ARNDT - EISTERT REACTION

* A carboxylic acid can be converted to a higher homologue with one additional carbon

atom by employing Arndt-Eistert reaction.

SOCl2 CH2N2 Ag+

R-

COOH ------------->  R-COCl   --------------> 

 R-COCHN2 

 ----------------------

>  R-CH2CONu 

ether, Et3N Nucleophile (Nu)

* The steps involved in this reaction are:

1) Conversion of carboxylic acid to an active compound like acid chloride or an

anhydride.

R-COOH + SOCl2 --------> R-COCl + SO2 + HCl

2) Conversion of acid chloride to a diazoketone. A base like Et3N is employed to

neutralize HCl liberated in this step.

R-COCl + CH2N2 --------> R-COCHN2 + HCl

3) Wolff rearrangement of diazoketone into a ketene and subsequent conversion of it to a

higher carboxylic acid or its derivative by using a nucleophile.

Ag2O, Δ or hν, -N2 Nucleophile (Nu)

R-COCHN2

 ---------------------------------

----> 

 R-

CH=C=O

 ----------------------------

----> R-CH2CONu

diazoketone Wolff rearrangement ketenehigher

homologue

* Nucleophile (Nu) = water or alcohol or amines

MECHANISM

1) Initially the diazomethane is acylated by the acid chloride to give a diazoketone.

* The HCl liberated in the first step must be neutralized by a suitable base to avoid the

formation of chloromethyl ketone.

 

2) Thus formed diazoketone is rearranged to a ketene. This is called Wolff-

rearrangement.

* Silver salts like PhCO2Ag, Ag2O along with heat or light catalyze the Wolff

rearrangement.

* The configuration of 'R' group during Wolff rearrangement is retained.

 

3) The ketene is immediately attacked by an appropriate nucleophile in the solution.

 

ILLUSTRATIONS

1)

* Nitro group is not affected in the above reaction.

 

 

2)

* The carboxylic group is first converted to an anhydride using ethyl chloroformate,

ClCO2Et.

* Also note that the configuration at stereogenic carbon is retained

ARNDT - EISTERT REACTION - EXERCISES

I) Carry out the following conversions using appropriate reagents.

II) Write the products formed during the following reaction sequences.

BAEYER VILLIGER OXIDATION

* In Baeyer villiger oxidation, the ketones are oxidized to esters by using peroxy acids. 

Note: The reaction is regioselective and in above case, the R' group is assumed to possess

greater migratory aptitude and hence only one product is formed preferentially.

* The reagents which can be employed include:

Metachloroperbenzoic acid (MCPBA), 

Peroxyacetic acid (PAA), 

Peroxytrifluoroacetic acid (TFPAA) 

Hydrogen peroxide/BF3 , 

Caro's acid buffered with disodium hydrogen phosphate

Sodium percarbonate (Na2CO3.1.5H2O2), 

Magnesium salt of monoperoxyphthalic acid (MMPP), 

Potassium peroxomonosulphate (potassium caroate) supported on hydrated silica also

known as "reincarnated caro's acid".

Baeyer villiger monooxygenase (an enzyme abbreviated as BVMO).  

* Cyclic ketones furnish lactones (cyclic esters). 

* The aldehydes may give carboxylic acids or formates. In the latter case, alcohols are

finally formed due to hydrolysis of unstable formates under the reaction conditions. 

* This reaction involves the oxidative cleavage of C-C bond.

 

MECHANISM

 * Initially the peroxy group is added to the carbonyl carbon to give a Criegee like

intermediate. Then one of the group attached to carbonyl carbon is migrated on to the

electron deficient oxygen atom in a concerted step, which is the rate determining step. 

* The substituents which can stabilize the positive charge can migrate readily. The

migratory aptitude of various substituents is approximately:

 3o-alkyl > cyclohexyl >  2o- alkyl > benzyl > aryl > 1o - alkyl > methyl.

* The electron withdrawing groups (-I groups) on peroxy acids enhance the rate of the

reaction.

* Thus the Baeyer villiger oxidation of unsymmetrical ketones is regioselective.

* As the rearrangement is a concerted process, the configuration of the migrating chiral

substituent is retained.

