Experiment 3A: Properties of Alkane (LAB)

Question 1

What alkane was used in the experiment to test alkane properties?

Answer 1

Hexane or C6H14

Question 2

Because alkanes are inert, they usually require _____ to undergo _____

Answer 2

Catalysts; Substitution reactions

Question 3

I. Baeyer’s Test for Unsaturation:

Answer 3

Materials:

• 1 mL of Permanganate solution (KMnO4)
• 1st vial of 1mL C6H14

Procedure:

1. Add 1 mL of KMnO4 to the first vial of C6H14.
2. Stopper and shake.

Question 4

Results of Baeyer’s Test:

Answer 4

The purple color of KMnO4 remained, no other physical changes

Therefore, it is a NEGATIVE / NO RESULT

Question 5

Why did Baeyer’s Test yield negative results in presence of alkane?

Answer 5

Baeyer’s Test for Unsaturation yields positive results only upon detecting an unsaturated organic compound.

No visible change was detected in testing the C6H14 because it was a saturated, single-bond hydrocarbon; thus being non-reactive to Baeyer’s test.

Question 6

II. Testing with Bromine Water:

Answer 6

Materials:

• 1mL Bromine water (Br2)
• 2nd vial of C6H14

Procedure:

1. Add 1mL of Br2 to the second vial with C6H14.
2. Stopper and shake while exposing the tube under sunlight.

Question 7

Results for Testing w/ Bromine Water:

Answer 7

• Two separate liquid layers formed

This is due to difference in densities between C6H14 and Br2

• Bubbles formed at the bottom; these are called polybromohexanes

Question 8

Why was no reaction yielded in the hexane-bromine mixture until it was exposed to sunlight?

Answer 8

This is because alkanes are relatively inert and inactive in aqueous reagents, like Bromine water.

However, the mixture’s exposure to sunlight caused the Br2 to lose its reddish color. This is because when catalyzed by sunlight, or specifically UV light, a H atom was substituted for a Br atom.

Question 9

Reaction in testing w/ Br2 is called:

Answer 9

Halogenation

Question 10

Hexane + Bromine Water formula:

Organic compound formed:

Answer 10

C6H14 + 3 Br2 –>[UV light] C6H13Br + 3 HBr

Organic compound formed is C6H13Br or 1-bromohexane

Question 11

III. Testing with Alcoholic Iodine

Answer 11

Materials:

• Alcoholic iodine or Alc. I2
• Third vial of C6H14

Procedure:

1. Add two drops of Alc. I2 to vial of C6H14.
2. Stopper and shake.

Question 12

Results of Alc. I2 Test:

Answer 12

• Alc. I2 remains yellow
• NO REACTION

Question 13

Why did Alc. I2 not react with C6H14?

Answer 13

This is because iodine does not react with alkanes.

In the order of halogen reactivity in halogenation, I2 is placed last behind F2, Cl2, and Br2, making it the least reactive halogen.

Therefore, its weak bond dissociation energy causes no substitution reaction to occur between I2 and Hexane

Question 14

IV. Ignition Test

Answer 14

Materials:

• 5 drops of C6H14
• Ignited flame

Procedure:

1. Place 5 drops of C6H14 to an evaporating dish.
2. Ignite using match or flame torch.

Question 15

Results of Ignition Test:

Answer 15

Appearance of luminous, yellow flame

Question 16

What process occurred during Ignition Test?

Answer 16

Combustion

Question 17

What caused C6H14 to combust?

Answer 17

Hexane’s exposure to its open surroundings, thus to oxygen, caused it to combust.

Question 18

What occurs in Combustion? What are formed?

Answer 18

Combustion is a process wherein hydrocarbons are rapidly oxidized.

Oxygen-containing compounds, specifically H2O (water) and CO2 (carbon dioxide) are formed.

Question 19

The appearance of the yellow flame, however, indicates that the combustion process was _____

Answer 19

Incomplete; happens when there is a lack of continuous and sufficient oxygen supply

Question 20

Ignition Test Formula:

Answer 20

2 C6H14 + 19 O2 –> 12 CO2 + 14 H2O * C6H14 + C