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Mechanism of elimination of HCl from chlorobenzene

Chemistry

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Answer

Styrene\text{Styrene}

Step-by-step solution

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Huu ni utaratibu wa kuondoa hidrojeni-halojeni (dehydrohalogenation) wa 1-chloro-1-phenylethane ili kutengeneza styrene, unaofanywa na besi kali (KOH) katika ethanoli. Utaratibu huu unafuata njia ya E2 (Elimination, bimolecular).

  1. Uundaji wa besi kali

    Potassium hydroxide (KOH) katika ethanoli (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}) huunda ioni ya ethoxide (CH3CH2O−\text{CH}_3\text{CH}_2\text{O}^-), ambayo ni besi kali.

    KOH+CH3CH2OH⟶CH3CH2O−K++H2O\text{KOH} + \text{CH}_3\text{CH}_2\text{OH} \longrightarrow \text{CH}_3\text{CH}_2\text{O}^- \text{K}^+ + \text{H}_2\text{O}

    Ioni ya ethoxide ndiyo itakayotoa protoni.

  2. Utaratibu wa E2

    Ioni ya ethoxide (CH3CH2O−\text{CH}_3\text{CH}_2\text{O}^-) hutoa protoni (hidrojeni) kutoka kwa kaboni ya beta (kaboni iliyo karibu na kaboni yenye klorini). Wakati huo huo, elektroni kutoka kwa dhamana ya C-H huhamia kuunda dhamana mpya ya C=C, na klorini huondoka kama ioni ya kloridi (Cl−\text{Cl}^-). Hii hutokea katika hatua moja iliyoratibiwa.

    C6H5CH(Cl)CH3+CH3CH2O−⟶C6H5CH=CH2+CH3CH2OH+Cl−\text{C}_6\text{H}_5\text{CH(Cl)CH}_3 + \text{CH}_3\text{CH}_2\text{O}^- \longrightarrow \text{C}_6\text{H}_5\text{CH=CH}_2 + \text{CH}_3\text{CH}_2\text{OH} + \text{Cl}^-
  3. Bidhaa za mwisho

    Bidhaa kuu ya kikaboni ni styrene (phenylethene). Ioni ya kloridi huungana na ioni ya potasiamu (K+\text{K}^+) kuunda potassium chloride.

    K++Cl−⟶KCl\text{K}^+ + \text{Cl}^- \longrightarrow \text{KCl}

    Bidhaa za mwisho za mmenyuko ni:

    • Styrene (C6H5CH=CH2\text{C}_6\text{H}_5\text{CH=CH}_2)
    • Ethanoli (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}) (kama bidhaa ndogo na pia kutumika kama kiyeyushi)
    • Potassium chloride (KCl\text{KCl})

    Bidhaa kuu ya mmenyuko huu ni:

    Styrene\text{Styrene}
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