Question

Part A of experiment:

1. Working in the fume hood, combine 4.0 mL of 10% NaOH solution and 4 mL EtOH in a 50 mL round bottom flask.

2. Dissolve 1.0 mL (8.6 mmol) of acetophenone into your solution.

3. Add 1.0 mL (9.8 mmol) of benzaldehyde and a magnetic stir bar to your flask. Clamp the flask

   above a stir plate.

4. Monitor the reaction for 45 minutes by TLC. Acetophenone needs to be diluted prior to TLC, and

   cannot be spotted neat.

5. Observe the formation of a yellow product. If no reactivity is apparent, heat the sample. Avoid

   the formation of a red oil from excess heating.

6. Pour the reaction mixture into an Erlenmeyer flask. Rinse the flask with a small amount of ethanol

   to ensure complete transfer of the reaction mixture.

7. Cool the reaction in an ice bath with rapid stirring for five minutes. Stop the stirring and let the

   flask stand. Crystals should crash out, but sometimes it may take as long as 30 minutes. Scratch the sides of the flask to induce crystallization if necessary. You may also add a small volume of cold water to induce precipitation.

8. Filter the crystals by suction filtration using a Buchner funnel (with an initial gravity filtration as performed in previous experiments) and wash the crystals with 20 – 30 mL of cold water and then with 5 mL cold EtOH. Dry the crystals by aspiration

9. Determine the melting point and yield of the crude product

Recrystallization

1. Dissolve the crude product in a minimum amount of boiling ethanol.

2. Allow the solution to cool to room temperature and allow crystallization to proceed for 10

   minutes at room temperature.

3. Cool the solution in an ice bath for up to 5 minutes to complete the process.

4. Filter and dry the crystals by aspiration using a Buchner funnel (with an initial gravity filtration as

   performed in previous experiments).

5. Dissolve a small amount of your pure product in ethanol and take another TLC to verify the purity.

6. Determine the melting point and yield of your pure product. The product should appear as pale

   yellow needles. The literature melting point for benzalacetophenone is 57 – 58 °C.

Question

Mr. White, a well-respected chemist, is studying the reactivity of benzaldehyde and acetophenone at room temperature using similar conditions to what you used in Part A of the experiment. Using some of his famed recrystallization techniques, Mr. White (or Walter, as his friends call him) was able to isolate two compounds, both of which are over 99.1% pure. As part of his characterization routine, Walter obtained NMR spectra for both compounds, which are included below along with zoom-ins on the regions of interest. While synthetically strong, Walter could use some help for the characterization part.

1. a) Based on what you know from the reaction performed in Part A, draw out the two compounds that Walter has isolated.

2. b) Help Walter assign each NMR spectra to the corresponding compound that you identified in a).

3. c) Provide multiple arguments to explain your assignment in b) (Based on such pieces of data like chemical shifts, multiplicity, integrations, etc.)

4. d) (BONUS)OnehydrogeninoneofthetwocompoundsdoesnotappearintheNMR provided. Identify which one it is and propose a reason why

.

Figure 1 -Compound M- Composé M 8om 100 76 72 68 p

Answer 1

Here I have given the answer with more or less detail analysis of the Spectra. Please I have marked the H atoms as a,b,x,y,z to denote different H-atom and gave their approximate peak values with the integration and pattern.

"CH3 to + NaOH (door NaOH (107) > Etон xaß unsaturated product (A) aldor product B) Based on the reaction form. these two products would ~6.7ppm, d, 1H & oth Not visible in the spectra ~7.5ppm, atto a Analysis of H NMR N2.5 (2, 4) u. 7ppm (t, 1 14) from the NMR spectra of figure 2. It is clear this spectre belongs to aldol product (B) • It has 10 aromatic proton between 7-8 ppm region. from one of the spectra ie Figure 1. We see lot protons in the aromatic region. tone (7-8) pom To 1 proton in the alkene region 6-7 ppm. 1. We will get to protons - in the aromatic region E as it is clear from the stenecture, lo 1 extra proton is the proton marked Ho. This proton is bit desheilded than Ha as it is • He gives a tots doublet in the region 2-3 ppm. with two proton • Hy gives a triplet with 1 ptroton in region 4.5-5.0 ppm this proton is more deshielded as it has an alcohol (OH) group attached to the same carbon. adjacent to the keto- group hence in the region (2-8ppm) Ha gives a doublete with 1 proton at 6.7 pom I The proton Hz is not visible in iH spectra since it is a replaceable hydrogen.