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1, 3-Dichloropropane one reaction with Zn and NaI gives:
- a)Propane
- b)1-chloropropane
- c)Cyclohexene
- d)Cyclopropane
Correct answer is option 'D'. Can you explain this answer?
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1, 3-Dichloropropane one reaction with Zn and NaI gives:a)Propaneb)1-c...
Reaction of 1,3-Dichloropropane with Zn and NaI
The reaction of 1,3-Dichloropropane with Zn and NaI is a reductive dehalogenation reaction.
Reductive dehalogenation reactions involve the reduction of a halogen atom to a hydrogen atom, resulting in the formation of an alkane or an alkene.
In the reaction of 1,3-Dichloropropane with Zn and NaI, the halogen atoms (chlorine) are reduced to hydrogen atoms, resulting in the formation of a cyclic alkane, cyclopropane.
The mechanism of the reaction involves the following steps:
1. Formation of an organozinc intermediate: The reaction of Zn with NaI results in the formation of an organozinc intermediate.
Zn + NaI → R-Zn-I
2. Formation of an alkyl iodide: The organozinc intermediate reacts with 1,3-Dichloropropane to form an alkyl iodide.
R-Zn-I + CH2Cl-CH2-CH2Cl → R-CH2-CH2-CH2-I + ZnCl2
3. Reductive dehalogenation: The alkyl iodide undergoes reductive dehalogenation in the presence of excess Zn and NaI to form cyclopropane.
R-CH2-CH2-CH2-I + 2Zn + 2NaI → R-CH2-CH2-CH2-H + 2ZnI2
Therefore, the correct answer is option D, cyclopropane.
The reaction of 1,3-Dichloropropane with Zn and NaI is a reductive dehalogenation reaction.
Reductive dehalogenation reactions involve the reduction of a halogen atom to a hydrogen atom, resulting in the formation of an alkane or an alkene.
In the reaction of 1,3-Dichloropropane with Zn and NaI, the halogen atoms (chlorine) are reduced to hydrogen atoms, resulting in the formation of a cyclic alkane, cyclopropane.
The mechanism of the reaction involves the following steps:
1. Formation of an organozinc intermediate: The reaction of Zn with NaI results in the formation of an organozinc intermediate.
Zn + NaI → R-Zn-I
2. Formation of an alkyl iodide: The organozinc intermediate reacts with 1,3-Dichloropropane to form an alkyl iodide.
R-Zn-I + CH2Cl-CH2-CH2Cl → R-CH2-CH2-CH2-I + ZnCl2
3. Reductive dehalogenation: The alkyl iodide undergoes reductive dehalogenation in the presence of excess Zn and NaI to form cyclopropane.
R-CH2-CH2-CH2-I + 2Zn + 2NaI → R-CH2-CH2-CH2-H + 2ZnI2
Therefore, the correct answer is option D, cyclopropane.
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1, 3-Dichloropropane one reaction with Zn and NaI gives:a)Propaneb)1-chloropropanec)Cyclohexened)CyclopropaneCorrect answer is option 'D'. Can you explain this answer?
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