New Way Chemistry for Hong Kong A-Level Book 3A1 Alkenes 30.1Introduction 30.2Nomenclature of...

New Way Chemistry for Hong Kong A-Level Book 3A

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Alkenes

30.130.1 IntroductionIntroduction

30.230.2 Nomenclature of AlkenesNomenclature of Alkenes

30.330.3 Physical Properties of AlkenesPhysical Properties of Alkenes

30.430.4 Preparation of AlkenesPreparation of Alkenes

30.530.5 Reactions of AlkenesReactions of Alkenes

Chapter 30Chapter 30


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30.1 Introduction (SB p.122)

Functional group of alkenes:

The C = C double bond in ethene


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Alkenes show geometrical isomerism

30.1 Introduction (SB p.122)


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30.2 Nomenclature of Alkenes (SB p.123)

1. Determine the stem name by selecting the longest possible straight chain containing the C = C double bond and use the ending ‘-ene’

2. Number the parent chain so as to include both carbon atoms of the double bond, and begin numbering with the end of the chain nearer the C = C double bond

3. Designate the position of the C = C double bond by using the number of the first atom of the double bond

4. Designate the positions of the substituents by using the numbers obtained by application of rule 2

Nomenclature of AlkenesNomenclature of Alkenes


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Examples:


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5. If two identical groups are present on the same side of the C = C double bond, the compound is designated as cis; if they are on opposite sides, the compound is designated as trans.

e.g.


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Example 30-1Example 30-1Give the IUPAC names for the following alkenes:

(a)

(b)

Answer


Solution:

(a) trans-3,4-dichlorohept-3-ene

(b) cis-3,4-dimethyloct-3-ene


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Check Point 30-1 Check Point 30-1

Draw the structural formula for each of the following alkenes:

(a) cis-hex-3-ene

(b) trans-2,3-dihydroxybut-2-ene

(c) cis-1,2-dichloroethene Answer


(a)

(b)

(c)


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30.3 Physical Properties of Alkenes (SB p.124)

Name Formula Boiling point (°C)

Melting point (°C)

Density at 20 °C (g cm-3)

Ethene CH2 = CH2 -104 -169 —

Propene CH3CH = CH2 -47.7 -185 0.514

But-1-ene CH3CH2CH = CH2 -6.3 -185 0.595

Pent-1-ene CH3(CH2)2CH = CH2 30 -165 0.641

Hex-1-ene CH3(CH2)2CH = CH2 62.9 -140 0.673

cis-But-2-ene CH3CH = CHCH3 (cis) 4 -139 0.621

trans-But-2-ene CH3CH = CHCH3 (trans)

1 -106 0.604

2-Methylbut-1-ene

CH3CH3C(CH3) = CH2 31 -138 0.650


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30.4 Preparation of Alkenes (SB p.125)

• alkenes can be prepared industrially by cracking of high molecular mass alkanes

CrackingCracking

422C600

33 CH2CHCH CHCH2

242223C600

323 HCHCHCHCHCHCH CHCHCH2


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• Dehydrohalogenation is the elimination of a hydrogen halide molecule from a haloalkane

Elimination ReactionsElimination Reactions

Dehydrohalogenation


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Examples:


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The ease of dehydrohalogenation of haloalkanes decreases in the order:

tertiary > secondary > primaryhaloalkane haloalkane haloalkane


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Dehydrohalogenation of 2° and 3° haloalkanes can take place in more than one way and a mixture of alkenes is formed

alc. KOHCH3CH2CHClCH3 CH3CH = CHCH3 + CH3CH2CH = CH2

heat

2-chlorobutane But-2-ene But-1-ene

(80%) (20%)

Note: the more highly substituted alkene is formed as major product


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The relative stabilities of alkenes decrease in the order:


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Dehydration is the removal of a water molecule from a reactant molecule

Dehydration of Alcohols


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The experimental conditions of dehydration depend on the structures of alcohols

e.g.


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The relative ease of dehydration of alcohols generally decreases in the order:

tertiary > secondary > primaryalcohol alcohol alcohol

Like dehydrohalogenation, the more highly substituted alkene is formed as the major product


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Example 30-2Example 30-2Classify the following alcohols as primary, secondary or tertiary alcohols.

(a) CH3CHOHCH2CH3

(b) CH3CH2CH2OH

(c) (CH3)2COHCH2CH2CH3 AnswerSolution:

(a) Secondary alcohol

(b) Primary alcohol

(c) Tertiary alcohol



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Classify the following haloalkanes as primary, secondary or tertiary haloalkanes.

