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Which of the following compounds exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O ?
  • a)
    3-methyl-1,2-cycloheptanedione
  • b)
    2-methyl-1,3-cycloheptanedione
  • c)
    5-methyl-1,3-cycloheptanedione
  • d)
    6-methyl-1,4-cycloheptanedione 
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Which of the following compounds exchanges the largest number of hydro...
It has eight α-H which can undergo exchange with deuterium.
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Which of the following compounds exchanges the largest number of hydro...
Answer:

To determine which compound exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O, we need to analyze the structures of the given compounds and identify the potential exchangeable hydrogens.

Analysis of Compound Structures:

a) 3-methyl-1,2-cycloheptanedione
b) 2-methyl-1,3-cycloheptanedione
c) 5-methyl-1,3-cycloheptanedione
d) 6-methyl-1,4-cycloheptanedione

Explanation:

In order to exchange hydrogens with deuterium, the compound must have acidic hydrogens. In this case, the acidic hydrogens are the α-hydrogens, which are next to the carbonyl group.

Analysis of Compound a:
3-methyl-1,2-cycloheptanedione
This compound has only one α-hydrogen, which can potentially exchange with deuterium.

Analysis of Compound b:
2-methyl-1,3-cycloheptanedione
This compound also has only one α-hydrogen, which can potentially exchange with deuterium.

Analysis of Compound c:
5-methyl-1,3-cycloheptanedione
This compound has two α-hydrogens, which can potentially exchange with deuterium.

Analysis of Compound d:
6-methyl-1,4-cycloheptanedione
This compound has three α-hydrogens, which can potentially exchange with deuterium.

Conclusion:

Among the given compounds, compound d (6-methyl-1,4-cycloheptanedione) has the largest number of exchangeable hydrogens (three α-hydrogens). Therefore, when treated with KOD in D2O, compound d will exchange the largest number of hydrogens from deuterium.
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Which of the following compounds exchanges the largest number of hydro...
It has eight α-H which can undergo exchange with deuterium.
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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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Which of the following compounds exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O ?a)3-methyl-1,2-cycloheptanedioneb)2-methyl-1,3-cycloheptanedionec)5-methyl-1,3-cycloheptanedioned)6-methyl-1,4-cycloheptanedioneCorrect answer is option 'D'. Can you explain this answer?
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Which of the following compounds exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O ?a)3-methyl-1,2-cycloheptanedioneb)2-methyl-1,3-cycloheptanedionec)5-methyl-1,3-cycloheptanedioned)6-methyl-1,4-cycloheptanedioneCorrect 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 Which of the following compounds exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O ?a)3-methyl-1,2-cycloheptanedioneb)2-methyl-1,3-cycloheptanedionec)5-methyl-1,3-cycloheptanedioned)6-methyl-1,4-cycloheptanedioneCorrect 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 Which of the following compounds exchanges the largest number of hydrogens from deuterium when treated with KOD in D2O ?a)3-methyl-1,2-cycloheptanedioneb)2-methyl-1,3-cycloheptanedionec)5-methyl-1,3-cycloheptanedioned)6-methyl-1,4-cycloheptanedioneCorrect answer is option 'D'. Can you explain this answer?.
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