ALKENES NOTES

Chapter 12

Nomenclature Of Alkenes

ALKENES

12.1: Structure of Alkenes Describe alkenes as unsaturated hydrocarbon with general

formula of CnH2n, n>2,(C2- C10). 12.2: Nomenclature of alkenes Draw the structures and name the compounds according to

the IUPAC nomenclature for:

i. Straight chain (C2- C10)ii. Branched alkenes (C4- C10)iii. Cyclic alkenes (C4-C6)

LEARNING OUTCOMES

Contain at least one carbon–carbon double bond.

Unsaturated hydrocarbon.

Old name: olefin.

General formula CnH2n

Name derived from alkane by changing the suffix from ‘ane’ to ‘ene’.

INTRODUCTION TO ALKENES

NOMENCLATURE OF ALKENES

STEP 1

Find the longest carbon chain containing C═C.

Name the parent hydrocarbon.

correctC C

H

H

CH3CH2

CH3CH2CH2

C C

H

H

CH3CH2

CH3CH2CH2

named as pentene

not as hexene

NOMENCLATURE OF ALKENESSTEP 2

Begin at the end nearer the C═C.

Number the C atoms in the chain.

If C═C is at same distance from the two ends, begin at the end nearer to the first branch point.

CH3CH2CH2CH CHCH3

123456

CH3CHCH CHCH2CH3

CH3

1 2 3 4 5 6

Make sure C═C get the lowest possible number!

NOMENCLATURE OF ALKENESSTEP 3

Number the substituents according to their position in the chain, and list them alphabetically.

Write the full name.

CH3CH2CH2CH CHCH3

123456

2–hexene or hex–2–ene

CH3CHCH CHCH2CH3

CH3

1 2 3 4 5 6

2–methyl–3–hexeneor

2–methylhex–3–ene


3- heptene

C C

CH3

CH3

CH3CH2

CH2CHCH3

CH3

1

3

2

6

4

5 7

Name : 3,4,6 - trimethyl – 3 - heptene

3-methyl 6-methyl4-methyl


CH2C

CH2CH3

CH2CH2 CH2 CH2CH3

1

3

2

64 5 7

Name : 2 – ethyl - 1 - heptene

2-ethyl

CH2C

CH2CH3

CH2CH2 CH2 CH2CH3

12

3

4 5 6 7 8

not as 3 -octene

named as

correct

1-heptene1-heptene

CH2C

CH2CH3

CH2CH2 CH2 CH2CH3


1- hexene

1 32 64 5

Name : 3 - isopropyl - 2,5 – dimethyl-1-hexene

2-methyl 3-isopropyl5-methyl

CH2 C

CH3

CH2CH2CHCH3

CH

CH3

CH3

CH3

NOMENCLATURE OF CYCLOALKENES

Cycloalkenes are named similarly as acyclic alkenes, but number the cycloalkenes so that the C═C is between C1 and C2 and the first substituent has the lowest number possible.

CH31

2

3

4

56

1–methylcyclohexene

If there is only one C═C, no need to indicate the position of C═C !

NOMENCLATURE OF CYCLOALKENES

1

2

3

4

56

1,4– dimethylcyclohexene

CH3

CH31

23

4

5

1,5–dimethylcyclopentene

CH3

CH3

Cycloalkenes are named similarly, but number the cycloalkenes so that the C═C is between C1 and C2 and the first substituent has the lowest number possible.


cyclobutene

1

3 2

4

Answer : 1,3 - dimethylcyclobutene

1 - methyl 3 - methyl

CH3

CH3


propene

13 2

Name : 3 - cyclopropylpropene

3 - cyclopropyl

CHCH CH2

EXERCISE - 1

Give IUPAC names for the followingcompounds:

a)

H2C CCCH3

H3C CH3

CH3

b)

CH3CH2CH CCH2CH3

CH3

c)

CH3CHCH

CH3

CHCHCH3

CH3

d)CH3CHC CCH2CH3

CH3

CH3

CH3

EXERCISE - 2

Give the structures and the IUPAC namesfor all alkenes with molecular formula of C6H12,ignoring cis-trans isomer.(Hint: There are 13.)

EXERCISE - 3

Name the following compounds:

a) b)

c)

CH3CH CCH2CH3

CH2CH3CH3

CCH2CH3

CCH2CHCH3H2C

CH3CH3

CCH2CH2CH3

CH2

CH2CH2

EXERCISE - 4

Name the following compounds:

a) CH3 b) c)

d) e)

CH3

CH3CH3

CH3

CH3

CH3

CH3

CH2CH3

Restricted rotation of carbon-carbon bonds due to:

presence of C═C.

Two different atoms or group of atoms attached to each of C atoms which form double bond.

