Topic 10 · page 9 of 27

Mechanism: hydroxide and a primary halogenoalkane

The hydroxide ion donates a lone pair to the δ+ carbon while the C–Br bond breaks and bromine leaves with the bonding pair. Two curly arrows, one from a lone pair and one from the bond, describe the whole mechanism.

12 min read10.910.9(i)

Read first: Hydroxide: substitution or elimination, Nucleophiles, electrophiles and bond polarity

The mechanism 10.910.9(i)

The reaction is the one from page 6: a primary halogenoalkane and aqueous potassium hydroxide.

CHX3CHX2Br+OHX−→CHX3CHX2OH+BrX− \ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}

The potassium ion is a spectator and does not appear in the mechanism. Three facts decide where the arrows go:

The mechanism of bromoethane with hydroxide. On the left is a hydroxide ion with three lone pairs on oxygen and a negative charge, and bromoethane with delta plus on carbon 1 and delta minus on bromine. A first curly arrow starts at a lone pair on the oxygen and points to the carbon. A second curly arrow starts in the middle of the C-Br bond and points to the bromine. On the right are the products, ethanol and a bromide ion with four lone pairs and a negative charge.
Arrow 1 starts at a lone pair on O and ends at the δ+ carbon. Arrow 2 starts in the C–Br bond and ends at Br. The products are ethanol and Br⁻.
  1. Arrow 1 starts at a lone pair on the oxygen of OH⁻ and points to the δ+ carbon. The pair becomes the new C–O bond.
  2. Arrow 2 starts at the middle of the C–Br bond and points to the bromine. Both bonding electrons go to bromine, which becomes Br⁻ with a lone pair.

The two arrows happen together, in a single step. For a moment the carbon is partly bonded to both O and Br. This is sometimes drawn as a transition state, in square brackets, with dashed partial bonds. You will not be asked to draw it.

The transition state of the reaction in square brackets. The carbon has dashed partial bonds to both oxygen and bromine, the oxygen and the bromine each carry a partial negative charge, delta minus, and the overall minus charge is written outside the bracket. The double dagger symbol marks a transition state.
The transition state, for interest. In square brackets: the C–O bond is forming and the C–Br bond is breaking at the same time.

MethodDrawing the mechanism for a primary halogenoalkane with aqueous KOH

  1. Draw the halogenoalkane with the C–X bond clear. Mark δ+ on the carbon and δ− on the halogen.
  2. Draw the hydroxide ion as O with its lone pairs and a minus charge.
  3. Arrow 1: from a lone pair on O to the δ+ carbon.
  4. Arrow 2: from the C–X bond to the halogen.
  5. Draw the products: the alcohol and the halide ion with its lone pairs and a minus charge.
  6. Name the mechanism: nucleophilic substitution.

Other primary halogenoalkanes 10.910.9(i)

The mechanism is the same for any primary halogenoalkane. Only the R group and the halogen change. The leaving halide ion is Cl⁻, Br⁻ or I⁻.

Worked example1-chloropropane with aqueous KOH

Draw the mechanism for 1-chloropropane, CH₃CH₂CH₂Cl, with aqueous potassium hydroxide.

  1. Draw CH₃CH₂CH₂–Cl and mark δ+ on the carbon bonded to Cl, and δ− on Cl.
  2. Draw OH⁻ with a lone pair on O.
  3. Arrow 1: from the lone pair on O to the δ+ carbon.
  4. Arrow 2: from the C–Cl bond to Cl.
  5. Products: propan-1-ol, CH₃CH₂CH₂OH, and Cl⁻ with four lone pairs.
The mechanism of 1-chloropropane with hydroxide, drawn like the bromoethane mechanism: a curly arrow from a lone pair on the oxygen of OH minus to the carbon bonded to chlorine, and a curly arrow from the C-Cl bond to chlorine. The products are propan-1-ol and a chloride ion.
The same two arrows for 1-chloropropane. Only the R group and the halogen are different.

Check yourself

  1. Draw the mechanism for the reaction of 1-chloropropane with aqueous potassium hydroxide.
  2. State where curly arrow 1 starts and where it ends in the reaction of bromoethane with hydroxide.
  3. State where curly arrow 2 starts and where it ends, and say what it shows.
  4. Name the mechanism and give the role of the hydroxide ion.
  5. Why is a δ+ and δ− shown on the C–Br bond in the mechanism?
  6. A student draws arrow 1 from the H of the OH⁻ ion to the carbon. Explain why this earns no mark.

Answers cover these until you have tried the questions

  1. See the last figure above: arrow 1 from a lone pair on O of OH⁻ to the δ+ carbon; arrow 2 from the C–Cl bond to Cl; products CH₃CH₂CH₂OH and Cl⁻ (with lone pairs and a minus charge).
  2. It starts at a lone pair on the oxygen of OH⁻ and ends at the δ+ carbon (where the new C–O bond forms).
  3. It starts at the middle of the C–Br bond and ends at the bromine atom. It shows the bonding pair going to bromine in heterolytic fission, so Br⁻ forms.
  4. Nucleophilic substitution. OH⁻ is a nucleophile.
  5. Bromine is more electronegative than carbon, so the bond is polar. The δ+ carbon is the atom attracted to the nucleophile's lone pair.
  6. An arrow must start at a lone pair or a bond. The H of OH⁻ is not where the electrons are, and an arrow from H would show the wrong pair moving.