Question Description
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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. 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 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer?.

Solutions for 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice 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?a)Same rate of reactions of C6H6, C6D6 and C6T6b)C6T6 >C6D6 >C6H6c)C6H6 >C6D6 > C6T6d)C6H6 > C6D6 = C6T6Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.