Guaicol ( 2 - methoxyohenol, d. 1.112 g/ml.) 1.1 mL is dissolved in 6 mL. 95% et
ID: 921829 • Letter: G
Question
Guaicol ( 2 - methoxyohenol, d. 1.112 g/ml.) 1.1 mL is dissolved in 6 mL. 95% ethanol in a fume hood, and a solution of 0.5 g of crushed NaOH pellets in 2 mL. water is added. The mixture is heated under reflux for 10 min. Then a mixture of 1.0 mL of (+_-) - 3 - chloro- 1, 2- propanediol, (d. 1.32 g/mL) in 1.0 mL 95 % ethanol is added drop wise to the phenoxide anion and the reflu is continued for 1 h. 1.2 g of guaifenesin was obtained from this reaction. Calculate the present yield of giaifenesin. How will you use IR and NMR to identify the product for the above reactionExplanation / Answer
Solution :-
Lets calculate the moles of the each reactant using the given information
Guaiacol density = 1.112 g /ml
Volume = 1.1 ml
Mass = v*d
= 1.1 ml * 1.112 g per ml
= 1.2232 g
Moles of Guaiacol = mass / molar mass
= 1.2232 g / 124.14 g per mol
= 0.009853 mol
Moles of NaOH = 0.5 g / 40.0 g per mol = 0.0125 mol
3-chloro-1,2-propanediol = 1.0 ml * 1.32 g per ml = 1.32 g
Moles of 3-chloro-1,2-propanediol = 1.32 g / 110.54 g per mol = 0.0119 mmol
So the moles of the Guaiacol is the limiting because all have 1 :1 mole ratio therefore the theoretical moles of the product that can be formed are 0.009853 mol
Now lets calculate the theoretical yield using this moles of the product
Mass = moles * molar mass
Mass of product = 0.008953 mol * 198.216 g per mol
= 1.953 g
Percent yield = (actual yield / theoretical yield )*100%
= (1.72 g / 1.953 g) * 100%
= 88.07 %
So the percent yield is 88.07 %
Q2) IR spectrum shows the specific peaks for the specific functional groups
In the reactant it will show the 1600 cm-1 peak for the C=C in the ring
Also it will show the peak at about 3300 cm-1 which is for the OH group this peak will also be present in the product.
In the product the C-C alkyl group peaks are obtained at about 750 to 850 cm-1
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