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A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A is
- a)o-toluic acid
- b)o-chlorotoluene
- c)o-bromo toluene
- d)m-toluic acid
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) ...
°C gave a compound (D) which on heating with Cu gave benzene. Deduce the structure of the compound (A).
The reaction sequence can be summarized as follows:
A + PCl5 → B
B + Br2 + KOH → C
C + NaNO2 + HCl → D
D + Cu → benzene
First, we need to determine the structures of B, C, and D in order to work backwards and deduce the structure of A.
B is the product of the reaction between A and PCl5 followed by NH3. This is a classic reaction for converting alcohols to alkyl chlorides via an SN2 mechanism. Therefore, B is likely an alkyl chloride. The exact structure depends on the structure of A and the specific conditions of the reaction.
C is the product of the reaction between B, Br2, and KOH. This reaction is known as the Hell-Volhard-Zelinsky reaction and is used to convert carboxylic acids to α-bromo acids. Therefore, C likely contains a carboxylic acid group. The α-bromo acid product would then be deprotonated by the KOH to form the corresponding α-bromo carboxylate.
D is the product of the reaction between C, NaNO2, and HCl at 0°C. This is the Sandmeyer reaction, which is commonly used to convert primary aromatic amines to aryl diazonium salts. Therefore, D is likely an aryl diazonium salt.
Finally, the fact that D can be heated with Cu to produce benzene indicates that D contains a phenyl group. When aryl diazonium salts are heated with Cu, they undergo a coupling reaction to form an aryl-aryl bond, which in this case would be a bond between two phenyl groups to form benzene.
Putting all of this together, we can deduce that A is likely a compound containing a phenyl group and a carboxylic acid group. One possible structure for A is benzoic acid (C6H5COOH), which would give the following reaction sequence:
benzoic acid + PCl5 → benzoyl chloride
benzoyl chloride + NH3 → benzamide
benzamide + Br2 + KOH → 2-bromobenzoic acid
2-bromobenzoic acid + NaNO2 + HCl → 2-bromo-1-nitrobenzene
2-bromo-1-nitrobenzene + Cu → benzene
The reaction sequence can be summarized as follows:
A + PCl5 → B
B + Br2 + KOH → C
C + NaNO2 + HCl → D
D + Cu → benzene
First, we need to determine the structures of B, C, and D in order to work backwards and deduce the structure of A.
B is the product of the reaction between A and PCl5 followed by NH3. This is a classic reaction for converting alcohols to alkyl chlorides via an SN2 mechanism. Therefore, B is likely an alkyl chloride. The exact structure depends on the structure of A and the specific conditions of the reaction.
C is the product of the reaction between B, Br2, and KOH. This reaction is known as the Hell-Volhard-Zelinsky reaction and is used to convert carboxylic acids to α-bromo acids. Therefore, C likely contains a carboxylic acid group. The α-bromo acid product would then be deprotonated by the KOH to form the corresponding α-bromo carboxylate.
D is the product of the reaction between C, NaNO2, and HCl at 0°C. This is the Sandmeyer reaction, which is commonly used to convert primary aromatic amines to aryl diazonium salts. Therefore, D is likely an aryl diazonium salt.
Finally, the fact that D can be heated with Cu to produce benzene indicates that D contains a phenyl group. When aryl diazonium salts are heated with Cu, they undergo a coupling reaction to form an aryl-aryl bond, which in this case would be a bond between two phenyl groups to form benzene.
Putting all of this together, we can deduce that A is likely a compound containing a phenyl group and a carboxylic acid group. One possible structure for A is benzoic acid (C6H5COOH), which would give the following reaction sequence:
benzoic acid + PCl5 → benzoyl chloride
benzoyl chloride + NH3 → benzamide
benzamide + Br2 + KOH → 2-bromobenzoic acid
2-bromobenzoic acid + NaNO2 + HCl → 2-bromo-1-nitrobenzene
2-bromo-1-nitrobenzene + Cu → benzene
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Community Answer
A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) ...
Catch the question from the stage where C is being produced.
Br2+KOH combination is used in only Hoffmann bromamide reaction. In this reaction the reactant is an amide. That means B amide hai.
Chalo ab dekhte hain ki yeh amide kaise bana.
Here NH3 is used which is used to make amide from a carboxylic acid. That means ki A is a carboxylic acid.
That means ki answer is A or D.
Final product o-cresol hai. o-Cresol me ortho position par CH3 group hota hai. Is group par reaction ka koi asar nahi hoga. That means ki A is the correct answer bcoz it has both the groups i.e. COOH and CH3 at ortho position.
Br2+KOH combination is used in only Hoffmann bromamide reaction. In this reaction the reactant is an amide. That means B amide hai.
Chalo ab dekhte hain ki yeh amide kaise bana.
Here NH3 is used which is used to make amide from a carboxylic acid. That means ki A is a carboxylic acid.
That means ki answer is A or D.
Final product o-cresol hai. o-Cresol me ortho position par CH3 group hota hai. Is group par reaction ka koi asar nahi hoga. That means ki A is the correct answer bcoz it has both the groups i.e. COOH and CH3 at ortho position.
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A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer?
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A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer?.
A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound (A) when reacted with PCl5 and then with NH3 gave (B), (B) when treated with Br2 and KOH produced (C). (C) on treatment with NaNO2 and HCl at 0°C and the boiling with water forms o-cresol. Compound A isa)o-toluic acidb)o-chlorotoluenec)o-bromo toluened)m-toluic acidCorrect answer is option 'A'. Can you explain this answer?.
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