GE 4 [SEM 4]

CHEMISTRY THEORY

Alcohols, Phenols and Ethers

Alcohols

Alcohols are compounds in which a hydroxyl (– OH) group is attached to saturated carbon

atom.

The hydroxyl group is the functional group of alcohols. Alcohols containing one

hydroxyl group are called Monohydric Alcohols. Alcohols with two, three or more hydroxyl

groups are known as Dihydric Alcohols, Trihydric Alcohols, and Polyhydric Alcohols

respectively. For example,

(Monohydric)

2

2Monohydric

(Two OH group)

CH OH

CH OH

|

Di

CH 2 – OH

|

CH – OH

|

CH 2 – OH Tr ih y d r ic

(Th r ee O H g r ou p )

Monohydric alcohols contain one –OH group attached to saturated carbon. They may be

represented as (R–OH)

Primary (1°) Secondary (2°) and Tertiary (3°) Alcohols

Monohydric alcohols are classified as primary (1°) secondary (2°), or tertiary (3°), depending

upon whether the –OH group is attached to a primary, a secondary, or a tertiary carbon.

Methods of Preparation of Alcohols

1. Reduction of Aldehydes and Ketones

Aldehydes and ketones can be reduced with H2/Ni or lithium aluminium hydride (LiAlH4) to

form the corresponding alcohols. Aldehydes give primary alcohols. Ketones give secondary

alcohols.

C OH

Functionalgroup

General formula

R OH

3 2(One OH group)

CH CH OH

CR

Primary carbon

OH

H

HPrimary alcohol

CR

Secondarycarbon

OH

HSecondary alcohol

R

CR

Tertiarycarbon

OH

Tertiary alcohol

R

R

2. Reduction of Acids

Carboxylic acids can be reduced with strong reducing agents like sodium borohydride

(NaBH4) or lithium aluminium hydride (LiAlH4) to form the corresponding alcohols.

3. Hydrolysis of Esters

Alcohols may be prepared by base or acid-catalysed hydrolysis of ester.

4. Using Grignard reagent

Primary (1°), secondary (2°) or tertiary (3°) alcohols can be prepared using Grignard reagent

(R-Mg-X) and suitable carbonyl compounds, where R is alkyl group. Formaldehyde

(HCHO) and R-Mg-X yield primary alcohol. Similarly, higher aldehydes and ketones form

secondary and tertiary alcohols respectively.

Reactions

Reaction with Active Metals

Alcohols react with sodium or potassium to form alkoxides with the liberation of hydrogen

gas.

The above reaction shows that alcohols are acidic in nature. The reason for this is that the

O–H bond in alcohols is polar and allows the release of the hydrogen atom as proton (H+).

However, alcohols are weaker acids (Ka = 10–16 to 10–18) than water. This is because the

alkyl groups in alcohols have a +I effect. They release electrons towards the oxygen atom so

that it becomes negatively charged. This negative charge on oxygen makes the release of the

positive proton more difficult.

Tertiary alcohols are less acidic than secondary alcohols. The secondary alcohols are less

acidic than primary alcohols. This is because the +I effect would be maximum in tertiary

alcohols, as they contain three alkyl groups attached to the carbon bearing the –OH group.

Alcohols are not acidic enough to react with aqueous NaOH or KOH.

R – OH + NaOH → No reaction

Oxidation reaction

Alcohols can be oxidised. The nature of the product depends on the type of alcohol and the

conditions of the reaction. Most widely used oxidising agents are KMnO4 + H2SO4 or

Na2Cr2O7 + H2SO4. Oxidation of alcohols can be used to distinguish between primary

secondary and tertiary alcohols.

Primary alcohols are first oxidised to aldehydes and then to acids.

Notice that the acid and the alcohol contains the same number of carbon atoms. The

reaction can be stopped at the aldehyde stage by removing it from the oxidising medium as it

is formed (e.g., by distillation).

Secondary alcohols are oxidised to the corresponding ketones.

Further oxidation under very drastic conditions breaks up the ketone molecule, producing

carboxylic acids containing fewer carbon atoms per molecule.

Tertiary alcohols are stable under normal oxidation conditions.

Under drastic conditions, tertiary alcohols give ketones and acids, each containing less

carbon than the alcohols.

