Organic Chemistry Peer Tutoring Department Chem 51LCUniversity of California, Irvine Professor LinkIbrahim Thein (ithein@uci.edu) http://sites.uci.edu/ochemtutorsDaniel De La Cruz (dcdelacr@uci.edu)

Final Review Packet Answer Key1.Extraction:

Assume you have a mixture of compounds 1,2, and 3 dissolved in ether and poured into aseparatory funnel. Then, you pour 1M NH3 (aq) into the separatory funnel to perform anacid-base extraction. Determine which compounds will end up in organic and aqueous layersonce the extraction is complete. Note that the pKa of ammonium ion is around 9.

Compound 1 is a neutral molecule and will remain in the organic layer. Compound 1 will notengage in any acid-base chemistry for the scenario described above.

Compound 2 will attempt to react with ammonia to produce the following equilibrium:

However, the equilibrium favors the side of the reaction containing the weaker acid (higher pKa).Therefore, the equilibrium favors the left side of the reaction. As a result, the salt ions will not beproduced in significant amounts. Compound 2 will remain in the organic layer.

Compound 3 will react with ammonia to produce the following equilibrium:

The equilibrium favors the right side of the reaction because ammonium is a weaker acid thanbenzoic acid. Therefore, the salt ions will be produced which will shift the benzoic acid into theaqueous layer in the form of benzoate anion.

2. Recrystallization: You are trying to recrystallize crude compound X. As part of your proceduralworkup, you must choose which solvent(s) to use for the recrystallization. Based on thefollowing information regarding the solubility of the crude compound X in various solvents,determine the appropriate solvent(s) for the recrystallization of compound X.

Solvent A: Crude compound X is soluble at all temperatures

Solvent B: Crude compound X is soluble at low temperatures but not at high temperatures

Solvent C: Crude compound X is not soluble at all temperatures

Solvents A and C; When performing recrystallization, the compound of interest should besoluble in the solvent at high temperatures but not at low temperatures. If one solvent does notwork, a solvent pair can be used. Minimum amount of Solvent A can be used to initially dissolvethe crude compound. Then, a minimum (dropwise) amount of Solvent C can be added until aslight cloudiness persists. Afterwards, Solvent A can be added dropwise until the cloudinessdisappears, thereby creating a supersaturated solution. The solution can be cooled initially onthe lab bench, then placed in an ice bath to completely cool.

3. TLC: You are running a TLC trial for a mixture of two compounds using 90% ethylacetate:10% hexane as your eluting solvent. After staining to visualize the spots, the TLC plateappears in the following way:

What is the problem with your TLC plate and what can be done to remedy this issue?The spots are too high up on the TLC plate (Rf values are too big) and there is not muchseparation between the spots. You should use a less polar eluting solvent mixture to reduce the

Rf values; put more hexane and less ethyl acetate when making your eluting solvent. Rf valuesshould ideally be between 0.2 and 0.8.

In a separate experiment, you run a TLC trial to monitor the progress of your reaction. The TLCplate appears as below (starting material-left lane, co-spot-middle lane, reaction mixture-rightlane):

Given the TLC plate above, has your reaction reached completion?No, there is still starting material present in the reaction mixture (lane on the far right).

Is your reaction product more or less polar than your starting material?Less polar because it has a larger Rf

What is a reason why the TLC plate must be stained to visualize the spots instead of using a UVlamp?The sample compounds used are not conjugated. Conjugation is required for a compound to beUV-active.

4. 1H NMR: Which of the following molecules corresponds to the 1H NMR spectrum depictedbelow?

A)

B)

C)

D)D; The molecule contains three equivalent alkyl proton groups on the far right. The moleculehas two equivalent allylic proton groups on the far left. The molecule has a single vinyl proton inthe middle. The molecule has two sets of equivalent aromatic hydrogens on the aromatic ring.