* In case of aldehydes, usually the hydrogen atom is migrated preferentially and thus by

furnishing carboxylic acids. But formates are also produced when the migrating group is

other than the hydrogen. This is possible when the other substituent is a tertiary alkyl group

or electron rich vinyl or aryl group (e.g. Dakin reaction).

* One of the competing reaction is the formation of epoxide when a double bond is

present in the molecule especially at low temperatures in neutral solvents.

 

ILLUSTRATIONS

1) In the following Baeyer villiger oxidation, the preferential migration of more

substituted secondary alkyl group is observed (regioselective) without disturbing its chiral

integrity i.e., the configuration at the chiral carbon is retained in the product

(stereospecificity).

 

2) Cyclic ketones furnish lactones as illustrated below.

 

3) Cyclohexyl group migrates preferentially over methyl group as illustrated below:

 

4)  In the following example, the subsequent hydrolysis of the ester gives the desired

alcohol. Another example of regioselectivity and stereospecificity.

 

5) Baeyer villiger oxidation is preferred over the epoxidation of double bond by the

peracid as illustrated in the following reaction.

6) The lactone formed can be reduced to a dihydric alcohol.

 

7) The bacteria strain variants which produce BVMO can be employed in Baeyer villiger

oxidation.

 

8) In the following reaction, the aldehyde is oxidized to formate due to preferential

migration of aryl group. But it undergoes hydrolysis under the reaction conditions to yield a

phenol. This reaction is similar to that of Dakin reaction.

 

9) As illustrated below, the aldehyde group is oxidized to carboxylic acid due to

preferential migration of the hydride ion. The aryl group with electronegative halogen groups

has less migratory aptitude. Remember the groups which can stabilize positive charge possess

greater migratory aptitude.

 

10) The greater migratory aptitude of aryl group over the -CH2 group can be observed in

the following example.

 

11) The -CH2 group is migrated preferentially in the following reaction. The -CH-CF3

group has less migratory aptitude due to electron withdrawing nature.

BAEYER VILLIGER OXIDATION - EXERCISES

I) Write the products formed in the following reactions.

BAMFORD-STEVENS REACTION

* In the Bamford-Stevens reaction, the tosyl hydrazones (p-Toluenesulfonyl hydrazones)

of aliphatic aldehydes or ketones furnish more substituted alkenes when treated with strong

bases like NaOMe, NaH, LiH, NaNH2 etc.

* The reaction may be performed either in protic solvents like glycols or in aprotic

solvents like ethylene glycol dimethyl ether. 

* Both the Bamford-Stevens reaction and the Shapiro reaction afford alkenes from tosyl

hydrazones.

* In case of Bamford-Stevens reaction, the more substituted alkene is formed as the

thermodynamic product. 

* Whereas in Shapiro reaction, the less substituted alkene is formed as the kinetic

product. This reaction employs bases such as alkyllithiums and Grignard reagents.

 

MECHANISM

 * The mechanism involves two steps. Initially, the reaction of tosyl hydrazone with a

strong base leads to a diazo compound, which can be isolated in some cases.

* The diazo compound may follow either one of the two pathways depending on the

reaction conditions. In protic solvents, the reaction proceeds via formation of carbenium ion,

whereas in aprotic solvents, the reaction proceeds via a carbene.

 

In protic solvents:

* In protic solvents, the diazo compound abstracts a proton from the solvent and thus by

forming a diazonium ion, which subsequently loses dinitrogen to give a carbenium ion.

Finally, a mixture of E & Z alkenes is formed from the carbenium ion through loss of a

proton.

* However, carbenium ions can easily undergo a Wagner–Meerwein rearrangement, and

hence the corresponding rearranged alkenes may be formed as side products in protic

solvents. 

Note: A carbenium ion is a trivalent carbocation. Whereas, the carbocation with five

coordinated carbon is nowadays referred to as a carbonium ion.

In aprotic solvents:

* In aprotic solvents, the diazo compound loses dinitrogen and gives a carbene, which

undergoes a faster 1,2-hydrogen shift to furnish a Z-alkene predominantly.

* The desired alkene is obtained in high yield in aprotic solvents.

