Halogen compounds (AS)：练习题

• a.

  Diethyl ether

• b.

  Ethene

• c.

  Ethanoic acid

• d.

  Ethanol

• a.

  The benzene ring in chlorobenzene provides steric hindrance, preventing nucleophilic attack.

• b.

  Chlorobenzene is less polar than chloroethane, making it less susceptible to nucleophilic attack.

• c.

  The C-Cl bond in chlorobenzene is stronger due to sp² hybridisation of carbon and partial double bond character.

• d.

  The delocalisation of electrons in the benzene ring destabilises the transition state for nucleophilic substitution.

• a.

  Bromoethene

• b.

  Ethene

• c.

  Ethanoic acid

• d.

  Ethanol

• a.

  Both compounds undergo S_N2 mechanism, but the carbon-bromine bond in 2-bromo-2-methylpropane is weaker due to inductive effects.

• b.

  2-bromo-2-methylpropane undergoes S_N2 mechanism due to steric hindrance, while bromoethane undergoes S_N1 mechanism.

• c.

  Both compounds undergo S_N1 mechanism, but the tertiary carbocation formed from 2-bromo-2-methylpropane is more reactive.

• d.

  2-bromo-2-methylpropane undergoes S_N1 mechanism, forming a stable tertiary carbocation, while bromoethane undergoes S_N2 mechanism.

• a.

  I. \( \text{KCN(ethanolic), heat} \)
  II. \( \text{LiAlH}_4, \text{then H}_2\text{O} \)

• b.

  I. \( \text{KCN(aq), heat} \)
  II. \( \text{H}_2\text{O/H}^+, \text{heat} \)

• c.

  I. \( \text{NaOH(aq), heat} \)
  II. \( \text{KMnO}_4/\text{H}^+, \text{reflux} \)

• d.

  I. \( \text{KCN(ethanolic), heat} \)
  II. \( \text{H}_2\text{O/H}^+, \text{heat} \)

先自己写一遍答案，再对照解题步骤。