First, we write the reaction equation:
NiCO₃ + 2HBr → NiBr₂ + H₂CO₃
Now, writing this in ionic form:
NiCO₃ + 2H⁺ + 2Br⁻ → NiBr₂ + 2H⁺ + CO₃⁻²
(NiCO₃ is insoluble so it does not dissociate in to ions very readily)
Overall equation:
NiCO₃ + 2Br⁻ → NiBr₂ + CO₃⁻²
Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
Hi, you have not provided structure of the aldehyde and alkoxide ion.
Therefore i'll show a mechanism corresponding to the proton transfer by considering a simple example.
Explanation: For an example, let's consider that proton transfer is taking place between a simple aldehyde e.g. acetaldehyde and a simple alkoxide base e.g. methoxide.
The hydrogen atom attached to the carbon atom adjacent to aldehyde group are most acidic. Hence they are removed by alkoxide preferably.
After removal of proton from aldehyde, a carbanion is generated. As it is a conjugated carbanion therefore the negative charge on carbon atom can conjugate through the carbonyl group to form an enolate which is another canonical form of the carbanion.
All the structures are shown below.