 

ILLUSTRATIONS

1) Bamford-Stevens reaction of tosyl hydrazone of 2-methylcyclohexanone affords more

substituted 1-methylcyclohexane.

Whereas, the Shapiro reaction conditions lead to less substituted 3-methylcyclohexane.

 

2) The Bamford-Stevens reaction of tosyl hydrazone of cyclopropane carbaldehyde

furnishes bicyclobutane: a special case.

BECKMANN REARRANGEMENT

* The Beckmann rearrangement is an acid catalyzed rearrangement of an oxime  to an N-

substituted amide. 

* Conc.H2SO4, HCl, PCl5,  PCl3, SOCl2, ZnO, SiO2, PPA (Poly phosphoric acid) etc., are

commonly employed in Beckmann rearrangement.

* Mostly applied for ketoximes. 

* Aldoximes are less reactive. 

* Cyclic oximes give lactams (cyclic amides).

 

MECHANISM

 * Initially the -OH group of the oxime is protonated. Then 1,2 shift of alkyl group (R1)

onto electron deficient nitrogen and the cleavage of N-O bond occurs simultaneously. 

Always the alkyl group which is 'anti' to the -OH group on nitrogen undergoes 1,2 shift

which indicates the concerted nature of the beckmann rearrangement.

Comment: The migration of hydrogen is seldom observed. Hence the N-unsubstituted

amides cannot be obtained by this reaction.

 

ILLUSTRATIONS

1) Industrial conversion of cyclohexanone to caprolactam, which is used in the

manufacture of Nylon-6, involves Beckmann rearrangement.

 

 

2) The relative migratory aptitudes of different groups in Beckmann rearrangement is

illustrated below.

The 1,2 shift  of phenyl group is faster than that of alkyl groups. It is due to formation of

phenonium ion. Hence the anti isomer reacts faster than the syn isomer. 

 

3) Abnormal Beckmann rearrangement occurs when the migrating group departs from the

intermediate and thus by furnishing a nitrile. 

BECKMANN REARRANGEMENT - EXERCISES

I) Mention the products formed in the following reactions.

 

II) How do you carry out the following conversions?

 

III) Propose a mechanism for the following reaction.

BIRCH REDUCTION

* In Birch reduction, aromatic rings are reduced to 1,4-dienes by alkali metals in liquid

ammonia.

* Commercial ammonia often contains iron as impurity. Therefore, it is often necessary to

distill the ammonia before using it in the reaction.

* The reaction is carried out at -33oC (boiling point of ammonia). Co-solvents like Et2O,

THF, DME etc., are added to improve the solubility of organic compounds at this

temperature.

MECHANISM

 

Comments:

* The alkali metals dissolve in liquid ammonia by forming solvated electrons which are

responsible for the reduction.

* The second protonation is almost always occurs at a site para to the first protonation

site.

* Hence in the final step of protonation, thermodynamically less stable but kinetically

favored 1,4-diene is formed predominantly (about 80%). But thermodynamically more

stable 1,3-diene(conjugated) is not formed predominantly as it is not favored kinetically.

* The formation of 1,4-diene can also be explained by "the principle of least motion",

which states that the reaction that involves the least change in atomic positions or electronic

configuration is favored

* Alcohols (EtOH or t-BuOH) are used as protonating agents. They also suppress the

formation of amide, NH2- ion, which may otherwise isomerize the 1,4-diene to more stable

1,3-diene.

* Under the reaction conditions (below 240 K), the alcohols do not react with the metals. 

* Relatively the reductions using Li metal are very faster.

 

ILLUSTRATIONS

1) Naphthalene is reduced to 1,4,5,8-tetrahydronaphthalene.

 

Regioselectivity: The positions of protonation on substituted benzenes depend on the

nature of the group. i.e., whether it is electron withdrawing group (EWG) or electron

donating group (EDG). 

 

* EWG: The electron-withdrawing groups promote ipso & para reduction. These groups

activate the ring towards birch reduction. Initially the protonations occurs para to the EWG.

E.g. -COOH, -CONH2, aryl group etc.,

 

* EDG: The electron-donating groups promote ortho & meta reduction. They deactivate

the ring for overall reduction compared to the EWG.

