1. Dehydrohalogenation of Haloalkanes(Alkyl...

1

Chemistry of ALKENES

Dr. S. S. Tripathy

ALKENES

General Methods of Preparation:

1. Dehydrohalogenation of Haloalkanes(Alkyl halides):

* E2 β-elimination of HX from alkyl halides occurs giving more substituted alkene as themajor product(Saytzeff’s rule) and less substituted product as the minor product(Anti-Saytzfeffor Hoffman elimination). Hence elimination is always regioselective, as both the products areformed with one being the dominant product.

* The order of reactivity : 30 halide > 20 halide > 10 halide* Dehydrohalogenation of alkyl halide giving alkene major product, takes place by E2

mechanism. Anti-periplanarity(the two leaving groups to remain anti w.r.t each other in the TS)is the stereoelectronic requirements for E2 mechanism. Refer GCOC-III for details.

* Strong bases like OH–(NaOH/KOH), OR–( NaOEt type), CN–(KCN) etc. are used toabstract a β-proton for E2 reaction. Weak bases like H

2O, ROH, NH

3 etc. cannot abstract a β-

proton efficiently and hence do not favour elimination reaction.* Usually alcoholic KOH is used to bring about elmination of HX in tert- and sec-

halides.* Primary halides with the above strong bases, give substitution as the major product(S

N2

mechanism. However, strong bulky base like t-BuOK is used for 10 halide to get elimination asmajor product.

* For for 30 and 20 halides, if elimination can occur in more than one directions, thenSatyzeff product is the major one. However, by using strong bulky base like t-BuO–(t-BuOK),anti-Saytzeff product becomes major.

* 30 halide gives nearly 100% elimination product while 20 halide gives elimination asthe majpr product and substitution(S

N2) as the minor product.

CH3 C

CH3

CH2

H

Br

OH(alcoholic)CH3 C

CH3

CH2

100%E2 (no substitution product)

CH2 C

CH3

CH

H

Br

CH3

HOH(alcoholic)

E2 CH3 C

CH3

CH CH3 CH2 C

CH3

CH2 CH3+major minor

(no substitution product)

CH3 CH

Cl

CH2 CH3alc. KOH CH3 CH CH CH3 CH2 CH CH2 CH3

CH3 CH

OH

CH2 CH3

+

major minor

E2 : MAJOR

SN2: Minor

(trans+cis)

+

20 halide

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Dr. S. S. Tripathy

You found that in the first example, there is one product i.e elimination, as it is a symmetricalhalide(tert-butyl bromide). The 2nd example(tert-pentyl bromide) also gives only eliminationproducts, with Saytzeff product(more substittuted) as major( > 80%). No substitution productsare formed in these cases as both are tert- halides.In case of a the third example(sec-butyl chloride), which is a 20 halide, both elimination andsubstitution products are formed, with the former being the major one(>80%). E2 elminationgives more than 80% but-2-ene as a mixture of trans(E) and cis(Z) isomers out of which transisomer is major(more stable). S

N2 gives the minor product butan-2-ol(<20%) which is a chiral

molecule exist as either d or l, as SN2 involves inversion of configuration. So there are altogether

3 products(including stereoisomers) namely trans and cis-but-2-enes and d or l- buta-2-ol.* If strong but bulky base like t-BuO–(t-BuOK) is used, a 20 halide gives anti-

Saytzeff(Hoffman elimination) as the major product among the elimination products.

+CH3 CH CH

Br

CH2

HH

O C

CH3

CH3

CH3

(t-BuOK)CH3 CH2 CH CH2 CH3 CH CH CH3

but-1-ene(major)

but-2-ene(minor)

In the above example, the very minor substitution product(sec-butyl tert-butyl ether) has not beenshown. S

N2 attack to a 20 halide is very less probable and is almost non-existent.

Primary halides:

KOEt(alc.)CH3 CH2 CH CH2 Br

HOEt

CH3 CH2 CH2 CH2 OEt CH3 CH2 CH CH2

SN2 product(major)90%

E2 product(minor)10%

+

10 halide

In case of 10 halide like n-butyl bromide, the major product with a strong base like OEt–, OH,CN– etc. is substitution(S

N2) and the minor product is elimination(E2). But for obtaining elimination

as the major product, strong bulky base like t-BuO–(t-BuOK) is used, which does not favour SN2

attack due to steric hindrance.

CH3CH2CH2CH2Br(CH3)3CO-K+

CH3CH2CH CH2 CH3CH2CH2CH2OC(CH3)3but-1-ene(E2; 85%)

n-butyl t-butyl ether(SN2; 15%)

+

N.B: Often in school level chemistry, it is told that alcoholic KOH gives elimiation product andaqeous KOH gives substitution product. Even a school level student will say that 1-bromobutaneon treatment with ethanolic KOH will give but-1-ene and with aqueous KOH will give ethylalcohol. But this is a wrong. Alcoholic(ethanolic) solution is made to enhance the S

N2 rate(less

polar than water) and make the organic substrate soluble(as most substrates are sparinglysoluble in water. Even with alcoholic KOH, 1-bromobutane will give butan-1-ol as major product,not but-1-ene. Yes, with 20 and 30 halide, the major product will be elimination with alcohlicKOH. In these cases too, use of aq. KOH will give elimination as major products, not thesubstitution product. Only using a weak base like H

2O, ROH(solvolysis reaction), the major

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Dr. S. S. Tripathy

product becomes substitution for 30 and 20 halides by SN1 mechanism. 10 halide will be very slow

with weak bases as both SN1 and SN2 are not favoured. In any case, there will be negligibleelimination product in this case. (Refer GCOH-III for details).SAQ: Give the products indicating the mechanism.

(a) 2-methyl-2-bromobutane reacts with KCN(alcoholic).(b) Ethyl bromide reacts with t-BuOK(alcoholic).(c) 1-bromopentane with KOH(alcoholic)(d) Tert-butyl chloride reacts with sodium ethynide.(e) n-butyl bromide reacts with sodium ethynide

SAQ: Show how many products will be obtained(include stereoisomers) when (1R,2S)1-bromo-2-methylcyclohexane is treatd with EtOH ?Solution: 6 products.

CH3

BrS

R EtOH

CH3

H

E1

CH3

- HCH3

E1

EtOH

- H CH3

OEt

(1R,2S) and (1S,2S)2 diastereomers

S

R/S

SN1hydride shift

CH3CH3

E1

(repeated)

CH2

E1CH3 OEt

SN1 EtOH

- H

Since the substrate is a 20 halide, and we are using a weak base EtOH, the major product is theS

N1 product(both the ethers) and the minor product is the E1 product(all elimination products).

Since we took a fixed stereoisomer(1R,2S) and since the configuration of C-2 will remain thesame, two diastereoisomers will be formed from the carbocation as the C-1 will have 50% R and50% S.

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Dr. S. S. Tripathy

(2) Dehydration of Alcohols:

* When a mixture of alcohol(containing at least 2 carbon atoms) and acid catalyst likeH

2SO

4 or H

3PO

4 is heated and distilled, alcohol gets dehydrated to form alkene. Acid acts as

catalyst.

C

H

C

OHH

+

heatC C H2O+

alcohol alkene

* The vapours of alcohol also on passing over heated alumina(Al2O

3) gets dehydrated

to form alkene.

Acid Catalysed Dehdyration:* Reactivity Order : 30 alcohol > 20 alcohol > 10 alcohol

(same order as for the carbocation stability)* 10 alcohol will require conc. H

2SO

4 for dehydration. In this case, the reaction is

complicated by the formation of unwanted byeproducts. Since conc. H2SO

4 is a strong oxidising

agent, it will oxidise some alcohol to CO2 and itself get reduced to SO

2, in addition to the

formation of some solid carbon(charring reaction). In such case, the gaseous mixture is passedthrough caustic soda(NaOH) solution to remove SO

2 and CO

2 from alkene.

* 20 and 30 alcohols dehydrate in prensence of dil. H2SO

4 and do not encounter with the

formation unwanted byeproducts.* HEATING of the mixture is absolutely required for the removal of a proton from the

β-position w.r.t the –OH group. If the mixture is not heated, the alcohol remains in the form ofa salt (alkyl hydrogen sulfate) due to esterification with acid.

R C OH H OSO3H R C OS3H

sulfuric acid alkyl hydrogen sulfate

H2O-alkene H2SO4+

* Regioselectivity: When β-H atoms are available on either side of –OH, then the majorproduct is formed according to Saytzeff’s rule.

* Stereoselectivity: If the alkene formed exhibits geometrical isomerism, then it will beformed as a mixture of E and Z forms, with one dominating the mixture. For simple alkenes likebut-2-ene, E is more stable and abundant than Z. [ E(trans) > Z(cis) ].

* Mechanism:For 10 alcohol, E2 and for 20 and 30 alcohols E1 mechanism operates.

* For preparing alkene, H2SO

4 is taken in excess of alchol. If alcohol is taken in excess,

then ether is formed instead of alkene at a lower temperature that that required for alkeneformation. S

N reaction occurs in steat of Elimination.

* Since the reaction goes via carbocation formation, often rearranged alkenes are formeddue to hydride or alkyl 1,2-shift.

* Acids like HCl, HBr, HI are not used for dehydration, as there will be some substitutionproducts proeduced in such case. The halide ions from these acids are good nucleophiles formessing up the reaction.

Examples:(1) A mixture of ethyl alcohol and excess conc. H

2SO

4 is heated at 1700C ethene gas is

produced which can be collected by the downward displace of water. Ethylene gas is contaminated

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Dr. S. S. Tripathy

with some CO2 and SO

2 gases(already discussed before) which are separated by passing the

mixture through caustic soda, which removes the acidic gases.

CH2 CH2

OHH

1700C

H2SO4(conc.)CH2 CH2 H2O+

Caution: If excess ethyl alcohol is taken, the product obtained is diethyl ether (in stead of ethene)at 1400C due to dehydration between two ethanol molecules.

1400C

H2SO4(conc.)+CH3 CH2 OH H O CH2 CH3 C2H5 O C2H5 H2O

(2) Propan-2-ol is dehdrated with excess conc. H2SO

4 at 1000C to form only one product

i.e prop-1-ene. Since it is 20 alcohol, lower temperature of 1000C is required.

CH3 CH

OH

CH2

HH2SO4(conc)

1000CCH3 CH CH2 H2O+

N.B: Propan-1-ol also will give the same alkene(propene) but since it is a 10 alcohol will needhigher temperature(1700C).

(3) Butan-2-ol on dehydration will give a mixture of but-2-ene( trans and cis) being themajor product and but-1-ene, the minor product. In total three products are obtained.

CH CH

OH

CH2

H

CH3

HH2SO4(conc)

CH3 CH CH CH3 CH3 CH2 CH CH2major(trans and cis) minor

+

(4) Butan-1-ol also on dehydration gives the same mixture as given by butan-2-ol. Onydifference is, a higher temperature of dehydration is required for this. This is due to rearrangementof a less stable 10 carbocation to a more stable 20 carbocation by hydride shift. But-2-ene alsois the major product here. We shall explain this below through mechansim.

