UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.3
Substitution of Halogenoalkanes with Aqueous Sodium Hydroxide
FREE CHEMISTRY REVISION · TOPIC 58
Substitution of Halogenoalkanes with Aqueous Sodium Hydroxide
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.3
What you will learn
Predict, draw and name the alcohol formed when a halogenoalkane reacts with aqueous sodium hydroxide. Write full and net ionic equations and give the conditions. Pearson E1.7.3 requires product prediction and conditions; mechanisms and practical techniques are outside the required E1.7 scope.
1. The functional-group change
In this substitution, the halogen attached to a saturated carbon is replaced by an OH group. A carbon–halogen connection becomes a carbon–oxygen connection, and the oxygen is also bonded to hydrogen. The carbon skeleton is retained in the straightforward examples here. The reaction is commonly called hydrolysis.
R–X + NaOH(aq) → R–OH + NaX
R–X + OH⁻(aq) → R–OH + X⁻(aq)
X represents the halogen; NaX is the corresponding sodium halide. R is the rest of the organic group, not a new element. Replace C–X, not an arbitrary C–H bond. An alcohol contains C–O–H rather than C–Na or C–O–Na.
2. Reagent and reaction conditions
| Item | What to state | Why it matters |
|---|---|---|
| Reagent | Aqueous sodium hydroxide, NaOH(aq) | Provides hydroxide ions in a water-based mixture |
| Heating | Warm/heat; heating under reflux is a standard condition for these examples | Allows sufficient reaction time at elevated temperature without simply losing volatile material |
| UV | Not required for this substitution | UV alkane halogenation is a different reaction |
| Product | Corresponding alcohol and sodium halide | Check both the organic change and the displaced halogen |
The word aqueous is essential. It means dissolved in water. Do not replace the specified aqueous reagent with a hot ethanolic hydroxide condition: changing the solvent changes the reaction competition. This lesson covers the aqueous substitution route, not the later elimination topic.
There is no single universal numerical temperature for all halogenoalkanes. State aqueous NaOH and warming/heating under reflux for the standard examples; follow any more specific conditions supplied in the question. Reflux is a condition description here, not a requirement to memorise apparatus or a laboratory procedure. Real reactions can have competing pathways; the equations below identify the substitution product, not a guarantee of 100% selectivity.
3. Balanced-equation and naming examples
Worked example 1 — bromoethane
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr(aq)
Heat under reflux. Product: ethanol. Bromoethane is C₂H₅Br; adding OH replaces Br with OH to give C₂H₆O. The displaced Br is present as bromide in sodium bromide. The complete equation has C₂H₆BrNaO on each side.
Worked example 2 — 1-chloropropane
CH₃CH₂CH₂Cl + NaOH(aq) → CH₃CH₂CH₂OH + NaCl(aq)
Use aqueous NaOH with heat/reflux. The OH group occupies the terminal position previously bonded to Cl. Product: propan-1-ol, not propan-2-ol. Formula changes from C₃H₇Cl to C₃H₈O; the organic product gains one O and one H while losing Cl.
Worked example 3 — 2-bromopropane
CH₃CHBrCH₃ + NaOH(aq) → CH₃CH(OH)CH₃ + NaBr(aq)
The substitution product is propan-2-ol. Its middle carbon remains bonded to two carbon atoms and one H, and gains a bond to OH in place of Br. Do not move the functional group to the end. Use the stated aqueous heating/reflux conditions.
Worked example 4 — 1-iodobutane
CH₃CH₂CH₂CH₂I + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaI(aq)
Heat with aqueous NaOH. The alcohol is butan-1-ol and the salt is sodium iodide. Carbon count remains four. The product formula is C₄H₁₀O: nine original organic H plus the H in OH.
4. Branches and reverse prediction
Worked example 5 — a branched primary halogenoalkane
(CH₃)₂CHCH₂Br + NaOH(aq) → (CH₃)₂CHCH₂OH + NaBr(aq)
Starting compound: 1-bromo-2-methylpropane. Product: 2-methylpropan-1-ol. The branch stays on carbon 2 and OH replaces Br on the terminal CH₂. Carbon and hydrogen counts in the product: C=4; H=3+3+1+2+1=10, so C₄H₁₀O.
Worked example 6 — work backwards from an alcohol
To obtain butan-2-ol by this substitution route, choose a suitable 2-halobutane such as 2-chlorobutane:
CH₃CHClCH₂CH₃ + NaOH(aq) → CH₃CH(OH)CH₂CH₃ + NaCl(aq)
State aqueous NaOH and heat/reflux. Starting with 1-chlorobutane would instead place OH on carbon 1. This reverse task checks connectivity as well as formula.
| Starting halogenoalkane | Substitution alcohol | Sodium salt |
|---|---|---|
| Chloroethane | Ethanol | NaCl |
| 1-Bromopropane | Propan-1-ol | NaBr |
| 2-Iodobutane | Butan-2-ol | NaI |
| 1-Bromo-2-methylpropane | 2-Methylpropan-1-ol | NaBr |
5. Ionic equations and quantities
Worked example 7 — remove the spectator ion
Write NaOH as Na⁺ + OH⁻ and NaBr as Na⁺ + Br⁻. Cancel Na⁺ from both sides:
CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
Left charge −1; right charge −1. The alcohol is neutral. Do not cancel the Br atom while it is covalently attached in the starting molecule: it changes chemical environment and is a product ion.
Worked example 8 — a theoretical amount
The substitution equation uses one mole of OH⁻ for each mole of a monohalogenoalkane. If 0.125 mol NaOH is limiting, halogenoalkane is in excess, and all of that hydroxide follows the substitution route, the theoretical alcohol amount is 0.125 mol. For ethanol, using Mᵣ=46.0 gives m=0.125×46.0=5.75 g. This is a theoretical maximum, not a measured yield or a claim of complete selectivity.
Practice with answers
| Task | Answer |
|---|---|
| Name CH₃CH₂CH₂OH | Propan-1-ol; OH is on an end carbon. |
| Name the salt from a bromoalkane and NaOH | Sodium bromide, NaBr. |
| Write the ionic product charge for the displaced chlorine | Cl⁻, not neutral Cl or Cl₂. |
| Would C₂H₅ONa be the specified organic product? | No. Aqueous hydrolysis produces the neutral alcohol C₂H₅OH. |
Common misconceptions
- Hydroxide simply removes the halogen with no replacement: OH occupies the original carbon–halogen position.
- All products should be named ethanol: chain length, branching and OH position must be retained.
- The OH hydrogen can be ignored in a formula count: alcohols include that H.
- Sodium bonds to the organic chain: Na⁺ is a spectator in the net ionic equation.
- Aqueous and ethanolic are interchangeable: the specified solvent is part of the conditions.
- The displaced halogen becomes X₂: it leaves the substitution equation as halide X⁻, represented in NaX in the full equation.
Examination tips
- Write NaOH(aq), not only NaOH.
- Include warming/heating under reflux when stating conditions for the standard examples.
- Locate C–X before drawing C–OH at exactly that position.
- For a full equation include NaX; for a net ionic equation include X⁻ and check charge.
- Preserve the carbon skeleton and check each carbon has four bond units and O has two.
- If the question asks for a displayed formula, expand every grouped H into its own atom and bond.
Condition cross-check: RSC, reflux and hydroxide substitution; RSC, halogenoalkane hydrolysis. Practical instructions and mechanisms in references are not required for this lesson.
Check your understanding
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