Topic 10 · page 10 of 27

Mechanism: ammonia and a primary halogenoalkane

Ammonia is a neutral nucleophile, so the first product carries a positive charge on nitrogen. A second ammonia molecule removes an H⁺ in a second step, which is why two moles of ammonia appear in the equation.

11 min read10.910.9(ii)

Read first: Mechanism: hydroxide and a primary halogenoalkane, Ammonia and cyanide

Step 1: ammonia attacks the carbon 10.910.9(ii)

The overall reaction is the one from page 7:

CHX3CHX2Br+2 NHX3→CHX3CHX2NHX2+NHX4Br \ce{CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br}

Ammonia has a lone pair on nitrogen and no charge. Like hydroxide, it is attracted to the δ+ carbon. The first step has the same two arrows as the hydroxide mechanism.

Step 1 of bromoethane with ammonia. On the left is an ammonia molecule with a lone pair on nitrogen and bromoethane with delta plus on carbon 1 and delta minus on bromine. A first curly arrow starts at the lone pair on nitrogen and points to the carbon. A second curly arrow starts in the C-Br bond and points to bromine. On the right the products are CH3CH2NH3 with a plus charge on the nitrogen, and a bromide ion.
Step 1. Arrow 1 starts at the lone pair on N and ends at the δ+ carbon. Arrow 2 starts in the C–Br bond and ends at Br. The nitrogen now has four bonds and a + charge.
  1. Arrow 1 starts at the lone pair on nitrogen and points to the δ+ carbon. The pair becomes the new C–N bond.
  2. Arrow 2 starts at the C–Br bond and points to bromine. Br⁻ leaves with both electrons.

Nitrogen has now used its lone pair to make a fourth bond, so it carries a + charge. The product is the ethylammonium ion, CH₃CH₂NH₃⁺, with Br⁻ as the counter-ion.

Step 2: a second ammonia removes H⁺ 10.910.9(ii)

The ethylammonium ion is positive and has an N–H bond that can give up H⁺. A second ammonia molecule has a lone pair, so it acts as a base.

Step 2 of the mechanism. On the left is CH3CH2NH3 with a plus charge on nitrogen and an ammonia molecule with a lone pair. A curly arrow starts at the lone pair on the ammonia nitrogen and points to a hydrogen on the CH3CH2NH3 nitrogen. A second curly arrow starts in the N-H bond and points to the nitrogen. The products are ethylamine, CH3CH2NH2 with a lone pair on nitrogen, and the ammonium ion NH4 with a plus charge.
Step 2. Arrow 3 starts at the lone pair on the second NH₃ and ends at an H. Arrow 4 starts in the N–H bond and ends at N. The products are ethylamine and NH₄⁺.
  1. Arrow 3 starts at the lone pair on the second NH₃ and points to a hydrogen on the nitrogen of the ethylammonium ion.
  2. Arrow 4 starts at the N–H bond and points to the nitrogen. The pair stays with nitrogen and becomes its lone pair again.

The products are ethylamine, CH₃CH₂NH₂, and the ammonium ion, NH₄⁺. With the bromide ion they form ammonium bromide, NH₄Br, which is why the overall equation has 2NH₃ on the left and NH₄Br on the right.

MethodDrawing the mechanism for ammonia

  1. Draw the halogenoalkane with δ+ on carbon and δ− on the halogen. Draw NH₃ with its lone pair.
  2. Arrow 1: from the lone pair on N to the δ+ carbon. Arrow 2: from the C–X bond to X.
  3. Draw the CH₃CH₂NH₃⁺ ion and X⁻.
  4. Draw a second NH₃ with its lone pair. Arrow 3: from that lone pair to an H on the N⁺. Arrow 4: from the N–H bond to N.
  5. Draw ethylamine and NH₄⁺.

Check yourself

  1. Draw step 1 of the mechanism for 1-bromopropane with ammonia.
  2. Write the formula of the ion formed in step 1, and explain why it carries a positive charge.
  3. Describe step 2: what the second ammonia molecule does and what is formed.
  4. Explain why two moles of ammonia are needed for each mole of halogenoalkane.
  5. Name the type of mechanism for step 1.
  6. State where curly arrow 1 starts and ends.

Answers cover these until you have tried the questions

  1. A lone pair on N of NH₃ to the δ+ carbon of CH₃CH₂CH₂–Br; the C–Br bond to Br. Products: CH₃CH₂CH₂NH₃⁺ and Br⁻. (Same arrows as in the figure for bromoethane.)
  2. CH₃CH₂NH₃⁺ (ethylammonium). The nitrogen used its lone pair to form a fourth bond, and it kept all its protons, so it has four bonds and a + charge.
  3. A second NH₃ acts as a base. Its lone pair takes an H⁺ from the N⁺, and the N–H pair stays on nitrogen. Ethylamine, CH₃CH₂NH₂, and the ammonium ion, NH₄⁺, form.
  4. One NH₃ is the nucleophile that attacks the carbon, and a second removes the H⁺ from the product. Two NH₃ per halogenoalkane.
  5. Nucleophilic substitution.
  6. It starts at the lone pair on the nitrogen of NH₃ and ends at the δ+ carbon.