbon atoms adjacent to the double bond.
3. Attack of Bromide Ion:
In the next step, the bromide ion attacks the carbocation, leading to the formation of the bromoalkane product. The attack can occur from either face of the carbocation, resulting in the formation of two enantiomers.
4. Enantiomeric Products:
Since the carbocation can form on either of the two adjacent carbon atoms, two different enantiomers are formed. These enantiomers are mirror images of each other and cannot be superimposed. Therefore, the reaction results in the formation of a mixture of both enantiomers.
5. Racemic Mixture:
A racemic mixture is one that contains equal amounts of both enantiomers. In this case, the pure enantiomer of 3-methyl-1-pentene will react to form a 50:50 mixture of the two enantiomers of the bromoalkane.
6. Explanation:
The chiral carbon in 3-methyl-1-pentene has two different substituents (a methyl group and a hydrogen atom), which gives rise to two different enantiomers. When the reaction occurs, the initial chiral carbon becomes a new chiral center in the product. Since the reaction can occur on either face of the carbocation, both enantiomers are formed, resulting in a racemic mixture.
In conclusion, when a pure enantiomer of 3-methyl-1-pentene is treated with HBr, a racemic mixture is formed due to the formation of a new chiral center during the reaction.
1. Reaction Mechanism:
The reaction between 3-methyl-1-pentene and HBr proceeds via an electrophilic addition mechanism. The first step involves the formation of a carbocation intermediate. The HBr molecule acts as an electrophile, donating a proton to the alkene to form a secondary carbocation.
2. Formation of the Carbocation:
The carbon atom in 3-methyl-1-pentene that becomes chiral after the reaction is the one that initially forms the carbocation. The carbocation can be formed on either of the two sp² hybridized carbon atoms adjacent to the double bond.
3. Attack of Bromide Ion:
In the next step, the bromide ion attacks the carbocation, leading to the formation of the bromoalkane product. The attack can occur from either face of the carbocation, resulting in the formation of two enantiomers.
4. Enantiomeric Products:
Since the carbocation can form on either of the two adjacent carbon atoms, two different enantiomers are formed. These enantiomers are mirror images of each other and cannot be superimposed. Therefore, the reaction results in the formation of a mixture of both enantiomers.
5. Racemic Mixture:
A racemic mixture is one that contains equal amounts of both enantiomers. In this case, the pure enantiomer of 3-methyl-1-pentene will react to form a 50:50 mixture of the two enantiomers of the bromoalkane.
6. Explanation:
The chiral carbon in 3-methyl-1-pentene has two different substituents (a methyl group and a hydrogen atom), which gives rise to two different enantiomers. When the reaction occurs, the initial chiral carbon becomes a new chiral center in the product. Since the reaction can occur on either face of the carbocation, both enantiomers are formed, resulting in a racemic mixture.
In conclusion, when a pure enantiomer of 3-methyl-1-pentene is treated with HBr, a racemic mixture is formed due to the formation of a new chiral center during the reaction.
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3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer?
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3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer?.
3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer?.
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3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of 3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed isa)a pure enantiomer is formedb)racemic mixture is formedc)optically inactive product is formedd)meso-product is formedCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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