Alkene+Alkyne
Short Description
hydrocarbon alkene and alkyne...
Description
LECTURE NOTES Session - 2009-10
ORGANIC CHEMISTRY TOPIC :
ALKENE & ALKYNE
Contents : Alkene/Alkyne Lect.
Lect.
1 Physical properties of Alkenes/Alkyne 2 Relative Besicity 3 Lab test of alkene and alkynes 4 Stability of alkenes/cycloalkenes 5 Method of preparation
(H) Acid-Catalyzed Hydration
(A) By partial reduction of alkynes (B) Dehydration of alcohol
(i) For Alkene (a) Oxymercuration-Demercuration (b) Alkoxymercuration - demercuration (c) Hydroboration-oxidation (Syn addition) (ii) Acid-Catalyzed Hydration For Alkyne Examples
(C) By pyrolysis of esters Examples
Lect. (I) Oxidation Reaction
Lect.
(i) Ozonolysis (C = C, C C)
(D) By Dehalogenation of vicinal dihalides/
Examples
tetrahalides (E) Dehydrohalogenation of alkyl halides/ vicinal dihalides (F) Replacement of the acetylenic hydrogen
Lect. (ii) Hydroxylation Reactions (Alkenes, Alkynes)
atom of terminal alkynes to form higher
(iii) Epoxidation of Alkenes
alkynes
(iv) Oxidative cleavage (Alkenes, Alkynes)
(G) Miscellaneous
Examples
Hydrolysis of carbides Examples
Lect.
Lect.
7. Miscelleneous
6. Chemical Reactions
(B) Addition of NOCl
(A) Electrophilic addition reaction
(C) Halogenation. Allylic substitution
(B) Addition of halogen to alkenes
(D) Alkylation
(C) Addition of halogen to alkynes
Examples
(A) Addition of carbenes to Alkenes
(D) Addition of HOX to Alkene (E) Addition of HOX to Alkynes Examples
Lect. Discussion
Lect. (F) Addition of Hydrogen Halides (+ HX) to Alkene (G) Addition of Hydrogen Halides (+ HX) to Alkyne Examples
Page # 2
Physical properties of alkenes & General method of preparation of alkenes
1. Physical Properties of Alkenes TABLE : 1 Physical properties
Homologus series
Isomers
1.
Physical state
C1 – C3 gases C4 – C20 liquids > C20 : solids
2.
Dipolemoment ()
3.
Polarity
–
4.
Melting point
increases with M.W.
trans > cis (due to more packing capacity)
5.
Boiling point
increases with M.W.
cis > trans # branching decreases B.P.
cis > trans cis > trans (for Cab=Cab type of alkenes)
trans Polarity increases, solubility in polar solvents increases.
trans > cis (cis isomers has more VanderWaals repulsion)
TABLE-2 – COMPARATIVE STUDY OF ALKANES, ALKENES, ALKYNES Properties
Alkanes
Alkenes
Alkynes
1.
Bond length
1.54 (C – C)
1.32 (C = C)
1.20 (C C)
2.
Bond energy(KJmol-1)
415 (C – C)
615 (C = C)
835 (C C)
3.
Hybridization
sp3
sp2
sp
4.
% s character
25%
33%
50%
5.
pKa
50
44
25
6.
Electronegativity of ‘C’
sp3 (C)
7.
Polarity
R – CH2 – CH3
R – CH = CH2
R–CC
8.
Rate of hydrogenation
–
less
more
9.
Rate of electrophilic addition reaction
–
more
less
10.
Heat of combustion
C2H6 (– 373)
C2H4 (– 337 kcal)
C2H2 (– 317 kcal)
11.
Density (g/cm3)
C3H8 (0.50)
C3H6 (0.52)
C3H4 (0.67)
<
sp2 (C)
<
sp (C)
Page # 3
12.
H 121.2° H C=C H H
Structure
1.08Å
ethene
13.
Shape
Tetrahedral
Planar
Linear
2. Rel at i ve Basi ci t y (a) CH3 CH2 : H2C CH :– HC C :– (b)
-
OH < OR < C
-
CR <
NH2 <
CH2 < CH2CH 3
CH
3. Stability of Alkenes/Cycloalkenes Overall relative stabilities of Alkenes Studies of numerous alkenes reveal pattern of stabilities that related to the number of alkyl groups attached to the carbon atoms of the double bond. The greater the number of attached alkyl groups (i.e., the highly substituted the carbon atoms of the double bond), the greater is the alkene’s stability. This order of stabilities can be given in general terms as follows. Relative stabilities of alkenes >
>
R
H C=C
>
>
H
R
R
R C=C
H
>
>
H
Cycloalkenes trans-Cycloheptene has been observed spectroscopically, but it is a substance with a very short lifetime and has not been isolated. trans-Cyclooctene has been isolated, however. Here the ring is large enough to accommodate the geometry required by a trans double bond and still be stable at room temperature. Trans-Cyclooctene exists as a pair of geometrical isomers. You may wish to verify this using hand-held models.
CH2 – CH2 CH2 | CH2
CH || CH
CH2 – CH2 cis-Cyclooctene
7 6
2 3
8 1
5
4
trans-Cyclooctene
The way of measuring the stability of an alkene is the determination of its heat of hydrogenation or the heats of combustion. Both are exothermic reactions. (H = – ve) Hhyd
1 (– ve sign indicates the exothermic nature of reaction) stability
In general, the order of stability of alkenes is : R2C = CR2 > R2C = CHR > R2C = CH2 > RCH = CHR (trans > cis) > RCH = CH2 > CH2 = CH2
Page # 4
4.
Methods of preparation
(A)
BY PARTIAL REDUCTION OF ALKYNES (a) By Catalytic Hydrogenation of Alkenes in presence of poisoned catalyst (A Syn Addition of Hydrogen: Synthesis of cis-Alkenes : This is performed by)
(i) Lindlar’s catalyst : Metallic palladium deposited on calcium carbonate conditioned with lead acetate and quinoline.
(ii) P-2 catalyst (Ni2B nickel boride) General Reaction
H2 ,Pd / CaCO 3 (Lindlar 's catalyst ) quinoline
R – C C – R
Mechanism of hydrogenation : (1)
( 2)
Steps : The reactant alkyne molecules and hydrogen molecules get adsorbed at the surface of metal catalyst. It is chemical adsorption (chemisorption). In this state, the reactants lie very close to each other and so the hydrogen atoms start forming bond with carbon. Two hydrogen atoms are added to two triply bonded carbon atoms from the same side of bond and a cis or syn addition product is formed. The product alkene now escapes away from the surface of the catalyst. Quinoline occupies the metal surface inhibiting further reduction to alkanes. Quinoline therefore is called catalyst poison and such palladium is called deactivated catalyst or poisoned catalyst.
