Experiment 4: Reaction of Alcohols

Question 1

It is a colorless, volatile liquid with a characteristically pleasant odor. It is used as an excellent solvent of many substances.

Answer 1

Ethyl alcohol or C2H5OH

Question 2

AI. PREPARATION FOR ANHYDROUS ALCOHOL

Answer 2

1. Place 10 mL of ethyl alcohol in a test tube. Then add 1 gram of CaC2
2. Cork tube loosely and heat in 60 celsius water bath for 10 minutes.
3. Cool the mixture and filter. Set liquid filtrate aside.

Question 3

A2. What is an anhydrous alcohol?

Answer 3

It is a pure alcohol, meaning it contains 0% water

Question 4

A3. What is the use of CaC2 in the preparation of anhydrous alcohol?

Answer 4

It separates the water properties of C2H5OH, making it become 100% pure alcohol

Question 5

A4. Equation for the Preparation of Anhydrous alcohol:

Answer 5

95% C2H5OH + CaC2 [60 celsius]–> 100% C2H5OH + C2H2 + Ca(OH)2

Question 6

B1. Difference in color between Hydrous vs Anhydrous CuSO4 (Copper Sulfate)

Answer 6

Hydrous CuSO4 is blue

Anhydrous CuSO4 is white

Question 7

B2. II. TEST FOR WATER IN ALCOHOL

Answer 7

1. Place powdered anhydrous CuSO4 in two dry test tubes.
2. Add 2 mL of 95% C2H5OH in one test. Add equal amount of prepared anhydous C2H5OH in the other test tube.

Question 8

B3. What happened to the mixture with CuSO4 and 95% C2H5OH?

Answer 8

Powdered CuSO4 turned blue because it absorbed H2O from 95% C2H5OH

Equation:
95% C2H5OH + CuSO4 –> C2H5OH + CuSO4 * H2O

Question 9

B4. What happened to the mixture with both anhydrous CuSO4 and 100% C2H5OH?

Answer 9

No change in color, remained white

Equation:
100% C2H5OH + CuSO4 –> no change in color

Question 10

B5. Why was powdered CuSO4 used in testing for presence of H2O?

Answer 10

Powdered CuSO4 changes color when it detects water or moisure, turning blue when water H2O is detected

Question 11

C1. What is Esterification?

Answer 11

In the presence of an acid catalyst, it is the process of heating carboxylic acids (COOH) with alcohol (OH) to form an ester (COOR)

Question 12

C2. What is used as the acid catalyst and dehydrating agent in this experiment’s Esterification process?

Answer 12

Sulfuric acid or H2SO4

Question 13

C3. III. ESTERIFICATION

Answer 13

Ester A

1. Mix in a test tube 1 mL of C2H5OH and acetic acid (CH3COOH).
2. Add a little less than 1 mL of H2SO4
3. 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).

Question 14

C4. Results and Equation of Esterification of Acetic Acid and Ethyl Alcohol:

Answer 14

CH3COOH + C2H5OH [H2SO4]–> CH3COOC2H5 + H2O

Ethyl acetate

Odor is Plastic balloon-like

Question 15

C5. Results and Equation of Esterification of Salicylic Acid and Methyl Alcohol:

Answer 15

C6H4OHCOOH + CH3OH [H2SO4]–> C6H4OHCOOCH3 + H2O

Methyl salicylate

Odor is like oil of Winter Green

Question 16

D1. It is a yellow solid with a characteristic medicinal / antiseptic odor

Answer 16

Iodoform or CHI3

Question 17

D2. IV. IODOFORM TEST

Answer 17

1. Add drops of C2H5OH and Sodium Hydroxide (NaOH) in a test tube with water.
2. Add a solution of I2 until a faint yellow color persists.
3. Warm in a bath of 60 celsius, and then set aside.

Question 18

D3. Results and Equation of Iodoform Test:

Answer 18

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

Question 19

D4. How was Iodoform formed?

Answer 19

The NaOh + I2 mixture oxidized the C2H5OH.

The halogen I2 reacted and yielded CHI3

Question 20

E1. What three alcohols were used for the Oxidation and Lucas Test procedures?

Answer 20

• Ethyl alcohol / C2H5OH
• Isopropyl alcohol / C3H7OH
• Tert-butyl alcohol / C4H9OH

Question 21

E2. It is the addition of oxygen and the removal of hydrogen

Answer 21

Oxidation

Question 22

E3. What oxidizing agents were used in this experiment’s Oxidation?

Answer 22

Potassium dichromate or K2Cr2O7

Sulfuric acid or H2SO4

Question 23

E4. V. OXIDATION

Answer 23

1. The three alcohols are placed in three different test tubes.
2. K2Cr2O7 and H2SO4 are added into all test tubes.
3. Warm in a water bath.

Question 24

E5. Results of Oxidation Test:

Answer 24

C2H5OH and C3H7OH changed into deep blue-green / OXIDATION OCCURRED

C4H9OH remained brownish-yellow / OXIDATION DID NOT OCCUR

Question 25

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?

Answer 25

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

Question 26

E7. Equations of Oxidation Test:

Answer 26

Ethyl alcohol:

C2H5OH + K2Cr2O7 + H2SO4 –> CH3COOH + K2SO4 + Cr2(SO4)3 + H2O

Isopropyl alcohol:

C3H7OH + K2Cr2O7 + H2SO4 –> CH3COCH3 + K2SO4 + Cr2(SO4)3 + H2O

Question 27

F1. This test is used to distinguish primary, secondary, and tertiary alcohols

Answer 27

Lucas Test

Question 28

F2. These are the hydrogen halide and catalyst used during Lucas Test:

Answer 28

Hydrogen chloride (HCl); Zinc chloride (ZnCl2)

Question 29

F3. VI. LUCAS TEST

Answer 29

1. The three alcohols are placed in three different test tubes.
2. HCl and ZnCl2 are added into all test tubes.
3. Warm in a water bath.
4. Stopper the tubes, shake, and allow to stand.

Question 30

F4. What is the first sign of a reaction in Lucas Test?

Answer 30

Cloudiness in solutions

Question 31

F5. What was the order of reaction based on how fast solutions clouded over?

Answer 31

Tert > Iso > Ethyl

Question 32

F6. In Lucas Test, the replacement of the ______ with Halogen occurs

Answer 32

Hydroxyl (OH) group

Question 33

F7. Equations for Lucas Test:

Answer 33

Ethyl

C2H5OH + HCl [ZnCl2] –> C2H5Cl + H2O

Iso

C3H7OH + HCl [ZnCl2] –> C3H7Cl + H2O

Tert

C4H9OH + HCl [ZnCl2] –> C4H9Cl + H2O