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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
NOMENCLATURE OF ALKENES
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
NOMENCLATURE OF ALKENES
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
NOMENCLATURE OF ALKENES
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.
NOMENCLATURE OF ALKENES
cyclobutene
1
3 2
4
Answer : 1,3 - dimethylcyclobutene
1 - methyl 3 - methyl
CH3
CH3
NOMENCLATURE OF ALKENES
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
HYDROHALOGENATION OF ALKENE
EXAMPLE:
+
2-butene 2–bromobutane
H BrC CCH3 CH3
HH
C CCH3 CH3
HH
BrH
HYDROHALOGENATION OF ALKENE
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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