Pinacol-Pinacolone rearrangement

This is an acid catalysed 1, 2 shift rearrangement of vicinal diol (1, 2-glycol) to ketone or

aldehyde with the elimination of water and this rearrangement is called Pinacol-Pinacolone

rearrangement.

The migrating group may be alkyl, aryl, hydrogen and even ester function (e.g., CO2Et).

The catalyst is either mineral acids (dilute H2SO4 or H3PO4) or Lewis acids (AlCl3, BF3, and

ZnCl2) or any electrophilic reagent (PCl5).

Mechanism of Pinacol-Pinacolone rearrangement:

Phenols

Preparation

1. Cumene hydroperoxide synthesis

This method involves arial oxidation of cumene (Isopropylbenzene) followed by

treatment with dilute HCl. It is the industrial process for phenol synthesis. This process

accounts for 80% of the total world production of phenol. The success of this method is due

to the availability of benzene and propene from petroleum and because of the formation of

acetone, a valuable by product.

Steps:

1. Fridel crafts alkylation with propene and H3PO4 leading to production of cumene.

2. Oxidation of cumene to cumene hydroperoxide by arial oxygen.

3. Treatment of cumene hydroperoxide with dilute HCl to produce phenol and acetone.

2. Synthesis from diazonium salts

This reaction can be performed easily in the laboratory and simply requires warming a

solution of benzenediazonium chloride, prepared from aniline, on a water bath at 50°C. The

phenol is recovered by steam distillation and extracted with diethyl ether.

Reactions

Electrophilic substitution reactions

Halogenation

Phenol reacts with bromine water (aqueous bromine) to give precipitate of 2, 4, 6-

tribromophenol. Chlorine reacts in the same way.

If the reaction is carried in CS2 or CCl4 (nonpolar solvents), a mixture of o- and p-

bromophenol is formed.

Nitration

Phenol reacts with dilute nitric acid to give a mixture of o- and p-nitrophenol.

Reimer-Tiemann Reaction

This involves the treatment of phenol with chloroform in aqueous sodium hydroxide solution

followed by acid hydrolysis to yield salicylaldehyde, If carbon tetrachloride is used in place

of chloroform, salicylic acid is formed.

The Reimer-Tiemann reaction is also given by other phenols and introduces –CHO group in the

ortho position.

Mechanism

In the above reaction, chloroform first reacts with sodium hydroxide to produce dichlorocarbene.

The electron-deficient dichloricarbene then reacts with sodiumphenoxide to form a

dihalide. This is then hydrolysed to form the aldehyde.

Fries Rearrangement

The phenol is first treated with acetic anhydride in the presence aqueous sodium

hydrazide to give phenyl acetate. The ester is then heated with aluminium chloride catalyst

when the acyl group migrates from the phenolic oxygen to an ortho or para position of the

ring. The product is a mixture of o- and p-hydroxyacetophenone.

The fries rearrangement is also given by other phenols and introduces –COR group in the

ortho and / or para positions.

Claisen rearrangement

The thermal rearrangement of allyl-vinyl (aryl) ether with [3,3]-sigmatropic shift is called

Claisen rearrangement.

Claisen rearrangements involving aromatic rings often produce phenol derivatives.

i)

This is an aromatic ortho-ortho claisen rearrangement.

If ortho positions are blocked then reaction goes to para position via Cope-rearrangement.

This is an aromatic para claisen rearrangement.

ii)

Ether

Ethers are compounds in which two alkyl substituents are attached by an oxygen atom. If the

alkyl substituents are identical, the ether is symmetrical or simple ether.

The groups R and R' can be alkyl or aryl. When both the R groups are alkyl, they are called

alkyl ether and when at least one of the R group is an aryl, they are called aryl ethers.

Preparations

Williamson synthesis

When an alkoxide is treated with an alkyl halide, ether is produced. The method was

developed by Williamson and is useful for the preparation of symmetrical as well as

unsymmetrical ethers.

where R' can be same as or different from R.

The alkoxide can be prepared by the action of alcohols with sodium or potassium

This reaction involves SN2 mechanism, with alkoxide ion acting as nucleophile and

alkyl halide as the substrate. The complete mechanism follows as

For ether to be obtained as major product, reaction should follow substitution and

minimize elimination reaction. This can be achieved by taking only methyl halides or 1°

halides with lesser branching near the reaction site.