(b) Birch Reduction : (Anti Addition of Hydrogen: Synthesis of trans-Alkenes) General Reaction
Na / Li Liq. NH3
H R C C H R
Reagents Na(or Li,K) + liq NH3 Na+ + e¯ (solvated electron) R
-
H – NH 2
••
Mechanism :
R C C R
C=C •
R
NaNH2 +
Na+
••
Note : This process of reduction is not eligible when terminal alkynes are taken.(R–CCH) because terminal alkynes form sodium salt with Na metal. CH3 – C CH + Na / NH3 CH3 – C C¯ Na+ + H
(B)
BY DEHYDRATION OF ALCOHOLS Alcohols when heated in presence of following reagents undergo loss of water molecule and form alkenes. The elimination is elimination. (i) H2SO4 / 160°C (ii) H3PO4 / (iii) P2O5 / (iv) Al2O3 / 350°C undergo loss of water molecule and form alkenes
Page # 5
General Reaction
(C)
BY PYROLYSIS OF ESTERS Thermal cleavage of an ester involves formations of a six membered ring in the transition state leading to the elimination of an acid leaving behind an alkene.
500C
Remarks:-
+
(1) When an ester is heated at high temperature, it forms an alkene by elimination of an acid. (2) It is an intramolecular 1, 2 elimination (-elimination) (3) It passes through a cyclic T.S. (6 membered). (4) It is a syn elimination, i.e., both the eliminating groups lie on the same side. (5) The least crowded -Hydrogen is eliminated to give the major product (Hoffman alkene).
As a direct consequence of cyclic transition state, both the leaving groups namely proton and carboxylate ion are eliminated from the cis position. This is an example of cis elimination.
Examples : H2 / Ni2B(P 2 ) or H2 / Pd / CaCO3
1.
CH3CH2C CCH2CH3 ( syn addition)
3 Hexyne
2.
Na / NH ( )
CH3 – CH2 – C C – CH2 – CH3 3
CH2OH
CH2
3.
CH3
+ (I) Minor
(II) Major
CH2OH
4.
If the starting material is labelled with deuterium as indicated, predict how many deuterium will be present in the major elimination product ?
Page # 6
HO
HO
CD3
HO
(b)
CD3
H
H2SO4
(a)
+ H2O
CD3
H O
2
CH3
CH2
CH3
D
D
D
D
H2SO4
D
D
D
D
D
D
D
+
(C – D cleavage is about 7 times slower than C – H cleavage so the product from C – H cleavage should be formed about 7–times as fast) 5.
Write the intermediate species as indicated :
H 3, 3 – dimethyl – 2 – butanol .................. (A) oxonium ion
H2 O Migration .................... ..................... (B) (C) 20 carbocation 30 carbocation
-H+
-H+
(E) Alkene
CH3 Ans.
(A) =
CH3 C
CH3
CH CH3
CH3 OH2 +
CH3 (B) = CH3 C CH + CH3
(D) Alkene
CH3 CH3 (C) = CH3 C +
CH CH3
CH3
CH3 CH3 (D) =
CH3 C
C
CH3
(E) =
CH3 C CH CH2 CH3
Major product is (D) Minor product is (E) 6.
CH3 – CH2 – O – C – CH2 – CH3 ethene + propanoic acid O
7.
Ph – CH2 – C – O – CH2 – CH2 – CH3 propene + Ph – CH2 – COOH O
–
CH3
8.
– O – C – CH3
O
Page # 7
O
–
O – C – CH2 – CH3
9.
O
ethene +
+ CH3CH2COOH
COOH
C2H5 O Et
Et H
Me
H
Me CH3 COOH H OH H
10.
H
O – C – CH3
CH3
(D)
O
+ CH3 – COOH
CH3
Dehydrohalogenation of alkyl halides Dehydrohalogenation is the elimination of a hydrogen and a halogen from an alkyl halide to form an alkene. Base can be used. (i) Hot alcoholic solution of KOH EtO¯ / EtOH (ii) NaNH2 (iii) tBuO¯K+ in t-BuOH (Hoffmann) General reaction :
| | alcohol C C + KOH | | H X
Mechanism
| | C C KX H2O
+ B–H +
Undergo elimination of hydrogen halide (HX) leading to the formation of alkenes.
(E)
By Dehalogenation of vicinal dihalides There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atoms are attached to the same carbon atom and vicinal dihalides in which the two halogen atoms are attached to the adjacent carbon atoms. Dehalogenation of vicinal dihalides can be effected either by NaI in acetone or zinc in presence of acetic acid or ethanol. General Reaction
(i)
Br | | CC | | Br
NaI or Zn, CH3COOH
Br Mech.
ZnBr
C C
Zn
Br (ii)
C C
C
C
C
C
+ ZnBr2
Br Zn dust
CH3 – CHBr – CH2Br CH3 – CH = CH2 CH3 COOH or C2H5OH as solvent
Mech. With NaI in acetone :
Page # 8
+ IX
It involves an antielimination of halogen atoms Remarks (1) Both are E2 elimination. (2)
Both are stereospecific antielimination. NaI Acetone
e.g.
CH3 – CHBr – CHBr – CH3 CH3 – CH = CH – CH3
Method of preparation of alkynes
(A)
Dehalogenation of tetrahaloalkane & Trihaloalkanes General Reaction
R – C C – R + 2Zn X2
CHCl3 + 6Ag + Cl3 CH CH CH + 6 AgCl
(B)
By dehydrohalogenetion of gem and vic dihalide : Alkynes can be synthesized from alkenes. A vicinal dibromide can be synthesized by addition of bromine to an alkene. The vic-dibromide can then be subjected to a double dehydrohalogenation reaction with a strong base to yield an alkyne. H H |
General Reaction
|
2 NaNH
2 R C C R 2NaBr RCH CHR Br2 R C C R
|
|
Br Br A vic dibromide
The dehydrohalogenations occur in two steps, the first yielding a bromoalkene and the second, the alkyne. Mechanism :
••
••
•• ••
••
••
••
••
••
••
••
Bromoalkene
R
Br–
••
+ H—N—H +
C— —C Br
Br Br
Amide ion vic-Dibromide (The strongly basic amide ion brings about an E2 reaction.)
H Ammonia
Bromide ion
H
R C—C ••
Step 2
••
H
H
R
••
H
+
H—N + R—C—C—R –
••
H Step 1
••Br
+ R
••
Bromoalkene (A second E2 reaction produces the alkyne)
.. - .. R – C C – R + H – N – H + :Br: :N – H .. | | H H Alkyne Ammonia bromide Amide ion ion
From vicinal/Geminal Dihalides (–2HX) Step 1
Step 2
Page # 9
(1) Reactant (2) Nature of Elimination
E2
(3) Rate of E2
r1 > (anti elimination)
E2 r2 (anti elimination)
Reagents (1) NaNH2, or , LiNH2 , or , KNH2 (2) alc. KOH Depending on the conditions, these two dehydrohalogenations may be carried out as separate reactions, or they may be carried out consecutively in a single mixture. Sodium amide, a strong base, is capable of effecting both dehydrohalogenations in a single reaction mixture. (At least two molar equivalents of sodium amide per mole of the dihalide) must be used. If product is to be a terminal alkyne, then three molar equivalents must be used because the terminal alkyne is deprotonated by sodium amide as it is formed in the mixture, consuming the sodium amide otherwise needed for the remaining dehydrohalogenation steps) Dehydrohalogenations with sodium amide are usually carried out in liquid ammonia or in an inert medium such as mineral oil. Geminal dihalides can also be converted to alkynes by dehydrohalogenation. A geminal dihalide (abbreviated gem-dihalide) has two halogen atoms bonded to the same carbon (from the Latin geminous, twins). Ketones can be converted to gem-dichlorides through their reaction with phosphorus pentachloride, and these products can be used to synthesize alkynes. Ketones can be converted to gem-dichloride through their reaction with phosphorus pentachloride and these products can also be used to synthesize alkynes.
