Benzene (1 g, 12.5 mmol) is allowed to react with 1-chloropropane (1 g, 12.5 mmo

Question

Benzene (1 g, 12.5 mmol) is allowed to react with 1-chloropropane (1 g, 12.5 mmol) and AlCl3. The product (1.2g) is subjected to analysis on a GLC equipped with a thermal conductivity detector. The chromatogram shows two product peaks identified as n-propylbenzene (area=65 mm^2, Wf=1.06) and isopropylbenzene (area=113 mm^2, Wf=1.09). Calculate the percent yield of each of the two isomeric products obtained in this reaction. Note that since each of the products has the same molar mass of 120, the use of weight factors gives both weight and mole percent composition.

Explanation / Answer

reaction of benzene with 1-chloropropane&AlCl3 is Friedal-Crafts alkylation of Benzene.

Benzene + 1-chloropropane and AlCl3.----->n-propylbenzene +isopropylbenzene

corrected area for n-propylbenzene =(peak area x W.f)= 65 x1.06 =68.9mm^2

corrected area of isopropylbenzene =(peak area x W.f)= 113 x 1.09 =123.7

total area = 68.9+123.7= 192.6mm^2

% of n-propylbenzene = 68.9/192.6 x 100 = 35.77%

% of isopropylbenzene = 123.7/192.6 x100 = 64.226%

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