A symmetrical or simple ether

R — O — R

An unsymmetrical or mixed ether

R — O — R

12 2RO H K RO K H

But if we take 2° and 3° alkyl halides, which have more tendencies to undergo

elimination, will give very less yield of ethers.

Suppose, we want to prepare ethyl tert-butyl ether. There are two possible routes to

prepare it.

Route (A) is more suitable to prepare given ether as the alkyl halide involved in this

route is a primary one while in route (B), the halide being a 3°, it will lead to elimination

reaction forming an alkene, 2-methyl propene.

Cleavage by heating with acids

Ethers on reaction with acids (HX) in presence of heat undergo cleavage to form an

alkyl halide and an alcohol. Alcohol on further reaction with HX gives second molecule of

alkyl halide.

HI is more reactive towards ether than HBr and HBr is more reactive than HCl.

Reaction in second step can take the direction of SN1 or SN2 pathway, depending

upon the conditions employed and the structure of ether. When both the alkyl groups are

methyl or 1º, it will follow SN2 reaction and when atleast one of the alkyl group is 3º, the

reaction follows SN1 pathway. For all other cases, it can undergo SN1 or SN2 pathway,

depending, depending upon the reaction conditions.

Mechanism

In the first step, ether is portonated by HX to give protonated ether. In the second

step, halide ion acts as nucleophile and attacks protonated ether to undergo cleavage. This

step is favoured because the leaving group (alcohol) is weakly basic.

Step I.

R O R H X R X R OH

2R OH H X R X H O

|. .

protonated ether

H

R O: R H X R O R X

Step II.

Reaction in second step can take the direction of SN1 or SN2 pathway, depending

upon the conditions employed and the structure of ether. When both the alkyl groups are

methyl or 1º, it will follow SN2 reaction and when atleast one of the alkyl group is 3º, the

reaction follows SN1 pathway. For all other cases, it can undergo SN1 or SN2 pathway,

depending, depending upon the reaction conditions.

Carbonyl compounds

Aldehydes are the compounds which have general formula RCHO. Ketones are compounds

having general formula RR'CO. The groups R and R' may be aliphatic or aromatic, similar or

different alkyl/aryl groups.

Both aldehydes and ketones contain the corbonyl group, >C=O, and are often referred to

collectively as carbonyl compounds.

Preparation

Reduction of Acid Chlorides

Acid chlorides can be reduced to aldehydes, only by the use of bulky hydride reducing agent,

tri-t-butoxy lithium aluminium hydride. If LiAlH4 is used as a reducing agent, the product

isolated is an alcohol and not an aldehyde.

R–COCl or Ar–COCl 3LiAlH(OBu t) R–CHO or Ar–CHO

Acid chlorides can also be reduced to aldehydes by H2 gas in the presence of Pd supported on

BaSO4 in xylene, poisoned with quinoline or sulphur. This reaction is called Rosenmund’s

reduction, which is applicable for the preparation of aliphatic as well as aromatic aldehydes.

R–COCl or Ar–COCl 2 4H / Pd BaSO

Poisoned with quinoline or S

R–CHO or Ar–CHO + HCl

Synthesis from Grignard reagent

Aldehydes can be synthesized with the help of Grignard reagent starting from formyl chloride

or methyl/ethyl formate in dry solvent.

H

RC = O

An aldehyde

R

R

C = O

A ketone

Similarly, ketones can be synthesized by the reaction of Acyl chloride or esters with Grignard

reagent. Another process involves reaction of alkyl nitrile with Grignard reagent followed by

hydrolysis.

Nucleophilic Addition Reactions of Carbonyl Compounds

Due to the difference in electronegativity between carbon and oxygen, the -electrons

shift towards oxygen creating partial positive charge or carbon and making it electrophilic. So

this electrophilic carbon can be attacked by different nucleophiles.

(a) Addition of cyanide

(b) Addition of sodium bisulphite

Example:

(c) Addition of derivative of ammonia.

H2N – G PRODUCT

The rate of such reaction is maximum at some particular pH. These reactions are

catalysed by the presence of slightly acidic conditions. In slightly acidic conditions,

dehydration step is the RDS, whose rate is increased by the protonation of OH, leading to

overall increase in rate of the reaction. But, when the acidity increases for the rate of addition

step decreases because concentration of NH2 – Z reduces due to its conversion to conjugate

acid, 3NH Z (which cannot function as a nucleophile because of absence of lone pair).

Thus, at low pH or more acidity, the addition step becomes the RDS.

