Question

(4 pts) (4 pts) (4 pts) (1 pts) 5. Calculations 6. Conclusions 7. Post lab Questions 8. TA signature POST LAB QUESTIONS: 1. Why do we add methanol to the reaction if we see a purple ring? 2. What reaction is occurring with the methanol (show reaction)? Does this reactior affect your product? 3. Why do you get a precipitate when adding the HCl to make the solution acidic? Exp 8-5

Answer 1

1.

If purple ring appears then it indicates presence of unreacted KMnO₄ in reaction mixture. We add methanol to consume oxidizing capacity of KMnO₄ present in reaction mixture. If KMnO₄ remains in reaction mixture than it will interfere in further steps and will consume added reagents like HCl by oxidation.

2.Reaction of Methanol with   KMnO₄ : Methanol is oxidized to Formic acid, which may be further oxidized to Carbon dioxide.

CH₃-OH (Methanol)  $\rightarrow$ HCOOH  $\rightarrow$ CO₂

This reaction produces formic acid which is more acidic than Adipic Acid, this reaction has negligible effect on product.Any effect might be due to more acidic nature of Formic acid, which may effect solubility in water.

3. Adding NaOH brings adipic acid in form of Salt Sodium Adipate, making whole product soluble. Adding HCl (pH = 1) , changes whole salt in form of Adipic acid, it also bring product as precipitate since HCl (pH = 1 solution) is much more acidic than product , and prevents ionization thus rendering product insoluble in water : ppt. formation.

^-OOC-(CH2)₄-COO^- and ^-OOC-(CH2)₄-COOH + (HCl solution pH = 1 )  

= HOOC-(CH2)₄-COOH (precipitate).