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Which of the following reagents does not reduce the CO group of carboxylic acids to CH2 groups to form alcohols?
- a)LiAlH4-ether
- b)B2H6-THF
- c)H2-CuCr2O4
- d)NaBH4
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Which of the following reagents does not reduce the CO group of carbox...
Reagents that reduce the CO group of carboxylic acids to CH2 groups to form alcohols:
1. LiAlH4-ether
2. B2H6-THF
3. H2-CuCr2O4
Reagent that does not reduce the CO group of carboxylic acids to CH2 groups:
4. NaBH4
Explanation:
1. LiAlH4-ether:
LiAlH4 (lithium aluminum hydride) is a strong reducing agent that can reduce carboxylic acids to alcohols. It reacts with the carbonyl group (CO) of the carboxylic acid, breaking the double bond and replacing it with a single bond to form an alcohol. The reaction proceeds in the presence of an ether solvent.
2. B2H6-THF:
B2H6 (diborane) is another strong reducing agent that can reduce carboxylic acids to alcohols. Similar to LiAlH4, it reacts with the carbonyl group of the carboxylic acid, breaking the double bond and forming an alcohol. The reaction is usually carried out in the presence of THF (tetrahydrofuran) as a solvent.
3. H2-CuCr2O4:
H2 (hydrogen gas) in the presence of CuCr2O4 (copper chromite) is a mild reducing agent that can also convert carboxylic acids to alcohols. The reaction involves the hydrogenation of the carbonyl group to form an alcohol. This process is known as catalytic hydrogenation.
4. NaBH4:
NaBH4 (sodium borohydride) is a reducing agent that is commonly used to reduce aldehydes and ketones to alcohols. However, it is not capable of reducing carboxylic acids to alcohols. This is because the carboxylic acid group is more acidic and less reactive than the aldehyde or ketone group. NaBH4 is not strong enough to break the CO double bond in carboxylic acids and convert it into a CH2 group.
Therefore, the reagent NaBH4 does not reduce the CO group of carboxylic acids to CH2 groups to form alcohols.
1. LiAlH4-ether
2. B2H6-THF
3. H2-CuCr2O4
Reagent that does not reduce the CO group of carboxylic acids to CH2 groups:
4. NaBH4
Explanation:
1. LiAlH4-ether:
LiAlH4 (lithium aluminum hydride) is a strong reducing agent that can reduce carboxylic acids to alcohols. It reacts with the carbonyl group (CO) of the carboxylic acid, breaking the double bond and replacing it with a single bond to form an alcohol. The reaction proceeds in the presence of an ether solvent.
2. B2H6-THF:
B2H6 (diborane) is another strong reducing agent that can reduce carboxylic acids to alcohols. Similar to LiAlH4, it reacts with the carbonyl group of the carboxylic acid, breaking the double bond and forming an alcohol. The reaction is usually carried out in the presence of THF (tetrahydrofuran) as a solvent.
3. H2-CuCr2O4:
H2 (hydrogen gas) in the presence of CuCr2O4 (copper chromite) is a mild reducing agent that can also convert carboxylic acids to alcohols. The reaction involves the hydrogenation of the carbonyl group to form an alcohol. This process is known as catalytic hydrogenation.
4. NaBH4:
NaBH4 (sodium borohydride) is a reducing agent that is commonly used to reduce aldehydes and ketones to alcohols. However, it is not capable of reducing carboxylic acids to alcohols. This is because the carboxylic acid group is more acidic and less reactive than the aldehyde or ketone group. NaBH4 is not strong enough to break the CO double bond in carboxylic acids and convert it into a CH2 group.
Therefore, the reagent NaBH4 does not reduce the CO group of carboxylic acids to CH2 groups to form alcohols.
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Community Answer
Which of the following reagents does not reduce the CO group of carbox...
Carboxylic acids on reduction with LiAlH4, B2H6 or H2 in the presence of suitable medium, are reduced to alcohols. The COOH group is reduced to CH2OH group. Sodium borohydride cannot reduce the carboxyl group.
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Which of the following reagents does not reduce the CO group of carboxylic acids to CH2 groups to form alcohols?a)LiAlH4-etherb)B2H6-THFc)H2-CuCr2O4d)NaBH4Correct answer is option 'D'. Can you explain this answer?
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