For example,

H

3 2 3 2Acetaldoxime

CH CHO O H N OH CH CH N OH H O

(Capable of showing geometrical isomerism)

2 2Ph CH O H N NHPh Ph CH N NHPh H O

Benzaldehyde phenyl hydrazone

(Capable of showing geometrical isomerism)

H3C

H3CC = O + H2N – NHCONH2

H3C

H3CC = N – NHCONH2 + H2O

Acetone semicarbazone

Fehling’s & Tollen’s Test

Aldehydes are easily oxidised by Fehling’s solution producing the corresponding acid.

Fehling’s solution is made by mixing, Fehling A solution, which contains copper sulphate,

with Fehling B solution, which contains sodium hydroxide and Rochelle salt (Sodium

Potassium Tartarate). During the oxidation of aldehydes to acids, the cupric ions are reduced

to cuprous ions which are precipitated as red cuprous oxide.

2

2 2RCHO 2Cu 3OH RCO 2Cu 2H O

2 2Cuprous oxide (red)

2Cu 2OH Cu O H O

Tollen’s reagent contains diamminesilver (I) ion, which is obtained through ammoniacal

AgNO3 solution. Tollen’s reagent oxidizes aldehydes to acid salt and reduces itself to free

silver in the form of silver mirror.

3 2 2 3 2Silver mirrorColourless Solution

RCHO 2Ag(NH ) 3OH RCO 2Ag 4NH 2H O

Fehling’s & Tollen’s reagents are useful in differentiating aldehydes from ketones

because ketones do not react with them. Tollen’s reagent is a mild and selective oxidizing

agent, attacking only aldehydic group, keeping other groups untouched. Unsaturated

aldehydes can be converted to unsaturated acid salt using Tollen’s reagent.

Tollen 's

2reagent, unsaturated uldehyde , unsaturated acid salt

R CH CH CH O R CH CH CO Ag

Ketones are not easily oxidized, thus they do not reduce Fehling’s solution or Tollen’s

reagent. But α-hydroxy ketones (compounds containing the unit CH(OH) C R)

O

| | readily

reduced.

Iodoform reaction

Iodoform reaction is a chemical reaction in which a methyl ketone is oxidized to

a carboxylate by reaction with aqueous HO- and I2. The reaction also produces iodoform

(CHI3), a yellow solid which is precipitated out from the reaction mixture. The iodoform

reaction is often used as a chemical test for the presence of a methyl ketone moiety.

A keto methyl (COCH3) group must be present in a compound for a positive iodoform test.

Other functional groups, which can be oxidised to a keto methyl (COCH3) group,

also respond positively toward this test. After oxidisation RCH(CH3)OH is

converted to

which has a keto methyl group and hence gives positive iodoform test .

Aldol Condensation

Aldehyde having α-hydrogen(s) undergo self-condensation on warming with dilute or

mild base to give β-hydroxy aldehydes, called aldols (aldehyde + alcohol). This reaction is

known as aldol condensation. A typical example is the reaction of acetaldehyde with base

under mild condition.

CH3CHO + CH3CHO → CH3CH(OH)CH2CHO OH-

Various basic reagents such as dilute sodium hydroxide, aqueous alkali carbonate, alkali

metal alkoxides, etc., may be used. The reaction is not favourable for ketones.

Aldol condensation has broad scope. It can occur between

(i) two identical or different aldehydes,

(ii) two identical or different ketones and

(iii) an aldehyde and a ketone.

When the condensation is between two different carbonyl compounds, it is called

crossed aldol condensation.

Mechanism

The first step involves the formation of a resonance-stabilized enolate anion by the

removal of an α-hydrogen from the aldehyde by the base. In the second step the enolate anion

attacks the carbonyl carbon of the second molecule of the aldehyde to form an alkoxide ion.

The later then takes up a proton from the solvent to yield aldol in the third step.

OH + H CH2 C

O

H

H2O + CH2 – C

O

H

CH2 = C

O

H

CH3 – C + CH2CHO

O

H

CH3 – C – CH2CHO

O

H

H2O

–OH

CH3 – C – CH2CHO

OH

H

Crossed aldol condensation between two different aldehydes

When both the aldehydes have α-hydrogen(s) both can form carbanions and also can act as

carbanion acceptors. Hence a mixture of four products are formed which has little synthetic

use. If one of the aldehydes has no α-hydrogen then it can act only as a carbanion acceptor. In

such case two products are formed, e.g.