(C)
Replacement of the acetylenic hydrogen atom of terminal alkynes to form higher alkynes (a) General Reaction
(I) (II)
Ether
R'– X RC = C – MgBr RC C – H + CH3MgBr R – C C – R’
Sodium ethynide and other sodium alkynides can be prepared by treating terminal alkynes with sodium amide in liquid ammonia :
liq. NH
3 CH3C C – H + NaNH2 NH3 +
Na Metal CH3C C – H + ½ H2 +
H–CC–H
(D)
2CH4 + BrMg – C C – MgBr
CH3 – C C – CH2 – CH3 CH3 | CH3 C C CH CH3
H5 C2 – C C – C2 H5
By hydrolysis of carbides CaC2 + 2HOH C2H2+Ca(OH)2 MgC2 + 2HOH C2H2+Mg(OH)2 Mg2C3 + 4HOH CH3– C CH + 2Mg(OH)2
(E)
BY Kolbe's Electrolysis Formation of alkene
Page # 10
Formation of alkyne
EXAMPLES : Identify the product in the following reactions :
CH3
H Na I / acetone
1.
H
H3C
Na I / acetone
2.
3.
What are the various product due to loss of HBr from
CH3 CH3 Ans.
,
, (minor)
4.
What are the major products of dehydrohalogenation of : (i) CH3 – CH2 – CH2 – CH2 – CH2 Cl
(ii) CH3 CH2 CH2 CH CH3 | Cl CH3 |
(iii) CH3 CH2 CH CH2 CH3 | Cl
(v) CH3 CH CH CH3 | | Ans.
| Cl
Cl | (vi) CH3 CH C CH3 | | CH3 CH3
Cl CH3 (i) gives CH3CH2CH2CH = CH2, (ii) & (iii) = CH3CH2CH = CHCH3
(iv) & (v) = 5.
(iv) CH3 CH2 C CH3
(vi) =
Identify the major (satytzeff) and minor (Hofmann) product of dehydrochlorination of :
Page # 11
Ans.
NaNH2 CH CH
6.
7.
CH3 – CHCl2
NaNH2 CH3 – C CH
8.
9.
NaNH2 CH CH
NaNH2 CH3 – CH2 – CHCl2 CH3 – C CH
10.
NaNH2 CH3 – CCl2 – CH3 CH3 – CCl = CH2
11. 12.
NaNH 2 CH3 – CH2 – CH2 – CHCl2 CH3 – CH2 – C CH NaNH2 CH3 – CH2 – C CH
13.
NaNH 2 CH3 – CH2 – CCl2 – CH3 CH3 – CH2 – C CH NaNH 2 CH3 – CH2 – C CH
14.
15.
16.
Give the structure of three isomeric dibromides that could be used as starting materials for the preparation of 3, 3 – dimethyl – 1 – butyne. CH3
Sol. 17.
CH3 Br
CH3
CH3 C CH2 CHBr2 CH3 C CH CH3 CH3 Br (I) (II) Show the product in the following reaction CH3 EtOK CH CH2 ? EtOH Br Br
CH2 Br
CH3
C
C
CH3
CH3 Br (III)
Page # 12
CH3 C CH
Sol. 18.
1, 1 – dibromopentane on reaction with fused KOH at 470K gives 2 – pentyne
Br
Ans.
fused KOH CH3 CH2 C CCH3 Br CH CH2 CH2 CH 2 CH3 470 K 1, 1 – dibromo pentane 2 – pentyne Give the mechanism of this rearrangement. Mech. Br H H H -OH H C C CH2CH2CH3 C C -OH Br CH2CH2CH3 Br H
H
C C C CH2CH3
C
C C CH2CH3
-OH
H
H
H OH
C C C CH2CH3
H
H
H H
H
H C C C CH2CH3 H
H2O OH
H H
C C C CH2CH3
H
H
C C C CH2CH3
H OH OH
H
C C C CH2 CH3 H
H
2 – pentene 19.
Br2 NaNH2 CH3 CH2 C HCH2Br CH3CH2C CH CH3 CH2 CH CH2 CCl 4
|
Br
mineral oil 110 160 º C
20.
Page # 13
6. Chemical Reactions : Due to presence of weak electrons in alkene and alkyne, it well go for electrophilic reaction. Now question is electrophilic addition or electrophilic substitution ? In electrophilic substitution reaction, one bond is broken and a new - bond between one of the doubly bonded carbon atoms and the electrophilic is formed. Since the bond energes of the bond broken and the new bond formed are not much different therefore electrophilic substitution reaction are not accompnied by large energy changes. On the other hand in electrophilic addition reactions one weak -bond (251 KJ mol–1) is broken and two strong bonds (2 × 347 = 694 KJ mol–1) are formed. The overall reaction is accompnied by a release of about 694-251 = 443 KJ mol–1 of energy. In other words electrophilic addition reactions are energetically more favourable than electrophilic substitution reactions Thus the typical reactions of alkenes are electrophilic addition reaction and not the electrophilic substitution reactions.
(A)
Electrophilic Addition Reaction (Mechanism, Remarks, Reactivity order) (a) General Reaction of electrophilic addition :
............. Addition product.
(b) Mechanism :
(c) Remarks : Alkenes, Alkynes and Alkadienes are electron rich species. So they function as species (due to loosely bound – electrons) These compounds mainly give electrophilic addition reactions. Due to nucleophilic nature alkenes / alkynes have affinity for
.
The reaction is initiated by an attack of
.
(d) Reactivity of an Alkene:Factors: (1) Presence of e– releasing groups (+m, +I) at C = C increases nucleophilicity and reactivity.
Page # 14
(2) Presence of ERG stabilises the intermediate carbocation. (3) more stable C , more is reacetivity.. (4) better bridged C (e) Examples of Reactivity Orders:(i)
(ii)
>
>
>
>
>
>
..
G – CH = CH2 >> CH2 = CH2 (reactivity) ( m )
..
–
–
NH2 – CH = CH2 (enamine) .. R – .O. – C = C – (vinyl ether) – –
–
..
–
–
H – .O. – C = C – tauto. C – C – O
(enol)
(iii)
ERG – CH = CH2 > CH2 = CH2 > EWG – CH = CH2 (a) CH3 – O – CH CH2 > CH2 = CH2 > CH2 CH – CN (acrylonitr ile ) ( vinyl ether ) (b)
– CH = CH2 > CH2 = CH2
(c)
< CH2 = CH2
CH = CH2
–
–
–
>
–
>
–
(d)
CH = CH2
CH = CH2
ERG
EWG
(f) Electrophilic addition in – C C –
E Nu – C C –
E Nu
(i) + X2 (ii) + HOX (iii) + HX (iv) + H2O/H+
(B) Additon of halogen to alkenes (Halogenation) X2 (a) General Reaction:- R – CH = CH – R R – CHX – CHX – R Page # 15
C=C
X | | CC | | X usually anti addition
+ X2
(b) Mechanism Step-1 Formation of a halonium ion .. + : .X.
