Nucleophilic substitution + elimination

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What is the general equation for a nucleophilic reaction?

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1

What is the general equation for a nucleophilic reaction?

CH3CH2X + Nu- → CH3CH2Nu + X-

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2

What is a nucleophile?

An electron pair donor that is attracted to areas of electron deficiency. Has a lone pair.

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3

Which three nucleophiles can halogenoalkanes react with?

OH, CN, NH3

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4

What conditions are required for nucleophilic substitution with an OH nucleophile and what is produced?

  • Warm, aqueous KOH/NaOH

  • Reflux

  • An alcohol and a halogen ion are produced

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5

Draw the mechanism for the reaction between 1-bromopropane and sodium hydroxide (textbook pg. 213)

XXX.

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6

What conditions are required for nucleophilic substitution with a CN nucleophile and what is produced?

  • Warm, ethanolic KCN (as water could act as a competing nucleophile)

  • Reflux

  • Nitrile is produced

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7

Draw the mechanism for the reaction between 1-bromopropane and potassium cyanide (textbook pg. 213).

XXX.

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8

What conditions are required for nucleophilic substitution with an NH3 nucleophile and what is produced?

  • Warm, excess ethanolic ammonia

  • Amine is produced

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9

Draw the mechanism for the reaction between 1-bromopropane and ammonia (textbook pg. 214).

XXX.

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10

What are the three points summarising nucleophilic substitution?

  • Occurs between a halogenoalkane and a nucleophile.

  • Nu attacks the delta positive C, and breaks the C-X bond.

  • One new bond (C-Nu) is formed and one is broken (C-X)

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11

How can the conditions of nucleophilic substitution with a hydroxide nucleophile change to cause elimination?

Warm the halogenoalkane with hydroxide ions dissolved in ethanol, instead of water. Under reflux.

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12

What is produced by elimination of halogenoalkanes?

An alkene, water and halogen ion.

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13

Draw the mechanism for the elimination of 2-bromopropane with potassium hydroxide (textbook pg. 213).

XXX.

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14

What happens if a longer, unsymmetrical halogenoalkane is used in elimination?

Two products are formed as the double bond could go on either side of the carbon with the halogen attached, e.g., 2-bromobutane forms but-1-ene and but-2-ene.

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