E.g. -R, -OR, -NR2, -SR, PR2, -CH2OH, -CHO, -C(O)R, CO2R etc.,

The -CHO, -C(O)R, CO2R act as electron donating groups because they are reduced to -

CH2O- prior to the reduction of the ring.

 

2) In the birch reduction of benzoic acid, the protonation occurs at ipso and para positions

relative to -COOH group on the benzene ring.

comments:

* The reason is electron withdrawing groups stabilise the radical anion at ipso and para

positions.

* The carboxylate ion, -COO- formed during the reaction is electron rich and hence

cannot be reduced. 

* The reduction is also possible even without the presence of alcohol due to strong

activation by -COOH group.

 

3) Protonation occurs at ortho and meta positions of benzene ring incase of anisole and

thus by giving more substituted double bond. It is due to the fact that, electron donating

groups stabilise the radical anion at ortho and meta positions.

* The above product can be hydrolyzed to β,γ-unsaturated ketone in presence of mild

acid. However, &alpha,β-unsaturated ketone is formed due to isomerization in vigorous acid

catalytic conditions. 

 

4) The electron donating groups deactivate the ring towards birch reduction. This is

exemplified below.

 The reduction occurs in the unsubstituted ring of naphthalene.

 

5) But with aniline derivatives (even though electron donating), the conjugated enamines

are formed directly as the major products (No need of acid catalyst). 

 

6) In case of phenols, birch reduction is not possible. It is because, phenolic function

becomes phenolate ion under the reaction conditions (basic) and does not react further.

 

7) The carboxyl groups have a dominating directive influence than other groups.

 

 

8) The directive influence of alkoxy groups is greater than that of alkyl groups.

 

 

9) The carbanion formed during birch reduction can undergo alkylation.

 

10) Electron deficient heterocyclic aromatic compounds can be reduced.

E.g. Pyridine gives 1,4-dihydropyridine, which can be further hydrolyzed to 1,5-

dicarbonyl compound.

 

11) Nowadays alkali metals encapsulated in nano structured oxides like silica gel are used

instead of liquid ammonia-metal solutions. E.g. sodium in silica gel (Na-SG). In the

following example, phenanthrene is reduced to 9,10-dihydrophenanthrene

Suggest the products likely to be formed during the birch reduction of the following

compounds.

 

II) Write the products formed in the following reactions.

CANNIZZARO REACTION : EXPLANATION

* The disproportionation reaction of aldehydes without α-hydrogens in presence of a

strong base to furnish an alcohol and a carboxylic acid is called Cannizzaro reaction. One

molecule of aldehyde is reduced to the corresponding alcohol, while a second one is

oxidized to the carboxylic acid.

* The applicability of Cannizzaro reaction in organic synthesis is limited as the yield is

not more than 50% for either acid or alcohol formed.

* In case of aldehydes that do have α-hydrogens, the aldol condensation reaction takes

place preferentially.

* The α,α,α-Trihalo aldehydes undergo haloform reaction in strongly alkaline medium.

E.g. Choral will give chloroform in presence of an alkali.

 

MECHANISM

 

* The reaction is initiated by the nucleophilic attack of a hydroxide ion to the carbonyl

carbon of an aldehyde molecule by giving a hydrate anion. This hydrate anion can be

deprotonated to give an anion in a strongly alkaline medium. In this second step, the

hydroxide behaves as a base.

* Now a hydride ion, H- is transfered either from the monoanionic species or dianionic

species onto the carbonyl carbon of another aldehyde molecule. The strong electron donating

effect of O- groups facilitates the hydride transfer and drives the reaction further. This is the

rate determining step of the reaction.

 

 

* Thus one molecule is oxidized to carboxylic acid and the other one is reduced to an

alcohol.

* When the reaction is carried out with D2O as solvent, the resulting alcohol does not

show carbon bonded deuterium. It indicates the hydrogen is transferred from the second

aldehyde molecule, and not from the solvent. 

* The overall order of the reaction is usually 3 or 4.

* The Cannizzaro reaction takes place very slowly when electron-donating groups are

present. But the reaction occurs at faster rates when electron withdrawing groups are present.