CH2 CH2 CH2

OH

CH3H2SO4(conc)

CH3 CH CH CH3 CH3 CH2 CH CH2major(trans and cis) minor

+

(5) Tert-butyl alcohol on dehdyration gives one alkene i.e 2-methylprop-1-ene(isobutene),since it is a symmetrical alcohol

(6) However, tert-pentyl alochol(2-methylbutan-2-ol) on dehydration gives two alkenes,one Saytzeff product(2-methylbut-2-ene) and the other anti-Saytzeff product(2-methylbut-1-ene)..

H2SO4(conc)CH3 C

CH3

CH2 CH3

OH

CH3 C

CH3

CH CH3 CH2 C

CH3

CH2 CH3+major minor

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Dr. S. S. Tripathy

5) Neopentyl alcohol, on dehydration gives the same products as given by tert-pentylalcohol ( i.e 2-methylbut-2-ene as the major product along with 2-methylbut-1-ene as minorproduct). Even though the molecule does not have a β-H atom w.r.t –OH group, a methyl shiftduring rearrangement of carbocation does this magic.

H2SO4(conc)CH3 C

CH3

CH3

CH2 OH CH3 C

CH3

CH CH3 CH2 C

CH3

CH2 CH3+major minor

Detailed Mechanism:(A) For 30 and 20 Alcohols: (E1 mechanism).

CH3 C

CH3

CH3

OHslow

30 stable carbocation

HCH3 C

CH3

CH3

OH

HCH2 C

CH3

CH3

H fast

H3O+-

CH2 C

CH3

CH3

H2O

fastheat

You know that protonation of –OH enhances its leavability as the leaving group now is a weakbase, H

2O which is a good leaving group.

The detailed explanation given in GOC-III(mechanism).(B) For 10 alcohol like Ethanol (E2 mechanism)

+ +CH3 CH2 OH H+CH2 CH2

H

OH H

H2SO4 HO3SO-

-CH2 CH2 H2O

H2SO4-HO3SO-

Since 10 carbocation is unstable, one step E2 elimination is the mechanism operating in case ofethanol dehydration.

N.B: In other 10 alcohols like propan-1-ol or 2-methylpropan-1-ol(isobutyl alcohol) etc. althoughonly one product is possible from them, there is scope for hydride shift to form more stablecarbocations. Hence the mechanism should be E1.

Mechanism for dehydration of some other Alcohols:Dehdyrationof butan-1-ol:

CH2 CH2 CH2

OH

CH3

+

H+

CH2 CH CH2

OH2

CH3

H

CH3 CH CH CH2

H

HH2O-

CH3 CH CH CH3 CH3 CH2 CH CH2

(cis and trans)but-2-enemajor

but-1-ene(minor)

H+-

hydride shift

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Dr. S. S. Tripathy

The alcohol is protonated first. In the 2nd step, which is slow and rate determining, a hydrideshift followed by dehydration occurs to form a stable 20 carbocation. In the 3rd step, deprotonationoccurs from the adjacent carbon atoms by heating to form both Saytzeff(major) and anti-Saytzeff(minor) products. Note that H

2O, as a base, is resonsible for the removal of proton in the

form of H3O+ in the last step, which has been shown as -H+ for simplicity.

N.B: Student to note that, often in different texts it is given that , a 10 carbocation is formedwhich then undergoes rearrangement by a hydride shift to form a more stable 20 carbocation. Buti do not believe that to be correct concept. Hydride shift and dehydration processes take placesimultaneously to form a stable 20 carbocation. A 10 carbocation is NEVER formed.

Dehydration of neopentyl alcohol:

+

H2O-

H+-

CH3 C

CH3

CH3

CH2 OH CH3 C

CH3

CH3

CH2 OH2CH3

- shiftCH2 C

CH3

CH CH3

H H

CH3 C

CH3

CH CH3 CH2 C

CH3

CH2 CH3

major(Saytzeff product)

minor(Anti-Saytzeff)

Here also i have shown the CH3– shift and dehydrtion occuring simultaneously to form stable 30

carbocation.

SAQ: (1)Give the mechanism for the acid catalysed dehydration of propan-1-ol and isobutylalcohol. (You do these yourself).(2) How many alkene products will be formed(including stereoisomers) by the acid catalyseddehdyration of 2-ethylpentan-1-ol. Give mechanism.Solution: 3 + 2 = 5 alkene products will be formed. Work out the mechanism involving carbocationrearrangement yourself.

CH3–CH

2–CH

2–C(CH

3)=CH–CH

3(E and Z) + CH

3–CH

2–CH=C(CH

3)–CH

2–CH

3(E and Z)

+ CH3CH

2CH

2–CH(C

2H

5)=CH

2 = 5 alkenes

(3): Give the dehydration product(s) for (a) cyclohexanol (b) 1-methylcyclohexan-1-ol.Solution:

(a)

OH

H3PO4(conc.)

cyclohexene

(conc. H3PO

4 is often used for cycloalkanols)

(b)

OH CH3

H+

CH3 CH2

1-methylcyclohexene

methylidenecyclohexane

+

(major) (minor)

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Dr. S. S. Tripathy

Mechanism for formation of Ether with excess Alcohol:

CH3 CH2

OHH+

CH3 CH2

OH2HO Et

CH3 CH2 O Et

HH2O-

Et O Et-

SN2

Since alcohol is in excess, free alcohol, as nucleophile to carry out SN reaction to form ether. In

case of ethanol, a SN2 mechanism is proposed. But for other alcohols were stable carbocation(20

or 30) can be formed, SN1 mechanism will operate for the formation of ether.

SAQ: Write the products formed by the dehydration of 1-(1-methylcyclopentyl)ethanol. Alsopredict the major product.

OH

Hint: Ring expansion gives cyclohexene Saytzeff product as major.

(3) Dehalogenation of Vicinal dihalides

* When the substitutents occupy adjacent carbon atoms, it is called vicinal(vic-) compoundeg. vicinal dihalide, vicinal diol etc. If the two are bonded to the same carbon atom then it iscalled geminal(gem-) compound.

* When vicinal dihalide reacts with Zn/EtOH(or MeOH) or Zn/CH3COOH or KI/acetone,

E2 elimination (dehalogenation) occurs to form an alkene.

R CH

X

CH

X

R'Zn/EtOH

R CH CH R' ZnX2+

Mechanism:* With Zn, first Zn(0) inserts into one C–Br bond to form an organozinc intermediate.

This undergoes E2 elimination to form alkene.

C C

Br

Br

Zn

insertionC C

ZnBr

Br

C C ZnBr2+

organozinc intermediate

* With NaI/acetone, I– does abstracts X+ and the other leaves as X– in the E2 mechanism.

C C

Br

Br

C C + +

I

KI

acetoneIBr K

+Br

-

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Dr. S. S. Tripathy

*Stereospecificity of E2 elimination is maintained here. If we take meso-2,3-dibromobutane,on dehalogenation will give trans-but-2-ene and an active isomer will give cis-but-2-ene. Referthe GOC-III(mechanism) for details.Examples:

(i) +

Zn/EtOHCH2

Br

CH2

Br

CH2 CH2 ZnBr2

1,2-dibromoethane(ethylene dibromide)

(ii) +CH

Br

CH

Br

MeMemeso-2,3-dibromobutane

NaI/acetoneC C

CH3

CH3

H

H

IBr NaBr+

trans-but-2-ene

(3) Dehydrogenation of alkanes:Though it is not possible at ordinary temperature(ΔG0 = +ve ), but at high temperature

of 5000C, it takes place (ΔG0 = –ve). Alkanes are dehydrogenated to form alkenes and H2

alongwith other products. This is a thermal cracking reaction. For higher alkanes, a mixture ofsmaller alkanes and alkenes are obtained. Refer the “Thermal Cracking” in the chapter “Alkanes”for details.

(4) Kolbe’s Electrolytic Method:* This is similar to the Kolbe’s electrolytic method for preparation of alkanes.* Ethene is prepared by the electrolysis of sodium/ postassium succinate soution(salt of

dicarboxylic acid). Higher alkenes cannot be prepared by this method.

H2C CH 2 CC

O

OK

O

KOpotassium succinate

H2C CH 2 CC

O

O-

O

-O + 2 K+

H2C CH 2 CC

O

O-

O

-O H2C CH 2 + 2CO2 + 2e

At anode

Mechanism:At anode:

C

O

O CH2 CH2 C

O

O C

O

O CH2 CH2 C

O

O + 2ediradicalsuccinate ion

C

O

O CH2 CH2 C

O

O CH2 CH2 CO22+

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Dr. S. S. Tripathy

At anode, succinate ion loses two electrons(one from each carboxylate ion) to form a diradical,which undergoes decarboxylation with one-electron-shift processes in one step to form etheneand 2 moles of CO

2. Here there is no report of the formation of any other bye products, which

indicates that no free carbon radicals are formed in this case, as we found in the Kolbes’s processfor preparing alkanes from salt of monocarboxylic acids.At cathode:

H2 is liberated at cathode. (equation same as given in ‘alkanes’)

(5) Controlled hydrogenation of Alkynes:* Alkynes react with H

2 in presence of poisoned catalyst like Lindlar’s catalyst(Pd-CaCO

3

or Pd-BaSO4 poisoned by lead acetate and quinoline) to form alkenes. The mechanism of addition

is SYN. For details, refer the GOC-III(mechanism).* Ni

2B (P-2 catalyst) can also be used for controlled addition of H

2 with SYN mechanism.

R C C R' H2Pd CaCO3

lead acete/quinoline

C C

R

H

R'

HCis(Z)alkene

/

* Alkynes are hydrogenated with ANTI steroselectivity by Na/liq. NH3. For details refer

GOC-III(mechanism).

R C C R' C C

R

R'

H

HTrans(E)alkene

Na/ l iq. NH3

[2H]

(6) Wittig Reaction:* Organophosphorous ylides(Wittig reagent) reacts with an aldehyde or a ketone to form

a higher alkene. In this method, the position of C=C remains the same as the C=O in carbonylcompound and hence is one of the best methods of preparation of alkene. In other methods likedehydration of alchols or dehydrohalogenation of alkyl halides we get a mixture of isomericalkenes, which is not the case here.

* Ylides are compounds having opposite charges on adjacent atoms each of which is acompletely filled octet. In P-ylides, phosphours atom will carry +ve charge and the adjacent C-atom will carry –ve charge.

Ph3P CH R Ph3P CH R

Phosphorous Ylide

Alkylidenetriphenylphosphorane

Alkylidenetriphenylphosphorane is a P- ylide, in which P atom carries +ve charge and theadjacent C-atom carries –ve charge. This has another resonance structure in which there is a P=Cdue to lateral overlapping of a filled C-orbital with a vacant d-orbital of P- atom. The dipolarion(Zwitterion) is more stable and more contributing to the hybrid. However, while showing thereaction, sometimes we shall take the P=C structure. The suffix ‘idene’ is given for indicatingthe divalency of the carbon bearing group. P- ylides can be very easilty prepared from alkylhalides and tripheynylphosphine(Ph

3P) (just wait for amoment).

* Wittig reaction is the reaction of the phosphorous ylide with aldehyde or ketone to formalkene(both trans and cis) and triphenylphosphine oxide.

