answer questions 1 and 2 ) Guaiacol (2-methoxyphenol, d. 1.112 g/mL) 1.1 mL is d
ID: 542838 • Letter: A
Question
answer questions 1 and 2
Explanation / Answer
1.
The volume of guaiacol used for reaction is 1.1 mL. Density of guaiacol is 1.112 g/mL. Calculate weight of guaiacol.
Wt. of guaiacol used in reaction = Volume x density
Wt. of guaiacol used in reaction = (1.1 mL) x (1.112 g/mL)
Wt. of guaiacol used in reaction = 1.223 g
According to given reaction one mole of guaiacol gives one mole of guaifenesin. Molar mass of guaiacol is 124 g/mol and that of guaifenesin is 198 g/mol. Thus, calculate theoretical yield.
Theoretical yield = (wt. of 1 mol of guaifenesin / wt. of 1 mol of guaiacol) x 1.223 g
Theoretical yield = (198 g / 124 g) x 1.223 g
Theoretical yield = 1.95 g
Actual yield of guaifenesin is 1.72 g. Thus, calculate percent yield as follows:
Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (1.72 g / 1.95 g) x 100%
Percent yield = 88.2 %
Therefore, percent yield of guaifenesin is 88.2 %
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2.
IR spectrum of gauifenesin will show following prominent absorptions:
1. Strong absorption near 3500 to 3200cm-1 due to presence of two hydroxy groups.
2. Absorption near 3100-3000 cm-1 due to aromatic C-H stretching vibrations.
3. Absorption near 2950 cm1- due to aliphatic C-H stretching vibrations,
4. Absorption near 1200 cm-1 due to C-O stretching vibrations.
NMR (proton) spectrum will show following peaks characteristic of guaifenesin:
1. Multiplate near 7.0 ppm due to aromatic protons.
2. Singlet near 3.8 ppm due to methoxy group.
3. Multiplate near 3.5 ppm to 4.5 ppm due to five aliphatic protons attached to carbons which are further attached to oxygen.
4. Two D2O exchaneable protons in the region 3 to 6 ppm due to two hydoxy groups.
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