Simun Mishra posted an Question
June 29, 2020 • 15:18 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Can u explain it here written without peroxide br attach to more substituted end and with peroxide br attached to less substituted end ?

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    Dinesh khalmaniya 1 best-answer

    https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Hydrogen_Bromide_and_Aklenes%3A_The_Peroxide_Effect#:~:text=Because%20the%20HBr%20adds%20on,product%20predicted%20by%20Markovnikov's%20Rule.

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    open this link to understand peroxide effect

  • Simun mishra

    it is written opposite according to markownikoffs rule br attach to less no H and anti markownikoffs rule with peroxide Br attached to higher no h attaching carbon

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    yes br attached to carbon having more hydrogen in presence of peroxide

  • Suman Kumar

    https://www.google.com/url?sa=t&source=web&rct=j&url=https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Hydrogen_Bromide_and_Aklenes%253A_The_Peroxide_Effect&ved=2ahUKEwjQssCKlqbqAhXPe30KHbzbBrwQFjAXegQIAxAC&usg=AOvVaw0awiXQJuJvAeBH_1p_xc3B&cshid=1593404502946

  • Suman Kumar

    https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.chem.ucla.edu/~harding/IGOC/P/peroxide_effect.html%23:~:text%3DPeroxide%2520effect%253A%2520The%2520change%2520in,the%2520presence%2520of%2520a%2520peroxide.&ved=2ahUKEwjQssCKlqbqAhXPe30KHbzbBrwQFjANegQIChAu&usg=AOvVaw32OkJfAfPbWhQpiQbUnkI5&cshid=1593404454259

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    Priyanshu kumar best-answer

    This is based on Markownikov and Anti markownikov addition reactio....with peroxide anti-markownikov addition occurs and Br (the negatively charged species) attached to carbon which have more hydrogen opposite to markovnikov addition

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    When an alkyl halide is added to an unsymmetrical alkene, an addition reaction takes place. To predict the product of the reaction, we use the markovnikovs rule. Markovnikovs rule says that "The electronegative part (halogen) of the alkyl halide gets attached to the least substituted carbon (carbon having the lowest number of hydrogens attached to it) out of the double bonded carbons." Anti-Markovnikov addition (otherwise known as Peroxide effect or Kharasch effect) just says the opposite, "The halogen of the alkyl halide gets attached to the more substituted carbon out of the double bonded carbons. But remember, the reagent plays a important role here and the mechanism is different in both the cases. The intermediate formed in Markovnikovs addition is a carbocation which can undergo rearrangement which can also be expained via 1,2-hydride shift. The stability of carbocations increases as 3°>2°>1°, so rearrangement occurs whenever possible. In case of Anti Markovnikovs addition, the intermediate formed is a free radical which doesn't undergo rearrangement. Also note that the Anti Markovnikov addition only takes place in case of highly selective halide like HBr. If you add HCl or HI in presence of Peroxide, then the reaction will still proceed via the Markovnikov addition.

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    simun read this

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    which pic I send is wrong or write

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    see your pic is not visible to us...as app is not working properly since yesterday

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    got it simun??...read the differences between mark. addition and anti. mark addition...

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    in the pic written ...... No peroxide = Br goes to more substituted end and with peroxide = Br goes to less substituted end

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    yes same thing...read above it is written

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    ok then the topic is written wrong in this ...

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    may be as it is not visible to us.... see this also as reference http://leah4sci.com/markovnikov-vs-anti-markovnikov-in-alkene-addition-reactions/

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    see simun..there is little bit confusion so keep in mind these two things....In markovnikov addition the negative part of addendum react to more substituted carbon and proton attached to less subsituted carbon...and vice versa in anti-markovnikov addition

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    got simun??

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