(a)

(b)

(c)

Answer


(a) Secondary haloalkane

(b) Primary haloalkane

(c) Tertiary haloalkane


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• hydrogenation of alkynes using Lindlar’s catalyst produces alkenes

• prevent further hydrogenation of the alkenes formed to alkanes

Addition ReactionsAddition Reactions

Hydrogenation


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30.5 Reactions of Alkenes (SB p.129)

Alkenes are more reactive than alkanes

Reason: presence of the C = C double bond

alkenes undergo addition reactions and the reactions are exothermic

Energetically favourable!!


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Electrons of bond are more diffuse and less firmly held susceptible to attack by electrophiles

Electrophiles such as H+, neutral reagents such as bromine (can be polarized) react with C = C double bond


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Addition of hydrogen bromide to C = C double bond yields a bromoalkane

Electrophilic Addition ReactionsElectrophilic Addition Reactions

Addition of Hydrogen Bromide



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Examples:



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Propene reacts with HBr to give 2-bromopropane (majo

r product) and 1-bromopropane (minor product)


The formation of two possible products can be explained

by the reaction mechanism.


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The mechanism for the addition of HBr to an alkene involves 2 steps

Step 1:


Reaction Mechanism: Electrophilic Addition Reaction of Hydrogen Bromide to Alkenes

Step 2:


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If the alkene is unsymmetrical, two different carbocations

can be formed


Theoretical Explanation of Markownikoff’s Rule


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2-bromopropane is the major product because the more stable

secondary carbocation is formed in the first step



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Markownikoff’s rule states that in the addition of HX to an unsymmetrical alkene, the hydrogen atom adds to the carbon atom of the carbon-carbon double bond that already has the greater number of hydrogen atoms.


Example:


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Markownikoff’s rule is related to the stability of the carbocation intermediate formed in the electrophilic addition reaction.

The relative stabilities of carbocations:


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Alkenes react rapidly with Br2 in 1,1,1-trichloroethane

at room temperature and in the absence of light

Addition of Bromine


e.g.


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The behaviour of alkenes towards Br2 in CH3CCl3 is a useful

test for the presence of carbon-carbon multiple bonds


Add Br2 in CH3CCl3 to excess alkene

The reddish brown colour of Br2 is decolourized


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In an aqueous solution of Br2, the following equilibrium exists

Addition of Bromine Water


Br2 + H2O HBr + HOBr

Bromic(I) acid

The bromine atom bears a partial positive charge while

the oxygen atom bears a partial negative charge

∵ oxygen is more electronegative than bromine


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e.g.


When bromic(I) acid reacts with alkenes, bromohydrin i

s formed


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Alkenes react with cold and concentrated H2SO4 to for

m alkyl hydrogensulphates

Addition of Sulphuric(VI) Acid


e.g.


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The large bulky –OSO3H group makes the alkyl hydrogensulphate

very unstable. Two possible further reactions take place:

1. Regeneration of alkenes


2. Production of alcohols


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In the presence of metal catalysts (e.g. Pt, Pd or Ni), H2 is ad

ded to each atom of C = C double bond to form an alkane

Catalytic HydrogenationCatalytic Hydrogenation


e.g.


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• Catalytic hydrogenation is

used to convert liquid

vegetable oil to semi-solid fats

in making margarine and

solid cooking fats (known as

hardening of oils).

• Hydrogenation is useful in analyzing unsaturated hydrocarbons

• The number of double or triple bonds present in the unsaturated hydrocarbon molecule can be deduced by the number of moles of hydrogen reacted


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(a) What chemical tests would you use to distinguish between two unlabelled bottles containing hexane and hex-1-ene respectively?

Answer

30.5 Reaction of Alkenes (SB p.136)

(a) Either one of the following tests:

Hex-1-ene can decolourize bromine water or chlorine

water in the dark while hexane cannot.

Hex-1-ene can decolourize acidified potassium manganat

e(VII) solution while hexane cannot.


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(b) What is the major product of each of the following reactions?

(i)

(ii)

Answer


(b) (i)

(ii)


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(c) Give the reaction products for the following reactions:

Ni(i) CH3CH = CH2 + H2

conc. H2SO4(ii) CH3CH = CHCH3

(iii) CH3CH = CHCH3 + Br2 Answer


(c) (i) CH3CH2CH3

(ii)

(iii)


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(d) Arrange the following carbocations in increasing order of stability. Explain your answer briefly.


Answer

(d) The increasing order of stability of carbocations is :

Tertiary carbocations are the most stable because th

e three alkyl groups release electrons to the positiv

e carbon atom and thereby disperse its charge. Prim

ary carbocations are the least stable as there is only

one alkyl group releasing electrons to the positive c

arbon atom.