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Dr. S. S. Tripathy

Ph3P CH R

R'

C O R''aldehyde/ketone P-Ylide

C

R'

CH R

R''

Ph3P O+ +

alkene triphenyl phospine oxide

Example:

(1)

Ph3P CH CH3

CH3

C O CH3 P-Ylide

C

CH3

CH CH3

CH3

Ph3P O+ +

acetoneethylidenetriphenylphosphorane

2-methylbut-2-ene

(2)

Ph3P C CH3

CH3

C O H

C C CH3

H CH3

Ph3P O+ +

benzaldehydeisoproylidenetriphenylphosphorane

2-methyl-1-phenylprop-1-ene

N.B: Cyclic or aromatic aldehydes and ketones can also undergo Wittig reaction with P-Ylides.

Preparation of P-Ylides:* First the corrsponding alkyl halide is allowed to react with Ph

3P(triphenyl phosphine),

a wonderful nucleophile(soft base). SN2 reaction leads to an phosphonium salt.

* This salt is then allowed to react with a strong base like alkyl or aryl lithium(BuLi, orPhLi) to carry out deprotonation from the carbon atom to form a –ve charge, thus forming thestable P – ylide.

Ph3P R CH2 X R CH PPh3

H

X

Bu Li

R CH PPh3 R CH PPh3

P-Ylide

+

phoshonium salt

Mechanism:

Ph3P CH R

COR'

R''

Ph3P CH R

R''

R'

CO

Ph3P CH R

R''

R'

CO

oxaphosphetane

R CH C

R'

R'

Ph3P O

+alkene

triphenylphosphineoxide

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Dr. S. S. Tripathy

Wittig reaction involves three steps. First the P - Ylide makes a nucleophilic attack using itscarbanioninc centre onto the electrophlic carbonyl carbon of the aldehyde or ketone to form abipolar intermediate. In the second step the negative charge(lone pair) of O– atom forms a bondwith the electron deficient P– atom to produce a four membered cyclicintermediate(oxaphosphetane). This cyclic intermediate, in the third state undergoes intramolecularring opening to form the alkene and triphenylphosphine oxide.Example:

ClPh3P

BuLi

(i)

(ii)PPh3

O

butylidenecyclpentan

Wittig reacnP - Yliden-butyl chloride

(6) Pyrolysis of Xanthates, Amine Oxides and Acetate:(Ei / Pericyclic SYN Eliminations)

* These belong to a separate class of elimination occuring via a cyclic TS in onestep(pericyclic reactions) and the elimination mechansim is SYN(CIS) i.e the two leaving groupsremain on the same side for cyclic TS. For details, refer the GOC-III(mechanism).

* β-H are available on either side of the leaving group(xanthate, amine oxide or actategroups), then the Hofmann elimination becomes the major product. Pyrolysis means simplyheating the compound in the absence of air/oxygen.

* If more than one alkene is formed, then Hofmann product is the major product here.

(a) Pyrolysis of xanthate ester (Chugaev Reaction)

OC

S

MeS

H

Me Et

MeH

2000C

MeH

MeEt

Threo active E-alkene

COS MeSH+ +

(xanthate ester)

When a xanthate ester is heated we get an alkene alongiwth carbonyl sulfide(COS) and methylmercaptan(methanethiol). The two leaving groups H– on the β- carbon and –OCSSMe(methylxanthate group) on the α− carbon lie on the same side(eclipsed) in the TS for attaining the cyclicTS(not shown). Thus the stereospecificity can be observed. A threo active isomer on heatinggives the E-alkene(shown) while a erythro active isomer will form Z-alkene(not shown).

* Xanthate ester used here is ROCSSMe. Xanthic acid(unstable) is ROCSSH and we takeits methyl ester. This is easily prepared from the corresponding alochol(ROH) by treating firstwith Na or NaH to form alkoxide ion and then with CS

2 to form xanthate salt(sodium xanthate).

Finally this salt carrying the xanthate nucleophile reacts with Me–I (SN2) to form xanthate ester.

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Dr. S. S. Tripathy

Et CH

Me

CH

Me

OH NaH

Et CH

Me

CH

Me

O NaH2

_

CS S

Et CH

Me

CH

Me

O C

S

S Na

CH3 I- NaI

Et CH

Me

CH

Me

O C

S

S Me

Threo- alcohol alkoxide xanthate salt

xanthate ester

(b) Pyrolysis of Amine oxide (Cope Reaction):

* N-oxides of a 30 amine on heating at 80 - 1600C, forms an alkene with the same Eimechanism explained before.

NR'2

H

R1R2

R3 R4

O

R3R4

R1

R2

+

(particular diastereoisomer)30-amine N-Oxide

80-1600C R'2NOH

dialkyl hydroxyl amine

(particular diastereoisomericalkene)

N-oxide is first prepared from the amine very easilty by treating with m-CPBA(m-chloroperoxybenzoic acid) or H

2O

2 in MeOH.

(R1)(R2)CH C(R3)(R4) N

R'

R'

mCPBA(R1)(R2)CH C(R3)(R4) N

R'

R'

O

30 amine Amine N-oxide

Alkene will be formed if one of the alkyl groups bonded to N- atom in amine contains a miniumof 2 carbon atoms. There can be more than one alkene if the alkyl groups are different and eachcontains 2 or more number of carbon atoms. The least substituted alkene(Hofmann elimination)is the major product in all these cases.

(c) Pyrolysis of acetates:

* Heating alkyl acetates ROCOCH3 at 400 - 5000C, Ei elimination occurs to form the

respective alkene.

O

H

R1 R2

R3

R4

CO

CH3

heat

R3R4

R1

R2

CH3COOH+

Diastereomeric acetate Diastereomeric alkene

14


Dr. S. S. Tripathy

(7) Heating of Tetraalkyl ammonium hydroxide :

* Quaternary ammonium hydroxide(QAH) (tetralkyl ammonium hydroxide) forms alkenewith Hofmann elimination i.e least substituted alkene, as major product(Hofmann Rule), if alkylgroup(s) contains two or more carbon atoms. If there is one ethyl group, then the exclusiveproduct is ethene(>95%). For more, refer GOC-III(mechanism).

* Here the mechanism is E2 (NOT Ei, though it is a thermal decomposition reaction).* The 40 ammonium hydroxide gives an alkene and a 30-amine. There can be more than

one pairs of alkene and tert- amine if more than one alkyl group contains more than two carbonatoms and are different. See these cases.(1)

Me N

Me

CH2 CH CH3

H

CH2

CH2H

OH

CH2 CH2 Me2N CH2CH2CH3

CH3 CH CH2 Me2N Et

+

+

98%

2%Quaternary ammonium hydroxide

heat

In the above case( one alkyl group is ethyl and the other n-propyl which undergo E2 elimination.So we get two pairs i.e ethene with dimethy n-propyl amine and propene with ethyl dimethylamine. But you see that the the first pair is 98% as ethene is purely unsubstituted. In fact sterichindrance is the greatest factor in this bulky compound for undertaking elimination. Hence theelimination of β-H occurs from the least hindered site.(2)

CH3 N

CH3

CH3

CH

CH2

CH

CH3

CH3

H

OH

CH2

CH CH

CH3

CH3 (CH3)3N+

Hoffmann product

less hindered carbon

more hindered carbon

heat

trimethyl 3-methylbutan-2-ylammonium hydroxide(3-methylbut-1-ene)

In this example, only one alky group has more than 2 carbon atoms which undergoes elimination.Here also we find that the exclusive product is the less branched alkene(Anti-Saytzeff eliminationor Hofmann elimination). The other alkene is formed to a very minor extent(not shown).

* QAH can be easily prepared by Hofmann Exhaustive methylation of a 10 amine. Weshall discuss this in the chapter “Amines”. Please wait for that.(6) From Geminal Dihalide

* Gem. dihalides undergo Wurtz type of coupling with sodium in anhydrous ether to formakene containing double the number of carbon atoms.

15


Dr. S. S. Tripathy

CH3 CH2 CH

Br

Br

4 Na CHBr

Br

CH2 CH3

ether+ +

CH3 CH2 CH CH CH2 CH3

hex-3-ene

1,1-dibromopropane(propylidene bromide)

Properties of Alkenes

(A) Physical :* C

2 – C

4 are gases, C

5 – C

17 (unsubstituted) are liquids and C

18 onwards solids at room

temperature.* They burn in air with luminous smoky flame.* The physical properties of alkenes are very similar to alkanes

(a) BP and MP :BP increases with increase in number of carbon atoms while MP varies in a irregular manner.

Alkene MP(0C) BP(0C)Ethene –169 –104Propene –185 –47But-1-ene –6.5Trans-but-2-ene –106 +1Cis-but-2-ene –139 +42-methylpropene(isobutylene) –141 –7Pent-1-ene –165 +30Trans-pent-2-ene –135 +36Cis-pent-2-ene –180 +373-methylbut-1-ene -135 +252-methylbut-2-ene –123 +39Hex-1-ene –138 +63.5Hept-1-ene –119 +93

* Unlike alkanes ( in which there is a systematic trend in BP and MP in the homlogousseries), alkenes do not show any systematic trend in BP/MP. While trans-but-2-ene has a MPof –1060C, hex-1-ene and hept-1-ene have lower MPs i.e –138 and –1190C respectively. SinceMP involves packing of molecules in solid state, it is not solely dependent on the number ofcarbon atoms. However, BP increases with the number of carbon atoms, though, not as regularlyas in alkanes. In general, alkenes have lower BPs than the respective alkanes having samenumber of carbon atoms.

* Even among structural isomers of alkenes, it is not worth comparing, because, we haveboth position as well as chain isomerism here. Trans-pent-2-ene has higher MP than pent-1-enebut cis-pent-2-ene has lower MP than pent-1-ene. So MP trend is very wild.

* Even the decrease of BP with increase in degree of branching is also not strictly validhere. Isobutylene has a lower BP than but-1-ene and both the GIs of but-2-ene. However, in

16


Dr. S. S. Tripathy

C5H

10 isomeric alkenes, it is not the case. 2-methylbut-2-ene has a higher BP than pent-1-ene,

though 3-methylbut-1-ene has a lower BP. So i feel, further discussion on the trend of MP, BPof alkenes will be monotonous. Forget it.

BP/MP of GIs:

For Geometrical Isomer: Cis > Trans (Boiling point) due greater polarity of CisTrans > Cis (Melting point) due more efficient packing in

solid state for its greater symmetryThis you can have a look the data for but-2-enes and pent-2-enes.(b) Polarity and Dipole Moment:

Alkanes are non-polar irrespective of whether they are linear or branched. However,unsymmetrical alkenes(structurally) are polar, though weakly. Even bu-2-ene having symmetricalstructure is polar for Cis- isomer and non-polar for Trans- isomer.