 

ILLUSTRATIONS

1) Formaldehyde is disproportionated to formic acid and methyl alcohol in strong alkali.

 

2) Benzaldehyde can be converted to benzoic acid and benzyl alcohol.

 

3) Furfural gives furoic acid and furfuryl alcohol in presence of strong alkali.

 

 

4) When a mixture of formaldehyde and a non enolizable aldehyde is treated with a

strong base, the later is preferentially reduced to alcohol while formaldehyde is oxidized to

formic acid. This variant is known as crossed Cannizzaro reaction.

E.g. Benzyl alcohol and formic acid are obtained when a mixture of benzaldehyde and

formaldehyde is treated with alkali.

The reason may be: the initial nucleophilic addition of hydroxide anion is faster on

formaldehyde as there are no electron donating groups on it.

The preferential oxidation of formaldehyde in crossed Canizzaro reactions may be

utilized in the quantitative reduction of some aldehydes. 

 

5) α-keto aldehydes can be converted to α-hydroxy carboxylic acids by an intermolecular

Cannizzaro reaction.

E.g. Phenylglyoxal undergoes intramolecular cannizzaro reaction by giving Mandelic

acid (α-hydroxyphenylacetic acid or 2-Hydroxy-2-phenylethanoic acid)

 

6) Phthalaldehyde can undergo intramolecular Cannizzaro reaction by giving (o-

hydroxymethyl) benzoic acid.

 

CLEMMENSEN REDUCTION : EXPLANATION

* The carbonyl compounds, which are stable to strongly acidic conditions, can be reduced

to alkanes conveniently by this method. Amalgamated zinc in HCl is used as reducing agent.

The C=O group is converted to CH2 group.

* This reaction is complementary to Wolff-Kishner reduction, which can be used to

reduce acid sensitive compounds.

 

MECHANISM

 

* The reduction occurs over the surface of zinc catalyst. The probable mechanism is

shown below. 

 

* There is a net flow of electrons from zinc to the carbonyl compound.

* As there is no formation of alcohol during the reaction, this method is not useful to

reduce alcohols to alkanes.

 

ILLUSTRATIONS

1) 

  Zn-Hg  

CH3CHO + 4(H)  -----------------> CH3CH3 + H2O

   HCl    

2) 

3) In the following reaction, along with the reduction of carbonyl group, the -OH group is

substituted by the -Cl group (side reaction). However the side reaction can be avoided by

employing Wolff-Kishner method.

4) But the phenol group is not affected. (Why? Ans: Nucleophilic substitution is not easy

on sp2 carbon of benzene ring!)

 

FAVORSKII REARRANGEMENT

* The Favorskii rearrangement is the base catalyzed rearrangement of enolizable α-

haloketones or cyclopropanones to carboxylic acids or their derivatives.

The α-haloketones must contain acidic α'-hydrogens.

 

 

* Esters are formed if alkoxides are used as bases. Amides are formed when amines are

used as bases.

* The Favorskii rearrangement of α-halo cyclic ketones results in ring contraction.

 

MECHANISM

* Initially, the α'-carbon is deprotonated to generate an enolate ion which is followed by

intra molecular nuclophilic substitution to give a cyclopropanone by ring closure.

* Thus formed cylcopropanone undergoes nucleophilic addition by a base at the carbonyl

carbon which is followed by the cleavage of the CO-Cα bond. Usually the cleavage occurs so

as to give less substituted and more stable carbanion.

 

ILLUSTRATIONS

1) 2-Chlorocyclohexanone can be converted into ethyl cyclopentanecarboxylate by

treating with sodium ethoxide. The six membered ring is contracted to the five membered

one.

 

2) 3-Bromobutan-2-one yields 2-methypropanoic acid as major product when treated with

alkali. The ring opening of cyclopropanone derivative occurs to give less substituted

carbanion. 

 

3) The α-haloketones will rearrange to amides when the Favorskii reaction is catalyzed by

amines. 

 

4) Synthesis of cubane involves favoroskii rearrangement step as shown below.

 

5) α,α-dihaloketones or α,α'-dihaloketones rearrange to give α, β-unsaturated carboxylic

acid derivatives under Favorskii conditions. For example, 1,3-Dibromobutan-2-one will give

ethyl but-2-enoate when treated with sodium ethoxide.