5. For the reaction shown below, 128 mg of phenol, 165 mg of 1,2-dibromoethane, andexcess sodium hydroxide are used to obtain 93 mg of 1,2-diphenoxyethane. What isthe percent yield of 1,2-diphenoxyethane?

phenol MW = 94.11 g/mol; 1,2-dibromoethane MW = 187.86 g/mol;sodium hydroxide MW = 40.0 g/mol; 1,2-diphenoxyethane MW = 214.26 g/mol

By analyzing the reactants and products, we see that the mole ratio of the reactants is2 moles of phenol: 1 mole of 1,2-dibromoethane. Our most important step is to find the limitingreactant, and the rest of our calculations will be based off of that limiting reactant. Recall that thepercent yield is (actual yield/theoretical yield) x 100.

6. Draw the arrow-pushing mechanism for the following Wittig reaction. Show all intermediatestructures.

The driving force in a Wittig reaction is the formation of the strong O-P double bond. TheWittig reaction is used in organic synthesis to generate a carbon-carbon double bond.Both (E) and (Z) alkene isomers will be produced but the (E) isomer will be the majorproduct because the ylide is stabilized (has an electron-withdrawing group, namely thecarbonyl).

7. a) Calculate the cis/trans percent product ratio given the following 1H NMR spectrum.

Cis: [1.00/(1.00+5.41)] x 100 = 15.6

Trans: [5.41/(1.00+5.41)] x 100 = 84.4

15.6%:84.4%

Recall that the 1H NMR peak for the cis product is more downfield (around 4.0 ppm) than thetrans product (around 3.5 ppm)

b) Given the cis:trans percent ratio calculated above, what are the possible reducing agents thatwere used in the reduction reaction (circle all that apply)?

A) NaBH4B) L-selectrideC) Aluminum isopropoxide (MPV reaction)

A & C: Both the NaBH4 and Aluminium isopropoxide reduction reactions produce the transproduct as the major product. NaBH4 reduction will proceed by axial attack (same face as thetert-butyl group), selectively producing the trans product as the major product. The MPVreaction is reversible which indicates that the reaction is thermodynamically controlled and willproduce the lower energy product as the major product, namely the trans product.

8. A classmate is designing an experiment to determine the effect of steric or electronic effectson the overall rate of a Diels-Alder reaction. Would using the following list of compounds be anappropriate way to design the desired experiment?

Yes. When designing an experiment, it is important to keep as many variables as constant aspossible. The substituents on the dienes vary based on their steric character, going from H(least sterically hindered) to tert-butyl (most sterically hindered). Meanwhile, the dienophile isheld constant, and there aren’t any significant differences in electronic character of thesubstrates used which could confound the results.

9. Which IR descriptions best correlate to the structure of the given organic molecule?

A) Two peaks around 3400 cm-1, One sharp peak around 1675 cm-1; One strong peak to theright of 3000 cm-1

B) One peak around 3400 cm-1, One sharp peak around 1675 cm-1; One strong peak to theright of 3000 cm-1

C) Two peaks around 3400 cm-1, One sharp peak around 1800 cm-1; One strong peak to theright of 3000 cm-1

A; The N is only bonded to one C, which means that there will be two distinct N-H peaks on theIR spectrum around 3400 cm-1. The carbonyl is conjugated which means that the IR peak will beslightly less than 1700 cm-1. The methyl substituent on the aromatic ring contains Csp3-H’swhich have an IR peak to the right of 3000 cm-1 (remember the line at 3000 cm-1).

10. In three different trials of the same reaction, Compound X has been shown to have anexperimental melting range of 1) 72.8-75.9 C 2) 73.2-76.3 C 3) 73-76.1 C. Would it bereasonable to say that the identity of Compound X has a literature value melting point of 80 C?Yes, it would be reasonable to say that Compound X has a literature value melting point of 80 C.Recall that when a mixture contains impurities (as is often the case in experiments), the meltingrange lowers and broadens when compared against the standard melting point of the compoundof interest. The literature melting point of any given compound is referring to the melting point ofthe standard compound with no impurities.

