Substitution reactions of benzene 12-13 Flashcards

1
Q

Draw the general reaction mechanism of electrophilic substitution of benzene.

A
  • The electron-dense ring attracts an electrophile- the electrophile accepts a pair of the pi electrons from the delocalised ring to form a covalent bond.
  • An intermediate forms that that contains both the electrophile and the hydrogen atom that is being substituted. The delocalised pi-electron cloud has been disrupted and the intermediate is less stable than benzene.
  • The unstable intermediate rapidly loses the hydrogen as an H* ion. The delocalised ring of electrons reforms and stability is restored.
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2
Q

Write the equation for the formation of NO2+.

A

HNO3 + H2SO4 —–> NO2+ + HSO4- + H2O

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3
Q

What are the reagents needed for the nitration of benzene and why are they needed in the reaction?

A
  • A mixture of concentrated nitric acid and concentrated sulfuric acid is used.
  • The sulfuric acid is needed to generate an electrophile from the nitric acid.
  • The electrophile is the nitryl cation, nitronium ion, and has the formula NO2+.
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4
Q

Draw the mechanism for the nitration of benzene.

A
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5
Q

Explain how the sulfuric acid acts as a catalyst in the nitration of benzene.

A

The H+ reacts with the HSO4- from the first step to reform H2SO4. So the sulfuric acid is acting as a catalyst.

H++ HSO4- —–> H2SO4

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6
Q

Write how the equation for the formation of Br+(bromonium ion, reacts with benzene).

A

Br2 + FeBr3 ——> Br+ + FeBr4-

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7
Q

Draw out the mechanism for the bromination of benzene.

A
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8
Q

Show how the iron(III) bromide used as a halogen carrier in the bromination of benzene is a catalyst.

A

The H+ reacts with FeBr4- from the first step to form the other product of the overall reaction, HBr, and to reform FeBr3. So the iron(III) is acting as a catalyst.

H+ + FeBr4-——-> FeBr3 + HBr

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9
Q

For bromination what halogen carriers could you use?

A
  • For bromination, the halogen carrier is AlBr3, FeBr3or Fe.
  • You can use Fe for bromination because Fe reacts with the halogens to form the iron (III) halide.
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