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Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?
  • a)
    Only heat
  • b)
    NaNO2/Cu
  • c)
    Cu2O/H2O
  • d)
    NaF/Cu
Correct answer is option 'A'. Can you explain this answer?
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Fluorination of an aromatic ring is easily accomplished by treating a ...
Answer:

The correct condition for the fluorination of an aromatic ring is only heat (option A). Let's understand this reaction in detail:

Fluorination of Aromatic Ring:
Fluorination of an aromatic ring involves the introduction of a fluorine atom into the ring. This can be achieved by treating a diazonium salt with a fluorine source. In this case, the fluorine source used is HBF4 (hydrogen tetrafluoroborate).

Reaction:
The reaction can be represented as follows:
C6H5-N2+ + HBF4 -> C6H5-F + N2 + HBF3

Here, the diazonium salt (C6H5-N2+) reacts with HBF4 to form the desired product, which is an aromatic compound with a fluorine atom attached to the ring (C6H5-F). Nitrogen gas (N2) is also evolved as a byproduct, along with the formation of HBF3 (hydrogen fluoroborate).

Conditions:
The correct condition for this reaction is only heat (option A). Heating the mixture of the diazonium salt and HBF4 helps in the generation of the reactive species required for the reaction to occur. The reaction is typically carried out at elevated temperatures, usually around 80-100°C.

Other Options:
The other options mentioned are not correct for the fluorination of an aromatic ring:
- NaNO2/Cu: This combination is used for the Sandmeyer reaction, which involves the conversion of a primary aromatic amine to a diazonium salt. It is not suitable for the fluorination of an aromatic ring.
- Cu2O/H2O: This combination is used for the reduction of a diazonium salt to yield an aromatic compound. It is not suitable for the fluorination reaction.
- NaF/Cu: This combination is not commonly used for the fluorination of an aromatic ring. HBF4 is a more commonly used fluorine source in this reaction.

In conclusion, the correct condition for the fluorination of an aromatic ring is only heat (option A).
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Can you explain the answer of this question below:On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic.The increasing order of stability of these compounds are is under:Anti-aromatic compound non-aromatic compound aromatic compounds. Compounds to be aromaticfollow the following conditions (according to valence bond theory)(i) The compounds must be be cyclic in structure having (4n + 2) e, where n = Hckels number = 0, 1,2, 3 et.c(ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of thecompounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.(iii) There must be a ring current of electrons in the ring or cyclic structure i.e. cyclic structure mustundergo resonance .Compounds to be anti-aromatic, it must have 4ne where n = 1, 2 and it must be planar and undergoresonance. Non-aromatic compounds the name itself spells that compounds must be non-planarirrespective of number of electrons. Either it has 4ne or (4n + 2) electrons it does not matter.The rate of reaction of any aromatic compounds depends upon the following factors:(i) Electron density(ii) stability of carbocation producedHigher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilicsubstitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after theattack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a greateffect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher theatomic weight or, molecualr weight, higher will be the van der Waals force of attraction or, bond energy.Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.)otherwise there will be no effect of kinetic labelling.Q. Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towardsnitrations?A:C6H6 C6D6 C6T6B:C6H6 = C6D6 = C6T6C:C6H6 C6D6 = C6T6D:C6T6 C6D6 C6H6The answer is b.

On the basis of aromaticity, there are three types of compounds i.e. aromatic, non-aromatic and antiaromatic. The increasing order of stability of these compounds are is under: Anti-aromatic compound < non-aromatic compound < aromatic compounds. Compounds to be aromaticfollow the following conditions (according to valence bond theory)(i) The compounds must be be cyclic in structure having (4n + 2)π e–, where n = Hückel’s number = 0, 1, 2, 3 et.c(ii) The each atoms of the cyclic structure must have unhybridised p-orbital i.e. the atoms of the compounds have unhybridised p-orbital i.e. usually have sp2 hybrid or planar.(iii) There must be a ring current of π electrons in the ring or cyclic structure i.e. cyclic structure mustundergo resonance .Compounds to be anti-aromatic, it must have 4nπe– where n = 1, 2… and it must be planar and undergo resonance. Non-aromatic compounds the name itself spells that compounds must be non-planarirrespective of number of π electrons. Either it has 4nπe– or (4n + 2) π electrons it does not matter.The rate of reaction of any aromatic compounds depends upon the following factors:(i) Electron density(ii) stability of carbocation producedHigher the amount of electron density of the ring, higher will be its rate towards aromatic electrophilic substitution and vice-versa. Similarly, higher will be the stability of the produced carbocation after the attack of electrophile, higher will be its rate towards aromatic electrophilic substitution. There is a great effect of kinetic labelling on the rate of aromatic electrophilic substitution. As we known that higher the atomic weight or, molecualr weight, higher will be the van der Waal’s force of attraction or, bond energy. Since there will be effect of kienetic labelling if the 2nd step of the reaction will be the slow step, (r.d.s.) otherwise there will be no effect of kinetic labelling.Q. Which of the following is correct order of the rate of reaction of C6H6, C6D6 and C6T6 towardssulphonation?

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Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?a)Only heatb)NaNO2/Cuc)Cu2O/H2Od)NaF/CuCorrect answer is option 'A'. Can you explain this answer?
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Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?a)Only heatb)NaNO2/Cuc)Cu2O/H2Od)NaF/CuCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?a)Only heatb)NaNO2/Cuc)Cu2O/H2Od)NaF/CuCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?a)Only heatb)NaNO2/Cuc)Cu2O/H2Od)NaF/CuCorrect answer is option 'A'. Can you explain this answer?.
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