Chapter 11 - Alkenes

Question 1

How does the physical properties of alkenes change down the group?

Answer 1

1) boiling and melting point: increases
2) density: less dense than water but increase slightly down the group
3) viscosity: increases

Question 2

Why does the boiling point of alkenes increase with increasing number of carbon atoms in the chain?

Answer 2

The increase in number of carbon atoms leads to an increase in number of electrons, resulting in an increase in the size of the electron cloud, leading to stronger dispersion forces between molecules.

Question 3

Why does the density of alkenes increase with increasing number of carbon atoms in the chain?

Answer 3

The increase in number of carbon atoms leads to an increase in number of electrons, resulting in an increase in the size of the electron cloud, leading to stronger dispersion forces between molecules. The increase in the strength of intermolecular dispersion forces cause alkene molecules to attract more closely together, resulting in a slightly smaller volume of liquid. Since p=m/V, with a larger relative molecular mass, the density of liquid increases.

Question 4

Does straight-chain alkenes or branched alkenes have higher viscosity?

Answer 4

straight-chain alkenes

Question 5

Why does the viscosity of alkenes increase with increasing number of carbon atoms in the chain?

Answer 5

Strength of dispersion forces between molecules increase

Question 6

How do boiling and melting points in cis-trans isomers of alkenes differ?

Answer 6

• cis-isomer has a higher boiling point (less noticeable in alkenes with less polar substituent groups)
• trans-isomer has a higher melting point

Question 7

Why does the cis isomer tend to have higher boiling points?

Answer 7

In cis-isomers, since the same molecules are on the same side of the double bond, the polar bonds between C and that molecule (e.g. C-Cl) will result in dipole moments that do not cancel out, hence producing an overall dipole moment for the molecule. Thus, the cis-isomer is polar and pd-pd interactions exist between molecules (on top of dispersion forces). In the trans-isomer, the two dipole moments will cancel out since they are on opposite sides, such that the overall dipole moment for the molecule is zero. Hence, the trans isomer is non-polar and only dispersion forces exist between molecules. More energy is needed to overcome the stronger forces of attraction between cis isomers, leading to a higher boiling point.

Question 8

Why does the trans isomer tend to have higher melting point?

Answer 8

In order for the intermolecular forces of attraction to work well in a solid, the molecules must be able to pack together efficiently in the solid state. The trans isomer pack better than the cis isomer due to the higher symmetry of the trans isomer. The “U” shape of the cis isomer does not pack well as compared to the straighter shape of the trans isomer. Since the trans isomer packs better in the solid state, the dispersion forces work more efficiently in holding the molecules together when compared to the cis isomer. Hence, more energy is required to overcome the forces of attraction between the trans isomer, leading to a higher melting point.

Question 9

What is the solubility of non-substituted alkenes in non-polar solvents?

Answer 9

Non-substituted alkenes can form favourable dispersion forces with non-polar solvents, hence they are soluble.

Question 10

What is the solubility of non-substituted alkenes in water?

Answer 10

Non-substituted alkenes can only form pd-id intermolecular forces with water, which is much weaker than hydrogen bonding between water molecules and the dispersion forces between the alkenes themselves. Hence, they are insoluble in water.

Question 11

How can alkenes be prepared?

Answer 11

1) Elimination of hydrogen halide from halogenoalkenes

2) Dehydration of alcohol

Question 12

What are 3 chemical reactions alkenes can partake in?

Answer 12

1) Electrophilic addition of: hydrogen halides, halogens, bromine water, water/steam
2) Reduction
3) Oxidation: Mild oxidation, oxidative cleavage

Question 13

What is electrophilic addition to alkenes?

Answer 13

It is when the weaker pi bond is broken instead of the sigma bond in the C=C bond. In place, two strong bonds are formed in the product, leaving carbon atoms joined by a single bond. The electron rich C=C, containing the pi bond, attracts electrophiles.

Question 14

Describe the general mechanism of electrophilic addition to alkenes. (2)

Answer 14

Step 1: Initial attack by the pi electrons of the C=C bond on the electrophile, forming a positively-charged carbocation intermediate. (slow)
Step 2: Subsequent atack of the unstable by nucleophile to generate a stable product. (fast)

Question 15

Describe the structure of a carbocation intermediate. (3)

Answer 15

1) Central carbon atom is positively charged and sp2 hybridised.
2) 3 substituents are arranged in a trigonal planar geometry around central carbon atom - nucleophiles may attack from either side of the plane
3) unhybridised p orbital is empty as an electron is lost

Question 16

How is the stability of carbocations determined? (2)

Answer 16

1) Since alkyl groups are electron donating groups, the more alkyl substituent the carbocation has, the more stable it would be, as the electron donating groups disperse the positive charge on the central carbon atom.
2) Conversely, electron withdrawing substitutents like halogens would destabilize the carbocation as they intensify the positive charge on the central carbon atom.

Question 17

What are the 2 reagents and 1 condition needed for electrophilic addition of hydrogen halides to alkenes to occur?

Answer 17

Reagents: Gaseous HCI & HBr/HI
Condition: Room temperature

Question 18

Describe the mechanism of the electrophilic addition of hydrogen halides to alkenes. (2)

Answer 18

Step 1: (Rate determining step): As the polar hydrogen halide approaches the pi electron cloud of the alkene in the correct orientation, the high electron density of the pi electron cloud “attacks” the electron deficient H atom on the hydrogen halide molecule. The pi bond in the C=C bond breaks, forming a new C-H bond. the H-X bond undergoes heterolytic fission, producing a carobocation intermediate and a halide ion.
Step 2: The halide ion acts as a nucleophile and attacks the unstable carbocation intermediate quickly, as the negatively charged halide ion in attracted to the positively charged carbon, forming a new C-X bond. A stable product is formed.

