the hemiacetal below is treated with 18o-labeled methanol (ch3o*h) and acid. where will the label appear in the products?
1 answer:
The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
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Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles =
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = = 0.03mole
Number of moles of H₂O = = 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2