.. + : .X. :
X
: .X. ..
C=C
C
+
C
Halonium ion
Step-2
Opening of the halonium ion.
X attacks from the back side of halonium ion. The nucleophile attacks the electrophilic nucleus of one halogen atom, and the other halogen serves as the leaving group, departing as halide ion. Many reactions fit this general pattern. (c) Reagents:- Br2 > Cl2 (F2, I2 not used) Solvents:- CCl4 , CHCl3 (H2O, MeOH, EtOH, CH3COOH) best (d)Remarks : (i) F2 is not added because F+ is never generated. Morever reaction is explosive giving CO2 & H2O (ii) I2 is not added because reaction is reversible with equilibrium in backward direction. (iii) Reaction with bromine is basis for test of alkenes. (iv) Halogen addition is stereospecific anti addition. (v) Halogens can also be added in presence of sun light and give free radical addition. (Reactivity of halogen addition in sunlight is F2 (explosive) > Cl2 > Br2 > I2) (e) Stereospecific Nature:- With longer E (longer than H ion) the reaction is highly stereospecific. If a bridged C is formed, then it is anti-addition or trans addition of two species at C = C. i.e., one stereoisomer of alkene (cis/trans) will give only one stereoisomer of product. (either d/ or meso) Alkene Cis Cis Trans Trans
Product meso type d type d type meso type
Addition of halogen to alkynes (– C C –) – –
– –
–
Br Br Br Br2 (leq.) Br2 (leq.) (a) R – C C – R R – C = C – R R – C – C – R (1) r1 ( 2) r2 Br Br Br (Tetrahalide) (Trans-dihalide) –
(C)
Addition Syn. Anti Syn. Anti
(b) Rate:- r2 > r1 (Criteria : Stability of carbocation)
Page # 16
–
–
Br R – C ––– C – R > R – C = C – R + + Br Br Br (Vinyl C+ ) (Alkyl C + ) (c) Nature of Addition:- Anti in both step
X (leq.) D R – C C –R 2 2 (1) Anti ( 2) Syn.
D / Pd / CaCO 3 R – C C –R 2 (1) Syn. [lindlar , catalyst ]
(D)
Addition of HOX to Alkene
(a) General Reaction
OH | | | | C C + HOX C C | | X
(b) Reagents (i) X2 + H2O HOX + HX (ii) HOX Remarks :
Anti adition No rearrangement No carbocation Bridged Carbocation
As H2O attacks the bromonium ion, positive charge develops on the carbon from which the bromine departs. The transition state has some of the character of a carbocation. More substituted carbocation are more stable than less substituted ones; therefore, when the bromonium ion ring opens, it does so by breaking the bond between bromine and the more substituted carbon.
Addition of HOX to Alkynes –
X HOX HOX(leq.) R – C = CH R – C C – H (1 eq.) OH enol
(a)
–
(E)
–H2O
Page # 17
R – C – CH2X
R – C – CHX2
O 10% (minor) [-Haloketone]
O 90% (major) [-Dihaloketone]
(b) Remakrs : # Two molecules of HOX can be added, the end product is -Dihaloketone. # The intermediate product is an enol which gives a minor product -haloketone. Extention : –H2 O KOH R – C – CH(OH)2 R – C – CHO R – C – CHX2 O (major product)
O
O
Examples : Ex.1
Br / CCl 4 CH3 – CHBr – CH2Br CH3 – CH = CH2 2
Ex.2
IBr / CCl 4 CH3 – CH = CH2
Ex.3
CH3 – CH = CH2
BrCl
Ex.4
CH3 – CH = CH2
Br2
Ex.5
Me Ex.6
ICl
H Br
Br2
Br H Br H
Me
Br
Me Me CH 3
Ex.7
CH3 C=C
H
cis
H
Ex.9
H
(d + mixture)
(cis)
H
H
H
H C=C trans
CH3
Me Me
Br2
X H
Me H
CH3 Ex.8
X
X
Me H
Br2
(meso)
H
(trans) Me
H
X Me
Br
2 Ph – CHBr – CHBr – Ph
(d + ) mix
Br Ex.10
Ph Ph
IBr
H
I H Page # 18
Ex.12
HOBr / H CH3 – CH – CH2 CH3 – CH = CH2 OH Br
Ex.13
–
–
–
HOCl / H CH2 – CH2 CH2 = CH2 Cl OH
–
Ex.11
Br | Br2 / H2 O CH3 – CH = CH2 CH3 CH CH2 NaCl, KI | impurities G 4 products are formed. G = OH– , Br–, I–, Cl–
Br / H O 2 2
Ex.14
+
OH H
–
Me
Me Br
–
Ex.17
–
Ex.16
––
Ex.15
–
–
- 18 18 OH 18 OH Cl NaO H Me Me Me HOCl / H ––– CH – CH3 Ph –––– CH – CH3 50 – 60 º C Ph –––– CH – CH3 Ph O OH OH
Cl 2 Ph – C C – CH3 Excess
Cl2 D / Ni (Q ) 2(R ) CH3 – C CH (leq.) (leq.)
Cl HOCl (excess) Ph – C C – CH3 Ph – C – C – Cl O CH3
Ex.19
CH3 CH3 HOBr KOH CH3 – C – C – CHO CH3 – C – C CH (Excess) H H O – –
– –
– –
Ex.18
Ex. 20
Ex.21
Ex. 22
Page # 19
Ex. 23
Ex. 24
CH 2Cl2 + Br2
Ex. 25
CH 2Cl2 + Br2
Stereo Chemistry of addition to Alkene : Since Cis & Trans isomers form differnet products which are stereoisomers the reaction is stereospecific as well as stereoselective.
(F)
Addition of Hydrogen Halides (+ HX) to Alkene (a) General Reaction
+ H – X
Page # 20
Markovnikov’s Rule : When reagent (like: HX, H2O) adds to asymmetrical alkene eg, propene isobutene etc. the addition occurs such that the nucleophilic attaches itself to the carbon atom of the alkene bearing the least number of hydrogen and electrophilic adds to the sp2 carbon that is bonded ot the greater number of hydrogen. (b) Remarks : (i) Non stereoselective (ii) Markovnikob additon (iii) Rearrangement
A. Reaction in which two or more constitutional isomers could be formed but one of them predominates is called a regioselective reaction. Regioselective reaction can be 1.
Moderately Regioselective :-
eg.
2.
Highly Regioselective : -
CH3 – CH = CH – CH2 – CH3 + HCl Not Regioselective because intermediate is only sec. carbocation.
