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One Integer Value Correct Type
Direction (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).
Q.
If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?
Correct answer is '5'. Can you explain this answer?
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One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section c...
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One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section c...
Explanation:
Step 1: Analyzing the given compound
- The given compound is trans-3-methylchlorocyclopentane.
- It contains a methyl group (CH3) and a chlorine atom (Cl) attached to a cyclopentane ring.
- The trans configuration indicates that the methyl group and the chlorine atom are present on opposite sides of the ring.
Step 2: Understanding the reaction conditions
- The compound is treated with KOH solution, which is a strong base.
- Under these conditions, two types of reactions can occur: SN2 (substitution nucleophilic bimolecular) and E2 (elimination bimolecular).
Step 3: Identifying possible reaction sites
- The compound contains two reactive sites: the methyl group and the chlorine atom.
- Both of these sites are susceptible to nucleophilic attack or elimination.
Step 4: Considering the SN2 reaction
- In an SN2 reaction, the nucleophile displaces the leaving group, resulting in substitution.
- The chlorine atom is a good leaving group, and the methyl group can act as a nucleophile.
- Therefore, the chlorine atom can be substituted by the methyl group, resulting in the formation of trans-3-methylcyclopentane.
Step 5: Considering the E2 reaction
- In an E2 reaction, a strong base abstracts a proton adjacent to the leaving group, resulting in elimination.
- The chlorine atom is a good leaving group, and the methyl group can act as a base.
- Therefore, the methyl group can abstract a proton from the adjacent carbon, resulting in the formation of cyclopentene.
Step 6: Identifying the total products
- From the SN2 reaction: trans-3-methylcyclopentane (1 substitution product)
- From the E2 reaction: cyclopentene (1 elimination product)
Step 7: Considering the possibility of additional products
- In some cases, the E2 reaction can also result in the formation of a different elimination product, such as trans-3-methylcyclopentene.
- Additionally, the SN2 reaction can result in the formation of a different substitution product, such as cis-3-methylcyclopentane.
Step 8: Counting the total products
- Therefore, in principle, a total of 5 products can be formed: trans-3-methylcyclopentane, cyclopentene, trans-3-methylcyclopentene, cis-3-methylcyclopentane, and the starting compound itself.
Conclusion:
- In principle, a total of 5 substitution and elimination products can be formed when trans-3-methylchlorocyclopentane is treated with KOH solution.
Step 1: Analyzing the given compound
- The given compound is trans-3-methylchlorocyclopentane.
- It contains a methyl group (CH3) and a chlorine atom (Cl) attached to a cyclopentane ring.
- The trans configuration indicates that the methyl group and the chlorine atom are present on opposite sides of the ring.
Step 2: Understanding the reaction conditions
- The compound is treated with KOH solution, which is a strong base.
- Under these conditions, two types of reactions can occur: SN2 (substitution nucleophilic bimolecular) and E2 (elimination bimolecular).
Step 3: Identifying possible reaction sites
- The compound contains two reactive sites: the methyl group and the chlorine atom.
- Both of these sites are susceptible to nucleophilic attack or elimination.
Step 4: Considering the SN2 reaction
- In an SN2 reaction, the nucleophile displaces the leaving group, resulting in substitution.
- The chlorine atom is a good leaving group, and the methyl group can act as a nucleophile.
- Therefore, the chlorine atom can be substituted by the methyl group, resulting in the formation of trans-3-methylcyclopentane.
Step 5: Considering the E2 reaction
- In an E2 reaction, a strong base abstracts a proton adjacent to the leaving group, resulting in elimination.
- The chlorine atom is a good leaving group, and the methyl group can act as a base.
- Therefore, the methyl group can abstract a proton from the adjacent carbon, resulting in the formation of cyclopentene.
Step 6: Identifying the total products
- From the SN2 reaction: trans-3-methylcyclopentane (1 substitution product)
- From the E2 reaction: cyclopentene (1 elimination product)
Step 7: Considering the possibility of additional products
- In some cases, the E2 reaction can also result in the formation of a different elimination product, such as trans-3-methylcyclopentene.
- Additionally, the SN2 reaction can result in the formation of a different substitution product, such as cis-3-methylcyclopentane.
Step 8: Counting the total products
- Therefore, in principle, a total of 5 products can be formed: trans-3-methylcyclopentane, cyclopentene, trans-3-methylcyclopentene, cis-3-methylcyclopentane, and the starting compound itself.
Conclusion:
- In principle, a total of 5 substitution and elimination products can be formed when trans-3-methylchlorocyclopentane is treated with KOH solution.
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One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer?
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One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer?.
One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer?.
Solutions for One Integer Value Correct TypeDirection (Q. Nos. 22-25) This section contains 4 questions. When worked out will result in an integer from Ot oQ (both inclusive).Q.If trans-3-methylchlorocyclopentane is treated with KOH solution, both SN2 and E2 occur simultaneously. How many total substitution and elimination products are formed in principle?Correct answer is '5'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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