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Below is a scheme of some reactions of propanol. Study it and answer the questions that follow.

Chemistry

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f(ii)

CH3CH=CH2\text{CH}_3\text{CH=CH}_2

[-CH(CH3)-CH2-]n\text{[-CH(CH}_3\text{)-CH}_2\text{-]}_n

CH3CHBrCH3\text{CH}_3\text{CHBrCH}_3

Huu hapa ni ufumbuzi wa maswali yako:

e) Mlinganyo wa kemikali kwa ajili ya utayarishaji wa oksijeni kwa kutumia manganese (IV) oksidi na hidrojeni peroksidi:

2H2O2(aq)→MnO2(s)2H2O(l)+O2(g)\text{2H}_2\text{O}_2\text{(aq)} \xrightarrow{\text{MnO}_2\text{(s)}} \text{2H}_2\text{O(l)} + \text{O}_2\text{(g)}

f)

i) Nini kitatokea katika mirija mitatu ya majaribio R, S na T baada ya siku saba:

  • Katika mrija R, misumari itatua kutu. Mrija R una hewa (oksijeni) na maji, ambazo zote ni muhimu kwa kutu kutokea.
  • Katika mrija S, misumari haitatua kutu. Mrija S una hewa (oksijeni) lakini hakuna maji, kwa sababu kloridi ya kalsiamu isiyo na maji inafyonza unyevu wote. Maji ni muhimu kwa kutu.
  • Katika mrija T, misumari haitatua kutu. Mrija T una maji yaliyochemshwa (ambayo yameondoa oksijeni iliyoyeyuka) na safu ya mafuta inayozuia oksijeni kutoka hewani kuingia kwenye maji. Oksijeni ni muhimu kwa kutu.

ii) Sababu moja kwa nini baadhi ya metali huwekewa tabaka la metali nyingine kwa njia ya elektrolisisi (electroplated):
Metali huwekewa tabaka la metali nyingine kwa njia ya elektrolisisi ili kuzuia kutu au kuboresha mwonekano (kutoa umaridadi).

  1. Mpango wa athari za propanol:
  • Y: Propanol (CH₃CH₂CH₂OH) hupitia upungufu wa maji (dehydration) kwa kutumia asidi ya sulfuriki iliyokolea (Conc. H₂SO₄) kwenye 160°C ili kuunda alkene.

    CH3CH2CH2OH→Conc. H2SO4,160∘CCH3CH=CH2+H2O\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, 160^\circ\text{C}} \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O}

    Hivyo, Y ni propene (CH3CH=CH2\boxed{\text{CH}_3\text{CH=CH}_2}).

  • Q: Y (propene) hupitia upolimishaji (polymerisation) kuunda polima.

    nCH3CH=CH2⟶[-CH(CH3)-CH2-]n\text{nCH}_3\text{CH=CH}_2 \longrightarrow \text{[-CH(CH}_3\text{)-CH}_2\text{-]}_n

    Hivyo, Q ni poly(propene) ([-CH(CH3)-CH2-]n\boxed{\text{[-CH(CH}_3\text{)-CH}_2\text{-]}_n}).

  • Z: Y (propene) hugusana na HBr (hidrojeni bromidi) katika mmenyuko wa kuongeza. Kulingana na sheria ya Markovnikov, bromini huongezwa kwenye kaboni yenye hidrojeni chache.

    CH3CH=CH2+HBr⟶CH3CHBrCH3\text{CH}_3\text{CH=CH}_2 + \text{HBr} \longrightarrow \text{CH}_3\text{CHBrCH}_3

    Hivyo, Z ni 2-bromopropane (CH3CHBrCH3\boxed{\text{CH}_3\text{CHBrCH}_3}).

  • Step I: Y (propene) hubadilishwa kuwa CH₃CH₂CH₃ (propane). Huu ni mmenyuko wa haidrojeni (hydrogenation), ambapo hidrojeni huongezwa kwenye alkene.

    CH3CH=CH2+H2→Ni/Pt/Pd catalyst, heat/pressureCH3CH2CH3\text{CH}_3\text{CH=CH}_2 + \text{H}_2 \xrightarrow{\text{Ni/Pt/Pd catalyst, heat/pressure}} \text{CH}_3\text{CH}_2\text{CH}_3

    Hivyo, Step I ni Hydrogenation (au kuongeza hidrojeni).

  • Step III: CH₃CH₂CH₃ (propane) hubadilishwa kuwa CH₃CH₂CH₂Cl (1-chloropropane). Huu ni mmenyuko wa ubadilishaji wa radikali huru (free radical substitution) ambapo atomi ya hidrojeni hubadilishwa na atomi ya klorini. Inahitaji mwanga wa UV.

    CH3CH2CH3+Cl2→UV lightCH3CH2CH2Cl+HCl\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{HCl}

    Hivyo, Step III ni Free radical chlorination (au ubadilishaji wa klorini kwa radikali huru).

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