You use 1mL of p-anisaldehyde and 1mL of acetaphone (along with other reagent, n

ID: 868096 • Letter: Y

Question

You use 1mL of p-anisaldehyde and 1mL of acetaphone (along with other reagent, not important in this case) to make p-anisalacetophenone.

What is the molar mass(g/mol), volume, density, and moles of p-anisaldehyde?

What is the molar mass(g/mol), volume, density, and moles of acetophenone?

What is the limiting reagent? Theoritical yield of p-anisalacetophenone?

Hint: molar mass and density can be obtained via a credible internet source. Use that to help you do some of the calculations.

Given an IR Spectrum of p-anisalacetophenone, what will be the significant peaks and their assignments?

The calculations are as below,

a) p-anisaldehyde

given, volume = 1 mL

density = 1.119 g/cm3

thus, grams of p-anisaldehyde = 1 x 1.119 = 1.119 g

MW of p-anisaldehyde = 136.15 g/mol

moles = 1.119/136.15 = 8.2 mmol

b) acetophenone

given, volume = 1 mL

density = 1.028 g/cm3

thus, grams of acetophenone = 1 x 1.028 = 1.028 g

MW of acetophenone = 120.15 g/mol

moles = 1.028/120.15 = 8.6 mmol

c) the limiting reagent detection is by finding whichstarting material would give least amount of product.

If p-anisaldehyde to p-anisalacetophenone

8.2 mmol = 0.0082 mol x MW of p-anisalacetophenone = 0.0082 x 238.28 = 1.96 g

If acetophenone to p-anisalacetophenone

8.6 mmol = 0.0086 x MW of p-anisalacetophenone = 0.0086 x 238.28 = 2.05 g

Since, we get less amount of product from p-anisaldehyde, this will be the limiting reagent.

p-anisaldehyde is the limiting reagent.

Theoretical yield of p-anisalacetophenone = 1.96 g

d) IR spectrum will show peaks at,

965 cm-1 (strong peak) for trans-disubstituted alkene

800-860 cm-1 for para-disubstituted benzene

1675 cm-1 (medium) for trans-1,2-disubstituted alkene

1625 cm-1 (strong) conjugated C=C with bezene ring

1500-1600 cm-1 (weak to strong) for aromatic C=C

1685 cm-1 for alpha,beta-unsaturated ketone

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