Here we have to explain and get the product of the reaction between 3-methyl-1,3-diol and chromium-tri-oxide (CrO₃)
The reaction between 3-methyl-1,3-diol with chromium-tri-oxide (CrO₃) form 3-methyl-3-hydroxybutanal.
The CrO₃ is used as a mild oxidant to the alcohol. It only produces aldehyde or ketone depending upon the position of the -OH group in a secondary or primary alcohol.
Here we can see there are two hydroxy (-OH) group in the reactant. However one of the hydroxy is at terminal carbon atom, thus primary alcohol and other in between as tertiary alcohol.
As the reagent CrO₃ does not reacts on the tertiary -OH group thus the -OH group present at 3 position will not be oxidized. Thus only the terminal hydroxy group will oxidize to form the corresponding aldehyde.
Henceforth the final product of the reaction is 3-methyl-3-hydroxybutanal as shown in the figure.
Answer:
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The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
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