Cannizzaro Reaction

[TigerLily]

Hello.

 

Can someone please help me with these organic chemistry lab questions? Thanks in advance.

 

Here's the reaction:

 

4-Chlorobenzaldehyde + KOH,H+ >> 4-Chlorobenzyl alcohol + 4-Chlorobenzoic acid

 

Purpose: This experiment illustrates the simultaneous oxidation and reduction of an aromatic aldehyde to form the corresponding benzoic acid and benzoyl alcohol.

 

1. In this experiment, the alkaline phase was diluted with water. Then it was acidified by concentrated hydrochloric acid. Discuss the acidification of the alkaline phase. How does it help in the isolation of the carboxylic acid product? Use an equation to help explain this point.

 

2. Think about the possibility of the alcohol product reacting with methanolic KOH. How would this effect the purity of the acid product? How could you tell if your acid product was contaminated in this way? Use an equation to help explain this.

 

3. Why is CO2 generated in the sodium bicarbonate test. Explain the evolution of CO2 in the bicarbonate test using an equation. Also, why do we perform chemical tests?

Edited July 20, 2009 by TigerLily

[John Cuthber]

Before we do your homework for you perhaps you would like to tell us what you think the answers are?

[TigerLily]

I'm sorry about this. I wasn't asking you to do the homework for me, I was asking for help.

 

I think I know #3.

 

When a carboxylic acid comes in contact with a solution containing bicarbonate ion, carbon dioxide is generated. Would the formula for this be

 

Cl-C6H5-COOH + NA+HCO3- >> Cl-C6H5-COO- NA+ + H2CO3

 

H2CO3 >> CO2 + H20 ?

 

I don't really understand #1 and #2.

 

For #1: Acidification protonates the carboxylic acid, but that's all I can think of so far. I'm not really sure what the equation would be for that.

 

I don't know what to say about #2.

[Theophrastus]

Simply put, in your third response, you are correct that:

 

[ce] Cl-C6H5-COOH + NaHCO3 -> NaOOC-C6H5-Cl + H2O + CO2 [/ce]

 

or simplified:

 

[ce] H+ + HCO3 <-> H2CO3 <-> H2O + CO2 [/ce]

 

For number one, you are right to say that the acidic conditions, result in the protonation of the acid, however its more important that if the conditions remained alkaline during the formation of 4-chlorobenzoic acid, the acid would simply react with the basic potassium hydroxide. The reaction you have described, in fact, can be performed in absence of a solution, with powdered potassium hydroxide, however, the lack of regulation, would mean the final product would be benzyl alcohol and an organic potassium salt, ruining your benzoic acid yield:

 

[ce] Cl-C6H5-COOH + KOH -> Cl-C6H5COOK + H2O [/ce]

 

The second question, I'm not really sure about, however, I would guess that the [ce] KOH [/ce] would, in forming an organic potassium salt, in a similar fashion to the reaction above, only with the alcohol's hydroxyl group, in place of the carboxylic, may further react with the benzenoic acid, however, as I'm not sure what pathway such a reaction might take, it might be more sincere to say I have no idea. Best of luck; hope this helps.

Edited July 29, 2009 by Theophrastus

[Fuzzwood]

Was the benzoic acid, generated by reacting with the acid in the first step, still soluble in water? Was it easy to extract when it wasnt water-soluble anymore?

[Theophrastus]

Ah Fswd has a good point, refering to the solvents. (As well as his actual intent of refering to possible precipitation, due to protonation) In the first question, you state that the alkaline phase is diluted with water, before adding the acid (which generally implies using water as the solvent), however in your second question, you refer to methanolic potassium hydroxide. If I'm not mistaken, doesn't methanolic mean, dissolved in methanol? If so, where does the methanol come into play? If so, was the water allowed to intermix with the methanol, or was there formation of layers? (I'm slowly beginning to confuse myself )

Edited July 29, 2009 by Theophrastus