(a) OOH

3 3 3 2Crossed product

R C CHO CH CHO R C CH(OH) CH CHO

(b) OOH

3 3 3 2Simple product (normal)

CH CHO CH CHO CH CH(OH)CH CHO

However, a good yield of the crossed product is obtained by slowly adding the aldehyde

having α-hydrogen to a mixture of the aldehyde having no α-hydrogen and the catalyst, e.g.

3 2

OCH CHO H O

6 5 6 5 2 6 5Slow additionCinnamaldehyde

C H CHO OH C H CH(OH)CH CHO C H .CH CH CHO

Formaldehyde having no α-hydrogen is a reactive carbanion acceptor due to the absence of

steric hindrance and +I effect. Hence, when acetaldehyde is treated with excess of

formaldehyde in the presence of Ca(OH)2, crossed aldol condensation continues (three times)

until trihydroxymethyl acetaldehyde, (HOCH2)3CCHO is formed. The latter having no α-

hydrogen undergoes crossed Cannizzaro reaction to form pentaerythritol.

Cannizzaro Reaction

In the presence of a strong base, aldehydes without α-hydrogens, i.e., nonaldolizable

aldehydes undergo self-oxidation-reduction i.e., disproportionation reaction. This is known as

Cannizzaro reaction. Thus, aromatic aldehydes (ArCHO), formaldehyde (HCHO), trialkyl

acetaldehydes (R3CCHO), heterocyclic aldehydes,

O

CHO

etc., undergo Cannizzaro

reaction, e.g.

6 5 6 5 2 6 5Benzyl alcohol Sodium benzoate

2C H CHO NaOH C H CH OH C H COONa

3SodiumformateMethyl alcohol

2HCHO NaOH CH OH HCOONa

The reaction best proceeds with aromatic aldehydes. Although the reaction is

characteristic of aldehydes without α-hydrogen, a few aldehydes with α-hydrogen are known

which undergo Cannizzaro reaction (exception).

200 C

3 2 3 2 2 3 2Dimethyl acetaldehyde 2 Methylpropanol 1 Sodium 2 methyl propanoate

2(CH ) CHCHO (CH ) CHCH OH (CH ) CHCOONa

The reaction can also occur between two different aldehydes having no -hydrogen

when it is called crossed Cannizzaro reaction.

NaOH

6 5 6 5 2C H CHO HCHO C H CH OH HCOONa

When formaldehyde undergo crossed Cannizzaro reaction with other aldehydes without

α-hydrogens, it is seen that formaldehyde is oxidized and the other is reduced. This is because

the nucleophilic attack occurs more readily on formaldehyde than on other aldehydes.

Mechanism

Rapid addition of O

OH

to one molecule of aldehyde results in the formation of a hydroxyl

alkoxide ion which like aluminium-isopropoxide acts as a hydride-ion donor to the second

molecule of aldehyde. In the final step of the reaction the acid and the alkoxide ion exchange

proton for reasons of stability.

C H – C + OH6 5

O

H

C H – C – OH6 5

O

H

C H – C – OH + C – C H6 5 6 5

O

H

O

SlowC H – C – OH + H – C – C H6 5 6 5

O

H

proton exchange

(fast)

H

O

C H – CH OH + C H – C – O6 5 2 6 5

O

C H – C = O6 5

O

Benzoin Condensation

Alcoholic

KCNBenzoin

2PhCHO PhCH(OH)COPh

When aromatic aldehydes is treated with alcoholic KCN, the product is not a

cyanohydrin but α-hydroxy aromatic ketone called benzoin. The product of aromatic

aldehydes with KCN is different than aliphatic aldehydes because after the attack of CN–, the

intermediate (I) in aromatic aldehyde has sufficient acidity (due to – I effect of Ph) so that

intramolecular proton exchange takes place to form a carbanion, which is resonance

stabilized. This carbanion then attacks another molecule of aromatic aldehyde, which

undergoes intramolecular proton exchange and then ejection of CN– to give final product i.e.

benzoin. The rate-limiting step of the reaction is attack of carbanion on second molecule of

aromatic aldehyde.

Mechanism

Ph – C + CN

O

H

Ph – C – H

O

CN

IMPEPh – C – C N

OH

....

Ph – C = C = N

OH

......