Question 19

What is the Markovnikov’s rule?

Answer 19

The Markonikov’s rule states that in the addition of HX to an alkene, the H atom adds to the C atom of the double bond that holds a GREATER number of H atoms. The Markonikov product would thus be the minor product.

Question 20

Why are there 2 possible products when a hydrogen halide adds across the double bond of an unsymmetrical alkene?

Answer 20

2 possible intermediates would be formed, one being a secondary carbocation and one being a primary carbocation. Since the secondary carbocation has more alkyl groups that are electron donating, it is more stable as the charge is dispersed around the central carbon atom.
The more stable product, known as the major product, would be preferred over the minor product that is less stable.

Question 21

In what instance is it possible for electrophilic addition of hydrogen halides to alkenes to form a racemic mixture? Why does it occur?

Answer 21

When a hydrogen halide adds across the double bond of an alkene that produces an intermediate carbocation with 3 different groups attached to it, it produces equal amounts of 2 enantiomers, forming a racemic mixture.
This occurs because the geometry about the positively charged carbon is planar, hence it has equal chances of being attacked from the top or bottom by the bromide ion in the second step. Thus, in an unsymmetrical alkene, it has equal chances of producing the 2 enantiomers.

Question 22

What is the 1 reagent and 1 condition needed for electrophilic addition of halogens to alkenes to occur? What is observed during the reaction?

Answer 22

Reagent: chlorine gas/bromine liquid/Cl2 or Br2 dissolved in CCl4
Condition: Room temperature, absence of UV light (to prevent free radical substitution)
Observations: Greenish yellow Cl2/reddish brown Br2 decolourized

Question 23

Describe the mechanism of the electrophilic addition of halogens to alkenes. (2)

Answer 23

Step 1: (rate determining) as the non polar halogen molecule approaches the pi electron cloud of the alkene, the high electron density of the pi electron cloud polarizes the electron cloud of the halogen molecule. The halogen undergoes heterolytic fission, resulting in an anion with the other atom bonded to the alkene to form a positively charged carbocation intermediate.
Step 2: The anion acts as a nucleophile to attack the carbocation intermediate to form the product.

Question 24

What is the 1 reagent and 1 condition needed for electrophilic addition of bromine water to alkenes to occur? What is observed during the reaction?

Answer 24

Reagent: aqueous bromine (bromine water)
Condition: room temperature
Observation: yellow-orange aqueous Br2 decolourise

Question 25

Describe the mechanism of the electrophilic addition of bromine water to alkenes. (2)

Answer 25

Step 1: (rate determining) as the non polar bromine molecule approaches the pi electron cloud of the alkene, the high electron density of the pi electron cloud polarizes the electron cloud of the bromine molecule. Bromine undergoes heterolytic fission, resulting in an bromide ion with the other bromine atom bonded to the alkene to form a positively charged carbocation intermediate.
Step 2: the carbocation intermediate can be attacked by water and bromide ion (that act as nucleophiles). The major product usually depends on which nucleophile is present in excess, thus bromoalcohol is usually the major product instead of the dibromo compound, since water is present in a much higher concentration compared to the bromide ion.

Question 26

What are the 2 reagents and 2 conditions required for electrophilic addition of water/steam for the industrial method?

Answer 26

Reagents: steam, H3PO4 catalyst
Conditions: High temperature, high pressure

Question 27

What are the 2 reagents required for electrophilic addition of water/steam for the lab method?

Answer 27

Concentrated H2SO4 and cold H2O

Question 28

Briefly describe what is electrophilic addition of water/steam to alkenes.

Answer 28

Alkenes react with H2O to form alcohols. It is similar to that of addition of hydrogen halides.

Question 29

What are the 2 reagents and 2 conditions required for reduction in alkenes (via catalytic hydrogenation)?

Answer 29

Reagents: H2 gas, Ni catalyst (or Platinum/Palladium at rtp)
Conditions: Room temperature and high pressure

Question 30

Briefly describe what is reduction in alkenes.

Answer 30

Reduction via catalytic hydrogenation converts unsaturated alkenes into saturated alkanes. This is a form of heterogeneous catalysis.

Question 31

What are the 2 reagents and 1 condition required for mild oxidation in alkenes? What is observed?

Answer 31

Reagents: KMnO4, dilute NaOH
Condition: cold
Observation: Purple KMn)4 decolourises and a brown precipitate of MnO2 is formed.

Question 32

Briefly describe what is mild oxidation in alkenes.

Answer 32

Cold KMn)4 can oxidise alkenes into corresponding diols (an alcohol containing two hydroxyl groups in its molecule).

Question 33

What are the 2 reagents and 1 condition required for oxidative cleavage in alkenes? What is observed?

Answer 33

Reagents: KMnO4, dilute NaOH
Condition: heat
Observation: Purple KMn)4 decolourises. effervescence of carbon dioxide may be observed for alkenes with 2 or less hydrogen atoms. (terminal C=C double bonds)

Question 34

What are 2 用处s of oxidative cleavage?

Answer 34

1) establish location of the alkene double bond in a hydrocarbon chain/ring
2) distinguishing test between alkenes with terminal C=C double bonds (=CH2) and those without. Alkenes with terminal C=C double bonds would evolve hydrogen gas.

Question 35

Briefly describe what is oxidative cleavage in alkenes.

Answer 35

Hot KMnO4 can cleave the C=C in alkenes to give carbonyl compounds and/or carboxylic acids as products. (carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O)