Crossed Cannizzaro Reaction the Reduction of an Aldehyde to a Primary Alcohol by Hydride Transfer: the Crossed Cannizzaro Reaction of P-Chlorobenzaldehyde with Formaldehyde

Experiment One: Crossed Cannizzaro Reaction

The Reduction of an Aldehyde to a Primary Alcohol by Hydride Transfer: The Crossed Cannizzaro Reaction of p-Chlorobenzaldehyde with Formaldehyde

Abstract:

The purpose of this experiment was to use a Crossed-Cannizzaro reaction to convert aldehydes into carboxylic acids and alcohol. This reaction occurs when the reacting aldehydes are not exactly identical. In this experiment formaldehyde will be used as a reducing agent for aldehydes which have no alpha hydrogens. In the end of the experiment, the product should be formic acid and benzyl alcohol. This product is left out for a week to dry and evaporate, and then the weight and its melting point is taken. At the end of the experiment, it is expected that the product will contain 157 mg of p-Chlorobenzyl alcohol, however, in my experiment I only recovered 39 mg of p-Chlorobenzyl alcohol. The percent yield of this experiment was 25.8%.

Procedure:

The procedure used was titled Experiment One: Crossed Cannizzaro Reaction

The Reduction of an Aldehyde to a Primary Alcohol by Hydride Transfer: The Crossed Cannizzaro Reaction of p-Chlorobenzaldehyde with Formaldehyde and can be found on blackboard. No changes were made to the procedure.

Reactions:

4-chloro-benzaldehyde 4-Chlorobenzyl alcohol

Melting Point: 46 degrees Celsius Melting Point: 69- 72 degrees celsius

Boiling Point: 214 degrees Celsius Boiling Point: 234 degrees Celsius

Density: 1.196 g/cm3 Molecular Weight: 142.6 mg

Data/Calculations:

starting p-Chlorobenzaldehyde: 155 mg → 1.10 mmol

melting point of p-Chlorobenzyl alcohol: 70 °C

Limiting reagent:

(155 mg p-Chlorobenzaldehyde) x (mmol/140.57mg) = (1.10 mmol p-Chlorobenzaldehyde)

(330μL Formalin) x (1.08 mg Formalin/ 1 μL) = (356 mg Formalin)

(356 mg Formalin) x (0.37 mg CH2O/ 1 mg Formalin) = (132 mg Formalin)

(132 mg Formalin) x (mmol/30.03 mg) = (4.40 mmol Formaldehyde)

(400μL KOH) x (11μmol KOH/ 1 μL KOH) = (4400...