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Shweta Thakur posted an Question
June 08, 2020 • 03:54 am 30 points
  • IIT JAM
  • Chemistry (CY)

Plzexpalin that in first transition state here that why negative charge is on carbon although h is removing and why in second neg. charg. not appear

sec-Butyltrimethy lammoniunm ans hydroxide the developing double bond and shows little resemblanee with the final alkene. In contrast, the state for Saytzeff elimination is stabilized by the developing d

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  • Shweta thakur

    they are

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    Zaitsev product me alkene like transition state hai. That's why stable alkene is formed i.e more Substituted.

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    Where are you facing problem

  • Suman Kumar best-answer

    Because in zaitsev transition state have alkene like transition state and that alkene is stable which has more no. of hyperconjugation.

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    Dinesh khalmaniya 1 best-answer

    it is E2 elimination reaction. base abstract the most acidic proton. so H gets partially positive charge whereas Carbon atom gets partially negative charge. Here H and leaving group must be antiperiplanar and H and leaving group eliminate simultaneously

  • Suman Kumar Best Answer

    In ist case notice that C-N bond is not broken till hydrogen is abstracted by base. So there is negative charge on carbon. But in 2nd case hydrogen is abstracted by base at the same time C-Br bond breaking. So there is simultaneous abstraction of proton & elimination of leaving group. So there alkene like transition state.

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