Alkene (μ)Dipole Moment(D)Propene 0.35But-1-ene 0.37Cis-but-2-ene 0.33Trans-but-2-ene 0Cis-1,2-dichloroethene 1.85Trans-1,2-dichloroethene 01,1-dichlorethene 1.30Cis-1,2-dibromoethene 1.35Trans-1,2-dibromoethene 0Cis-1,2-diiodoethene 0.75Trans-1,2-diiodoethene 0

* Because, pi-bond is made up of loosely held electrons, they can be easily pushed orpulled by a attached group or atom. Therefore, alkenes(unsmmetrical) are polar(though weakly)while alkanes are nonpolar. For example. in propene, the CH

3_ group produces +I effect on the

C=C and hence forms a bond dipole vector as shwon in the figure. The opposite bond dipolevector is a H–C bond which has a different magnitude(though opposite in direction) and cannotnullify the CH

3–C bond diopole and hence there is a small resultant moment (0.35D). The other

two H atoms(anti w.r.t each other) cancel out their moments. Ethene is a symmetrical moleculeand hence the resultant moment is zero.* In Cis-but-2-ene, there is a resultant of the two CH

3–C bond dipoles (as CH

3– groups lie

on the same side) which cannot nullify the the resultantn due to C–H moments. Hence it is polar.In trans-but-2-ene the CH

3–C bond dioples nullify each other and so also the C–H moments.

Hence it is nonpolar.* However, in Trans-pent-2-ene, the two vectors CH

3–C and C

2H

5–C moments, though lie

in opposite directions, but their magnitudes are different. Hence trans-pent-2-ene is polar, unliketrans-but-2-ene which is nonpolar. In this case, both cis and trans isomers are polar, though theformer has greate dipole moment.

17


Dr. S. S. Tripathy

C C

CH3 C2H5

H H

C C

CH3 H

H C2H5

Cis-pent-2-ene Trans-pent-2-enemore polar( higher) less polar(-lower)

Conclusion: Cis-alkene has a greater dipole moment than its Trans isomer. Hence the boilingpoint of a Cis isomer is greater than its Trans isomer.(Exception: Trans-1,2-diiodoethene, being nonpolar has a greater boiling point(1920C) than thecis isomer(1880C). The dichloro and dibromo ethenes, however, follow the above rule of thumb.)(c) Solublity:Being weakly polar or non-polar they are too sparingly soluble in water and highly souble inorganic solvents.(d) Densities: They are lighter than water and their densities lies between 0.6 - 0.7 g/cc(liquidalkene).

CHEMICAL

(A) Addition Reactions* Since C=C is nuclophilic, most addtion reactions to alkenes are electrophilic in

nature(AdE). A few additions are free radical type and a few others take place via cyclic

intermediate.

1) Addition with H2:

* This is SYN addition mechansim which occur in the surface of catalyst(Ni/Pd/Pt). For details, refer GOC-III(mechansim). In presence of catalyst like Ni or Pd or Pt at highertemperature of 200 – 3000C, alkenes are hydrogenated to form alkanes.

meso-3,4-dimethylhexaneZ-3,4-dimethylhex-3-ene

MESOH

Me EtH

Me Et

Me Et

H

HMe Et

H

H

H

H

MeEt

MeEt

Ni

3000C

The Z-isomer shown above gives the MESO alkane and hence E-isomer(not shown here) willgive the d/l pair.2) Addition with X

2 :

* Cl2 and Br

2 in CCl

4 solvent easilty adds onto the double bond at room temperature

or in dark to form vicinal halide. Cl2 > Br

2. Iodiation is very slow and is not recommened.

* The discharge of red colour of Br2 in CCl

4 is often taken as +ve test for the

presence of unsaturation (CC multiple bond)* It is a A

dE mechanism and often free carbocation is not formed. In stead a 3-

membered cyclic halonium ion interemediate if formed. For details of mechanism , refer GOC-III(mechanism).

* The stereospecificity of reaction is ANTI addition. The two Br atoms join theC=C from the opposites sides.

18


Dr. S. S. Tripathy

+

E-but-2-enemeso-2,3-dibromobutane

C C

H

HBr

Br

CH3CCl4Br2C CCH3

CH3H

H CH3

E-but-2-ene with anti addition of Br2 gives the meso product, hence Z-but-2-ene would give d/

l pair(not shown).* Addition of Cl

2/Br

2 dissolved in nuclophilic solvents like H

2O or ROH will give

halohadrin as the major product due to the competing solvent nucleophile.

+C C Cl2H2O C C

Cl

OH

C C

Cl

Cl

majorvic. dichlorideminor

+

chlorohydrin

If EtOH is taken as solvent, we get ethoxy chloro compound(–OR in place of –OH) in the firstproduct shown.

* In presence of nucleophilic salts like NaCl, addition of Br2 gives vicinal dibromide as

the major product as Br– is more nucleophilic than Cl–. The minor product is chlorobromoproduct.

CH2 CH2

Br2/NaClCH2 CH2

BrBr

CH2 CH2

ClBr

+major minor

3) Addition with HX :

* HI > HBr > HCl >> HF* The mechanism is AdE. For details refer GOC-III(mechanism).* Markonikoff Rule is vaid for predicting the product for an unsymmtrical alkene.

The negative part of HX adds on the carbon of C=C which bears less number of H atoms(stericallymore hindered carbon). Hence addition is regiospecific for unsymmetrical alkenes.

R CH CH2 HX+ -

R CH

X

CH3+

Markonikoff additionunsymmetrical alkene

CH3NO2

* If the alkene is unsymmtrical structurally but symmetrical w.r.t C=C (lile CH3–

CH=CH–CH2–CH

3), then the addition is regioselctive. The major product can be predicted from

the relative stability of the carbocations on the basis of relative hyperconjugation effect.

+CH3 CH CH CH2 CH2 CH3 HI CH3 CH CH3CH2CH2CH2

I

CH3 CH2 CH3CH2CH2CH

I

2-iodohexae

3-iodohexane+

(major)hexa--2-ene

19


Dr. S. S. Tripathy

The 20-carbocation at C-2 is more stable( 5 α-H atoms) than the 20 carbocation at C-3(4 α-Hatoms).If there is a resonance stabilisation for the above type of substrates, then the reaction is sweptin favour of almost one product.

CH3 CH CH OCH3 HCl CH3 CH2 CH

Cl

OCH3

(only product)+

In this case, a carbocation adjacent to –OCH3 group is stablised by R-effect. Hence there is one

product(regiospecific).

CH CH CH3HCl

CH CH3CH2

Cl

In this case too, the 20-benzylic carbocation is more stable than the 20–alkyl, hence one product.* Non-nuclophilic polar solvents like nitromethane or acetonitrile is preferred in

this case, as nucleophilic solvents like water or alcohol will compete in the 2nd step to givealochol or ether products alongwith alkyl halide.

* Since carbocation formation occurs in the first rate determinin step, often theirrearrangement gives a mixture of products, major being different from that predicted byMarkonikoff’s rule.

CH3 CH

CH3

CH CH CH3HCl

CH3 CH

CH3

CH2 CH3CH

Cl

CH3 CH

CH3

CH CH3CH2

Cl

CH3 C

CH3

CH2 CH3CH2

Clrearranged product(major)

+

+

Hydride shift produces the most stable 30-carbocation andhence the rearranged chloro-product isthe major product in this case.

* Sometimes anti-Markonikoff product becomes major particularly when thecarbocation according to Markonikoff addition is unstable.

+CF3 CH CH2 HBr CF3 CH2 CH2

Br

In this case, 20 carbocation is not formed as it is destabilised by a strong –I group(–CF3), hence

the anti-Markonikoff product is formed.* Cyclic compounds often udergo ring expansion(rarely ring contraction) due to

carbocation rearrangement.

(1) HCl

Cl

(2) HBr

CH3

Br

In the example 1) a four member unstable ring expands to more stable 5-membered ring. Whilein example 2), a 5-membered ring exapands to still a more stable 6-membered ring. In the 2ndexample, we shall also get the normal products from teh 5-membered ring(not shown), 5-membered

20


Dr. S. S. Tripathy

ring has small angle strain so as to have some stability.Mechanism for the ring expansion:For 4 – 5 expansion, the mechanism is shown below. You practise yourself for the second one.

Cl

H

CH2H1

23

4 5

1

23

4 5

Cl

4) Addition with HOX or with X2/H

2O : (Halohydrin formation)

* HOX(hapohalous acid) adds onto an unsymmetrical C=C following Markonkoff’srule. The product, in common system, is called halohydrin of the respective alkene.

* HOCl or HOBr are very weak acids and are poor proton donors. In stead, theycan donate the electrophile chloronium ion(Cl+), if need be.

* The mechanism is AdE(like addition of HX). The formation of carbocation

occurs in the first step followed by addition with the nucleophile(OH–).

R CH CH2 (HO) X+-

R CH

OH

CH2

X

alkene halohydri+

Mechanism:

CH3 C

CH3

CH2 Cl OH CH3 C

CH3

CH2

Cl

- OH

OHCH3 C

CH3

CH2

Cl

OHisobutyleneisobutylene chlorohydrin(1-chloro-2-mehylpropan-2-ol)

+

If we use pure HOCl or HOBr, we get the halohydrin according to the above mechanism. Thereaction is almost regiospecific, driven by the stability of carbocation.* HOCl usually reacts in presence of strong acids like conc.H

2SO

4. In such case the

chloronium ion is produced in situ and the carbocation in the last step is attacked by nucleophileH

2O followed by deprotonation to form chlorohydrin.

Cl OH H+

Cl+ H2O+ +

* Usually we take X2/H

2O to get halohydrin in situ. In halogen/water mixture, there must

be some pure HOX, but most of it remain as a unreacted mixture of X2 and water. With pure

HOX, the reaction happens as shown above. But with X2 and water, the halohydrin is the major

product and the vicinal dihalide is the minor product. Again in this case, the reaction goes viathe cyclic halonium ion intermediate. Hence the addition mechanism is ANTI. However, themajor product is the one in which the the nucleophile H

2O, attacks to the more hindered carbon

atom(carbon bearing less number of H atoms), as that carbon has greater partial +ve charge(morecarbocationic character). Hence the Markonikoff product is the major product.

21


Dr. S. S. Tripathy

R CH CH2

Br Br

R CH CH2

Br

- Br

H2O

preferred attackR CH CH2

OH

Br

- Hbromonium ion intermediate

Here the other product is also formed in which H2O attacks the less hindered carbon and hence

the -Br is bonded to the more hindered carbon. But that product is the minor one. Since the morehindered carbon has a greater partial +ve charge(RSs not shown) i.e more carbocationic character,the nucleophile H

2O preferably attacks that position. Hence the Markonikoffr’s rule decides the

major product here. Note that the competing Br– also makes nucleophilic attack to give vicinaldibromide as the other minor product. Since water is taken in large excess of halogen, theprobability of H

2O is much greater. Since both are weak bases and poor nucleophiles by nature.

Note also that the ANTI sterospecificity is seen in this case in the major product, if properstereochemical alkene is taken. So the major product in this case, is not only, Markonikoff, butalso anti addition product like halogenation of alkenes.

5) Addition with Water(Hydration of Alkenes) :

* Addition of H2O to alkenes to form alcohols is catalysed by non-nucleophilic

acids like H2SO

4 and H

3PO

4.

* The mechanism of addition is AdE like addition with HX. H+ from acid adds

onto the alkene in the rate determining step to form carbocation, which reacts the nucleophileH

2O to form alcohol.