(G)
Addition of Hydrogen Halides (+ HX) to Alkyne –
–
– –
Br Br HBr (dark ) HBr (a) General Reaction R – C C – H R – C = C – H r R – C – CH 3 r1 2 Br H (MK)
–
–
+ + H R – C = CH2 R – CH – CH2 + R – C – CH3 r2 Br Br Br Vinyl halide Less stable more stable –
H+ (b) Mechanism:- R – C C – H r [R – C CH2 ] Br 1 Vinyl C
Br -
(c) Remarks:Page # 21
(1) MK Addition in both steps. (2) r2 > r1 criteria : carbocation stability I < II. (3) If two moles of HX one added the final product is Gemdihalide. (4) Electrophilic addition to terminal alkyne is regioselective. C=C
VS
CC
addition
(1) Reactivity order
C=C
>
–CC–
(2) C (Stability)
+ C–C
>
+ –C=C–
E (alkyl C + )
e– cloud is cylindrical and difficult to break
E (vinyl C + )
(3) For H2/Ni (R.O.)
C=C
<
–CC–
(4) Hhydrogenation (overall)
C=C
<
–CC–
<
–CC–
# Heat of hydrogenation per -bond ?
C=C (d) Reactivity Order:-
R – CH CH – R >
> R–C C–R
Free Radical Addition of HBr:- Kharasch Effect or Peroxide Effect General Reaction
e.g.
HBr / R O 2 / h R – CH2 – CH2 – Br R – CH = CH2 2
BrCCl3 / peroxides n C 6H13 CH CH2 n C 6H13 CH CH2 CCl3 1 Octene | Br 3 Bromo 1,1,1 trichlorononane
n C 6H13 CH CH2 1 Octene
HBr / R O
2 2 nC6H13CH2 – CH2 – Br
Mechanism : Steps:-
Remarks:(1) When HBr is added to an unsymmetrical alkene in presence of sunlight and peroxide. Then an Anti MK Addition Product is obtained. (2) It is a free radical chain reaction. (3) In presence of peroxide and sunlight Brº is formed in chain initiation step. (4) Brº forms more stable alkyl radical by homolysis of C = C bond. (5) In the last step alkyl radical abstracts Hº from HBr and AMK product is obtained.
Page # 22
Examples : 1. Ans.
Why do we get AMK product ? In the chain propagation Step (3), more stable alkyl radical (3º > 2º > 1º) is formed which gives the major product.
2.
If HF, HCl, HI are used even in sunlight electrophilic addition takes place. Why only HBr give free radical addition ? In chain propagation (iii) steps: R – CH = CH2 + X R – CH – CH2 – X + HR
Ans.
R – CH = CH2 + F R – CH = CH2 + Cl
R – CH = CH2 + Br R – CH = CH2 + I B.E.:- CF > CCl > CBr > CI
R – CH – CH2 – F R – CH – CH2 – Cl R – CH – CH2 – Br R – CH – CH2 – I
In step (iv) R – CH – CH2 – X + H – X R – CH – CH2 – X + H – F R – CH – CH2 – X + H – Cl R – CH – CH2 – X + H – Br R – CH – CH2 – X + H – I
exo exo exo endo
R – CH2 – CH2 – X + X + HR R – CH2 – CH2 – X + F + endo R – CH2 – CH2 – X + Cl + endo R – CH2 – CH2 – X + Br + exo R – CH2 – CH2 – X + I + exo
B.E.:- HF > HCl > HBr > HI Chain propagration steps indicate that addition of I is endo. is step 3 and is unfavourable. Addition of F, Cl is endothermic, unfavourable in step 4. Addition of HBr is energetically favourable [in both chain propagation steps-3 & 4]
–
– –
CH3 C HBr CH3 – C = CH2 – C–C–C ( dark ) Br
–
–
–
CH3 C HBr CH3 – C = CH2 – C – C – C (light ) Br
CH3 C HBr CH3 – C – CH = CH2 – C – C – C – C (R 2 O 2 ) Br C CH3 – –
– –
4.
# Light can also cleave HBr bond.
# R2O2 accelerates cleavage of HBr bond.
–
3.
HBr – (light ) HBr – ( dark )
–
– Br
HBr (R O ) Ph – CH = CH2 – 22 Ph – CH2 – CH2 Br
–
6.
Br
–
5.
– –
– –
CH3 C HBr CH3 – C – CH = CH2 – C–C–C–C ( dark ) CH3 Br C
Page # 23
–
HBr Ph – CH = CH2 – Ph – C – C Br
–
–
–
– –
7.
CH3 CH3 HBr CH3 – C – CH = CH2 CH3 – C – CH – CH3 CH3 Br CH3 HBr
8.
Br
–
H
9.
10.
HCl
Cl
–
11.
HBr
Me
Br –
CH3 CH3 – O – CH2 – CH = CH – CH – CH3
HCl (dry)
CH3 .. CH3 – O – CH – CH2 – CH2 – CH – CH3 –
12.
Cl
If water is taken, ether will be hydrolysed. Dry HCl is used.
13.
14.
+ HCl
Give the reactants (alkene) of the following products.
or
The planar carbocation (x) is responsible for non–stereoselectivity. The bromide nucleophile can attack from the top or bottom, leading to a mixture of stereoisomers. The addition is therefore a mixture of syn and anti addition.
Page # 24
17.
– –
–
HCl(leq.) CH2 = CH – CH2 – C CH CH3 – CH – CH2 – C CH Cl HCl(leq.) CH2 = CH – C CH CH2 = CH – C = CH2 Conjugated product Cl Conjugated System Neoprene
–
16.
Cl HCl(leq.) H – C = CH 2 H – C – CH3 Cl Cl gem dihalide Vinyl Chloride –
15.
HCl(leq.) H – C C – H
more stable product
(H)
Acid-Catalyzed Hydration
(i)
For Alkene Alkenes add water in the presence of an acid catalyst to yield alcohols. The addition takes place with Markovnikov regioselectivity. The reaction is reversible, and the mechanism for the acid-catalyzed hydration of an alkene is simply the reverse of that for the dehydration of an alcohol. The carbocation intermediate may rearrange if a more stable carbocation is possible by hydride or alkanide migration. Thus, a mixture of isomeric alcohol products may result. General Reaction
Mech. Step 1 :
+ H2 O
Protonation of the double bond forms a carbocation
+
Step 2 :
Nucleophilic attack by water
Step 3 :
Deprotonation to the alcohol
+
(a) Oxymercuration-Demercuration (ANTI – ADDITION) Alkenes react with mercuric acetate in a mixture of water and tetrahydrofuran (THF) to produce (hydroxyalkyl) mercury compounds. These can be reduced to alcohols with sodium borohydride and water. Oxymercuration
Page # 25
O OH O || | | || THF + H2O + Hg OCCH – C – C – + CH3 COH O 3 2 | | || Hg – OCCH3
General Reaction
OH | | –C–C– O + | | || Hg – OCCH3
| | + NaBH4 C C + Hg + | | HO H
(Demercuration)
In the oxymercuration step, water and mercuric acetate add to the double bond ; in the demercuration step, sodium borohydride reduces the acetoxymercury group and replaces in with hydrogen. The net addition of H – and – OH takes place with Markovikov regioselectivity and generally takes place without the complication of rearrangements.
