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A colourless liquid, at room temperature, reacts with soda-lime to form sodium salt of a carboxylic acid and ammonia gas. The liquid is
- a)propanoic acid
- b)formamide
- c)propanamide
- d)methyl enthanoate
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A colourless liquid, at room temperature, reacts with soda-lime to for...
Understanding the Reaction
The reaction of the colourless liquid with soda-lime indicates that the compound must have specific functional groups that can interact with soda-lime, which is primarily a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO).
Identifying the Compound
1. Formamide:
- Formamide (HCONH2) is a simple amide and contains a carbonyl group (C=O) attached to an amine (NH2).
- When treated with soda-lime, formamide undergoes a reaction that leads to the formation of sodium formate (the sodium salt of formic acid) and ammonia gas (NH3).
- The reaction typically involves the dehydration of the amide followed by hydrolysis.
2. Other Options:
- Propanoic Acid: This is a carboxylic acid and would not release ammonia upon reaction with soda-lime.
- Propanamide: Similar to formamide, but it has a larger alkyl group which may not directly produce ammonia when treated with soda-lime.
- Methyl Ethanoate: An ester that does not react with soda-lime to produce ammonia.
Conclusion
The key to understanding why formamide is the correct answer lies in its ability to decompose under the conditions provided by soda-lime, resulting in the formation of the sodium salt of a carboxylic acid (sodium formate) and the release of ammonia gas. Thus, the colourless liquid that reacts as described is indeed formamide (option B).
The reaction of the colourless liquid with soda-lime indicates that the compound must have specific functional groups that can interact with soda-lime, which is primarily a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO).
Identifying the Compound
1. Formamide:
- Formamide (HCONH2) is a simple amide and contains a carbonyl group (C=O) attached to an amine (NH2).
- When treated with soda-lime, formamide undergoes a reaction that leads to the formation of sodium formate (the sodium salt of formic acid) and ammonia gas (NH3).
- The reaction typically involves the dehydration of the amide followed by hydrolysis.
2. Other Options:
- Propanoic Acid: This is a carboxylic acid and would not release ammonia upon reaction with soda-lime.
- Propanamide: Similar to formamide, but it has a larger alkyl group which may not directly produce ammonia when treated with soda-lime.
- Methyl Ethanoate: An ester that does not react with soda-lime to produce ammonia.
Conclusion
The key to understanding why formamide is the correct answer lies in its ability to decompose under the conditions provided by soda-lime, resulting in the formation of the sodium salt of a carboxylic acid (sodium formate) and the release of ammonia gas. Thus, the colourless liquid that reacts as described is indeed formamide (option B).
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Community Answer
A colourless liquid, at room temperature, reacts with soda-lime to for...
Only amides (but not acids and esters) undergo hydrolysis in presence of soda-lime to form sodium salt of a carboxylic acid and ammonia gas. Further, since the given compound is a liquid, it must be formamide, because propanamide is a solid.
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A colourless liquid, at room temperature, reacts with soda-lime to form sodium salt of a carboxylic acid and ammonia gas. The liquid isa)propanoic acidb)formamide c)propanamided)methyl enthanoateCorrect answer is option 'B'. Can you explain this answer?
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A colourless liquid, at room temperature, reacts with soda-lime to form sodium salt of a carboxylic acid and ammonia gas. The liquid isa)propanoic acidb)formamide c)propanamided)methyl enthanoateCorrect answer is option 'B'. 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 colourless liquid, at room temperature, reacts with soda-lime to form sodium salt of a carboxylic acid and ammonia gas. The liquid isa)propanoic acidb)formamide c)propanamided)methyl enthanoateCorrect answer is option 'B'. 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 colourless liquid, at room temperature, reacts with soda-lime to form sodium salt of a carboxylic acid and ammonia gas. The liquid isa)propanoic acidb)formamide c)propanamided)methyl enthanoateCorrect answer is option 'B'. Can you explain this answer?.
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