Ph – C + C – OH

O

H

Ph – C – C – Ph

O¯

H

IMPEPh – C – C – Ph

OHO – H

H

O¯

CN

– CN¯Ph – C – C – Ph

OH

H

O

Benzoin

CN

Ph CN

IMPE = Intra molecular proton exchange

Reduction reactions (Clemmensen & Wolff-Kishner reduction)

Aldehydes and ketones can be reduced to hydrocarbons by the action of (a) amalgamated zinc

and concentrated hydrochloric acid (Clemmensen reduction) or (b) hydrazine (N2H4) and a

strong base like KOH or potassium tertiary butoxide (Wolff-Kishner reduction).

C = O

N2H4, KOH

Zn– Hg/

Conc. HCl

C HH

C HH

Clemmensen reduction forcompounds sensitive to bases

Wolff-Kishner reduction forcompounds sensitive to acids

For example,

CH3CH2CH = CHCHOZn– Hg/

Conc. HClCH3CH2CH = CHCH3

N2H5 OH–

Carboxylic Acids and their derivatives

Organic compounds which contain the carboxyl functional group (–COOH) are called the

Carboxylic Acids. The general formula is

R – C – OH or R – COOH or R – CO H2

O

where R is an alkyl group.

Carboxylic acids are further classified as monocarboxylic acids, dicarboxylic acids,

tricarboxylic acids etc. according as the number of –COOH groups present in the molecule is

1, 2, 3 or more. The long-chain monocarboxylic acids are commonly called Fatty Acids

because many of them are obtained by the hydrolysis of animal facts or vegetable oils.

Acidity of Carboxylic Acids

Carboxylic acids are acidic in nature. They can donate a proton and form salts with bases.

| | | | –

2

O O

R C OH NaOH R C O N a H O

Carboxylic acid Base Salt

Acidity Constants

Strong acids (e.g., HCl or H2SO4) completely ionize in aqueous solution. Carboxylic acids

are weak acids. They are only partially in aqueous solution and an equilibrium exists between

the ionized and un-ionised forms.

| |

2 2

O

R C OH H O R C O H O

It is defined as the concentration of the products of ionization in moles per litre divided by

the concentration of the un-ionised acid.

3a

[RCOO ] [H O ]K

[RCOOH]

The acidity constant describes the relative strength of a weak acid. Stronger acids

will have higher numerical value of acidity constants.

Explanation of Acidity

Carboxylic acids are acidic and lose a proton

readily because the carboxylate ion formed by

ionization or reaction with a base is stabilized

by resonance.

X-Ray studies support the fact that

carboxylate ion exists as a resonance hybrid. For

example, in formic acid the carbon-oxygen bonds

have different lengths, whereas in sodium

formate the two carbon-oxygen bond lengths are

identical and intermediate in length between

those of normal double and single carbon-oxygen

bonds.

The stability of carboxylate ion can

also be explained on the basis of its

molecular orbital structure. The carbon

atom of carboxylate ion is sp2-

hybridized. it is bound to each oxygen

atom by a bond. The unused carbon p

orbital overlaps with p-orbitals of both

oxygen atoms to form stable delocalized

molecular orbital.

Note that the four electrons are bound to three atoms (1 carbon + 2 oxygens). This

delocalization of α electrons is responsible for the extra-stability of the carboxylate ion.

Effects of Substituents of Acidity

The most important factor affecting the acidity is the Inductive Effect of substituents

on the α-carbon atom.

(i) Electron-releasing alkyl groups decrease the acidity. This is because the electron-releasing

groups increase the negative charge on the carboxylate ion and destabilize it. The loss of

proton becomes more difficult. Also, as the size of the alkyl group increases, acidity

decreases. For example,

R – CO

OR – C

O

OR – C

O

O

–

Resonance forms of carboxylate ion

Resonancehybrid

R – COH

O

1.23Å

1.36ÅFormic acid

H – C

O

O

+

1.27Å

1.27Å

Na

Sodium formate

CR

-MO

sp Carbon2

· Delocalised

R – C

O

O

| |

a

5

O

H C OH

K Formic acid

17.7 10

| |

3

5

O

C H C OH

Acetic acid

1.76 10

| |

3 2

5

O

CH C H C OH

Pr opionic acid

1.34 10

(ii) Electron-withdrawing substituents (Cl, Br, F, OH, CN) increase the acidity. This is

because the electron-withdrawing substituents decrease the negative charge on the

carboxylate ion and stabilize it. The loss of proton becomes relatively easy. For example,

chloroacetic acid is about 100 times stronger than acetic acid.

