ssume caffeine is a base which can be represented by the general formula R,N: a)

Question

ssume caffeine is a base which can be represented by the general formula R,N: a) Write the reversible reaction of RiN: with nitric acid, HNOstaq), b) If caffeine were present in the "salt" form in the tea leaves and in the aqueous extract, explain why it would not be well extracted into dichloromethane. c) If the aqueous solution were quite alkaline, use the reaction in part a) to explain why caffeine would mostly be present as the "neutral" molecule R3N: and, thus, would be more soluble in dichloromethane. Ko for caffeine-4.1.3 Assume the 100 mL aqueous tea extract contained 300 mg of caffeine. Calculate how much caffeine remains unextrac dichloromethane. Repeat th dichloromethane is broken up into two separate 20 mL extraction portions ted if the extraction is made with a single 40 mL portion of e calculation assuming that the 40 mL total volume of

Explanation / Answer

For cafeeine

a) reversible reaction,

R3N + HNO3 <==> R3NH + NO3-

b) Salts are ionic compounds which are also polar in nature. Therefore, they will not dissolve in less-polar organic solvents such as dichloromethane. Neutral caffeine however will dissolve in dichloormethane.

c) If the solution is alkaline, the reaction shown in a) would have its equilibrium towards left handside, that is neutral caffeine would be formed exclusively. This is according to the LeChatellier's principle. R3NH would react with the alkali present to form R3N until the equilibrium is reestablished to its initial state.

d) Kd for caffeine in (water/dichloromethane) is 4.636

Extraction with one 40 ml dichloromethane

let x amount is extracted

4.636 = [((300-x)/100)/(x/40)]

0.116x = 3 - 0.01x

x = 3/0.126 = 23.81 mg

--

with 2 extraction with 20 ml dichloromethane

first,

4.636 = [((300-x)/100)/(x/20)]

0.232x = 3 - 0.01x

x = 3/0.242 = 12.4 mg

remaining in water layer = 300-12.4 = 287.6 mg

Second extraction,

4.636 = [((287.6-x)/100)/(x/20)]

0.232x = 2.876 - 0.01x

x = 2.876/0.242 = 11.9 mg

Total amount extracted = 24.3 mg

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.