 

 

6) The rearrangement of α,α-dihaloketones to α-hydroxycarboxylic acids, followed by

oxidative decarboxylation to the ketones is known as Wallach degradation.

 

7) The rearrangement of non enolizable α-haloketones in presence of bases is called

quasi-Favorskii rearrangement. The synthesis of powerful pain killer Pethidine (also known

as Demerol or Meperidine) makes use of this rearrangement. 

The quasi-Favorskii rearrangement involves no cyclopropanone intermediate and  follows

a semibenzilic mechanism as shown below.

 

FAVORSKII REARRANGEMENT - EXERCISES

I) Write the products formed in the following reactions.

FITTIG REACTION

* In Fittig reaction, two aryl halides are coupled in presence of sodium metal in dry ether

or tetrahydrofuran to furnish biaryls.

Dry Ether

Ar-X  +  2Na  +  X-Ar ----------------->  Ar-Ar + 2NaX  

Where Ar = aryl group, X = halogen

* The yields will be improved by using ultrasound, especially in two-phase reactions.

* A modification of reaction which involves, an alkyl halide and an aryl halide is called

Wurtz-Fittig reaction.

* Refer Wurtz reaction for the reaction conditions and the detailed mechanism.

ILLUSTRATIONS

1) Biphenyl can be prepared by Fittig method as follows:

MICHAEL ADDITION : EXPLANATION

* The Michael reaction is the conjugate 1,4-addition of a resonance stabilized carbanion

(michael donor)  to an activated α,β-unsaturated  compound (michael acceptor). 

Michael donors: 

The Michael donors contain active -CH2 (methylene) group or -CH group. The acidic

nature of methylene group is enhanced by the electron withdrawing groups (EWG) like: keto,

cyano, nitro, carboxylic acid derivatice etc.

E.g. Active methylene compounds like  Malonates (e.g.Diethyl malonate), β-keto esters

(e.g.Acetoacetic ester) etc., are some of the examples for Michael donors.

Michael acceptors:

Not only  α,β-unsaturated ketones and also esters, nitriles, sulfones and compounds with

activated double bonds act as Michael acceptors. Vinyl ketones, alkyl acrylates, acrylo nitrile,

fumarates etc., are some examples.

* A strong base is employed to generate enolate ion from Michael donor. The bases used

in the reaction include: alkoxides, LDA, NaOH, KOH  etc., in protic solvents like alcohols.

* Michael addition is a thermodynamically controlled conjugate 1,4 addition reaction and

competes with kinetically controlled 1,2 addition to C=O. At low temperatures, 1,2 additon

occurs predominantly. But at higher temperatures, the michael addition is the preferred

route.

MECHANISM

 

* Initially a resonance stabilized  enolate ion (nucleophile) is produced from Michael

donor in presence of a strong base. 

It is added to the 4th carbon of  α,β-unsaturated carbonyl compound and thus by giving a

new enolate, which usually yields the final compound upon hydrolytic workup.

Note: The final product formed in protic workup appears to be a 3,4-addition product.

 

* The conjugate 1,4-addition product is thermodynamically controlled and occurs

predominantly at relatively higher temperatures. But a kinetically controlled 1,2-addition is

also possible which is preferred at low temperatures as shown below.

 

ILLUSTRATIONS

1) Michael addition of diethyl malonate with methyl vinyl ketone followed by protic

workup yields a 1,5-dicarbonyl compound. 

Note: Hydrolysis of ester groups and decarboxylation occurs in the final step.

 

2) Ethyl acetoacetate (acetoacetic ester) can be used as michael donor.

 

3) In the following reaction, Michael addition of  diethyl malonate to mesityl oxide yields

a 5-Oxocarboxylic acid.

 

4) Usually more substituted alpha-carbon of Michael donor is involved in the addition.

 

5) Robinson annulation involves, the Michael addition followed by intramolecular aldol

condensation as illustrated below.

 

6) Both cis and trans isomers are possible in Michael addition involving alkynes.

 

7) Michael additions to extended conjugate systems is also observed. Following is an

example of 1,6-addition.