* Hence for unsymmetrical alkenes, Markonikoff alcohol is formed regiospcifically.* Branched alkenes, often give rearranged alochols due to carbocation

rearrangements.* Alkene and acid-water mixture is to be heated for getting alkene. However, the

temperatue should be lower than that required for dehdyration(1700C). 20-alochol requries lessthan 1000C while 30 alcohol needs less than 250C. Usually we do not put a heat(Δ) symbol forhydration of alkene, as often we get get 30 or at the least 20- alcohol. Only ethene gives 10-alcohol(ethyl alcohol) which needs higher temperature( but < 1700C).

* When a chiral alcohol is formed by the addition of H2O to an alkene, we get

a rac. mixture product.

R CH CH2 H OHH2SO4

+ -R CH

OH

CH3+

Mechanism:

H2SO4

CH3 C

CH3

CH2 H+

H2SO4

HSO4--

CH3 C

CH3

CH3

H2OCH3 C

CH3

CH3

OH

H

HSO4-

CH3 C

CH3

CH3

OH-

+

22


Dr. S. S. Tripathy

Alkene accepts a H+ from the acid to form a stable carbocation in the first step. Then nucleophileH

2O attacks onto the carbocation to form protonated alcohol(oxonium ion) in the 2nd step.

Deprotonation of this species by the conjugate base of the acid (HSO4–) in the third step produces

alcohol.Rearrangement:

CH3 CH

CH3

CH CH2H2O/H+

CH3 C

CH3

OH

CH2 CH3

major product

Mechanism:

CH3 CH

CH3

CH CH2H

+

CH3 C

CH3

CH CH3

H

hydride

shiftCH3 C

CH3

CH2 CH3

H2O

H+-

CH3 C

CH3

CH2 CH3

OH

H2O

H+-

CH3 CH

CH3

CH CH3

OH

most stable carbocation

major product

+

minor product

* Like addition of HX, in this case too, there can be ring expansion.

CH CH CH3

H2O/H+

OH

CH2CH3CH CH3CH2

OH

CH2 CH3CH

OH

+ +

normal productsring expansionproduct

Besides the normal products from the hydration of prop-1-enylcyclopentane, we get the 2-ethylcyclohexan-1-ol as major product. Ring expansion to a more stable six membered ring occurshere.

CH CH CH3

H+

CH CH3CH2 CH2CH31

23

4

5 6

123

4

5 6

H2O

H+-

OH

CH2CH3

Interestingly, there are two chiral centres developed in the cyclohexanol product(hence fourstereoisomeric products namely RR, SS, RS and SR). Each of the normal products has one chiralcentre, hence exists as R/S pair. So altogether 8 alcohols, including stereoisomers, are formedin this reaction from any stereoisomer of the reactant(E or Z).

23


Dr. S. S. Tripathy

6) Peroxide Effect/Kharasch Effect(Anti Markonikoff Addtion) :* Unsymmetrical alkenes add onto some selected addendum like HBr, RSH, BrCl

3, ICF

3

etc. in presence of a peroxide like benzoyl peroxide, di-tert-butyl peroxide, H2O

2 etc in anti-

Markonikoff way. The negative part of the addendum adds onto carbon bearing more number ofH atoms.* This occurs in Free Radical Addtion mechanism(Ad

F). For details, please refer GOC-

3(mechanism).

R CH CH2 HBr H2O2 R CH2 CH2

Br

+

7) Hydroxylation of alkenes to vicinal diols(Reaction with Baeyer’s reagent OROsO

4.

* Cold dilute alkaline KMnO4 (Baeyer’s reagent) reacts with alkene at pH> 8 to

form vicinal diols(glycols). The purple colour of permanaganate changes to green colour ofmanganate in this redox reaction. Generally green colour is not observed, in stead discharge ofpurple colour of permanganate in this reaction is used as another test for unsaturation.

* The reaction takes place via an cyclic manganate ester intermediate.* The addition mode is SYN i.e the two -OH groups bond on the same side of

C=C. Hence E-but-2-ene gives d/l product while Z-but-2-ene gives meso-product.* OsO

4/pyridine can be used as a substitute for Baeyer’s reagent for the SYN

hydroxylation. A cyclic osmate ester intermediate is formed in this case, which is broken downby NaHSO

3 or H

2S to form

For details of mechanism, refer GOC-III(mechanism).

R CH CH R'

cold di l. alk.KMnO4(Baeyer's reagent)

H2O [O]+ +R CH R'CH

OH OH

(vicinal diol)

R CH CH R' R CH R'CH

OH OH

(vicinal diol)

OsO4 /pyridine

NaHSO3

(i)

(ii)

* Ethylene give ethylene glycol(glycol).8) Epoxidation and its conversion to vicinal diol:

* Alkene first is converted to alkene oxide(oxirane or epoxyalkane) by eitherreacting with (a) O2 in presence of Ag at 4000C OR with (b) m-CPBA(m-chloroperoxybenzoicacid) or any other peroxy acid.

* In the second step, the alkene oxide is hydrolysed by alkali(OH–) or H2O( in

presence of acid) to form vicinal diol.* Here the mechanism of addition is ANTI i.e the two -OH groups join on the

opposite side of C=C like halogenation of alkenes. Thus Z-but-ene gives d/l product while E-but-2-ene gives meso product.

For details of mechanism, refer GOC-III(mechanism).

R CH CH R' O2

Ag

4000C

12

+ R CH R'CH

O

H2O/H+

Na+ OH

-or

R CH R'CH

OH

OHalkene oxidevic. diol

* Ethene gives ethene oxide(oxirane) which on hydrolysis to form ethylene glycol.

24


Dr. S. S. Tripathy

9) Hydroboration-Oxidation (Formation of Anti-Markonikoff alcohol)

* Alkenes react with diborane in THF to form tiralkyl borane (R3B), which on oxidation

by H2O

2/OH– gives anti-markonikoff alcohol.

* It is essentially anti-Markonikoff addition H2O to an unsymmetrical alkene, though reallysuch addition does not take place.* The addition mechanism is SYN i.e -H and -OH add up from the same side of the C=C.Please refer GOC-III(mechanism) for details of the mechanism of HBO.* No rearrangement product is found here (like acid catalysed hydration of alkenes).

R CH CH2

BH3 / THF3 R CH2 CH2

3B

H2O2/OH-

R CH2 CH2

OH

3 H3BO3+

3

* It involves two steps. In the first step, borane(BH3) or diborane(B

2H

6) reacts with alkene

to form trialkyl borane. In the subsequent step, the borane is oxidised by H2O

2/OH– to form to

form alcohol.In brief, it can be represented as follows :

R CH CH2

BH3 / THF

H2O2/OH-

(i)

(i i)R CH2 CH2

OH

Some other Reactions of Tralkyboranes:Hydrobroation Amination, Reduction, Halogenation and Coupling(a) Trialkyl borane reacts with chloramine(Cl–NH

2) to form primary amine.

R CH CH2

BH3 / THF(i)

(i i)R CH2 CH2

NH2

Cl NH2

(b) Trialkyl bornae reacts with acetic acid to form alkane. This Hydroboration Reduction.

R CH CH2

BH3 / THF(i)

(i i)R CH2 CH3CH3COOH

(c) Trialkyl borane reacts with Br2 or I

2 to form bromide or iodide.

R CH CH2

BH3 / THF(i)

(i i) R CH2 CH2

Br

Br2

(d) Trialkylborane will undergo self coupling to form higher alkane in presence of AgNO3/

OH–.

R CH CH2

BH3 / THF(i)

(i i)R CH2 CH2 CH2 CH2 R

AgNO3 / OH-

(N.B: These above four reactions of trialkylboranes are not widely studied andhence are not popular)

25


Dr. S. S. Tripathy

SAQ:

(i) CH3 CH CH2

(i)

(i i)

BD3/THF

CH3COOH?

(ii) CH3 CH CH2

(i)

(i i)

BH3/THF

CH3COOD?

Ans (i) CH3 CH

D

CH3(ii) CH3 CH2 CH2

D

10) Oxymercuration-Demurcuration:• Alkene first reacts with Hg(OAc)

2 followed by water to form an organomercuric alcohol

intermediate(oxymercuration). This is electrophilic addition as alkene(nucleophile abstracts theelectrophile (HgOAc)+ from the reagent in the r.d.s.• This intermediate is reduced by NaBH

4/OH- to form a Markonikoff alcohol(Demercuration)

• Essentially it is Markonikoff addition of H2O.

• No rearranged product is formed here.• It is not stereospecific like HBO.• If ROH reacts in the first step in place of water, we get a Markonikoff ether.

R CH CH2

Hg(OAc)/THF

H2O

(i)

(i i)R CH CH2

HgOAc

OH

NaBH4/OH-

R CH CH3

OHOrgano Mercury alcohol

Mechanism:

R CH CH2

AcO Hg OAc

R CH CH2

Hg OAc

- OAc

H2O

R CH CH2

Hg OAc

OH H

OAc-

- AcOH

R CH CH2

Hg OAc

OH

BH3

H

BH3

R CH CH3

OHHg

AcO-Na

+

Na+

- -

-

mercurinium ion

organo mercury alcohol

26


Dr. S. S. Tripathy

* If R’OH is used in stead of water in the hydrolysis step. ether is formed instead of thealcohol.

R CH CH2Hg(OAc)2

R'OHR CH

OR'

CH2

HgOAc

NaBH4/OH-

R CH2

OR'

CH3

ether

11) Ozonolysis: (Additon with Ozone)* Addition of ozone to C=C of an alkene to form ozonide adduct and its subsequent

break-down to form simpler products(aldehydes/ketones) is taken as the important method toassign the position of C=C in a organic structure.

* Alkene first forms the unstable MOLOZONIDE which rearranges to form stableOZONIDE addition product. This on further treatments produce simpler products due to break-down at C=C.

* Essentially, ozonolysis is the break-down of the C=C to two carbonyl compoundsdepending on the structure of alkene. Two carbonyl compounds(aldehyde + aldyde OR ketone+ aldehyde OR ketone + ketone) are formed in the reaction depending the original structure ofalkene.

R C

R'

CH R''O3

Zn/CH3COOH

(i)

(i i)R C

R'

O O CH R'+ketone aldehydealkene

(a) Reductive Ozonolysis:The ozonide reacts with Zn/CH

3COOH or Zn/H

2O or (CH

3)

2S to form aldehyde/

ketone mixture due to cleavage at the ozonide site(previously C=C site). The other product H2O

2

is reduced by Zn so as to prevent further oxidation of aldehyde to carboxylic acid.

R CH C R' + O3 R CH C R'

O O

O

molozonide

R CH CR'

O O

O

ozonide

R'' R''

R''

R CH CR'

O O

O

ozonide

Zn/CH3COOHR CH

O

+ C R'

O

+ H2O2

R''

R''

aldehyde ketone

27


Dr. S. S. Tripathy

In fact H2O

2 should have been deleted from RHS as it really is reduced by Zn/CH

3COOH or Zn/

H2O.

(b) Oxidiative Ozonolysis:The ozonide is broken down in presence of an oxidising agent like Ag

2O or peroxy acids

or even with additional H2O

2 to form a mixture of carboxylic acid/ketone. Aldehyde in this case

converts to carboxylic acid.