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(e) Based on your answer in (d), arrange the following molecules in the order of increasing rate of reaction with hydrogen chloride.


Answer

(e) Upon the reaction with hydrogen chloride, it involves the

formation of carbocations. Therefore, the order of reaction rates

follows the order of the ease of the formation of carbocations, i.e.

the stability of carbocations:

Therefore, the rates of reactions of the three compounds

with hydrogen chloride increase in the order:


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Ozonolysis is a widely used method for locating the double bond of an alkene

OzonolysisOzonolysis


(unstable)

The unstable ozonide is reduced directly by treatment with Zn and H

2O


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Overall process of ozonolysis:


e.g.


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Example 30-3Example 30-3Predict the structures of the following hydrocarbons A, B and C using the information given below:

Answer


HydrocarbonMolecular formula

Products after ozonolysis

A C3H6

B C6H10

C C10H16


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Solution:

A: As C3H6 can be expressed as CnH2n, the hydrocarbon is a molecule with one C = C double bond. When A

undergoes ozonolysis, and are formed.

∴The possible structure of A is CH3CH = CH2.



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Solution:

B: As C6H10 can be expressed as CnH2n-2 and only one dicarbonyl compound is formed on ozonolysis, the hydrocarbon is an alicyclic molecule with one C = C double bond.

∴ The possible structure of B is .



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Solution:

C: As C10H16 can be expressed as CnH2n-4. Two products with totally five carbon atoms are formed. So the original compound is an acyclic molecule with three C = C double bonds.

∴ The possible structure of C is CH3CH = CHCH2CH = CHCH2CH = CHCH3.



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Draw the structures of the alkene molecules that give the following products on ozonolysis.

(a) CH3CH2CH2CHO and CH3CHO

(b) CH3CH2CHO and CH3COCH3

Answer


(a)

(b)


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Polymers: Compounds that consist of very large molecules made up of many repeating units

Monomer: Each repeating unit

Polymerization:The reaction by which monomers are joined together

Addition polymerization: alkene monomers are joined together without the elimination of small molecules

Addition polymer: The polymer produced by addition polymerization

PolymerizationPolymerization



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Monomer: ethene

Depending on the conditions, two kinds of poly(ethene) are

formed

Poly(ethene)



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Low density poly(ethene) (LDPE):

• Molecular mass: 50 000 to 3 000 000

• Light, flexible and low melting temperature

• Uses: make soft items like wash bottles, plastic bags and food wraps

High density poly(ethene) (HDPE):

• Molecular mass: up to 3 000 000

• Tougher and higher melting temperature

• Uses: make more rigid items like milk bottles and water buckets


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Some products made of poly(ethene)


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Reaction Mechanism: Free Radical Addition Polymerization of Ethene


1. Chain initiation

The diacyl peroxide molecule undergoes homolytic bond fission to generate free radicals

The radical reacts with an ethene molecule to form a new radical


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2. Chain propagation


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3. Chain termination

The radicals react to give a stable molecule and the reaction stops.


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Poly(propene)


Properties: more rigid than HDPE, high mechnical strength, strong resistance to abrasion

Uses: make moulded furniture; make crates, kitchenware, food containers; make ropes and hard-wearing carpets

The helmet is made of poly(propene)


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Preparation of monomer (phenylethene):

Poly(phenylethene) (or Polystyrene)


Formation of poly(phenylethene):


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Poly(phenylethene):

Properties:

• Transparent, brittle and chemically inert

Uses:

• Make toys, specimen containers and cassette cases



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Expanded poly(phenylethene):

Properties:

• Extremely light, while solid foam

Uses:

• Make light-weight ceiling tiles in buildings, food boxes and shock absorbers for packaging

Some products made of expanded poly(phenylethene)


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Name and structural formula

of monomer

Name and structural formula of polymer

Uses

Ethene

CH2 = CH2

Poly(ethene) LDPE: plastic bags, wash bottles, food wraps, pipes and tubing

HDPE: tough bottles and jugs, buckets, washing-up trays, toys, pipes and tubing

Propene

CH2 = CHCH3

Poly(propene)Moulded furniture, crates, kitchenware, food containers, fibres for making ropes and hard-wearing carpets and athletic wear

Phenylethene

CH2 = CHC6H5

Poly(phenylethene) Poly(phenylethene): moulded objects (combs, toys, cups, brush and pot handles), refrigerator parts and insulating materials

Expanded poly(phenylethene): good shock absorbents in packaging, light-weight ceiling tiles in building, disposable foam cups and food boxes


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New Way Chemistry for Hong Kong A-Level Book 3A1 Alkenes 30.1Introduction 30.2Nomenclature of...

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