Advantage over acidic hydrolysis (1) no requrement of acidic condition, which are harmful to many organic molecules. (2) No rearrangement occur. Mechanuism
There are two step
Step (A) : -
Step A is called oxymercuration i.e. addition of H2O & Hg(CH3COO)2.
Step (B)
NaBH4 OH–
NaBH4 converts the carbon mercury bond into a carbon- Hydrogen bond. Because the reaction result in the loss of mercury , it is called demercuration.
If solvent taken is alcohol product will be ether & process is called alkoxymercuration demercuration.
(b)
Alkoxymercuration - demercuration
General reaction
RO OR | | | | NaBH4 ROH + Hg(OAc)2 –C–C– C C | | | | H HgOAc (Markovnikov orientation)
Page # 26
(c)
Hydroboration-oxidation (SYN ADDITION) General Reaction
| | ROH + BH3.THF C C | | H B H | H
An alkene reacts with BH3 ; THF or diborane to produce an alkylborane. Oxiation and hydrolysis of the alkyborane with hydrogen peroxide and base yields an alcohol. (1) HBO leads to hydrogen of an alkene, however H2O is not a reactant. (2) During process of addition H comes from BH3 & OH comes from H2O2 (3) Net result is addition of H2O by anti markov. rule without rearrangement because no inermediate is formed. (4) There are two steps(a) Hydroboration: - Addition of borane on alkenes. In this reaction, the addition of electrophilic Borane & the nucleophilic H– take place in one steps i.e. it is concerted reaction. - No intermediate is formed.
In this way one BH3 react with three molecules o alkene successively to form trialkyl borane.
There are two reasons for the borane, alkylborane and dialkylborane to add to the sp2 carbon that is bonded to the most hydrogen . (A) To achieve most stable carbocation like T.S. (B) There is moree room at this carbon for the bulky group to attach itself i.e stearic factor.
Page # 27
(b)
Oxidation :- Replacement of B from OH ap. H2O 2 3ROH + BO 3– (CH3CH2CH2)3 B OR R3B 3 OH– Mechanism HOOH + HO–
HOO– + H2O
Re peat the two Pr eceding steps two time
3 ROH + BO33–
Re peat the three Pr eceding steps ROH + two times
.
(ii)
Acid-Catalyzed Hydration For Alkynes
G.R.
H SO / HgSO 4 R – C – CH3 R – C C – H 2 4 O
–
–
OH OH H SO 4 R – C = CH2 R – C = C – H 2 Hg + HgSO4 –
:
Hg 2 R – C = C – H Mechanism:- R – C C – H H2O: + Hg
O R – C – CH3
tauto.
Remarks:-
(1) Alkynes add one molecule of water. (2) The product enol tautomerises to a carbonyl compound (aldehyde or ketone) and further addition does not take place. (3) The reaction is catalysed by Hg+2 ions. (4) The product is M.K. Addition of water. (5) If HBO oxidation method is used, then AMK addition of water takes place. Important:- The product does not appears to be addition product. Arrange the following alkenes in order of their reactivity towards acid – catalysed hydration (a)
(b)
CH3 – (CH2)3 – CH = CH2 < CH3 – (CH2)2 – CH = CH – CH3 < H3 C (CH2 )2 C CH2 | CH3
H
H C=C
CH3 H3C cis-2-butene
(c)
<
H3C
H C=C
H CH3 trans-2-butene
CH2 – CH = CH – CH3 <
< H3 C C CH2 | CH3
CH = CH – CH2 – CH3 < H3C – C = CH – CH3
Page # 28
(Greater the stability of carbocation (intermediate), greater the hydration.)
EXAMPLES 1.
CH3 CH CH2 Pr opene
H O, H
2 CH3 CHCH3 | OH Isopropyl alcohol
2.
CH3 | CH3 C CH CH2 | CH3 3,3 dim ethyl 1 butene
CH3 H | | 50 % H SO 4 2 CH3 – C – C – CH3 | | OH CH3 2, 3-dimethyl-2-butanol (major product)
CH3 H
CH3 3.
D O/ D C CH3 2 CH3 C
CH3 C
C CH3
OD D
H
CH2
CH3
CH3
0
4.
CH3 C
+ H2SO4
CH3
5.
CH3 | H3 C C CH CH2 | CH3
8.
dil. aqueous H SO
4 2
H2 C CHCH2 CH3 1 butene
7.
CH3 C OSO3H CH3 t – Butyl hydrogen sulphate
3,3 dim ethyl 1 butene
6.
83 C H2O
CH3 C OH . CH3 t – butyl alcohol, bp 830C
H | H3 C C | CH3
CH3 | C CH3 | OH
+ H3 C C C CH3
2, 3 dim ethyl 2 bu tanol (major )
NaBH4
| | CH3 CH3 2, 3 dim ethyl 2 butene (min or )
OH | CH3 CHCH2CH3 2 butanol
CH2 CH CH2 CH3 1 butene
+ enantiomer + dialkyl-and trialkylborane
+ enantiomer
Page # 29
In the first step, boron and hydrogen undergo syn addition to the alkene in the second step, treatment with hydrogen peroxide and base replaces the boron with – OH with retention of configuration. The net addition of – H and – OH occurs with anti Markovnikov regioselectivity and syn stereoselectivity. Hydroboration-oxidation therefore, serves as a useful regiochemical complement to oxymercuration demercuration.
9.
C3H7CH = CH2
Hg(OAC)2 THF / H2O
NaBH 4 / NaOH C3H7CH(OH)CH2 – HgOAC C3H7CH(OH)CH3
(An organomercurial alcohol)
Hg(OAC)2
10.
NaBH
4
CH3OH
OCH3 OH
11.
12.
Hg( OAC )
2
NaBH4
O
NaOH
CH3 CH3 | | NaBH4 Hg(OAc )2 , H2O CH3 C CH CH2 CH3 C CH CH3 | | | CH3 CH3 OH 3,3 Dimethyl 1 butene 3, 3 Dimethyl 2 bu tan ol (No rearrangement )
13.
14.
CH3 CH3 CH3 + | | | H (dil. H2SO4) + + CH3 ¯ shift CH3 – C – CH – CH3 CH3 – C – CH = CH2 CH3 C CH CH3 | | | 3°carbocation CH3 CH3 CH3 2°carbocation (– H+ ) H2O CH3 OH CH3 | Hydroboration | | CH3 – C – CH2 – CH2OH oxidation | CH3 – C – CH – CH3 | CH3 CH3 OH CH3 Hg(OAc)2 | | CH3 – C – CH – CH3 NaBH4 | CH3
(i) Hydration with dil. H2SO4 proceeds via carbocation rearrangement (ii) Hydration with Hg(OAc)2, H2O, followed by NaBH4 proceeds via Markonikov’s rule. (iii) Hydration with (BH3)2 followed by H2O2 / OH¯ proceeds via Anti Markonikov’s rule.
Page # 30
15.
H
H 16.
BH / THF
H H2O2 / OHO
Me
3
BH2
Me
H
HBr Br
OH
H trans – 2 – cyclopentanol
1 – methylcyclopentene
17.
Me
BH3/THF
H SO OH 24
OH
H2O2/OH-
HBr
Br
18.