R CH CR'

O O

O

ozonide

R C

O

+ C R'

O

+ H2O2

R''

R''

ketone

Ag2OOH

c. acid

(c) Ozonolysis with Reduction:The intermediate ozonide reacts with RAs like LiAlH

4 or NaBH

4 to form alchols(10 and

20 alcohols)

CH3 C

CH3

O O

O

CH CH3

LAH

6[H}CH3 CH

OH

CH3 CH3 CH2 OH+

2-methylbut-2-ene ozonide

20-alcohol 10-alcohol

N.B: Usual meaning of ozonlysis is reducttve ozonolysis i.e to give a mixture of aldehdes andketones discussed in case (a) before.

Examples:1.

CH2 CH2 + O3 H2C CH2

O O

OAg

2 O

Zn/CH3COOH

H C

O

H2H C

O

OH2

(ethylene ozonide)

2.

CH CH2 + O3 CH CH2

O O

O

H3C C

O

H H C

O

H

H3C H3C

Zn/CH3COOH

+

28


Dr. S. S. Tripathy

3.

C CH + O3 C CH

O O

O

H3C C

O

CH3 H C

O

CH3

H3CH3C

Zn/CH3COOH

+

CH3

H3C

CH3 CH3

Mechanism:* In the first step, alkene undergoes 1,3-dipolar cycloaddition reactions resulting the

primary ozonide i.e molozonide. The molozonide is unstable and quickly rearranges into secondaryozonides(three from the unsymmetrical alkene shown below), which are stable. This takes placevia the formation of carboyl and carbonyl oxide pairs.

MOLOZONIDE (Two pairs of carbonyl and carbonyl oxide)

* The carbonyl oxides are analogous to ozone as 1,3-dipolar compound. These undergocycloaddition again with the carbonyl products to from three secondary ozonides called ALKENEOZONIDES.

Do you see above the most interesting aspect in the formation ozonides ?? We took one alkenebut got three different ozonides. These, in fact, could be separted as their physical properties arewidely different, but seldom it is done. The purpose is to break-down these ozonides to form twocarbonyl products.

I am putting the scheme for the formation of moloznide from alkene and its subsequentconversion to ozonide in a simplistic manner without invoking the possibliities of formation ofmore than one ozonides.

29


Dr. S. S. Tripathy

In the first step, O3 undergoes 1,3 - cycloaddition with alkene to form unstable molozonide(also

called Criegee intermediate) which breaks in a retro-(1,3) cycloaddition reaction to form acarbonyl compound and a carbonyl oxide. In the second step, the carbonyl oxide adds onto thecarbonyl compond in an inverted manner(carbonyl compound has been flipped) to form ozonide.(N.B: There might [3+3] cycloaddition of two carbonyl oxides to form dangerously explosivetetraoxane product. Beware of it !!!!!

* Decomposition of ozonides:

R CH

O

O O

C

R'

R''

OH2 R CH

O

O O

C

R'

R''

OH2

R CH O O C

R'

R''

O O

H

H+ +

H2O2

H2O or (CH

3)

2S will initiate ring opening which leads to break-down to form two carbonyl

compounds and H2O

2 or (CH

3)

3SO (as the case may be). In the absence of Zn or Zn/CH

3COOH,

or in the presence of Ag2O or additional H

2O

2, the aldehyde formed first is further oxided to

c.acid.

30


Dr. S. S. Tripathy

Ozonlysis of Dienes and Cycloalkenes:

H3C CHHC CH CH2 + 2 O3

H3C CH CH CH CH2

O O

O

O O

O

Zn/CH3COOH

H3C C

O

H H C

O

C

O

H H C

O

H+ +

gyloxal(oxalaldehyde)

acetaldehyde formaldehyde

diozonide

Exactly analogous to alkenes, dienes will give two monofunctional carbonyls and one difunctionalcarbonyl. Penta-1,3-diene gives a mixture of formalldehyde, acetaldehyde and ethanedial(glyoxal).

+ O3

O

OO

Zn/AcOH

H2C

CH2

H2C

CH2

O

H

O

H

hexanedial

Cyclohexene breaks down at C=C to give a acyclic difunctional dicarbonyl compound i.ehexanedial.Determination of Structure of alkene or alkadiene from ozonolysis products:

The ozonolyis products are joined up with a C=C at the two carbonyl functions.

Xozonolysis

CH3COCH3+ CH3CH2CHOLet us tie up acetone and propanal at their carbonyl functions by replacing them by a commnC=C bridge.

C

H3C

H3C

O O CH CH2 CH3

C

H3C

H3C

CH CH2 CH3

+

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Dr. S. S. Tripathy

Example 2:

Yozonolysis

CH3CHO + CH3COCH2CHO + HCHOSince two monofunctional and one difunctional compounds have been formed, it is a diene. Thedifunctional is to be kept the middle and the two monofunctional compounds to be kept the twoterminals so as to tie up with two C=C.

C

H3C

H

O O C

CH3

CH2 CH O C

H

O

H

HCH3C C

CH3

CH2 CH CH2

+ +

Example 3:

Z5-oxohexanal

CH3COCH2CH2CH2CHOozonolysis

Since the product is a difunctional compound, the alkene must be a cycloalkene.

CH3

C OCH2

CH2CH2

CH

O

CH3

1-methylpent-1-ene

12) Epoxidation and Formation of Vicinal Diols from Alkene oxides :* Alkenes are easily converted to alkene oxides(epoxyalkanes) by either (a) reacting with

O2 in presence of Ag/4000C or (b) reacting with a peroxy acid such as m-CPBA(m-

chloroperoxybenzoic acids)

R CH CH R' + 1/2 O2 R CH CH R'

O

epoxy alkane(oxirane)

Ag

4000C

CH 2 CH 2 1/2 O2 CH 2 CH 2

OAg

4000C+

ethylene oxide(epoxyethane or oxirane)

* Alkene oxides are the starting materials for vicinal diols. When alkene oxide is hydrolysedby water catalysed by acid or base, vicinal diol is formed. The overall addition mechanism foralkene to form the diol is ANTI, as the two -OH groups will join from the opposite sides(analgousto bromination of alkenes). Thus cis-but-2-ene gives rac. mix of diols and trans-but-2-ene givesmeso-diol.

32


Dr. S. S. Tripathy

R CH CH R'm-CPBA

[O]R CH R'CH

O

H2O/H+

( OH-)or

R CH

OH

CH

OH

R'

vicinal diolalkene oxide

(N.B: You shall get more details on this in GOC-III(mechanism).

13) Addition with Carbenes :* Alkene reacts with carbenes to form cyclopropane derivaties.* Carbenes can be obtained by the photolysis of pyrolysis of diazomethane(CH

2N

2) or

ketenes(RCH=C=O) or from carbenoids(I-Zn-CH2-I).

*Stereospecificity: The reaction is stereospeicific i.e cis-alkene give cis-cyclopropane andtrans-alkene give trans-cycloalkane i.e the configuration of alkene is retained in cyclopropane.This is the case, when singlet carbene is made to react and this happens if the reaction is carriedout as such in liquid phase(with ether solvent). This is a cheletropic concerted cycloadditionreaction taking place in one step via a TS.

C C

CH3 CH3

H H

CH2N2

h/ether

CH2

CH3 CH3

Cis-1,2-dmethylcyclopropane

- N2

carbene

Cis-but-2-ene

C C

CH3 H

H CH3

CH2N2

h/ether

CH2

CH3CH3

Trans-1,2-dmethylcyclopropane

- N2

carbene

Trans-but-2-ene

Example 2:

CHCl3/KOH

CCl2

C

Cl

Cl;;

C

OMe

H

MeOCH2Cl

BuLi

MeOCH

CHCl3 and KOH ( Or CHBr3 / t-BuOK) will produce dichloro(or dibromo) carbene in situ due

to α-elimination which adds upto C=C. MeO-CH2-Cl in presence of BuLi can also give MeOCH2:

carbene to form the product.

* Loss of Stereospecificity: When the reaction is carried out in gaseous phase and in thepresence of excess of inert gas like N

2, then the singlet carbene fans out some energy to the

bombarding N2 molecules and converts to triplet(ground state) carbene. Since triplet carbene is

a diradical, concerted reaction will not take place, rather the reaction takes place in two steps viathe formation of a diradical intermeidate. Since C-C rotation is allowed, the net result after the2nd step of cyclisation is the loss of stereospecificity. Each of cis and trans alkene gives amixture of cis and trans cyclopropane.

33


Dr. S. S. Tripathy

C C

CH3 CH3

H H

CH2N2

h(gas phase)

CH2

CH3 CH3

Cis-1,2-dmethylcyclopropane

N2

carbene

Cis-but-2-ene

excess

CH3CH3


+

+


CH3CH3

excess

Trans-but-2-ene

carbene

N2Cis-1,2-dmethylcyclopropane

CH3 CH3

CH2

h(gas phase)

CH2N2C C

CH3 H

H CH3

Mechanism:

14) With Alkanes (Alkylation of Alkanes)

Alkene reacts with alkane to form an alkylated alkane.

CH3 C

CH3

CH2 + CH3 CH

CH3

CH3

conc. H2SO4 or HF

0-10 C ?

Product: CH3 CH

CH3

CH2 C

CH3

CH3

CH3

(See carefully how alkane is added to alkene)

34


Dr. S. S. Tripathy

Mechanism:

CH3 C

CH3

CH2

H+

CH3 C

CH3

CH3+

CH3 C

CH3

CH2

CH3 C

CH3

CH2 C

CH3

CH3

CH3

+

CH3 C

H

CH3

CH3

CH3 CH

CH3

CH2 C

CH3

CH3

CH3

CH3 C CH3

CH3

++

Isobutene reacts with isobutane in presence of acidic catalyst to form 2,2,4-trimethylpentane(isooctane), which is an important component in petrol(gasoline).* This is electrophilic addition mechanism. In the first step, alkene is protonated to formthe most stable 30-carbocation. This adds onto another molecule of alkene to form the bigcarbocation(30). Then the role of alkane comes. The big carbocation abstracts a hydride ion(H–) from isobutane at the 30-postion to produce neutral alkane(alkylated alkane) and a tert-butylcarbocation. This ion again adds onto alkene to form big carbcation which abstracts a hydrideion from isoubtane to repeat the cyclic process.Note that if you apply Markonikoff rule for theaddtion, you have to take (CH

3)

3C of isobutene as +ve part and H as the –ve part, as you found

that a hydride ion is stransferred from alkane. Since electronegativity wise H- should be the +vepart, but since 30 carbocation is very stable and 30-carbanion is very unstable, the C–H bondheterolytically breaks towards the H atom to from a stable 30 carbocation and H–. The lattertransfers to the carbocation already formed from alkene.SAQ: What product will be obtained when propene reacts with propane in presnce of H+ ?Solution:

CH3 CH CH2 CH3 CH

CH3

HH

+

CH3 CH2 CH2 CH

CH3

CH3

+ -+

15) Dimerisation of Alkene:* An alkene can also add onto itself in presence of acid, if the solution is heated. The bigcarbocation formed in the previous case, undergoes elimiation of H+ to form two alkenes, onebeing major. This is dimerisation of alkene. Isobutene adds onto itself in prence of H

2SO4 when

heated to form 2,4,4-trimethylpent-2-ene as major product.

