Hydrohalogenation of alkenes and dehydrohalogenation of haloalkanes
Chemistry. Alkenes Session objectives 1.Kolbe’s method 2.Dehydration of alcohol...
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Transcript of Chemistry. Alkenes Session objectives 1.Kolbe’s method 2.Dehydration of alcohol...
Chemistry
Alkenes
Session objectives
1. Kolbe’s method
2. Dehydration of alcohol
3. Dehydrohalogenation
4. Miscellaneous method
General characteristics of alkene
•Olefins •General molecular formula CnH2n •C–C bond hybridization 1.34 A0
•sp2 hybridization •Show chain, positional and geometrical isomerism
General characteristics of alkene
Chain isomerism3 2 2 2CH —CH —CH —CH=CH
H3C — CH2 — C — CH3
CH22-methyl-1-butene
Positional isomerism
3 2 2 2CH - CH - CH - CH CH 3 2 3CH - CH - CH CH- CH1-pentene 2-pentene
General characteristics of alkene
Geometric isomerism
H H
CH3CC
CH3
Cis 2 – butene
HH
CCCH3
CH3
Trans 2–butene
Preparation of alkenesThe Kolbe’s electrolysis
CH2
CH2
CH2
CH2
.
.
Dimerization
(Minor)
(Major)
CH2COONa
CH2COONa+-
+-
sodium succinate
Electriccurrent
CH2COO
CH2COO-
-+ 2Na+
Anode Cathode
-2e2Na
H2O..
CH2COO
CH2COO-2CO2
+2e
2NaOH + H2
Dehydration of alcohols (E1 - elimination)
OH Conc. H2SO4H2O+
3 2 2 2CH CH CH CH OHAl2O3
300° C 3 2 2CH CH CH CH
1-butene
3 2 2CH CH CH OH 2 4Conc.H SO
160°
3 2 2CH CH = CH + H O
Mechanism
3 2 2CH - CH - CH - OHH+ +3 2 2 2CH - CH - CH - OH
2-H O 23 2CH CH CH
Oxonium ion
Carbocation
23 CHCHCH
H
H 3 2CH CH CH
1 propene
Features of dehydration
• A carbocation is formed as an intermediate.
• The ease of dehydration of alcohols is 3° > 2° > 1°
• 3° carbocation is most stable
R
R
OH
R
C R
H
OH
R
C R
H
OH
H
C
30 20 10
> >
Features of dehydrationThe stability of alkenes is governed by Saytzeff’s rule
2 2 2 2 2 2R C C R R C CH R R C CH ~ RCH CHR RCH CH
CH3
OH
CH3
1 2
3 CH2
CH3
CH3
CH3
CH2
CH3
CH CH3
C CH2
CH3
Cfrom 1, 2position
from 1, 3position
loss of H2O C
(Minor)
(major)
loss of H2O
Features of dehydrationDehydrohalogenation of alkyl halides: (E2 or 1,2-elimination or -elimination)
CH2
Alc.KOHCH2
Br
CH2+ KBr H2O+CH2
H12
Mechanism
H
HH Br
HO
H
HC C
HH
HHSlow
C C
H
H
H
H OH
HBr
+ Br- + H2OC C
–
Transition state –
Features of dehydration
• Stereochemically, the best conformation for elimination is the anti-coplanar conformation, i.e. the three bonds H—C, C—C and C—Br are in one plane.
• There is no formation of any intermediate in this reaction, rather it proceeds through a transition state.
• A trans- or anti- elimination in which the atoms leave from the opposite sides is observed in most cases
• The reaction is thus stereospecific, since a base pulls a hydrogen, a strong base will accelerate E2 elimination.
• predominant formation of a substituted alkene is formed according to Saytzeff’s rule
Features of dehydration
3 2 2 3CH - CH - CH - CH|Br
Alc. KOH
3 3 3 2 2CH - CH=CH- CH +CH - CH - CH=CHMajor Minor
Hoffman’s eliminationIf the leaving group is very strong electron-attracting, then elimination takes place contrary to Saytzeff’s rule. The formation of a less substituted alkene is observed.