Questions 11-13 correspond to the following reaction:

11. Which major product is expected, A or B?

a. Ab. Bc. Neitherd. A mixture of both

Organocuprate reagents preferentially add 1,4 to α-β-unsaturated compounds.

12. Provide an arrow-pushing mechanism for the above reaction that accounts for the finalproduct.

13. Given that 245 mg of aldehyde was reacted with 209 mg of Me2CuLi, calculate thepercent yield if 195 mg of the product was formed.

14. Use the spectra below to fill in the data tables and propose a structure for the moleculewith formula C9H13N.

1H shift (δ) Integration Splitting Assignment

8.5 2 multiplet aromatic

7.5 1 multiplet aromatic

7.2 1 multiplet aromatic

2.6 2 triplet methylene

1.6 2 quintet methylene

1.4 2 sextet methylene

0.9 3 triplet methyl

13C shift (δ) # of carbons Assignment

150 1 aromatic

149 1 aromatic

138 1 aromatic

137 1 aromatic

124 1 aromatic

33 1 aliphatic

26 1 aliphatic

22 1 aliphatic

14 1 aliphatic

Degrees of unsaturation: #C - #H - #X + #N + 1 (X = halogen)12

12

12

9 - 13/2 - 0 + 1/2 + 1 = 4 degreesTip: 4 or more degrees means look for an aromatic ring

IR: Peaks from 2850-3000 cm-1 indicate sp3 carbons, 3000-3200 cm-1 indicate sp2 carbons.

Structure:

15. Which separation method would be best to separate the following mixtures of two solids?If extraction, specify whether acid-base extraction would be best.

recrystallization

Acid-base extraction

Liquid-liquid extraction

16. Indicate whether each of the following compounds would be best visualized after TLCusing UV light, KMnO4 stain, or neither.

UV light can be used to visualize spots of compounds with aromatic or otherwiseconjugated systems. These spots will be shadows on the plate because the UV will beabsorbed while the rest of the plate appears greenish.

KMnO4 achieves visualization by oxidizing aldehydes/ketones and C-C double bonds,which leaves yellow spots after reaction, leaving the rest of the plate purple.

In this example, the compounds marked “UV” also happen to be visualizable withKMnO4, however, they both also happen to absorb UV. UV is favored because it is nondestructive and saves reagent.

Questions 17–20 correspond to a 1:1 mixture of the following solid compounds.

17. What method would be best to separate the two compounds?

Acid-base extraction. Recrystallization would be less than ideal because the two havesimilar solubility in different solvents. Using acid-base extraction would be ideal sincethere is a significant difference in pKa.

18. If any, which of the following reagents would be best used in this separation?

a. KOHb. NaHc. CH3OHd. HCl

The HCl, with a pKa of about -6 would be ideal. Because the left compound can act as abase, we want to add acid to protonate it and bring it into the aqueous layer. A and B areboth bases, which are the opposite of what we want. C is an alcohol with pKa ~16, so itwould also act as a base in a reaction with the amine. D is the only good answer and itwould be an ideal choice in the lab.

19. An extraction was performed on the mixture using ether and DI water. Which compoundshould be in each layer after adding the appropriate reagent, if needed?

The left compound (amine) should be in the aqueous layer after protonation. The rightcompound should be in the organic layer. The ether will form the organic layer above theaqueous layer in this case based on density.

20. Instead of pure compounds, the student found that a small amount of the othercompound was found in each layer. What steps can be taken to improve separation?

To ensure that the organic layer contains none of the left compound (protonated amine),it should be rinsed with DI water (or brine solution) and re-extracted. Then, the organiclayer should be dried over magnesium sulfate to ensure none of the amine is presentafter the ether is evaporated.

To minimize the amount of the neutral compound (right structure), the DI water can bereplaced with brine. The presence of salt ions in the aqueous layer will further reduce theneutral compound’s solubility in the water.