SANDMEYER’S REACTIONWhen solution of freshly prepared diazonium salt is mixed with cuprous chloride or cuprous bromide, then diazonium group is replaced by Cl or Br.

Where X = Cl, Br

Diazonium salt is inturn prepared by reaction by aniline with cold aqueous mineral acid & sodium nitrite at a temperature of 0 – 50C. This reaction is called Diazotisation.

Kinetics & MechanismSandmeyer’s reaction is an example for aromatic nucleophilic substitution.

This reaction is found to follow the rate law: Rate = The mechanism is analogous to SN1

The evidence for the reversible slow step was obtained by the following observation made

FINKELSTEIN REACTIONAlkyl iodides are often prepared by the reaction of alkyl chlorides / bromides with NaI in dry acetone. This reaction is called Finkelstein reaction.

Where X = Cl, Br NaI being more covalent is soluble in acetone, where as NaCl & NaBr are

relatively more ionic than NaI & hence get precipitated in acetone. This facilitates forwarded reaction according to Le-Chatelier’s principle.

SWART’S REACTIONThe synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride / bromide in the presence of AgF, Hg2F2, CoF3, SbF3. The reaction is termed as Swart’s reaction.

Similarly Allyl chloride Allyl fluoride

WURTZ FITTIG REACTIONHaloarenes when treated with alkylhalide in presence of sodium and dry ether form alkyl benzene

The reaction is said to proceed through free radical mechanism

FITTIG REACTIONHaloarenes when treated with sodium metal in the presence of dry ether produce diaryl compound

The reaction is said to proceed through free radial mechanism.

ULMANN REACTIONWhen iodoarene is heated with Cu powder, diaryl compound is formed.

The reaction is also said to proceed through free radical mechanismFRANKLAND REACTION

When an alkyl halide is heated with Zn powder, higher alkane is formedR – I + Zn + I – R R – R + ZnI2Some Examples

i)

ii)

SYNTHESIS OF GRIGNARD REAGENT

Grignard reagents can be obtained by the reaction of alkyl halide with Magnesium metal in dry ether.Ether solvates and thus stabilizes the Grignard reagentR – X + Mg RMgX Carbon – magnesium bond is covalent but highly polar and magnesium –

halogen bond is ionic The order of ease of formation of Grignard reagents is

R – MgI > R – MgBr > R – MgCl R – MgF is difficult to prepare from alkyl fluorides, because the latter don’t react

with Magnesium All mono halides except allyl and benzyl halides form Grignard reagents

Reason: The Grignard reagent of alkyl and benzyl halides readily react among themselvesEx: Ph – CH2 – MgX Ph – CH2 – CH2 – Ph

Grignard reagents are highly reactive and react with any source of proton to hydrocarbon. it is necessary to avoid traces of moisture from a Grignard reagentR – MgX + H2O RH + Mg(OH)X

alkane Viscinal dihalides undergo dehalogenation reaction with Mg instead of Grignard

reagent formation

1, 3 – dihalides undergo cyclisation to given cyclopropane when reacted with Mg

Cl – CH2 – CH2 – CH2 – Cl + Mg

HYDROBORATION-OXIDATIONThis procedure involves addition of BH3 to the double bond, with hydrogen being

attached to one doubly bonded carbon and boron to the other. The alkylborane is then oxidized during which ‘B’ is replaced by –OH.

Mechanism:Stage I:

Hydroboration is regioselective and it is anti-Markownikoff’s addition.

The two B-H bonds of alkyl borane undergo similar additions, finally leading to trialkyl borane.

Stage II: Oxidation of trialkyl boranes

CH2 – CH2 – R |

R – CH2 – CH2 – B – CH2 – CH2 – R R – CH2 – CH2 – O – B – O – CH2 – CH2 – R |

Trialkyl borate O – CH2 – CH2 – R

3OH-

3R – CH2 – CH2 – OH + Overall result is “SYN ADDITION” of H & OH groups on the double bond, i.e., the

reaction is stereospecific Alkanes can be prepared by heating trialkyl borane with CH3COOH.

The advantage of Hydroboration – Oxidation over acid – catalysed hydration of alkenes is that there is no possibility of any rearrangement.

ACID – CATALYSED HYDRATION OF ALKENESAlkenes react with water in the presence of dil acid as catalyst to form alcohols.

MechanismMechanism of the reaction involves three stepsStep I: Protonation of alkene to form “Cabocation”

Step II: Nucleophilic attack of water on carbocation

Step III:

All the steps are reversible and step I is rate determining Because carbocation is involved the overall addition of H2O onto the double

bond follows Markownikoff’s rule

Ex:

Major product Acids that have weakly nucleophilic anion such as dil H2SO4 are chosen as

catalyst

TRY TO ANSWERWhy dil HCl is not preferably used as catalyst in the acid catalysed hydration of alkenes? Can H3PO4 be used as the catalyst?

The reverse of this mechanism is dehydration (EI mechanism) of alcohol to get alkene, which takes place at higher temperature.