Hg2 / H2 SO 4 R–CC–H ( A ) (dil.)
BH3 – THF
R – CH2 – CHO
(B) H O / OH 2 2 19.
( A ) /( B ) H – C C – H CH3 – CHO
20.
Ph – C C – CH2 – CH3
(1) Hg 2 D 2 SO 4
(1) BH3 – THF
(2) D 2O 2 / OD
(1) BD 3 – THF (2) D 2 O 2 / OD
Common Sense :
D2SO4 – D2SO4 Present in D2O
OD – OD Present in D2O 21.
2
Hg / H 3 O Ph – C CH Acetophenone (Reactant)
OH | Ph – C = CH2 Retrosynthesis Ph – C CH Page # 31
Hg 2 / H SO
2 4
22.
23.
BH / THF
3 CH3 – C C – CH3
HOO
I / OH
2
+ CHI3
I / OH
2
NaOH / CaO ( s ) CH3 – CH2 – COOH Yellow ppt. CH3 – CH3
Alkadines (conjugated) : E+ addn :
Mechanism :
Page # 32
With a non-classical C+, the product is always a trans-alkene (anti-addition).
Ex. HBr
(1)
H2C = CH – CH = CH2 (r , t )
(trans) > (cis) .........classical.
Page # 33
HCl (1eq)
(2)
(r , t )
+
+
+
(1,2) (1, 6) > (1, 4) > (1, 2).
(3)
HCl H3C – CH = CH – CH = CH2
Alkenes = (I/I) ~ II > II. ...........on the basis of alkene stability.
HOX
(4)
6.
Miscellaneous
(A)
Addition of carbenes to Alkenes :
1eq
Methylene is the simplest of the carbenes : uncharged, reactive intermediates that have a carbon atom with two bonds and two nonbonding electrons. Like borane (BH3), methylene is a potent electrophile because it has an unfilled octet. It adds to the electrons rich pi-bond of an alkene to form a cyclopropane.
General Reactions Heating or photolysis of diazomethane (CH2N2) gives nitrogen gas and methylene. .. . . : N N CH2 : N N C H2 diazometha ne
Heat or ultraviole t light
H
N2 + :C
H Methylene
There are two difficulties with using CH2N2 to cyclopropanate double bonds. First, it is extremely toxic and explosive. A safer reagent would be more convenient for routine use. Second, methylene generated from CH2N2 is so reactive that it inserts into C – H bonds as well as C = C bonds.
Structure :
Page # 34
(a)
H2C
H2 C
(b) (c) (d) (e)
Singlet sp2/120º/bent Diamagnetic 6e– /e– deficient (electrophilic)
triplet paramagnetic (electrophilic)
Carbene is an E species. So it gives E addition reactions with alkenes/ alkynes CH N /
2 2
ex.
CH – CHCO /
3 CH3–CH=CH2
(B)
Addition of NOCl NaCl
CH3 – CH = CH2
(C)
Halogenation. Allylic substitution
General Reaction
X2 = Cl2, Br2
NBS = N-Bromosccinimide
(D)
NCS = N-Chlorosuccinimide
Alkylation : General Reaction
| | | | acid C C + R – H C C | | H R
Examples : 1.
.. C H2 N N: sh
Page # 35
Propene +
+
+
2.
Cl2 , 600C CH3 CH CH2 Cl CH2 CH CH2 Pr opylene (Pr opene )
3.
4.
Allyl chloride (3 Chloro 1 propene )
CH3 CH3 CH3 CH3 | | | | H SO 4 CH3 C CH2 + CH3 C H 2 CH3 C CH2 C CH3 Isobutylene | | | CH3 H CH3 Isobu tane 2, 2, 4 Trimethylpen tan e
(a)
CH3 CH CHCH3 + 2 Butene
light
Diazometha ne
NaOH / H2O + CHBr3
(b)
5.
CH 2N2
Show how the insertion of methylene into a bond of cyclohexene can produce the following. CH3
(a)
(b)
(c) norcarane, CH3
6. Ans.
Assertion (A):- Propene (CH3CH = CH2) undergoes allylic substitution. Reason (R) : CH2 = CHCH2 (allylic) free radical is stabilised by resonance. (A)
Br 7.
+ NBS Cyclohexene
1.
3-Bromocyclohexene
F.R Allylic substitution : H2C–CH=CH2
Br / r.t.
2 (1)
........... Eaddition (or F.R. addition)
Page # 36
Br / 400 º–500 ºC
2
.......... Allylic substitution
(2)
NBS / R O
2 2
(3)
X2 = Br2/Cl2 #
X2 [H –CH=CH ] + HX 2C 2
(2)
(500 ºC) – X2
............... (Chain reaction continues) At high temperature, an allyl free radical can be formed
#
NCS/NBS ...... N- Bromosuccinimide
(NBS)
(i)
R–O–O–R 2 R–Oº
(ii)
R–Oº + H–Br R–O–H + Brº
(iii)
(rds ) H2C–CH=CH2 H2C–CH=CH2CH + HBr Br º
(iv)
(v)
+ H–Br
+ Br2 (low conc.)
H2 C –CH=CH2 + Br2
+ Brº
NBS is used to provide very slow but steady supply of Br2 . If Br2 is used in excess then addition reaction takes place but if NBS is used then in one step only 1 molecule of Br2 is formed which is insufficient to give addtion product so the substitution product is the major product.
#
Addition always takes place when Br2 is in excess. Halolactonisation : [unsaturated acid + X2 + basic med.] NaHCO
3 H2C = CH – CH2 – CH2 – CH2 COOH
X2
Page # 37
#
Hydrolysis of vinyl ethers : Vinyl ethers = Alkoxy alkenes = Highly reactive alkene (Eaddition)
H
+ HOR
18
H3 O
Mechanism
H
H
H3C–CH=O18 ex.
– H
An alkene X (C5H10O) on hydrolysis with dil H2SO4 produces Y and Z. Both give iodoform test (+ve) X is 1015 times more reactive than ethene toward E
(I) (X)
(Y)
(Z)
(II) (X)
(Y)
(Z)
(I)
Oxidation Reaction
(1)
Oxidation of alkene & alkyne by acidic KMnO4 When alkene & alkyne heated with KMnO4 in acidic or in alkaline medium ; following changes takes place. [O] Terminal = CH2 CO2 + H2O Page # 38
[O] Terminal CH CO2 + H2O [O] Non Terminal alkene = CHR RCOOH [O]
=
[O] C – R RCOOH
(2)
Ozonolysis : Like permanganate ozone cleaves double bonds to give Ketones and aldehydes. However, ozonolysis is milder, and both Ketones and aldehydes can be recovered without further oxidation. O
R
R
R' C
(a) General Reaction
C
+ O3
H
R
R' C
C
H
R O
O
Ozonide (CH3)2S
R C
O+ O
R' C H
R Ketone (1) O 3 ( 2 ) H2 O
+ R'COOH (Oxidative ozonolysis)
(1) O 3 (2 ) Zn / H2 O or CH3 COOH or Me 2 S / H2 O
+
(1) O3 (2) LiAlH4/NaBH4
(reductive ozonolysis)
+ R'CH2OH (reductive ozonolysis)
(b) Mechanism
C
O
O
O
O
C
-
O+
O
C O
Molozonide (Primary ozonide) C
C
O
O
O
C
O
C
O
-
O+
Ozonide
Page # 39
O
R
R' C
C
H
R O
R
CH3 - S - CH3 dimethyl sulfide
C H
R
O
O
R'
O+ O
C
+ CH3 S
CH3
dimethyl sulyoxide (DMSO)
(3)
Ozonolysis of alkyne : Alkynes gives acid by ozonolysis ozonolysis of alkyne required neither oxidative nor reductive work up - it is followed only by hydrolysis. carbon dioxide is obtained from the CH gp. of a terminal alkyne.
