Halogenoalkana dan jalur kompetitif
Diperkenalkan| English | Makna |
|---|---|
| substitution/ˌsʌbstɪˈtjuːʃn/ | substitusi |
| elimination/ɪˌlɪmɪˈneɪʃn/ | eliminasi |
| nucleophile/ˈnjuːklɪɒfaɪl/ | nukleofil |
| hydrolysis/haɪˈdrɒləsɪs/ | hidrolisis |
Why can one bromide give an alcohol or an alkene?
- Warm aqueous KOH gives substitution 取代 to an alcohol. Hot ethanolic KOH favours elimination 消去 to an alkene.
- The solvent and conditions are part of the prediction, not optional labels.
Kondisi manakah yang menguntungkan pembentukan etena dari bromoetana?
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Alkali etanolik dengan pemanasan menguntungkan eliminasi; alkali akuatik menguntungkan substitusi menjadi etanol.
Identify the reactive carbon
- C–Br is polar: carbon is partly positive and accepts an electron pair from a nucleophile 亲核试剂.
- Primary, secondary and tertiary count the carbon groups attached to the carbon bearing the halogen.
Klasifikasikan CH₃CHBrCH₃ sebagai primer, sekunder, atau tersier.
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Karbon yang terikat pada Br terhubung ke dua gugus karbon lainnya.
Follow the electron pairs
- With OH⁻ and primary bromoethane, one arrow goes from the oxygen lone pair to the carbon bearing Br. Another goes from C–Br to Br.
- With NH₃, an alkylammonium ion forms first. A second NH₃ removes H⁺ to give the amine and NH₄⁺.
CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ with warm aqueous alkali. The carbon skeleton stays at two carbons; a cyanide route would make three.
Nukleofil mendonorkan pasangan elektron saat membentuk ikatan.
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Pasangan bermula dari pasangan bebas atau ikatan yang ada dan didonorkan ke situs yang kekurangan elektron.
Check the product carbon count
- Bromoethane with alcoholic KCN gives CH₃CH₂CN. The cyanide carbon adds one carbon, making propanenitrile.
- Excess alcoholic ammonia under pressure gives ethylamine. Hot ethanolic KOH instead removes H and Br overall, giving ethene.
Berapa banyak karbon dalam produk organik dari bromoetana dengan KCN alkohol?
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Dua karbon asli ditambah karbon sianida menghasilkan nitril tiga karbon.
Evaluate hydrolysis 水解 evidence
- In matched aqueous silver nitrate/ethanol tests, iodoalkanes hydrolyse faster than bromoalkanes, then chloroalkanes: C–I is weaker.
- Match temperature, concentrations and volumes. Visible AgX formation is an indirect endpoint with detection/solubility limitations.
“The most polar carbon–halogen bond always hydrolyses fastest” is false. Bond strength is central to the matched chloro/bromo/iodo comparison.
Identify the nucleophile/base, conditions, electron-pair source, leaving group and carbon count.
Kontrol manakah yang membantu membandingkan laju hidrolisis?
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Jaga variabel selain substrat pilihan tetap sama; batasan deteksi harus tetap dievaluasi.