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Write a balanced equation

What would be the balanced equation for the synthesis of benzoic acid using the Gignard reagent, phenylmagnesium bromide? 

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J.R. S. | Ph.D. in Biochemistry--University Professor--Chemistry TutorPh.D. in Biochemistry--University Profes...
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I can't draw phenyl magnesium bromide here but let's say it looks like øMgBr where ø represents a phenyl ring and then you have MgBr attached.  The reaction then would be 
øMgBr + CO2 ==> ø-C00MgBr ==H2O==>ø-COOH + Mg(OH)Br
Dave M. | Most Experienced, Helpful Organic Chemistry Tutor Available AnywhereMost Experienced, Helpful Organic Chemis...
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PhMgBr + CO2 → PhCO2MgBr the reaction is not complete however, and water will not workup the product to form benzoic acid. The balanced equation must include an acid-base reaction that turns the conjugate base to its corresponding acid. Water will never work. 
You must add dilute acid (like HCl) to the product to effect a "work-up" that neutralizes the product to the carboxylic acid. 
If you merely add water to the the magnesium carboxylate, you will never separate the product from the water layer. 
The yield will be poor or no recovery. I presume this is for a lab you did with the Grignard made in situ with bromobenzene with dry Mgo turnings which were then exposed to dry ice (CO(s)). Once the reaction was complete by whatever detection method you used (passage of time, TLC), you almost certainly did not just pour water over it and get benzoic acid. That is not possible.
Recovery by extraction cannot happen until the mixture is poured over water and the pH is brought at least 2 (litmus paper turns deep red) and such a pH cannot be reached without a strong acid (a pKa of less than -6 to obtain significant hydronium concentration--the strongest acid in aqueous solution). Only then can the neutral acid enter the organic phase, so the above answer is not suitable or correct for a lab. It is not correct in a paper-based problem either.  

If it sounds like I'm being too specific or even polemical, I am not. You cannot make protonated benzoic acid by the equation above. I've done it many times. It must precipitate out of water to be benzoic acid at all.