(1) O
3 CH COOH + CH CH COOH CH3 – C CCH2 CH2 ( 3 3 2 2) H2O
(1) O 3 CH3CH2 – CH2 – C CH CH3CH2CH2COOH + CO2 (2) H2O
Comparison of permanganate cleavage and ozonolysis : Both permanganate and ozonolysis break the carbon–carbon double bond and replace it with carbonyl C =0 groups. In the permanganate cleavage, any aldehyde products are further oxidized to carboxylic acids. In the ozonolysis-reduction procedure, the aldehyde products are generated in the dimethyl sulfide reduction step, and they are not oxidized.
Examples : CH3
CH3
O
O
O
O
CH3
Conc. KMnO4 - OH
1. H
H
O
(not isolated) CH3 CH3
2. H
O
(i) O3 (ii) (CH3)2 S
O H
3.
O3
+
KMnO4/OH -
+ R'COOH
KMnO / H 4
+ R'COOH
K Cr2O 7 / H / 2 or, CrO3 / H /
+ R'COOH
dil.KMnO 4 cold
HIO 4
+
Page # 40
O O
Lemienx Reagent (KMnO4 + NaO4)
4.
OH OH
H H
HO4
H HIO 4 no reaction OH
OH H O (1) TFPAA 2
5.
H
7.
H H
( 2) OsO 4 H O 2
6.
OH HIO 4 OH
O O
How will you distingnish cis, trans cyclohexane-1, 2-diol. H H
O + AgIO ppt. 3 O
OH HIO 4 / AgNO3 OH
cis H OH
OH H
HIO / AgNO3 4 No reaction
trans (i) O3 (ii) (CH3)2 S
8.
CH3CH2CHO + CH3(CH2)4CHO (65%)
3 - nonene
H
H (i) O3 (ii) (CH3)2 S
9.
H
C H
CHO
H
CHO
H O +
C H
(i) O3 (ii) (CH3)2 S
10.
H
O
CHO
CHO
O
(ii)
Hydroxylation Reactions of alkenes
(a) General Reaction
(b) Mechanism:(1) With Bayer's Reagent:-
Page # 41
2H 2O
+ H3MnO 4 ( 5 ) (colorless )
(2) With OsO4/H2O/NaHSO3:-
R – CH = CH – R
R – CH – CH – R
O
O
O
O Os
Os
(Cyclic osmate ester)
+ H2OsO 4 ( 6)
O
O
O
(48)
2H 2O
O
(3) Epoxidation/Hydrolysis:-
RCO3H = Peroxy acid HCO3H = PFA (Performic acid) CH3CO3H = PAA (Peracetic acid) Ph – CO3H = PBA CO H – 3 –
= MCPBA (Metachloro perbenzoic acid) Cl CF3CO3H = TFPAA (Trifluoro peracetic acid)
........... E addition + R'COOH
R – CH = CH – R .. O 18
O
H
C
O
(iii)
R
Epoxidation of Alkenes : An alkene is converted to an epoxide by a peroxyacid, a carboxylic acid that has an extra oxygen atom in a – O – O – (peroxy) linkage.
(a) General Reaction
C=C
O || +R– C –O–
–H
O || + R – C – O – H (acid)
The epoxidation of an alkene is clearly an oxidation, since an oxidation, since an oxygen atom is added. Peroxyacids are highly selective oxidizing agents. Some simple peroxyacids (sometimes called peracids) Page # 42
and their corresponding carboxylic acids are shown below :-
(b) Mechanism
(iv)
R
O
C O C
C
O H transition state
+
Oxidative cleavage of Alkenes (a) Cleavage by permanganate In a KMnO4 hydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur. Mixtures of Ketones and carboxylic acids are formed, depending on whether there are any oxidizable aldehydes in the initial fragments. A terminal = CH2 group is oxidized to CO2 and water. R
R' C
General Reaction
C
R
*
H
KMnO4 Warm conc. Glycol
R
R' O
C
+
H
R'
C
O O
OH
R
Aldehye (Oxidizable)
C
Ketone (Stable)
The same reagents that oxidize alkenes also oxodoze alkynes. Alkynes are oxidized to diketone by a basic solution of KMnO4 at room temp. KMnO 4 dil CH3 – C CCH2CH3 – OH
Examples
1.
+
Bayer ' s Re agent
2.
OsO 4 H2 O / NaHSO 3
3.
Me
OH
H H
Me H OH OH
–
Me
4.
OH
(1) PBA 18 (2) H 2O /OH -
Page # 43
–
H
R (1) OsO 4 (2) H2O H
–
–
R
R
R – CHOH – CHOH – R (d / mix )
H
C=C –
–
–
H
R
C=C –
–
R 5.
R H
R
H(meso isomer) OH OH
–H Bayer ' s C = C (d / mix ) – H R (1) Re agent
–
6.
–
R
–H Ag / O 2 / C = C (meso isomer) H– ( 2) H 2 O R
–
–
7.
What is the main utility of this reaction and why is it superior to KMnO4 cleavage for this purpose ?
Sol.
It locates the position of C = C's in molecules. KMnO4 cleavage is more vigorous and can oxidize other groups, i.e. OH.
8.
Give the products of the following reactions:(i)
+
KMnO 4 (aq. bassic )
(ii)
or Cis-1, 2-Cyclopentanediol OsO 4 in H2 O 2
KMnO (aq.) 4 mesoor OsO 4 in H2O 2
(iii)
KMnO (aq.) 4 rac-CH3CHOHCHOHCH3 or OsO 4 in H2O 2
9.
(i)
(ii)
10.
(a)
(b)
hot KMnO 4 (or K 2 Cr2 O 7 )
hot KMnO 4 (or K 2 Cr2 O 7 )
not KMnO 4 (or K 2 Cr2 O 7 )
not KMnO 4 HOOCCH2CH2COOH (or K 2 Cr2 O 7 ) Page # 44
O
KMnO4 Warm cond.
11.
+
13.
C
O
KMnO4 Warm cond.
12.
OH
+
COOH
CO2
COOH
O
+
+
Epoxycyclohexane 14.
Predict the products, including stereochemistry where appropriate, for the m-chloroperoxy-benzoic acid epoxidations of the following alkenes. (a)
H CH3
H
H C=C CH2CH2CH3
(c) Cis-cyclodecene
(b)
CH3
CH2CH2CH3 C=C H
(d) Trans-cyclodecene
Page # 45
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