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The reaction of ethanolic KOH on 1, 1-dichloropropane gives:
- a)Propanaldehyde
- b)Propyne
- c)Propan-1, 1-diol
- d)Propanone
Correct answer is option 'B'. Can you explain this answer?
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The reaction of ethanolic KOH on 1, 1-dichloropropane gives:a)Propanal...
Reaction of Ethanolic KOH on 1, 1-dichloropropane
The reaction of ethanolic KOH on 1, 1-dichloropropane is a type of elimination reaction known as dehydrohalogenation. In this reaction, the two chlorine atoms are removed from the 1,1-dichloropropane molecule to form a triple bond between the two carbon atoms. The resulting product is propyne.
Mechanism of the Reaction
The reaction mechanism for the dehydrohalogenation of 1,1-dichloropropane involves the following steps:
1. Deprotonation: The ethanolic KOH acts as a base and removes a hydrogen ion from the 1,1-dichloropropane molecule, forming a carbanion.
2. Elimination: The carbanion attacks one of the chlorine atoms and displaces it as a chloride ion. This forms an intermediate alkene.
3. Deprotonation: The ethanolic KOH acts as a base again and removes a hydrogen ion from the alkene intermediate, forming a second carbanion.
4. Elimination: The second carbanion attacks the remaining chlorine atom and displaces it as a chloride ion. This forms the final product, propyne.
Reaction Equation
The overall reaction equation for the dehydrohalogenation of 1,1-dichloropropane is:
1,1-dichloropropane + ethanolic KOH → propyne + potassium chloride + water
Conclusion
The reaction of ethanolic KOH on 1,1-dichloropropane is a dehydrohalogenation reaction that results in the formation of propyne. This reaction is an important process in organic chemistry and is used in the synthesis of many compounds.
The reaction of ethanolic KOH on 1, 1-dichloropropane is a type of elimination reaction known as dehydrohalogenation. In this reaction, the two chlorine atoms are removed from the 1,1-dichloropropane molecule to form a triple bond between the two carbon atoms. The resulting product is propyne.
Mechanism of the Reaction
The reaction mechanism for the dehydrohalogenation of 1,1-dichloropropane involves the following steps:
1. Deprotonation: The ethanolic KOH acts as a base and removes a hydrogen ion from the 1,1-dichloropropane molecule, forming a carbanion.
2. Elimination: The carbanion attacks one of the chlorine atoms and displaces it as a chloride ion. This forms an intermediate alkene.
3. Deprotonation: The ethanolic KOH acts as a base again and removes a hydrogen ion from the alkene intermediate, forming a second carbanion.
4. Elimination: The second carbanion attacks the remaining chlorine atom and displaces it as a chloride ion. This forms the final product, propyne.
Reaction Equation
The overall reaction equation for the dehydrohalogenation of 1,1-dichloropropane is:
1,1-dichloropropane + ethanolic KOH → propyne + potassium chloride + water
Conclusion
The reaction of ethanolic KOH on 1,1-dichloropropane is a dehydrohalogenation reaction that results in the formation of propyne. This reaction is an important process in organic chemistry and is used in the synthesis of many compounds.
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Community Answer
The reaction of ethanolic KOH on 1, 1-dichloropropane gives:a)Propanal...
(1) due to presence of Alcoholic medium the nucleophile starts acting as a base.
(2)E1 mechanism can't take place here due to lack of stability of carbocation(refer R.I).
(3)E2 is the only possible way so anti-elimitation occours over here.
(Hope you got it)
(2)E1 mechanism can't take place here due to lack of stability of carbocation(refer R.I).
(3)E2 is the only possible way so anti-elimitation occours over here.
(Hope you got it)
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The reaction of ethanolic KOH on 1, 1-dichloropropane gives:a)Propanaldehydeb)Propynec)Propan-1, 1-diold)PropanoneCorrect answer is option 'B'. Can you explain this answer?
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