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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: NorthWardVale on January 12, 2024, 04:02:18 PM

Title: What would the products of this reaction be?
Post by: NorthWardVale on January 12, 2024, 04:02:18 PM
C8H8O3 + CaH2 :rarrow:

I have tried solving it myself but what i get is C8H8O3 + CaH2 :rarrow: C8H10O3 + Ca(OH)2 + H2

But I then cannot balance it. Can you please help me? This is for a research project

Thanks all!  ;D
Title: Re: What would the products of this reaction be?
Post by: Hunter2 on January 12, 2024, 04:05:55 PM
What is C8H8O3? Some hundred  possibilities.
One could be vanilin?
Title: Re: What would the products of this reaction be?
Post by: Vidya on January 13, 2024, 03:20:50 AM
The balanced equation is with water-
C8H8O3 + 2H2O  + CaH2    
                             :rarrow: C8H10O3 +Ca(OH)2 + H2
Title: Re: What would the products of this reaction be?
Post by: Borek on January 13, 2024, 03:51:09 AM
Quote from: Vidya on January 13, 2024, 03:20:50 AM
The balanced equation is with water-
C8H8O3 + 2H2O  + CaH2    
                             :rarrow: C8H10O3 +Ca(OH)2 + H2

This is a complete nonsense, if you are coming here to get another series of links to your site at least give valid answers.

Not only you have added water which is not part of the original question, but you have balanced the equation ignoring presence of C8H8O3 (whatever it is). Why not

123C8H8O3 + 2H2O  + CaH2 :rarrow: 123C8H10O3 +Ca(OH)2 + H2

or just

2H2O  + CaH2 :rarrow: Ca(OH)2 + H2 ?

The latter is actually why CaH2 is occasionally suggested as a desiccant, assuming it doesn't react with the dried substance. But you can't be sure it doesn't. Some of the possible isomers of C8H8O3 include alcohols and acids, which definitely can get reduced.
Title: Re: What would the products of this reaction be?
Post by: Vidya on January 13, 2024, 04:03:26 AM
Whats wrong with the answer ??
Water is added as second step and not in first step
I know water can not be with CaH2  just like other other metal hydrides.
I have given the complete reaction taking into consideration of attack of H- as nucleophile on carbonyl carbon and then followed by protonation using water.

[edited by mod, discussion moved to PM]
Title: Re: What would the products of this reaction be?
Post by: Hunter2 on January 13, 2024, 04:05:08 AM
Quote
123C8H8O3 + 2H2O  + CaH2 :rarrow: 123C8H10O3 +Ca(OH)2 + H2

This is not any more balanced. Left side 990 H, right side 1234 H

Hydrids can be used to change carbonyl compounds to alcohol.

R'COR + H- => RR'HCO-

In this case Ca2+ will bond two of it.

2 R'COR + CaH2 => 2 RR'HCO- + Ca2+

This would be the basic reaction.
To get from the alcoholate the alcohol itself it has to be added water as suggested.

RR'HCO- + H2O =>  RR'HCOH + OH-

But clear we dont know what is the compound C8H8O3.
Title: Re: What would the products of this reaction be?
Post by: Vidya on January 13, 2024, 04:08:57 AM
Yes I also assumed there is carbonyl group attacked by H- as nucleophile and protonated by water in next step.
Title: Re: What would the products of this reaction be?
Post by: Babcock_Hall on January 13, 2024, 10:55:31 AM
@OP, Can you say something more about this research project?  I am having a difficult time understanding how such a project is going to get off the ground without this and other fundamental information.
Title: Re: What would the products of this reaction be?
Post by: Borek on January 13, 2024, 12:41:19 PM
My mistake with balancing the reaction, sorry about that. Missed the changing number of hydrogens in the molecule.

Still: you are making unjustified assumptions about what the compound is/can be and how it reacts. That's not a correct way of solving problems, nor a correct way of teaching anything.