CIS–TRANS ISOMERISM

C C

H

CH3

H3C

H

EXAMPLE: CH3CH═CHCH3

2–butene

C C

H

CH3

H3C

H

trans–2–butene

C CH

CH3H3C

H

cis–2–butene

Two groups on the same side cis

Two groups on opposite side trans

CIS–TRANS ISOMERISM ON C═C

EXERCISE - 5

Determine which of the following compoundsshow cis-trans isomerism.Draw and name the cis and trans isomers ofthose that do.

a) 3–hexeneb) 3–methyl–2–pentenec) 2,3–dimethyl–2–pentene

Which of the following compounds can exist as pairs of cis-trans isomers?Draw each cis–trans pair.

a) CH3CH═CH2

b) (CH3)2C═CHCH3

c) CH3CH2CH═CHCH3

d) (CH3)2C═C(CH3)CH2CH3

e) ClCH═CHClf) BrCH═CHCl

EXERCISE - 6

EXERCISE - 7

Give the structural formula of the following compound and indicate cis-trans isomer: • 2,3–dimethyl–2–butene• cis–2–methyl–3–heptene• 3,6–dimethyl–1–octene• 3–chloropropene• 2,4,4–trimethyl–2–pentene• trans–3,4–dimethyl–3–hexene

EXERCISE - 8

Menthene, a hydrocarbon found in mint plants, has the systematic name as 1–isopropyl–4–methylcyclohexene.Draw its structure.

EXERCISE - 9

Each of the following name is incorrect.Give the correct name

a) 4,7–dimethyl–4–octeneb) 5–bromo–3–methyl–3–hexenec) 2,3–dichlorocyclohexened) 4–ethyl–2–cyclohexenee) 7–bromo–5–ethyl–2–methyl–4–octenef) 3–bromo–6–chloro–5–methylcyclohexeneg) 3–propyl–1–cyclopenteneh) 3–chloro–2–methylcyclobutene

EXERCISE - 10

Each of the following name is incorrect.Draw the structure represented by the incorrectname (or a consistent structure if the name is ambiguous) and give your drawing the correct name:

a) cis–2,3–dimethyl–2–penteneb) 2–methylcylopentenec) 3,4–dimethylcyclohexened) cis–2,5–dibromo–3–ethyl–2–pentene

12.3: Preparation of alkenes

a. Show the preparation of alkenes through dehydration of alcohols.

b. State Saytzeff’s rule.

c. Predict the major product using Saytzeff’s rule

LEARNING OUTCOMES

By elimination reactions:

Dehydration

loss of H2O from alcohol

Latin: “de” means down, away, removal

PREPARATION OF ALKENE

C C

H OH

acid , heat+C C H2O

Reactant: alcohol.

Conditions:

Concentrated H2SO4 or other strong acids

Heat

H and OH eliminated!

DEHYDRATION OF ALCOHOLS

SAYTZEFF’S RULE

An elimination occurs to give the major product,which is the most highly substituted alkene.

EXAMPLE: dehydration of 2–butanol

CH3

OH

CH2 CH3CH – H+

2–butene

1–butene

major product

minor product

– H2O

CH3

CH

C

CH3

H

CH3CH2

CH

C

H

H

EXAMPLE: dehydration of 3,3–dimethyl–2–butanol

– H2O

OH

CH3–C–CH–CH3

CH3

CH3

– H+

major

minor

C

CH3H3C

C

H3C CH3

C

CH3HC

H CH(CH3)2

EXERCISE - 11

Predict the major products of dehydrationof the following alcohols:

a) 2–pentanolb) 1–methylcyclopentanolc) 2–methylcyclohexanol

Note: naming cycloalcohols:C where the OH group is attached shouldbe numbered as C1

EXERCISE - 12

Give the products that would be formedwhen each of the following alcohols issubjected to acid-catalyzed dehydration.If more than one products would be formed,designate the alkene that would be the majorproducts. (Neglect cis-trans isomerism).

a) 2–methyl–2–propanolb) 3–methyl–2–butanolc) 3–methyl–3–pentanold) 2,2–dimethyl–1–propanole) 1,4–dimethylcyclohexanol

12.4: Reaction of alkenes:

a) Explain the addition reactions of alkenes with:

i. Hydrogen in the presence of catalyst.

ii. Halogen (Cl2 or Br2) in inert solvent (CH2Cl2)

iii. Hydrogen halides (HCl and HBr)

b) Introduce Markonikov’s rule for reaction (iii)

LEARNING OUTCOMES

REACTION OF ALKENES

C C + A—B addition C C

BA

A–B can be:

hydrogen H2 in catalyst

hydrogen halides H–X

halogen in inert solvent X2 in CH2Cl2

HYDROGENATION OF ALKENE

alkane

C C + C C Ni, Pd or Pt

H H

alkene

The addition of H2 to a double bond

Alkane is formed

HH

HYDROGENATION OF ALKENE

EXAMPLE:

