preparation of flavanone (2-phenyl-4-chromanonel) 2-Hydroxyacetophenone (1.1mL)

Question

preparation of flavanone (2-phenyl-4-chromanonel)

2-Hydroxyacetophenone (1.1mL) with benzaldehyde 1.0mL and 160mL 0.1M sodium hydroxide solution.

I have recovered 1.43 g crystals after performing the recrystallization with ethanol.

I am not sure how do i find find the yield of product.

I started with 1.1mL 2-hydroxyacetophenone, sequence of aldol condensation elimination and michael addition , obtained 1.43g product=flavanone.

Explanation / Answer

2-hydroxyacetophenone

density of 2-hydroxyacetophenone = 1.13 g/mL

volume of 2-hydroxyacetophenone = 1.1 mL

Hence,

mass of 2-hydroxyacetophenone = density x volume = 1.13 g/mL x 1.1 mL  = 1.243 g

Then,

moles of 2-hydroxyacetophenone = mass /molar mass = 1.243 g/ 136.15 g = 9.13 x 10-3 mol

benzalehyde

density of benzalehyde = 1.04 g/mL

volume of benzalehyde = 1.0 ml

Hence,

mass of  benzalehyde   = density x volume = 1.04 g/mL x 1.0 mL  = 1.04 g

Then,

moles of benzalehyde =  mass /molar mass = 1.04 g/ 106.1 g = 9.8 x 10-3 mol

Therefore,

moles of 2-hydroxyacetophenone are less than moles of benzaldehyde.

Hence,

2-hydroxyacetophenone is the limiting reagent.

Yield is calculated based on  2-hydroxyacetophenone.

2-hydroxyacetophenone -------------->   flavanone

136.15 g 224.25 g

1.243 g ?

? = (1.243 g/ 136.1) x  224.25 g flavanone

= 2.05 g flavanone

This is theoretical yield of flavanone.

Given that actual yield of flavanone = 1.43 g

Therefore,

% yield of flavanone = actual yield / theoretical yield x 100

= (1.43 g/ 2.05 g) x 100

= 69.7 %

Therefore,

% yield of flavanone = 69.7 %

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