Experiment 48

Objective

To identify the presence of ethanol by its reaction with iodine and sodium hydroxide, forming a yellow precipitate of iodoform.

Introduction

When ethanol reacts with iodine (I₂) in the presence of sodium hydroxide (NaOH), it undergoes a series of reactions that yield a yellow precipitate of iodoform (CHI₃). This characteristic yellow precipitate confirms the presence of ethanol in the sample.

Reaction Steps

1.    Oxidation of Ethanol

CH3​CH2​OH + I2 ​→ CH3CHO + 2HI

2.    Formation of Triiodoacetaldehyde

CH3​CHO + 3I2 ​→ CCl3CHO + 3HI

3.    Formation of Iodoform (Yellow Precipitate)

CCl3CHO + NaOH → CHCl3 ​+ CH3COONa

Materials

Chemicals

●      Ethanol (CH₃CH₂OH)

●      Iodine solution (I₂)

●      Sodium hydroxide (NaOH)

Apparatus

●      Test tube

●      Bunsen burner

●      Thermometer

●      Pipette

Reagents

●      1.659 g NaOH and iodine solution; 5 mL of ethanol.

Procedure

1.    Preparation of Reaction Solution

o   Mix NaOH and iodine solution in a test tube.

2.    Addition of Ethanol

o   Add 5 mL of ethanol to the solution in the test tube.

3.    Heating

o   Heat the test tube gently with a Bunsen burner and wait for the reaction to complete.

4.    Observation of Precipitate

o   A yellow precipitate appears, indicating the formation of iodoform.

Observation

A yellow precipitate of iodoform confirms the presence of ethanol in the sample.

Questions

1.    What is methyl iodide?

2.    Why is iodine used in the ethanol test?

3.    What occurs during the ethanol test?