+Ni, Pd or Pt

C═C

H3C CH3

H H

H H H3C C C CH3

H H

cis-2-butene butane

HH

HALOGENATION OF ALKENE IN INERT SOLVENT

Alkenes reacts rapidly with Cl2 and Br2 in inert solvents such as CH2Cl2 or CCl4

EXAMPLE:

CH3CH═CHCH3 +CH2Cl2

2,3–dichlorobutane

C—CCH3CH3

HH

Cl ClCl Cl

2-butene

HALOGENATION OF ALKENE IN INERT SOLVENT

EXAMPLE:

+CH2Cl2

1,2–dichloropropane

+ CCl4

1,2–dibromocyclopentane

C—CH CH3

HH

Cl ClC—CH CH3

HH

Cl Cl

Br BrBr Br

HYDROHALOGENATION OF ALKENE

H–X: HBr and HCl

C C + H C CX

alkenehaloalkene

XH


EXAMPLE:

+

2-butene 2–bromobutane

H BrC CCH3 CH3

HH

C CCH3 CH3

HH

BrH


EXAMPLE:

+

2-methyl-2-butene2–chloro-2-methylbutane

H ClC CCH3 CH3

CH3HC CCH3 CH3

CH3H

ClH

For unsymmetrical alkene?

why H choose here ?

MARKOVNIKOV’S RULE

Vladimir Vassilyevich Markovnikov (1838-1904). Russian chemist.

In addition of HX to an unsymmetrical alkene, H atom adds to the C atom of the double bond that already has the greater number of H atoms.

CH2 CH3CH

C atom with the greater number of

H atoms

H Br

CH2 CH3CH

H Br

EXAMPLE:

+

2–chloro–2–methylbutane

CH3–CH–C═C

CH3

H

H H Cl

CH3–CH2–C

CH3

Cl

C H

H

H

C atom with the greater number of

H atoms

EXERCISE - 13

What product would you expect from reactionof HCl with:

a) 1–ethylcyclopenteneb) 1–methylcyclohexene

EXERCISE - 14

Suggest the possible alkene that can be used toprepare 3-chloro-3-methylhexane.There may be more than one possibility.

? CH3CH2CCH2CH2CH3

CH3

Cl

EXERCISE - 15

Predict the products of the following reactions:

a)

b)

+ HCl ?

(CH3)2C═CHCH2CH3+ HBr ?

c) CH3CH2CH2CH═CH2 + HBr ?

d) + HBr ?CH2

EXERCISE - 16

What alkenes would you start with to preparethe following compounds?

a) Bromocyclopentane

b) CH3CH2CHBrCH2CH2CH3

EXERCISE - 17

Predict the products of the following reactions:

a)

b)

c)

d)

CH3CH2CH═CCH2CH3

CH3

+ HCl ?

1–ethylcyclopentene + HBr ?

2,2,4–trimethyl–3–hexene + HBr ?

+ HBr ?

12.5: Unsaturation test for alkenes

Explain the unsaturation test for alkenes:

i. Baeyer’s test using dilute, alkaline solution of KMnO4 at room temperature.

ii. Reaction with bromine in CCl4.

LEARNING OUTCOMES

UNSATURATION TEST FOR ALKENE

to test the presence of C=C. to distinguish alkane and alkene (or alkyne).

BROMINE TEST BAEYER TEST

12.5

CCl4CCl4

Observation: Red–brown color of Br2 dissappear (colorless) almost instantly.

BROMINE TEST

BROMINE TEST

Alkanes do not react with Br2 or Cl2 at room

temperature and in the absence of light.

R—H

alkane

+ Br2room temperatureIn the dark , CH2Cl2

no appreciablereaction

Observation:The color bromine unchanged.

R—Halkane

+ Br2 bromoalkaneslight

Observation:The colour of bromin dissappears.

BAEYER TEST

C═C is oxidized dihydroxylation

C C

OH OH

+C C KMnO4

diol

dilute / cold OH-

EXAMPLE

CH3CH═CHCH3 + KMnO4dilute / cold OH-

decolouration: purple colourless

CH3CH—CHCH3

OH OH2,3–butanediol

EXAMPLE

CH3CH═CH2 + KMnO4

dilute / cold OH-

CH3CH—CH2

OH OH1,2–propanediol

propene

OH

OH

+ KMnO4

cyclohexene 1,2–cyclohexanediol

BAEYER TEST

dilute / cold OH-

EXERCISE - 17

Show the structures of alkenes that give the following products on reaction with cold, dilute alkaline solution of KMnO4:

CH3CH2CH═CH2 a)

b)

c) (CH3)2C═CHCH3

(CH3)2C═CHCHC(CH3)2

EXERCISE - 18

Write the structural formula for the productsthat is formed when 1–butene reacts with each of the following reagents:

a) HCIb) H2, Ptc) Br2 in CH2Cl2

d) cold dilute KMnO4, OH-

EXERCISE - 19

Repeat EXERCISE using cyclobutene instead of 1–butene.

END OF SLIDE SHOW

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ALKENES NOTES