What's the role of sodium carbonate and hydrochloric acid for the neutralization of disaccharides into monosaccharides?

A couple of days ago we conducted the following experiment at school:



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> Add 5mL of an aqueous solution of sucrose ($\ce{C6H22O11}$) to 1mL of concentrated hydrochloric acid and heat it for 10 minutes. Then, add sodium carbonate ($\ce{Na2CO3}$) until the foaming stops. Then, conduct the Fehling experiment by adding the Fehling reagent to the following solution and heat it.
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The results of the Fehling test on the solution was positive, meaning there was a compound with free aldehyde functional groups present.


The explanation to the above experiment was given as well:



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> $\ce{C12H22O11\_{(aq)} + H2O\_{(l)} -> C6H12O6\_{(aq)} + C6H12O6\_{(aq)}}$
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> Neutralisation:
>  $\ce{Na2CO3\_{(aq} + 2HCl\_{(aq)} -> 2NaCl\_{(aq)} + H2O\_{(l)} + CO2\_{(g)}}$
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> The neutralised solution performs a redox reaction on Fehling's reagent, so the result is the red precipitate.
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First of all, I am interested as to why the heating aqueous solution of sucrose and hydrochloric acid result in the occurrence of foaming. Why does the sodium carbonate stop the process of foaming?


Second of all, I'd like to know why do we need to neutralize the solution in question for the test to be positive? Glucose molecules ($\ce{C6H12O6}$) are aldehydes and they should color the Fehling's reagent all by themselves. Is neutralization therefore obsolete?




The foaming occurs due to carbon dioxide generation when you add the sodium carbonate to the sucrose and HCl solution as there is excess HCl. It stops when all the excess HCl is neutralised.


Fehling's solution B is strongly basic and when you add it to Fehling's A it forms the bis(tartrate) complex of Cu(II) which is deep blue. This is reduced by the aldehydes to Cu2O which is insoluble in neutral and basic aqueous solution - hence the characteristic red ppt. Cu20 is soluble in dilute HCl, so if you did not neutralise you would get no ppt.


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