OR OR R=CH3 O2N OCH O2N COOH OCH 1h, 1i, 1 OCH 1k, 11, 1m Compound 1h (1.25 g, 5
ID: 873596 • Letter: O
Question
OR OR R=CH3 O2N OCH O2N COOH OCH 1h, 1i, 1 OCH 1k, 11, 1m Compound 1h (1.25 g, 5.20 mmol) was dissolved in hot MeOH (20 mL), to which IN NaQH (10 mL 10.0 mmol) was added. The mixture was heated under reflux for 1 h and then more water (50 mL) was added. The aqueous layer was neutralized by addition of 1M HCI (15 mL). The precipitated crude product was collected by filtration, which was recrystallized from MeOH to give a yellow solid 1 k. Yield: 0.63 g, 53%. 1. What is the limiting reagent (LR) in the reaction above and why has it been chosen? If compound 1h is set to 1 equivalent (eq.), then how many eq. of NaQH are being used above? 2. What is the solvent in the reaction above? What is the solvent in your Exp. 4 reaction? Thinking about a solvent's purpose, why might these two solvents differ even though the overall reaction is similar? 3·Think of three important traits of an ideal reflux solvent? Approximately what temperature would the above reaction reach at reflux? 4. Find the ratio of moles of salicylic acid (10.0 mmol) and NaQH (15 mL of 6M) that you use in your Exp. 4 laboratory procedure. Then find the ratio of moles of the starting material 1h to NaQH above. Which procedure, the lab book or the journal procedure above, uses a greater ratio of NaQH to starting material? Why might this be? 5. If 6 mL of 1 N NaOi. instead of 10 mL, were used in the reaction procedure above, what other reaction detail(s) would possibly need to be modified and how?Explanation / Answer
1. The limiting reagent in above reaction is 1h as it lesser in amount (5.20 mmol) in comparision to NaOH (10 mmol). Thus it limits the reaction and hence limits the formation of the product. If compound 1 h is set to 1 equivalent then the 2 eq. of NaOH are being used above.
2. The solvent in the above reaction is methanol.
3.Ideal traits of refluxing solvent :
Free from dissolved impurities
High volatility, or low boiling point (for refluxing or evaporation it should have lesser b.p than H2O)
Unreactive to the solute
Readily dissolves the solute
Not easily oxidised at the temperatures used.
4. Ratio of moles of salicylic acid and NaOH = 10 mmol / (15 mL x 6 M )
= 10 / 90 = 0.11
Ratio of moles of starting material 1h to NaOH in above reaction = 5.20 mmol / 10.0 mmol
= 0.52
The above reaction has greater ratio than experiment 4 because in experiment 4 the quantity of NaOH taken is very large in comparision to that taken for the above reaction.
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