Question 6 of 39intermediate🔧 ApplyShort Answer2 marks

Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Correct Answer

The Williamson synthesis involves the reaction of a sodium alkoxide with an alkyl halide. For the preparation of 2-ethoxy-3-methylpentane from ethanol and 3-methylpentan-2-ol, the following reactions occur:

Step 1: 3-Methylpentan-2-ol reacts with sodium to form sodium alkoxide. CH₃—CH₂—CH(CH₃)—CH(CH₃)—OH + Na → CH₃—CH₂—CH(CH₃)—CH(CH₃)—ONa⁺

Step 2: Ethanol reacts with HBr to form ethyl bromide. C₂H₅OH + HBr → C₂H₅Br

Step 3: Sodium alkoxide reacts with ethyl bromide to form 2-ethoxy-3-methylpentane. CH₃—CH₂—CH(CH₃)—CH(CH₃)—ONa + C₂H₅Br → CH₃—CH₂—CH(CH₃)—CH(CH₃)—OC₂H₅ + NaBr

The reaction follows an Sₙ2 mechanism where the alkoxide ion attacks the primary alkyl halide.

Exercise: Intext Questions | Q: 7.10 | (Page Page 28)
For More Understanding

Explanation

The textbook context clearly shows the stepwise reactions for Williamson synthesis. The secondary alcohol (3-methylpentan-2-ol) is converted to its sodium alkoxide, while ethanol (a primary alcohol) is converted to ethyl bromide. This arrangement is important because Williamson synthesis works best when the alkyl halide is primary (to avoid elimination), which is why ethyl bromide is used rather than making a secondary alkyl halide from 3-methylpentan-2-ol.

Solution Steps

  1. Step 1: Convert 3-methylpentan-2-ol to sodium alkoxide using sodium metal

  2. Step 2: Convert ethanol to ethyl bromide using HBr

  3. Step 3: React sodium alkoxide with ethyl bromide via Sₙ2 mechanism to form the ether

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