Experiment 4: Reaction of Alcohols Flashcards
It is a colorless, volatile liquid with a characteristically pleasant odor. It is used as an excellent solvent of many substances.
Ethyl alcohol or C2H5OH
AI. PREPARATION FOR ANHYDROUS ALCOHOL
- Place 10 mL of ethyl alcohol in a test tube. Then add 1 gram of CaC2
- Cork tube loosely and heat in 60 celsius water bath for 10 minutes.
- Cool the mixture and filter. Set liquid filtrate aside.
A2. What is an anhydrous alcohol?
It is a pure alcohol, meaning it contains 0% water
A3. What is the use of CaC2 in the preparation of anhydrous alcohol?
It separates the water properties of C2H5OH, making it become 100% pure alcohol
A4. Equation for the Preparation of Anhydrous alcohol:
95% C2H5OH + CaC2 [60 celsius]–> 100% C2H5OH + C2H2 + Ca(OH)2
B1. Difference in color between Hydrous vs Anhydrous CuSO4 (Copper Sulfate)
Hydrous CuSO4 is blue
Anhydrous CuSO4 is white
B2. II. TEST FOR WATER IN ALCOHOL
- Place powdered anhydrous CuSO4 in two dry test tubes.
- Add 2 mL of 95% C2H5OH in one test. Add equal amount of prepared anhydous C2H5OH in the other test tube.
B3. What happened to the mixture with CuSO4 and 95% C2H5OH?
Powdered CuSO4 turned blue because it absorbed H2O from 95% C2H5OH
Equation:
95% C2H5OH + CuSO4 –> C2H5OH + CuSO4 * H2O
B4. What happened to the mixture with both anhydrous CuSO4 and 100% C2H5OH?
No change in color, remained white
Equation:
100% C2H5OH + CuSO4 –> no change in color
B5. Why was powdered CuSO4 used in testing for presence of H2O?
Powdered CuSO4 changes color when it detects water or moisure, turning blue when water H2O is detected
C1. What is Esterification?
In the presence of an acid catalyst, it is the process of heating carboxylic acids (COOH) with alcohol (OH) to form an ester (COOR)
C2. What is used as the acid catalyst and dehydrating agent in this experiment’s Esterification process?
Sulfuric acid or H2SO4
C3. III. ESTERIFICATION
Ester A
- Mix in a test tube 1 mL of C2H5OH and acetic acid (CH3COOH).
- Add a little less than 1 mL of H2SO4
- Warm in a water bath for 20 minutes. Then pour into 50 mL of cold water
Ester B
Repeat above process using methyl alcohol (CH3OH) and salicyclic acid (C4H6OHCOOH).
C4. Results and Equation of Esterification of Acetic Acid and Ethyl Alcohol:
CH3COOH + C2H5OH [H2SO4]–> CH3COOC2H5 + H2O
Ethyl acetate
Odor is Plastic balloon-like
C5. Results and Equation of Esterification of Salicylic Acid and Methyl Alcohol:
C6H4OHCOOH + CH3OH [H2SO4]–> C6H4OHCOOCH3 + H2O
Methyl salicylate
Odor is like oil of Winter Green
D1. It is a yellow solid with a characteristic medicinal / antiseptic odor
Iodoform or CHI3
D2. IV. IODOFORM TEST
- Add drops of C2H5OH and Sodium Hydroxide (NaOH) in a test tube with water.
- Add a solution of I2 until a faint yellow color persists.
- Warm in a bath of 60 celsius, and then set aside.
D3. Results and Equation of Iodoform Test:
I2’s yellow tint separated into precipitate, which settled at the bottom
Formed precipitate is CHI3
Equation:
C2H5OH + 6 NaOH + 4 I2 –> CHI3 + HCOONa + 5 NaI + 5 H2O
D4. How was Iodoform formed?
The NaOh + I2 mixture oxidized the C2H5OH.
The halogen I2 reacted and yielded CHI3
E1. What three alcohols were used for the Oxidation and Lucas Test procedures?
- Ethyl alcohol / C2H5OH
- Isopropyl alcohol / C3H7OH
- Tert-butyl alcohol / C4H9OH
E2. It is the addition of oxygen and the removal of hydrogen
Oxidation
E3. What oxidizing agents were used in this experiment’s Oxidation?
Potassium dichromate or K2Cr2O7
Sulfuric acid or H2SO4
E4. V. OXIDATION
- The three alcohols are placed in three different test tubes.
- K2Cr2O7 and H2SO4 are added into all test tubes.
- Warm in a water bath.
E5. Results of Oxidation Test:
C2H5OH and C3H7OH changed into deep blue-green / OXIDATION OCCURRED
C4H9OH remained brownish-yellow / OXIDATION DID NOT OCCUR
E6. What kind of alcohols were ETHYL & ISO, and what did they oxidize into?
What kind of alcohol was TERT, and why did it not oxidize?
C2H5OH was a PRIMARY ALCOHOL, and oxidized into a Carboxylic acid (COOH)
C3H7OH was a SECONDARY ALCOHOL, and oxidized into a Ketone (CO)
C4H9OH was a TERTIARY ALCOHOL, thus difficult to oxidize under normal conditions because of high electronegativity levels
E7. Equations of Oxidation Test:
Ethyl alcohol:
C2H5OH + K2Cr2O7 + H2SO4 –> CH3COOH + K2SO4 + Cr2(SO4)3 + H2O
Isopropyl alcohol:
C3H7OH + K2Cr2O7 + H2SO4 –> CH3COCH3 + K2SO4 + Cr2(SO4)3 + H2O
F1. This test is used to distinguish primary, secondary, and tertiary alcohols
Lucas Test
F2. These are the hydrogen halide and catalyst used during Lucas Test:
Hydrogen chloride (HCl); Zinc chloride (ZnCl2)
F3. VI. LUCAS TEST
- The three alcohols are placed in three different test tubes.
- HCl and ZnCl2 are added into all test tubes.
- Warm in a water bath.
- Stopper the tubes, shake, and allow to stand.
F4. What is the first sign of a reaction in Lucas Test?
Cloudiness in solutions
F5. What was the order of reaction based on how fast solutions clouded over?
Tert > Iso > Ethyl
F6. In Lucas Test, the replacement of the ______ with Halogen occurs
Hydroxyl (OH) group
F7. Equations for Lucas Test:
Ethyl
C2H5OH + HCl [ZnCl2] –> C2H5Cl + H2O
Iso
C3H7OH + HCl [ZnCl2] –> C3H7Cl + H2O
Tert
C4H9OH + HCl [ZnCl2] –> C4H9Cl + H2O