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