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Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4
(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are
(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are
[JEE Advanced 2014]
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider all possible isomeric ketones, including stereoisomers of Mol...
The general formula of isomeric ketone having molecular mass 100 is C6H12O [6×12+12×1+16].
The possible structure will be :
The number of ketones that gives racemic mixture with NaBH4 is 5 as the ketone with chiral center will give diastereoisomer with NaBH4
Most Upvoted Answer
Consider all possible isomeric ketones, including stereoisomers of Mol...
The general formula of isomeric ketone having molecular mass 100 is C6H12O [6×12+12×1+16].
The possible structure will be :
a) CH3−CH2−CH2−C∣∣O−CH3
b) CH3CH2CH2−C∣∣O−CH2CH3
c)CH3−CH3∣CH−CH2−C∣∣O−CH3
d) CH3−CH2−C∣HCH3−C∣∣O−CH3
e) CH3−C∣HCH3−C∣∣O−CH2−CH3
f) CH3−CH3∣C∣CH3−C∣∣O−CH3
The number of ketones that gives racemic mixture with NaBH4 is 5 as the ketone with chiral center will give diastereoisomer with NaBH
The possible structure will be :
a) CH3−CH2−CH2−C∣∣O−CH3
b) CH3CH2CH2−C∣∣O−CH2CH3
c)CH3−CH3∣CH−CH2−C∣∣O−CH3
d) CH3−CH2−C∣HCH3−C∣∣O−CH3
e) CH3−C∣HCH3−C∣∣O−CH2−CH3
f) CH3−CH3∣C∣CH3−C∣∣O−CH3
The number of ketones that gives racemic mixture with NaBH4 is 5 as the ketone with chiral center will give diastereoisomer with NaBH
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Consider all possible isomeric ketones, including stereoisomers of Mol...
Introduction:
To answer this question, we need to consider all possible isomeric ketones with a molecular weight of 100 and determine which ones will give a racemic product when reacted with NaBH4. Let's break down the problem step by step.
Step 1: Determine the possible isomeric ketones:
To find the possible isomeric ketones, we need to consider the different carbon chain lengths and functional groups. Here is a list of all the possible isomeric ketones with a molecular weight of 100:
- C6H14O: 2-hexanone, 3-hexanone, 4-methyl-2-pentanone
- C5H12O: 2-pentanone, 3-pentanone, 2-methyl-1-butanol, 3-methyl-1-butanol, 2-methyl-2-butanol
- C4H10O: 2-butanol, 3-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol
- C3H8O: 2-propanol, 1-propanol
Step 2: Determine the possible stereoisomers:
Now, we need to consider the possibility of stereoisomers for each ketone. Stereoisomers are molecules with the same molecular formula and connectivity but different spatial arrangements. In the case of ketones, we need to consider the possibility of different arrangements around the carbonyl carbon. Let's go through each ketone and determine the possible stereoisomers:
- 2-hexanone: No stereoisomers
- 3-hexanone: No stereoisomers
- 4-methyl-2-pentanone: No stereoisomers
- 2-pentanone: No stereoisomers
- 3-pentanone: No stereoisomers
- 2-methyl-1-butanol: One stereoisomer (enantiomer)
- 3-methyl-1-butanol: One stereoisomer (enantiomer)
- 2-methyl-2-butanol: One stereoisomer (enantiomer)
- 2-butanol: One stereoisomer (enantiomer)
- 3-butanol: One stereoisomer (enantiomer)
- 2-methyl-1-propanol: One stereoisomer (enantiomer)
- 2-methyl-2-propanol: One stereoisomer (enantiomer)
- 2-propanol: One stereoisomer (enantiomer)
- 1-propanol: One stereoisomer (enantiomer)
Step 3: Reacting with NaBH4:
When ketones are reacted with NaBH4, they undergo reduction to form alcohols. NaBH4 acts as a reducing agent and adds hydride (H-) to the carbonyl carbon. The hydride can add from either face of the ketone, leading to the formation of a racemic mixture (equal amounts of both enantiomers) of the corresponding alcohol.
Step 4: Identify ketones that give a racemic product:
Based on the information from Step 2 and Step 3, we can identify the ketones
To answer this question, we need to consider all possible isomeric ketones with a molecular weight of 100 and determine which ones will give a racemic product when reacted with NaBH4. Let's break down the problem step by step.
Step 1: Determine the possible isomeric ketones:
To find the possible isomeric ketones, we need to consider the different carbon chain lengths and functional groups. Here is a list of all the possible isomeric ketones with a molecular weight of 100:
- C6H14O: 2-hexanone, 3-hexanone, 4-methyl-2-pentanone
- C5H12O: 2-pentanone, 3-pentanone, 2-methyl-1-butanol, 3-methyl-1-butanol, 2-methyl-2-butanol
- C4H10O: 2-butanol, 3-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol
- C3H8O: 2-propanol, 1-propanol
Step 2: Determine the possible stereoisomers:
Now, we need to consider the possibility of stereoisomers for each ketone. Stereoisomers are molecules with the same molecular formula and connectivity but different spatial arrangements. In the case of ketones, we need to consider the possibility of different arrangements around the carbonyl carbon. Let's go through each ketone and determine the possible stereoisomers:
- 2-hexanone: No stereoisomers
- 3-hexanone: No stereoisomers
- 4-methyl-2-pentanone: No stereoisomers
- 2-pentanone: No stereoisomers
- 3-pentanone: No stereoisomers
- 2-methyl-1-butanol: One stereoisomer (enantiomer)
- 3-methyl-1-butanol: One stereoisomer (enantiomer)
- 2-methyl-2-butanol: One stereoisomer (enantiomer)
- 2-butanol: One stereoisomer (enantiomer)
- 3-butanol: One stereoisomer (enantiomer)
- 2-methyl-1-propanol: One stereoisomer (enantiomer)
- 2-methyl-2-propanol: One stereoisomer (enantiomer)
- 2-propanol: One stereoisomer (enantiomer)
- 1-propanol: One stereoisomer (enantiomer)
Step 3: Reacting with NaBH4:
When ketones are reacted with NaBH4, they undergo reduction to form alcohols. NaBH4 acts as a reducing agent and adds hydride (H-) to the carbonyl carbon. The hydride can add from either face of the ketone, leading to the formation of a racemic mixture (equal amounts of both enantiomers) of the corresponding alcohol.
Step 4: Identify ketones that give a racemic product:
Based on the information from Step 2 and Step 3, we can identify the ketones
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Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer?
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Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. 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 Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. 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 Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer?.
Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. 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 Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. 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 Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer?.
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Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are[JEE Advanced 2014]a)3b)4c)5d)6Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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