Because carbocation is formed in the mechanism, there is always a possibility of rearrangement.

Ring expansions also takes place(Demjanov rearrangement) in certain cases.

Ex;

DOW’S METHODPhenol is obtained by reacting chlorobenzene with fused NaOH at 523K and

300atm followed by acidification with HCl. This is called Dow’s method of preparation of phenol.

KOLBE’S REACTION( also called KOLBE’S-SCHMIDT REACTION)When sodium phenoxide is reacted with CO2 under pressure followed by hydrolysis then o-hydroxybenzoic acid is formed.

The reaction is an example for electrophilic aromatic substitution.Phenol is first converted to phenoxide because the later is more reactive towards

electrophilic aromatic substitution (especially when weak electrophiles like CO2

are used)Mechanism

Ortho substitution occurs majorly because of stabilization of intermediate formed in step-I due to chelation of Na+ ion.

When potassium phenoxide is used p-substituted product is formed in major amounts.Reason: Due to bigger size and low positive charge density on metal ion, chelation does not occur efficiently.

TRY TO ANSWER1. What happens when phenol is reacted with CCl4 in presence of KOH?2. Complete the following reactions:

RIEMER-TIEMANN REACTIONPhenol when refluxed with CHCl3 and alkali yields ortho and para-hydroxybenzaldehyde. The ortho product is the predominant one.

The reaction is electrophilic substitution where dichlorocarbene is the electrophile.MechanismStage I: Generation of carbene

Stage II: Electrophilic aromatic substitution

Pyrrole on Riemer-Tiemann reaction gives an abnormal product due to ring expansion along with normal product.

If one of the ortho positions is occupied, the para substituted product is formed predominantly.

Ex:

Other example,

TRY TO ANSWER1. Predict the structures of the products formed when indole undergoes Riemer- Tiemann reaction.

2. Complete the following reactions:

(i) (ii)

WILLIAMSON’S SYNTHESISIn this method, an alkyl halide is allowed to react with sodium alkoxide. As aresult, ethers are formed. Both symmetrical and unsymmetrical ethers can be prepared by this method.

Mechansim:Reaction involves SN2 attack of alkoxide ion on alkyl halide.

Better results are obtained when alkyl halide is primary. In case of secondary and tertiary alkyl halides elimination reaction competes with substitution. Consequently, when tertiary alkyl halide is taken only elimination product is formed.

2-methyl propene (almost exclusively formed)

In order to prepare phenyl ethers, phenoxide must be reacted with alkyl halide.

The otherwise is not feasible under normal conditions because phenyl halides donot undergo nucleophilic substitution reactions.

The reaction is more faster if good leaving group is present. Consequently, theorder of reactivity of various halides towards Williamson’s synthesis is

CLEAVAGE OF C – O BOND IN ETHERSEven though ethers have oxygen atom carrying two lone pairs, they are inert towards substitution reactions because (strong base) is not a good leaving group. But if the ethers are protonated, then they become reactive towards nucleophile, because now the leaving group is an alcohol which is a very weak base.Ether when reacts with cold conc. HI or HBr undergoes cleavage to form alcohol and corresponding alkyl halide.

The mechanism of the above reaction involves

(i) Formation of oxonium ion

(ii) Nucleophilic attack of on oxonium ion to produce alcohol and alkyl halide.Whether the nucleophilic substitution is SN1 or SN2 depends on (i) structure of the ether (ii) reaction conditions

Case-I If both the alkyl groups of ether are primary (or) one alkyl group is primary

and the other one is secondary, then the nucleophilic substitution takes the course of SN2 mechanism and the nucleophile attacks the less sterically hindered alkyl group.

Mechanism:Step-I: Formation of Oxonium ion

Step-II: Nucleophilic substitution.acts as nucleophile and attacks the less sterically hindered CH3 group.

Note: If excess conc HI is taken then the alcohol so formed further undergoes substitution to form alkyl halide.

Thus in such case the net reaction would be

Case-II When one of the alkyl groups is primary and the other is tertiary the nature of mechanism is governed by solvent.When aprotic solvent is used or anhydrous conditions are maintained SN2 mechanism is favoured.When protic polar solvent is used SN1 mechanism is favoured.

Case-III When one of the alkyl groups of ether is secondary and the other is

tertiary, then the mechanism is SN1.

Case-IV When phenyl alkyl ethers are taken then always phenol is formed because the bond between ‘C’ of phenyl ring and ‘O’ has double bond character due to resonance.

Similarly,

TRY TO ANSWER1. The C-O bond in ether is cleaved by HI/HBr and not by HNO3 or H2SO4 though all of them are strong acids. Why?2. Distinguish between ether and an alkane of comparable molecular weight.3. Complete the reaction (i) 4. Give the order of reactivity of halogen acids with ethers. Explain your answer.

ROSENMUND’S REDUCTIONAcyl halides when reacted with H2 in presence of Palladium supported over BaSO4partially poisoned by quinoline (Lindlar’s catalyst) is partially hydrogenated gives aldehyde.

