Exactly 2.00 g of an ester A containing only C, H, and O was saponified with 15.00 mL of a 1.00 M NaOH solution. Following the s
aponification, the solution required 5.30 mL of 1.00 M HCl to titrate the unused NaOH. Ester A, as well as its acid and alcohol saponification products B and C, respectively, were all optically active. Compound A was not oxidized by K2Cr2O7, nor did compound A decolorize Br2 in CH2Cl2. Alcohol C was oxidized to acetophenone by K2Cr2O7. When acetophenone was reduced with NaBH4, a compound D was formed that reacted with the acid chloride derived from B to give two optically active compounds: A (identical to the starting ester) and E. Propose a neutral structure for each compound that is consistent with the data. Note that the absolute stereochemical configurations of chiral substances cannot be determined from the data. Arbitrarily draw a configuration for an enantiomer (and be consistent with the derivatives); omit wedge/dash bonds for a racemate.
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Answer:
The empirical formula is =
The Valproic acid formula is =
Explanation:
Mass of the produced water = 0.166 g
Molar mass of water = 18 g/mol
The moles of are calculated as 0.166 g /18 g/mol = 0.00922 moles.
In 1 mole of water, there are 2 moles of hydrogen atoms.
Thus,
Moles of H = 2 x 0.00922 = 0.01844 moles
Each hydrogen atom's molar mass is 1.008 g/mol.
Hydrogen mass in the molecule = 0.01844 x 1.008 = 0.018588 g
Mass of produced carbon dioxide = 0.403 g
Molar mass of carbon dioxide = 44.01 g/mol
The moles of are calculated as 0.403 g /44.01 g/mol = 0.009157 moles.
Each carbon atom's presence is 1 mole in 1 mole of carbon dioxide.
So,
Moles of C = 0.009157 moles
The molar mass of carbon is 12.0107 g/mol.
Carbon mass in molecule = 0.009157 x 12.0107 = 0.11 g
Since Valproic acid comprises only hydrogen, oxygen, and carbon, the oxygen mass in the sample = Total mass - Carbon mass - Hydrogen mass.
The sample's overall mass = 0.165 g.
Oxygen mass in the sample = 0.165 - 0.11 - 0.018588 = 0.036412 g
The molar mass of oxygen is 15.999 g/mol.
Moles of O = 0.036412 / 15.999 = 0.002276 moles
Taking the simplest ratio for H, O, and C yields:
0.01844: 0.002276: 0.009157
= 8: 1: 4
The empirical formula becomes
While molecular formulas detail the precise count of atoms for each element, empirical formulas represent the simplest form or reduced ratio of these elements in the compound.
Consequently,
The molecular mass equals n × Empirical mass.
Here, n is a positive integer from 1, 2, 3...
Empirical mass = 4×12 + 8×1 + 16 = 72 g/mol.
Molar mass = 144 g/mol.
Thus,
The molecular mass = n × Empirical mass.
144 = n × 72
⇒ n = 2
The Valproic acid formula is =