 

8) Nitroalkanes with α-hydrogens are excellent Michael donors. The Michael addition of

nitromethane with acrylonitrile is one such example. In this case, three moles of acrylonitrile

are involved in the addition.

 

9) Enamines are excellent Michael donors and require no base.

 

10) Bicyclo[2.2.2]octane systems can be prepared by double michael addition as

illustrated below.

 

MICHAEL ADDTION REACTION - EXERCISES

I) Suggest the addition products likely to be formed between the following Michael

donor and Michael acceptor under the thermodynamic conditions.

 

PHILLIPS CONDENSATION REACTION : EXPLANATION

* Phillips reaction involves the condensation of ortho phenylenediamines with organic

acids in presence of dilute mineral acids to furnish benzimidazoles. 

* This method has the advantage that the benzimidazoles, which cannot be prepared by

heating the components together, are obtained easily  by using dilute acids at lower

temperatures.

* Good yields are obtained with aliphatic acids.

* However the yield can be improved by carrying out the reaction in sealed tubes with

aromatic acids also.

MECHANISM

 

* Initially one of the amine group is acylated with the organic acid in presence of mineral

acid to furnish an N-acylated compound. In the next step, the other nitrogen is also acylated

by making bond with the carbonyl carbon of the first acyl group leading to ring closure.

ILLUSTRATIONS

1) Benzene-1,2-diamine can be condensed with acetic acid in presence of 4N HCl to give

2-methyl-1H-benzimidazole.

2) 2-phenyl-1H-benzimidazole can be prepared by carrying out the reaction in a sealed

tube at 180oC

REFORMATSKY REACTION : EXPLANATION

* The Reformatsky reaction involves the treatment of an α-halo ester with zinc metal and

subsequent reaction with aldehyde/ketone to get β- hydroxy ester.  

* Usually inert solvents like diethyl ether or THF are used.

* Better yields are obtained by using Zn-Cu couple or in situ preparation of zinc by

reduction of zinc halides by potassium (also known as Rieke zinc).

MECHANISM

 

* Initially zinc reacts with α-halo ester to give an organozinc reagent called reformatsky

enolate. It is just like the Grignard reagent. It is added to the carbonyl group of aldehyde or

ketone to furnish β- hydroxy ester.

 

* The organozinc reagents are less reactive and hence the nucleophilic addition to the

ester group seldom occurs. Some of them are quite enough stable to be isolated and can be

elucidated for the structure by techniques like X-ray analysis.

ILLUSTRATIONS

1) 

  Zn   H3O+  

CH3CHO + Br-CH2-COOC2H5

  -----------------

> ----------->  CH3CH(OH)-CH2COOC2H5

   Diethyl ether    

2) Coumarin is formed from the β- hydroxy ester as final product.

I) Write the product(s) likely to be formed in the following reactions.

 

II) How do you carry out the synthesis of following compounds using Reformatsky

method?

WILLIAMSON'S SYNTHESIS

* In this method, an ether is prepared by the nucleophilic substitution (typically SN2) of

organic halide with alkoxide ion.

R-O-    +    R'-X  ----------------->  R-O-R'    +   X-

Alkoxide

ionHalide Ether

* The alkoxide ion is generated in situ by treating an alcohol with a metal or a strong

base.

* It follows bimolecular nucleophilic substitution (SN2) pathway. 

* Symmetrical or unsymmetrical ethers can be prepared.

* The nature of alkoxide ion is less important. It may be primary or secondary or tertiary.

* But due to strongly alkaline conditions, dehydrohalogenation (elimination) is a side

reaction. Hence the yields are relatively better with methyl or primary alkyl halides only.

The yields are affected when halides contain β-hydrogen. Elimination products are

formed exclusively with tert-halides.

MECHANISM

 

* It is a typical SN2 reaction. The alkoxide ion attacks the carbon atom containing the

halogen atom from the back side. The bond making and breaking occurs simultaneously in

the transition state. 

 

ILLUSTRATIONS

1) In the following reaction, initially, the sodium ethoxide is generated by treating ethyl

alcohol with sodium metal.

