Download presentation
Presentation is loading. Please wait.
Published byRolf Neal Modified over 9 years ago
1
Chemistry
2
Alkenes
3
Session objectives 1.Kolbe’s method 2.Dehydration of alcohol 3.Dehydrohalogenation 4.Miscellaneous method
4
General characteristics of alkene Olefins General molecular formula C n H 2n C–C bond hybridization 1.34 A 0 sp 2 hybridization Show chain, positional and geometrical isomerism
5
General characteristics of alkene Chain isomerism Positional isomerism 1-pentene2-pentene
6
General characteristics of alkene Geometric isomerism
7
Preparation of alkenes The Kolbe’s electrolysis
8
Dehydration of alcohols (E1 - elimination) 1-butene
9
Mechanism Oxonium ion Carbocation
10
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 3030 2020 1010 > >
11
Features of dehydration The stability of alkenes is governed by Saytzeff’s rule
12
Features of dehydration Dehydrohalogenation of alkyl halides: (E2 or 1,2-elimination or -elimination) Mechanism
13
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
14
Features of dehydration
15
Hoffman’s elimination If 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 C 1 is more acidic than and is preferably pulled by the base.
16
Hoffman’s elimination
17
Miscellaneous Methods Debromination of vicinal dibromides
18
Miscellaneous Methods Wittig reaction Corey-House method
19
Miscellaneous Methods Catalytic dehydrogenation
20
Class exercise
21
Class Exercise - 1 Acetone reacts with Ph 3 P + — C – H — CH 3 gives (a) (b) (c) (d) None of these Wittig reaction. Hence, the answer is (b). Solution:
22
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 CH 2 = CH — CH = CH 2 Hence, the answer is (c). Solution:
23
Class Exercise - 3 sp 2, sp 3 from the structure. Hence, the answer is (a). Solution: Which hybrid orbital/orbitals will form the following compound? (a) (b) Only (c) sp and (d) sp and
24
Class Exercise - 4 When 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:
25
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:
26
Class Exercise - 6 The IUPAC name of is (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:
27
Class Exercise – 7 Which of the following is a false statement? (a) There is sp 3 hydridization in propane (b) Ethyne has a linear structure (c) There is sp 2 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:
28
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:
29
Class Exercise - 9 Which of the following alkenes is least stable? (a) (b) (c) (d) No hyperconjugation. Hence, the answer is (c). Solution:
30
Class Exercise – 10 Convert the following reactant into product.
31
Solution
32
Thank you
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.