The – H at C1 is more acidic than and is preferably pulled by the base.
3H at C
Hoffman’s elimination
CHCH2 CH2 CH3
CH3
NaOHC 52
OHHC 52
+N(CH3)3
CH3CH2CH2CH CH2 + CH3CH2CH CHCH3
Majo Mino
Miscellaneous Methods
Na
H
H
Br
H
H
Br
H
H
H
HC C
H
H
H
HC C
Zn/HOAC
acetone+ I2 +2 NaBr
+ ZnBr2
C C
CH2CH
Br Br
CH2CH
ZnBr2
Zn+
Debromination of vicinal dibromides
Miscellaneous Methods
Cl (CH2 CH)2CuLi CH CH2 + LiCl + CH3CH2Cu+
Wittig reaction
(C6H5)3P CH2
O
CH2 + (C6H5)3P O
Methylene cyclohexane
Corey-House method
Miscellaneous Methods
3CrO3 2 2 3 3 2 2CH CH CH CH CH CH CH CH
Catalytic dehydrogenation
Class exercise
Class Exercise - 1Acetone reacts with Ph3P+ — C–H — CH3 gives
CH2 C — CH2 — CH3
CH3
CH2 — C CH2 — CH3
CH3
CH2 CH2(a)
(b)
(c)
(d) None of these
Wittig reaction. CH2 — C CH2 — CH3
CH3
Hence, the answer is (b).
Solution:
Class Exercise - 2
How many and bonds are present in 1,3-butadiene?(a) 6 and 2 (b) 2 and 2(c) 9 and 2 (d) 6 and 6
CH2 = CH — CH = CH2
9 2 and bonds.
Hence, the answer is (c).
Solution:
Class Exercise - 3
sp2, sp3 from the structure.
Hence, the answer is (a).
Solution:
Which hybrid orbital/orbitals will form the following compound?
3 2 3CH CH CH CH CH
(a) (b) Only
(c) sp and (d) sp and
2 3sp and sp
Class Exercise - 4When an aqueous solution of sodium acetate and sodium propionate is electrolyzed, one gets(a) ethane (b) propane
(c) butane (d) All of these
Along with normal butane we get unwanted product like ethane, propane, etc.
Hence, the answer is (d).
Solution:
Class Exercise - 5
Which has the least heat of hydrogenation?(a) 1, 3-butadiene (b) 1-butene
(c) Trans-2-butene (d) Cis-2-butene
trans-2-butene is maximum stable. Therefore, heat of hydrogenation will be least.
Hence, the answer is (c).
Solution:
Class Exercise - 6
The IUPAC name of
isCCH2
CH3
CH3
CH CH3
(a) 3, 3, 3-trimethyl-1-propene(b) 3,3-dimethyl but-1-ene(c) 1,1,1-trimethyl-2-propene(d) 2,2-dimethyl-3-butene
3,3-dimethyl but-1-ene
Hence, the answer is (b).
Solution:
Class Exercise – 7Which of the following is a false statement?(a) There is sp3 hydridization in propane(b) Ethyne has a linear structure(c) There is sp2 hybridization in ethylene(d) Alkynes show geometric isomerism
Alkynes do not show geometrical isomerism because it is a linear structure.
Hence, the answer is (d).
Solution:
Class Exercise - 8
Which of the following types of isomerism is not exhibited by alkenes?
(a) Position isomerism(b) Chain isomerism(c) Geometric isomerism(d) Metamerism
Metamerism will not be shown by alkenes.
Hence, the answer is (d).
Solution:
Class Exercise - 9Which of the following alkenes is least stable?
CH2 CH CH3C C
H
CH3
CH3
CH3
CH2 CH2 C CCH3
CH3CH3
CH3
(a) (b)
(c) (d)
No hyperconjugation.
Hence, the answer is (c).
Solution:
Class Exercise – 10Convert the following reactant into product.
CH2OH CH3
SolutionCH2 — OH CH2 — OH2
CH3
H
H
CH3
–H
CH2
H
+
H+ / H2O
–H2O
+
+
+
3° stable carbocation
Thank you