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Lab Report 10...

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Experiment F1: An Aldol Reaction: trans-4-Nitrochalcone Purpose The purpose of this experiment is to perform a base-catalyzed aldol reaction via the condensation of an aromatic aldehyde with an aryl alkyl ketone. This initial addition reaction will be followed by a dehydration reaction in order to yield a chalcone, or an ɑ,β-unsaturated ketone. The desired product will be isolated using a Hirsch funnel and analyzed by obtaining both an melting point and an IR spectrum. Experimental Procedure Experiment F1 was followed as described in Mayo, pages 351-354, with the modifications listed in the Canvas document. Reaction Scheme

Calculations Theoretical Yield of trans-4-Nitrochalcone Limiting Reactant [R] --- 4-Nitrobenzaldehyde 110 mg 110 mg ×

1𝑔 1000 𝑚𝑔

×

1 𝑚𝑜𝑙 𝑅 151.12 𝑔

×

1 𝑚𝑜𝑙 𝑃 1 𝑚𝑜𝑙 𝑅

×

253.25 𝑔 1 𝑚𝑜𝑙 𝑃

= 0.1843 grams --- 184 mg

Actual Yield of trans-4-Nitrochalcone Weighed directly on weighing paper --- 150 mg of trans-4-Nitrochalcone % Yield of Benzoic Acid (Experimental Yield / Theoretical Yield) x 100 = 150 𝑚𝑔 184 𝑚𝑔

x 100 = 81.52 ---- 81.5% Yield

Data and Results Compound

Mol. Weight

Amount

mmol

M.P. (℃)

B.P.

Density

(g/mol) Acetophenone

120.16

95% Ethanol 4-Nitrobenzaldehyde

0.86

20.7

4 mL 151.12

10% NaOH Trans-4Nitrochalcone

100 μL

110 mg

0.73

103-106

EY) 0.59 TY) 0.73

158-160

(℃)

(g/mL)

202.6

1.03

78.5

0.789

300

400 μL 253.25

EY) 150 mg TY) 184 mg

Observations ● Mixing all reactantants resulted in a clear liquid ● Upon dissolving the aldehyde, the solution turned a pale yellow ● The mixture turned orange and a solid precipitated after the addition of the NaOH solution ● Following the ethanol wash, the product appeared as pale yellow crystals ● Melting Point Range 153-158 ℃ Questions 1) Mechanism

2) A. 400 μL is equivalent to 1 mmol of NaOH B. 100 μL is equivalent to 0.25 mmols of NaOH

C. Given that NaOH serves as the base that catalyzes the production of the product, using a smaller amount will still make the reaction go to completion. However, the smaller quantity of the base will translate to a lower yield of the product as some of the initial reagents will remain unreacted. Therefore, in order to maximize yield and for the sake of their grades, the students should rerun the experiment with 400 μL. 3) A. 1658 cm-1 B. The carbonyl stretch is 23 cm-1 off from the starting ketone of acetophenone. This is due to the conjugation of the alkene next to the carbonyl. The double bond delocalizes electrons over a greater area of the molecule, resulting in a smaller wave number than seen in acetophenone (which lacks the alkene).

4) A. Intermediate B. Based on the NMR data, there is no intermediate that was contained within the isolated product. In the NMR, the signals for 11 Hydrogen are shown, which corresponds to the exact number of Hydrogens present in the desired product. In the intermediate, 13 Hydrogens are present which means the peak integration of the given NMR does not correlate to this molecule. Therefore, only the product present was Trans-4-Nitrochalcone

5) 1

H NMR Spectra for Trans-4-Nitrochalcone

Protons in Molecule

Peak Frequency (ppm)

Peak Multiplicity

Peak Integration

A

8.15

d, J = 8.5

2

B

7.98

J = 15.7 Hz

1

C

7.85

d, J = 8.5

2

D

7.51

m

5

E

6.98

J = 15.7 Hz

1

Works Cited

Mayo, D. W.; Pike, R. M.; Forbes, D. C. Microscale Organic Laboratory with Multistep and Multiscale Syntheses, 5th ed.; John Wiley & Sons, Inc., 2011; pp 7W38-7W41. Modifications for Experiment F1: Aldol Reaction: trans-4-Nitrochalcone, Canvas document....