The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0andordm;C gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer?

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The reaction of compound P with CH₃MgBr (excess) in (C₂H₅)₂O followed by addition of H₂O gives Q, The compound Q on treatment with H₂SO₄ at 0ºC gives R. The reaction of R with CH₃COCl in the presence of anhydrous AlCl₃ in CH₂Cl₂ f ollowed by treatment with H₂O produces compounds S. [Et it compounds P is ethyl group]

Q. The product S is -

Correct answer is option 'D'. Can you explain this answer?

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The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group ]Q.The reactions, Q to R and S to S, are

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The plot given below shows P-T curves (where P is the pressure and T is the temperatur e) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ______.Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product. Correct answer is '0.05'. Can you explain this answer?

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PARAGRAPH “X”Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product.Q. The compound Y is

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PARAGRAPH “X”Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K withmoist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product.QThe compound Z is

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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.

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The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer?

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The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. 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 The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer?.

Solutions for The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. 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 The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q, The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 f ollowed by treatment with H2O produces compounds S. [Et it compounds P is ethyl group]Q.The product S is -a)b)c)d)Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.

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