Ex: Rosenmund reduction offers a method for conversion of a carboxylic acid into the corresponding aldehyde via acid chloride.HCHO cannot be prepared by this method as HCOCl is unstable at room temperature.

TRY TO ANSWER

1) Complete the reaction

2) Can acyl halide be converted to ketones? If so how?3) Predict the product formed when acyl halides reacts with R – Mg - X

STEPHENS REACTIONNitriles when reacted with stannous chloride in presence of HCl give imine which on hydrolysis gives corresponding aldehyde. This is called Stephen’s reduction.

Alternatively, nitriles are selectively reduced by diisobutyl aluminium hydride (DIBAL – H) to imines followed by hydrolysis to give aldehydes

Similarly esters are also reduced to aldehydes with DIBAL – H

EXAMPLES Double bond is not reduced

TRY TO ANSWER1) How can you prepare ketones from nitriles?2) How can you prepare propiophenone from propane nitrile?3) Can Sn + HCl be used instead of SnCl2 + HCl for the reduction of nitriles to

aldehydes? Explain your answer.

ETARD REACTIONOxidation of Toluene with chromyl chloride in CS2 as solvent followed by hydrolysis gives benzaldehyde. This is a controlled oxidation method. This is called Etard reaction.

The above conversion can also be carried out by oxidation of Toluene with chromic oxide in acetic anhydride followed by hydrolysis

GATTERMANN – KOCH REACTIONWhen benzene or its derivative is treated with carbonmonoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted benzaldehyde.

MechanismStep – I : Generation of Electrophile

Step – II : Electrophilic substitution on benzene ring

GATTERMANN REACTIONWhen benzene is reacted with HCN in presence of HCl & AlCl3, aldimine is

formed which on further hydrolysis gives benzaldehyde

MechanismStep – I : Generation of Electrophile

Step – II : Electrophilic substitution on benzene ring.

Step – III : Hydrolysis of benzaldimine

Both Gatlermann – Koch and Gattermann reactions are electrophilic aromatic substitution reactions.

CLEMENSON’S REDUCTIONThe reduction of carbonyl functional group of aldehyde and ketones to methylene group with amalgamated zinc and conc HCl is known as Clemenson’s reduction.

Simultaneous dehydration of alcohol may take placeEx:

,-unsaturated ketones undergo reduction of both the olefinic and carbonyl groups.Ex:

and -keto acids are not reduced, but other keto acids are reduced.

WOLFF-KISHNER REDUCTIONIn this method, carbonyl group of an aldehyde or ketone are converted to methylene group by reacting with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol.

This method is not suitable for ,-unsaturated carbonyl compounds. Simultaneous dehydrohalogenation may take place because a strong base is

used

NOTE:

Both Clemenson’s and Wolff-Kishner reduction do not reduce double bonds(except in case of ,-unsaturated carbonyl compounds).

TRY TO ANSWER* Find out A,B,C,D and E in the following scheme of reactions.

TOLLEN’S TEST

When aldehyde is heated with Tollen’s reagent, a white silvery ppt is formed. Therefore this test is also called Silver Mirror Test.In this test aldehyde is oxidized to carboxylate ion by Tollen’s reagent. Silver of Tollen’s reagent gets reduced and precipitated as Silver metal.Tollen’s reagent is the ammoniacal solution of silver nitrate, where silver ions exists as complex.The overall reaction can be depicted as below:

FEHLING’S TESTWhen aldehyde is heated with Fehling’s reagent, a red ppt is formed.Fehling’s reagent is a solution of Fehling’s reagent ‘A’ and Fehling’s reagent ‘B’ mixed in equal amounts. Fehling’s reagent ‘A’ contains aqueous solution of CuSO4 while reagent ‘B’ contains alkaline solution of Rochelle salt(sodium potassium salt of tartarate).Function of Rochelle salt:

Cu+2 ions get precipitated in alkaline medium as Cu(OH)2. Therefore to keep Cu+2 ions in solution it is complexed with tartarate ion forming a soluble [Cu(tartarate)2]-2 complex.The overall reaction involved is Important points regarding Fehling’s and Tollens’s reagent:(i) Both the Tollen’s and Fehling’s reagent cannot oxidize ketones. Therefore ketones do not give these tests.(ii) Both the reagents cannot oxidize olefinic bonds if present in the aldehyde.(iii) Aromatic aldehydes are not oxidized by Fehling’s solution whereas they are oxidized by Tollen’s reagent.Ex: Benzaldehyde gives white ppt when heated with Tollens’s reagent and gets oxidized to benzoate ion, whereas it does not give ppt with Fehling’s reagent.(iv) Tollen’s test is also given by -hydroxy ketones.

(v) Tollen’s test is also given by alkyl and acyl hydroxylamines.

Ex: (vi) Formic acid gives positive test with both Fehling’s and Tollen’s reagent and itself is oxidized to CO2. Whereas other carboxylic acids do not give these tests.

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