Organic 1 Section 3 - Unit 14: Alkenes

Question 1

Explain how you would test for alkenes (2 marks)

Answer 1

• Shake with bromine water

- The colour will change from orange to colourless

Question 2

Explain why tertiary carbocations are more stable than primary carbocations (3)

Answer 2

• Alkyl groups on either side of the positive carbon are electron releasing
• So they reduce the charge
• And the carbocation is stable overall

Question 3

Change in functional group, reagent and conditions for reaction of alkenes with hydrogen halides (3)

Answer 3

• Alkene ➡ Halogenoalkane
• HBr or HCl (Hδ+)
• Room Temperature

Question 4

Change in functional group, reagent and conditions for reaction of alkenes with halogens (3)

Answer 4

• Alkene ➡ Dihalogenoalkane
• Br2 (Brδ+)
• Room Temperature

Question 5

Change in functional group, reagent and mechanism for reaction of alkenes with sulphuric acid (6)

Answer 5

Step 1:
- Alkene ➡ Alkyl Hydrogensulfate 
- Concentrated H2SO4
- Electrophilic Addition
Step 2:
- Alkyl Hydrogensulfate ➡ Alcohol
- Warm water
- Hydrolysis reaction

Question 6

Change in functional group and conditions for reaction of alkenes with water (4)

Answer 6

• Alkene ➡ Alcohol
• 300°C to 600°C
• 70 atm (high pressure)
• Concentrated H3PO4 (catalyst)

Question 7

Explain why alkenes have lower melting and boiling points than alkanes (2)

Answer 7

• Less electrons in alkenes so weaker Van der Waals forces

- So less energy is needed to overcome them

Question 8

Explain what is meant by the term ‘electrophile’ (1 mark)

Answer 8

• An electron pair acceptor

Question 9

Explain why, even though bromine molecules are non-polar, they react with the double bonds in alkenes (2 marks)

Answer 9

• Double bond have high electron densities

- So bromine forms an induced dipole

Question 10

Explain what is meant by an addition reaction (1 mark)

Answer 10

• Two reactants form one product

Question 11

Explain how a bromine molecule is able to act polar even though it is normally non-polar (1 mark)

Answer 11

• The electrons of the double bond polarise the Br-Br bond

- And Brδ+ becomes electrophilic