  Na   C2H5Cl  

C2H5OH  -------------->  C2H5O-Na+   ------------------>  C2H5OC2H5

      -NaCl  

2) A cyclic ether is formed in the following reaction.

 

3) In the following reaction, propene is also formed in good quantities due to elimination

side reaction. 

 

4) An epoxide can be synthesized from a halohydrin using this reaction.

 

5) Phenoxide ions can be employed to get aromatic ethers. 

 Note: Halobenzenes do undergo nucleophilic substitution and hence they cannot be not

used in Williamson's synthesis.

WILLIAMSON'S SYNTHESIS: EXERCISES

I) How do you get following ethers using Williamson's synthesis. Write the starting

materials and also mention any side products formed.

 

WURTZ REACTION

* Wurtz reaction is a coupling reaction, in which two alkyl halide molecules are made to

react with sodium metal in anhydrous ether or tetrahydrofuran to form a new carbon carbon

bond and thus by giving a symmetrical alkane.

Dry Ether

R-X  +  2Na  +  X-R ----------------->  R-R + 2NaX  

Where X = halogen

* The reaction must be performed under anhydrous conditions because the alkyl free

radical formed (see the mechanism) during the reaction is strongly basic and can abstract

proton from water.

* In case of alkyl and aryl fluorides as well as aryl fluorides, tetrahydrofuran is used as

solvent instead of ether. 

* This reaction is limited to synthesis of symmetrical alkanes with even number of carbon

atoms only. The number of carbons in the alkane is double that of alkyl halide (n ---> 2n type

reaction)

* If dissimilar alky halides are used, a mixture of alkanes is formed. It is usually difficult

to separate the mixture and hence is not a suitable method to synthesize unsymmetrical

alkanes.

E.g. The Wurtz reaction between R-X and R'-X yields not only R-R' but also R-R and R'-

R'. This mixture cannot be separated easily.

* Methane cannot be prepared by this method.

* A modification of this reaction involving alkyl and aryl halides is called Wurtz-Fittig

reaction. If only aryl halides are subjected to coupling, the reaction is called as Fittig reaction.

MECHANISM

 

* Initially an alkyl free radical is formed due to transfer of one electron from sodium

atom.

 R-X  +  Na ---------

> R.  +               X-

Alkyl free radical

* In the next step, one more electron is transferred from second sodium atom to the free

radical to give a carbanion.

 R.  +  Na ---------

> R-Na+     

* Thus formed alkyl anion displaces halide ion from the second molecule of alkyl halide.

It is an SN2 reaction.

 R- Na+  +  R-X ---------

> R-R  +            Na+X-

  

Symmetrical Alkane

Comment: Since the alkyl free radicals are formed, elimination side reactions leading to

alkenes is also possible, especially with bulky alkyl groups, which require more activation

energy during the nucleophilic substitution (SN2) step.

ILLUSTRATIONS

1) Ethane is formed when methyl chloride is treated with sodium metal in dry ether.

Dry Ether

CH3-Cl  +  2Na  +  Cl-CH3 ----------------->  CH3-CH3 + 2NaCl  

2) Strained carbon skeletons like bicyclobutane ( bicyclo[1.1.0]butane ) can be prepared

by an intramolecular Wurtz reaction as shown below.

 

 

3) When tert-butylhalides are subjected to wurtz reaction, isobutylene is formed as the

major product. It is because the elimination is favored over SN2 mechanism. The SN2 step

requires more activation energy due to steric hindrance. 

WURTZ-FITTIG REACTION

* The Wurtz-Fittig reaction is the modification of Wurtz reaction. It involves the coupling

of an aryl halide with an alkyl halide molecule in presence of sodium metal to furnish

alkylated aromatic hydrocarbons.

Dry Ether

Ar-X  +  2Na  +  X-R ----------------->  Ar-R + 2NaX  

Where Ar = Aryl group, R = alkyl group, X = halogen

 

* The more reactive alkyl halide will form the organosodium initially, which acts as a

nucleophile and attacks the aryl halide. 

* Usually the yields are very high.

* Refer Wurtz reaction for the reaction conditions and the detailed mechanism.

ILLUSTRATIONS

1) Toluene can be prepared by Wurtz-Fittig method as follows: