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What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?
- a)Methylation succeeds only if large excess of AICI3 is taken
- b)AICI3 reacts with phenol to give C6H5OAICI2
- c)Methylation occurs mainly at meta position
- d)Ortholpara methylation occur with para isomer as major product
Correct answer is option 'A,B,D'. Can you explain this answer?
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What is/are true regarding Friedel-Crafts methylation reaction of phen...
Methylation succeeds only if large excess of AICI3 is taken:
In the Friedel-Crafts methylation reaction, CH3Cl is used as the methylating agent and AlCl3 acts as a Lewis acid catalyst. The reaction proceeds through the formation of a complex between AlCl3 and CH3Cl, which activates the methyl group for attack on the aromatic ring of phenol. However, AlCl3 also reacts with phenol to form a complex, which is less reactive towards methylation. Therefore, it is necessary to use a large excess of AlCl3 to ensure that most of it reacts with phenol and leaves enough free CH3Cl for methylation to occur.
AICI3 reacts with phenol to give C6H5OAICI2:
AlCl3 reacts with phenol to form a complex called phenol-AlCl3 adduct. This adduct has the general formula C6H5OAlCl2. The reaction involves the Lewis acid-base interaction between the lone pair of electrons on the oxygen atom of phenol and the empty orbital of AlCl3. This complex is less reactive towards methylation, as the oxygen atom is already coordinated to the aluminum atom, making it less available for nucleophilic attack by CH3Cl.
Ortholpara methylation occurs with para isomer as major product:
The Friedel-Crafts methylation reaction can result in the substitution of a methyl group at either the ortho or para position of the aromatic ring. However, due to the steric hindrance caused by the bulky phenolic -OH group, the para substitution is favored. The methyl group is more likely to attack the para position, which is less hindered by the -OH group compared to the ortho position. Therefore, the para isomer is the major product in this reaction.
In summary, in the Friedel-Crafts methylation reaction of phenol using CH3Cl/AlCl3, a large excess of AlCl3 is taken to ensure sufficient availability of free CH3Cl for methylation to occur. AlCl3 reacts with phenol to form a less reactive complex, and methylation occurs primarily at the para position due to steric hindrance caused by the -OH group.
In the Friedel-Crafts methylation reaction, CH3Cl is used as the methylating agent and AlCl3 acts as a Lewis acid catalyst. The reaction proceeds through the formation of a complex between AlCl3 and CH3Cl, which activates the methyl group for attack on the aromatic ring of phenol. However, AlCl3 also reacts with phenol to form a complex, which is less reactive towards methylation. Therefore, it is necessary to use a large excess of AlCl3 to ensure that most of it reacts with phenol and leaves enough free CH3Cl for methylation to occur.
AICI3 reacts with phenol to give C6H5OAICI2:
AlCl3 reacts with phenol to form a complex called phenol-AlCl3 adduct. This adduct has the general formula C6H5OAlCl2. The reaction involves the Lewis acid-base interaction between the lone pair of electrons on the oxygen atom of phenol and the empty orbital of AlCl3. This complex is less reactive towards methylation, as the oxygen atom is already coordinated to the aluminum atom, making it less available for nucleophilic attack by CH3Cl.
Ortholpara methylation occurs with para isomer as major product:
The Friedel-Crafts methylation reaction can result in the substitution of a methyl group at either the ortho or para position of the aromatic ring. However, due to the steric hindrance caused by the bulky phenolic -OH group, the para substitution is favored. The methyl group is more likely to attack the para position, which is less hindered by the -OH group compared to the ortho position. Therefore, the para isomer is the major product in this reaction.
In summary, in the Friedel-Crafts methylation reaction of phenol using CH3Cl/AlCl3, a large excess of AlCl3 is taken to ensure sufficient availability of free CH3Cl for methylation to occur. AlCl3 reacts with phenol to form a less reactive complex, and methylation occurs primarily at the para position due to steric hindrance caused by the -OH group.
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What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer?
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What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer?.
What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer?.
Solutions for What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer?, a detailed solution for What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? has been provided alongside types of What is/are true regarding Friedel-Crafts methylation reaction of phenol using CH3CI/AICI3?a)Methylation succeeds only if large excess of AICI3 is takenb)AICI3 reacts with phenol to give C6H5OAICI2c)Methylation occurs mainly at meta positiond)Ortholpara methylation occur with para isomer as major productCorrect answer is option 'A,B,D'. Can you explain this answer? theory, EduRev gives you an
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