35


Dr. S. S. Tripathy

CH3 C

CH3

CH2H2SO4

80 C2

CH3 C

CH3

CH2

H+

CH3 C

CH3

CH3+

CH3 C

CH3

CH2

CH3 C

CH3

CH2 C

CH3

CH3

CH3

+

CH3 C

CH3

CH C

CH3

CH3

CH3

CH2 C

CH3

CH2 C

CH3

CH3

CH3

+

(major) (minor)

?

SAQ: What product is obtained when propene on dimiersed under acidic/heat conditions ?Soution:

CH3 CH CH2H

+

2 CH3 CH CH CH

CH3

CH3 CH2 CH CH3

CH3

CHCH

CH3 CH CH3

CH3

CCH CH3 CH CH2

CH3

CCH

+

+

+

major

(See the mechanism yourself. There will be rearrangement of carboncation from 20 to 20 and alsoto 30, resulting all the four products. Methyl shift also can take place to form 3-methylpent-2-ene and 3-methylpent-1-ene, as the very minor products. Excluding 10 carbocation, which neverforms, you take all possible 20 and 30 carbocation by hydride and methyl shifts, you will find theabove products plus the other two not shown.(Note that dimerization/oligomerization of alkenes by using TM catalysts to form many unusualproducts including isomerisation, never known in acid catalysts has revolutionised chemicalsynthesis. This is not discussed here.)

16) Addition of Alcohol and C. acids to Alkenes:(Alkylation of Alcohol and C. acid):

CH3 C CH2

CH3 C2H5OH/H+

?

36


Dr. S. S. Tripathy

Mechanism:

CH3 C CH2

CH3

H+

CH3 C

CH3

CH3

C2H5 OH

- HCH3 C

CH3

CH3

OC2H5

+

tert-butyl ethyl ether

isobutene

Isobutene adds on to EtOH to form tert-butyl ethyl ether.Alcohol and carboxylic acid adds onto alkene in presence of acid catalyst(H

2SO

4) to form ether

and ester respectively. Since it goes via carbocations, rearrangement is possible depending on thestructure of alkene. For simple alkenes, you have to take H as the +ve part and RO- or RCOO-as the –ve part respectively in order to apply Markonikoff rule to predict the product.

CH3 C CH2

CH3

H+

CH3 C

CH3

CH3- H

CH3 C

CH3

CH3

OCOCH3

+

tert-butyl acetate

isobutene

CH3COOH

Similary isobutene adds onto acetic acid to form tert-butyl acetate.

17) Hydroformylation of Alkenes(OXO Process) :* Alkenes are converted to aldehydes very economically by this method. It is an industrialmethod of synthesising aldehyde from alkene.* Alkene reacts with CO and H

2 gases in presence of the TM calayst eg [CoH(CO)

4] or

[RhH(CO)(PPh3)

3]at high presssure and temparature to form aldehyde containing one carbon

atom more.* It is essentially a hydroformylation process, i.e addtion of H–CHO to C=C. H atom andthe formyl group(CHO) are added to the two carbon atoms of C=C.* It is homogenous catalytic process where alkene is coordinated to the TM in the intermedatestep. H atom is bonded to the more hindered carbon of C=C and thus the formyl(CHO) part isbonded to the less hindered carbon. Thus the major product formed is the straight chain aldehydefrom a unbranched unsymmetrical alkene.

R CH CH2 H2 CO[CoH(Co)4]

10 - 100 atm40 - 2000C

R CH2 CH2 CHO

R CH

CHO

CH3

major

minor

+ + +

more hindered

Mechanism:The reaction of ethyene with the Co catalyst in presence of H

2 is shown with mechanism. The

H atom from the catalyst is added to the alkene at the more hindered position(in this case, it issymmetrical). The alkene forms first a π-complex(18VE) and then a σ-complex(16VE) with thecatalyst. In the latter complex H- atom from catalyst get added. The CO, then is inserted so asto convert into a carbonyl group in the carbon skeleton, joining at the less hindered position.Reductive cleavage of the sigma bond occurs by H

2 to from the less branched aldehyde. From

an unbranced alkene, straight chain aldehyde is the major product. In this example, ethene formpropanal.

37


Dr. S. S. Tripathy

(18) Addition with conc. H2SO

4:

* Alkenes are abosorbed by conc. H2SO

4 to form alkyl hydrogen sulfates. This

follows Markonikoff rule.* However, on hydrolysis of this compound, alcohol(s) is/are obtained. Again if the

hydrogen sulfate addition product is heated as such or heated after additing water, then isomericalkenes are again formed.

R CH CH2 H OSO3H+ -

R CH

OS3H

CH3

conc.

+

H2SO4

alkyl hydrogen sulfate

SAQ: What products are obtained when 3-methylbut-1-ene is absorbed by conc. H2SO4 and theproduct is (a) diluted by water (b) diluted to lesser extent and then heated.Solution: (a) 3-methylbutan-2-ol and 2-methylbutan-2-ol(major). The latter is formed due torearrangement of carbocation.(b) 2-methylbut-2-ene and 3-3-methylbut-1-ene and 2-methylbut-1-ene.Try with mechanism by yourself.

(B) Wacker’s Oxidation of Terminal Alkenes :* Alkene transforms to carbonyl compound when oxidised by of PdCl

2/CuCl

2 catalyst/

cocatalyst mixture in H2O/DMF to which O

2 gas is constantly bubbled. Ethene gives ethanal and

all other alkenes give ketones. Expecially terminal alkenes (alk-1-ene) are suitable for thisprocess resulting methyl ketones. In this reaction, H

2O plays an important role in the mechanism

(the first nucleophilc attack made by H2O onto the alkene bonded to PdCl

2) while O

2 plays

secondary role for converting Cu+ formed back to Cu2+ in the last step. So never consider this

38


Dr. S. S. Tripathy

as the additon of alkene with O2 molecule. Here the OA is Pd2+ which is reduced to Pd(0), which

is converted back to Pd2+ by Cu2+ which is reduced to Cu+. O2 ultimately converts Cu+ to Cu2+.

Thus Cu2+ acts as a co-catalyst with Pd2+.

CH2 CH2 PdCl2 H2O CH3CHO Pd HCl+ 2+ + +

+Pd CuCl2 PdCl2 Cu2Cl22 +

+2Cu2Cl2 O2 HCl+ +12 CuCl2 H2O

(note that the solution contains ample free HCl formed in the reaction steps.)

R CH CH2PdCl2/ CuCl2

O2 (H2O/DMF)60-700C

R C

O

CH3

* It is an industrial process of manufacturing ethanal and acetone from ethene and propenerespectively. It is a homogenous catalysis.

N.B: If alcohol(R’OH) is used in stead of H2O, then we get alkoxy alkene(CH

2=C(OR’)R in

stead of carbonyl compound)

R CH CH2PdCl2/ CuCl2

O2 (MeOH/DMF)60-700C

R C

OMe

CH2

(For detailed mechansim, please refer “Organic Name Reactions”)

(C) Substitution Reactions:(i) Allylic halogenation:

* NBS(N-bromosuccinimide) is effective reagent to carry out bromination at allylicposition.

R CH2 CH CH2

NBS/CCl4

allylic position

R CH CH CH2

Br

Structure of NBS and NCS:

C

N

C

O

O

Br

C

N

C

O

O

Cl

* NCS(N-chlorosuccinimide) / t-BuOCl / SO2Cl

2 is used for allylic chlorination.

39


Dr. S. S. Tripathy

CH2CH3 CH CH CH3 CHCH3 CH CH CH3

Cl

CH2CH3 CH CH CH2

Cl+

major

minor

20 allylic 10 allylcNCS/CCl4

In all these reagents are suitable for generation of allyl free raidcals, as the themechanism of substituion is S

F like we had in alkanes. Same initiation, propagation and termination

steps are also here. Since allyl radical is stable due to resonance, substituion occurs at thisposition.

Reactivity Order: 30 allyl > 20 allyl > 10 allyl* Cl

2/Br

2 with heat / light can also be used for the substitution. High temperature

chlorination is a possible alternative. If allylic position is absent, then halogenation occurs atvinylic position.

CH2 CH CH3Cl25000C

CH2 CH CH2

Cl

(allylic chlorination)

Cl2

5000CCH2 CH2 CH2 CH

Cl

vinyl chloride (vinylic chlorination)

Mechanism with NBS:A small amount of HBr is always present in NBS which slowly react with each other in

presence of heat or light to form Br2. Br

2 then carries out the free radical substitution as usual.

Please refer SF mechanism for details(initiation, propagation and termination)

CH2 CH CH3 NBr

O

O

CCl4heat or l ight

CH2 CH CH2

Br

NH

O

O

HBr+ + + HBr+

Allylic Rearrangement:3-methylcyclopen-1-ene on reaction with NBS gives gwo products, major product being

the rearranged product. Product (A) is kinetically controlled as the 30-allyl radical is most stable.Product (B) is thermodynamically controlled as more substituted alkene is more stable. This hasbeen formed by rerrangement of allyl radical.

40


Dr. S. S. Tripathy

CH3

H

Br

CH3CH3

NBS

CH3

Br

NBS

CH3

Br(A)

(B)

SAQ:

(ii) Allylic Oxidation:Alkenes are oxidised by SeO

2 followed by hydrolysis to form allylic alcohols.

R CH2 CH CH2

allylic position

R CH CH CH2

OHSeO2

41


Dr. S. S. Tripathy

N.B: SeO2 can do many interesting oxidation. It converts a carbonyl compound into a α-dicarbonyl

compound.

CH3 C

O

PhSeO2

C C

O

Ph

O

H

(D) Isomeization:Terminal alkenes isomerize to internal alkenes and branched alkenes when heated in presecne ofAlCl

3 at 200 - 3000C. The reverse process is not possible.

* But-1-ene on isomerisation gives a mixture of but-2-ene and isobutene alongwith but-1-ene.

H3C CH2 CH CH2

Anhy. AlCl3

or heatH3C CH CH CH3

H3C C CH2

CH3

+

but-2-ene

isobutylene

This reaction takes place via carboncations catalysed by the Lewis acid AlCl3.

Mechanism:

CH3 CH2 CH CH2

AlCl3CH3 CH CH CH2

AlCl3H

CH3 CH CH CH3

- AlCl3

CH3 CH2 CH CH2AlCl3 CH3 CH2 CH CH2

AlCl3

CH2 C

CH3

H

CH2

AlCl3

CH2 C

CH3

CH3

- AlCl3

(D) Polymerisation:* Alkenes and substituted alkenes are potential monomer to produce inumerable

polymers from which plastics, rubbers, fibers, paints, adhesives and many other commodities areprepared. The polymerisation belongs to addition type. The pi bonds of C=C open up and addonto each other consuming a large number of alkene molecules to form a macromolecule or apolymer molecule having large molecular mass.

H2C CH

Z

ncatalyst

H2C CH

Z

nmonomer polymer

42


Dr. S. S. Tripathy

Where Z= pendant group in the alkene monomer;n = degree of polymerisation which is very large (> 103). The formula inside parenthesis is calledREPEAT UNIT, which is repeated ‘n’ times in to build one macromolecule.The catalyst used for polyermerization may not be a catalyst, in stead it may be an intiator likea peroxide like benzoyl peroxide(BPO) or AIBN(azo-bis-isobutyronitrile) for radical polymerisationor a Lewis acid like BF

3/H

2O in cationic polymerisation or BuLi for anionic polymerisation. The

actual catalyst(which remains unchanged chemically) is the Zeigler-Natta catalyst of R3Al/TiCl

4

for the Ziegler Natta polymerisation.N.B: The details of polymerisation will be taken in POLYMER chemistry seprately.

H2C CH 2n H2C CH 2

npolyethylene(PE)ethylene

O2

High Pressure

H 2C CHn H 2C CH

npolyviny l chloride(PVC )vinyl chloride

Cl Clperoxide

CH2 CH

CN

n CH

CN

CH2

n

Bu-Li

+

acrylonitrile(AN) polyacrylonitrile(PAN)

CH2 CH

CH3

n CH

CH3

CH2

n

TiCl4 /

Et3Al

polypropylene polypropylene(PP)

The pi-bond of C=C is removed and the repeat unit is kept inside parenthesis with extendedbonds on either side. A subscrip ‘n’ after parenthesis will indicate a polymer moelcule.

ALKENE METATHESIS REACTIONS:* Exchange of C=C between alkenes to give alkene products with different substitutentsand carbon skeleton as compared to reactants and many other related reactions come under thiscategory. The following few examples demonstrates the metathsis between two molecules(sameor different) and hence is called Cross-Metathesis(CM).

R

R

+

R

R

Mo(CO)6+

(Homo CM)

43


Dr. S. S. Tripathy

* In 1964, the above metathesis was carried out using Mo(CO)6 catalyst. This reaction was

symmetry forbidden by W.H rule unlike photochemical dimerisation of ethylene. Even then, ithappend with the participation of catalyst. Catalyst plays the key role in bringing about thischange. Then came the dominance of Grubbs catalyst which was found to much more efficient.

Grubbscatayst

++

(Hetero CM)

R1 R1

Ru complex

R2 R2

R1 R2catalyst

+2

(Grubbs catalyst)

(ethene is not formed)

* Two different alkenes; one bonding to R1 on each side and the other bonding to R

2 on

each side of C=C. Reaction between the two in presence of the catalyst produces alkene bondedto R

1 on one side and R

2 on the other.

Me Me

Ru complex

Et Et

Me Etcatalyst

+2

but-2-ene

hex-3-ene

pent-2-ene

But-2-ene and hex-3-ene react in presence of a Ruthenium based complex catalyst to form pent-2-ene. Is it not interesting ??One can think of a possible [2+2] cycloaddition between the two C=C and ring opening thentaking place in the other direction(analogous to Wittig reaction) to give different alkene. But thereaction does not take place by photochemically method. The mechanism is really intriguing andtakes place by the participation of catalyst which enters into [2+2] cycloaddition with one alkeneand then does the rest of the magic. We shall not discuss the detailed mechanism of the reaction.Catalysts: First generation Grubbs catalyst: (Cy

3P)

2Cl

2Ru=CH-Ar

(Where Cy = cyclohexyl)2nd generation Grubbs catalyst : (Cy

3P)Cl

2(NHC)Ru=CH-Ar

(where NHC = N-heterocyclic carbene)

=

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Dr. S. S. Tripathy

Shrock Catalyst using Molybdenum complex is an improvent over the Grubbs catalyst. However,the second generation Grubbs catalyst is more widely used.Three Component CM:

+

+ +

O

CH2 CH22

Grubbscatayst

O

Isobutene and but-3-en-2-one independely do undergo CM, but in presence of au unjugateddiene(hexa-1,5-diene) forms a cross product alongwith two ethylene molecules.Other types of alkene metathesis reactions:

(a) Cross Metathesis:When two alkenes of same type or different type undergo metathesis to give two alkenes

with changed carbon skeleton, it is called cross metathsis(CM). The examples that we discussedbefore belong to this type. One more example is given below.

R1

R2

R1

R2

+ +

Grubbs Catalyst

Mechanism:

* Look the right side top. CH2=CH-R

1 cyclises with the catalyst L

4M=CHAr to form the

four membered metallocyclobutane ring(RHS middle). This on retro opening in the oppositedirection produces R

1CH=M and CH

2=CHAr(shown as CH

2=CH

2, as the catalyst has been shown

as M=CH2, which is not really true for the first step. It is M=CHR’/Ar). Then R

1-CH=M reacts

with R2–CH=CH

2 to form another metallocyclobutane intermediate(shown in the LHS middle).

This on retro opening of ring in reverse direction gives R1–CH=CH–R and M=CH

2. Now this

M=CH2 starts fresh cycle by reacting with CH

2=CH–R

1 to form cyclbutane interemediate whose

opening gives ethene and R1–CH=M. Only the inititaion step, we do not get ethene, in stead we

get CH2=CHAr but in subsequent propagation steps we get ethene. So we write the products of

this metathesis as R1–CH=CH–R

2 and CH

2=CH

2.

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Dr. S. S. Tripathy

(b) Ring Closing Metathesis(RCM):Terminal acylic diene undergoes RCN to form cycloalkene with elimination of ethene. Octa-1,7-diene, in presence of Grubbs catalyst, undergoes RCM to form cycloheexene. 5 to 30 memberedcyclic alkenes can be prepared by this method.

Mechanism:

(c) Ring Opening Metathesis Polymerisation(ROMP):Cycloakene undegoes ring opening polymerisation in a typical way by the participation of theGrubbs or Schrock catalyst.

Here the Schrock catalyst is used. I am tempted to give the mechanism for this.

46


Dr. S. S. Tripathy

Mechanism:

In the initiation step, a four membered ring is formed with the catalyst. Then retro opening inthe opposite direction gives two C=C, one bearing the metal at one terminal and the other bearingthe carbon with alkyl group(=C-R) from the catalyst at the other terminal. In the propagationstep, cyclization occurs between a new cycloalkene and the M=C at the left terminal and againon retro reopening forms the dimer and then with the third cycloalkene to form trimer and soon. till a polymer is formed. Chain growth takes place leftwards w.r.t the C=R present at theextreme right. In the termination step. the metal is delinked from the polymeric chain by usinga suitable doubly bonded agent(X=Y).

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Dr. S. S. Tripathy

DIENESPreparation of Dienes:

Types: (a) Conjugated (b) Isoloated (c) Cumlated (cumelenes)Conjugated dienes are more stable than isolated dienes. For that reason, often during prepartionof polyenes, conjugated diene is the major product. Cumlated dienes are least stable and hencemost reactive.Preparation of Conjugated and Isolated Dienes:* Dehydration of vic. diols and dehydrohalogenation of vic. dihalides produce conjugateddienes. However, the latter reaction also produce alkyne.* One can plan the synthesis of conjugated dienes many ways. A few examples givenbelow.

(a) Buta-1,3-diene can be prepared from but-1-ene as follows.

CH2 CH CH2 CH3NBS CH2 CH CH CH3

BrKOH(alc.)

CH2 CH CH CH2

(b) Penta-1,3-diene can be prepared from pent-2-ene as follows.

CH CH CH2CH3 CH3

cold dil alkKMnO4 CHCH3 CH3CH2CH

OH OHconc. H2SO4

heat

CHCH2 CH3CHCH

(c) Buta-1,3-diene can be prepared by dehydration of terminal diols as follows.

O3

H2O/ZnOHC CH2 CH2 CHO LiAlH4 HOH2C CH2 CH2 CH2OH

conc.H2SO4

heat

CH2 CH CH CH2

(d) Unconugated diene e.g penta-1,4-diene can be prepared as follows.

O3

H2O/ZnOHC CH2 CH2 CH2 CHO

LiAlH4HOH2C CH2 CH2 CH2 CH2OH

conc.H2SO4

heat

CH2 CH CH CH CH3

CH2 CH CH2 CH CH2 +(unconjugated or isolated diene)

(conjugated diene)

Due to carbocation rearrangement, occuring in the acidic medium, conjuagated penta-1,3-dieneis also formed.

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Dr. S. S. Tripathy

(d) Only penta-1,4-diene(isoated) can be prepared by dehydrohalogenation method.

O3

H2O/ZnOHC CH2 CH2 CH2 CHO

LiAlH4HOH2C CH2 CH2 CH2 CH2OH

CH2 CH CH2 CH CH2

PCl5

H2C CH2 CH2 CH2 CH2

ClClKOH(alc.)

(f) Cyclohexadiene can be prepared from cyclohexene by bromination followed bydehydrohalogenation.

(g) Enyne eg but-1-en-3-yne can be prepared by the dimerisation of acetylene.

CH CH Cu2Cl2NH4Cl

2 CH2 CH C CHbut-1-en-3-yne

Preparation of Cumulated Dienes:(1) From MAPP gas: Allene(propadiene) is present as a equilibrium mixture with propyne calledMAPP gas(methyl acytlene propadiene propane gas). MAPP gas also carries some amount ofpropane. Fractionation of these produces pure allene.

CH3 C CH CH2 C CH2

(2) Gem dihalocyclopropane reacts in presence of alkyl lithium with skeleton rearrangement toform allenes via carbene intermediate.

(3) Dehydrohalogenation of dihalides.1,3-dichloropropane with alcoholic KOH produces allene.Properties of Conjugated Dienes:1. Addition with X

2:

Electrophilic reagents like X2(halogen, Cl2 or Br2) produces two products with unequal yields.(a) 1,2- addition product or direct addtion(b) 1,4- addition product or conjugate addition

1,2-addition is always kinetically controlled product and occurs at lower temperature, while 1,4-addition product is thermodynamically controlled product and occurs at higher temperature.For details refer GOC-III(mechanism). The major product in each case is shown.

CH2 CH CH CH2 X2hexane CH CH CH2CH2

X X

+low temp.

(1,2-addition)

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Dr. S. S. Tripathy

CH2 CH CH CH2 X2acetic acid

higher tempCH CH CH2CH2

X X

+(1,4-addition)

2. Addition with HX:

CH2 CH CH CH2 HBr-80 C

CH2 CH CH

Br

CH3+

1,2-addition

CH2 CH CH CH2 HBr+

1,4-additionroom temp.

CH CH CH3CH2Br

(3) With water(acid catalysed):

CH2 CH CH CH2 H2O CH CH CH2CH3

OH

+low temp.

H+

CH2 CH CH CH2 H2Ohigher temp

CH CH CH2CH3

OH

+H

+

(4) Cycloaddition with a dienophile(Diels Alder Reaction): Refer “Name reaction”.

(5) Polymerisation: buta-1,3-diene on polymerisation with a free radical initiator giives a rubberypolymer polybutadiene(PB).

CH2 CH CH CH2n peroxideCH2 CH CH CH2 npolybutadiene(PB)

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Dr. S. S. Tripathy

E-CONCEPT IN CHEMISTRY

For Class XI( 1st Year +2)

Chemistryof

ALKENES

Dr. S. S. TripathyPresident, The Uranium

(Formerly Senior Reader in ChemistryRavenshaw College, Cuttack)

1. Dehydrohalogenation of Haloalkanes(Alkyl...

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