| |

2

5a

O

Cl – C H C OH

Chloroacetic acid

K 136 10

| |

3

5

O

CH C OH

Acetic acid

1.76 10

The strength of the electron-withdrawing substituents determines the magnitude of

the effect on acidity. For example, fluoroacetic acid is stronger than chloroacetic acid

since F is more electronegative than Cl.

| |

2

5a

O

F CH C OH

Fluoroacetic acid

K 260 10

| |

2

5

O

Cl CH C OH

Chloroacetic acid

136 10

As the number of electron withdrawing substituents increases, acidity also

increases. For example

| |

3

5a

O

Cl C C OH

Trichloroacetic acid

K 23,200 10

| |

2

5

O

Cl CH C OH

Dichloroacetic acid

5,530 10

| |

2

5

O

ClCH C OH

Chloroacetic acid

136 10

Key Points

(i) Strength of an acid is determined by the readiness with which they will donate a proton

(ii) Acid strength will be increased by any factor which increases stability of the anion or

which promotes proton loss.

(iii) Electron-releasing alkyl groups decrease the acidity.

(iv) Electron-withdrawing substituents (Cl, Br, F, OH, CN) increase the acidity. The strength

of electron withdrawing substituents determines the magnitude of its effect on acidity.

As the number of electron-withdrawing substituents increases, acidity also increases.

(v) The more halogen atoms attached to the acid molecule, the stronger the acid.

(vi) The closer the halogen atoms are attached to the carboxylic acid functional group, the

stronger the acid.

Preparation

Hydrolysis of Esters

When an ester is boiled with concentrated aqueous NaOH, sodium salt of the acid is formed.

This on treatment with dilute HCl gives the corresponding carboxylic acid. For example,

Reaction of Grignard Reagents with CO2

Grignard reagents (RMgX) react with carbon dioxide to form addition products that can be

hydrolysed to carboxylic acid.

ether

R X Mg RMgX

Carboxylic Acid derivatives

Several types of derivatives can be prepared with help of specific reagents.

Formation of Acid Halides

Carboxylic acids react with phosphorus halides (PCl3, PCl5) or thionyl chloride

(SOCl2), to form acid halides. For example,

Formation of Amides

Carboxylic acids react with ammonia to give salts, which on heating yield amides.

Formation of Esters

Carboxylic acids react with alcohols in the presence of a strong acid catalyst like

H2SO4 or HCl to form esters. The reaction is reversible and is called Esterification.

The equilibrium can be shifted to the right by using excess of alcohol or removal of

water by distillation.

Formation of Anhydrides

Carboxylic acids undergo dehydration with phosphorus pentoxide (P2O5) to form

acid anhydrides.

R – C – OH + H – OR

O

Carboxylicacid

Alcohol

H+

R – C – + H O2OR

O

Ester

R – C – OH + H – O – C – R

OP O2 5

(Two acid molecules)

R – C – + H O2O – R

O

Acid anhydride

OO

CH – C – OH + H – O – C – CH3 3

OP O2 5

Acetic acid

CH – C – + H O3 2O – C – CH3

O

Acetic anhydride

OO

Acetic acid

Anhydrides can also be prepared by treating sodium salts of acids with acid halides.

Reactions

Claisen condensation

The Claisen condensation is a carbon–carbon bond forming reaction that occurs

between two esters or one ester and another carbonyl compound in the presence of a strong

base, resulting in a β-keto ester or a β-diketone.

Perkin Reaction

The Perkin reaction gives an α,β-unsaturated aromatic acid by the aldol condensation of

an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. The

alkali salt acts as a base catalyst, and other bases can be used instead.

Cinnamic acid can be prepared by Parkin reaction.

Mechanism

Amines and Diazonium Salts

Amines are the derivatives of ammonia obtained by replacement of H atoms by any alkyl or

aryl groups. Replacement of one, two and three hydrogen atoms of ammonia produces

primary, secondary and tertiary amines respectively.

Aliphatic Amines

According to Lewis theory, a base is a substance which donates a lone pair of

electrons e.g. ammonia, water etc. Like ammonia, amines contain a lone pair of electrons on

nitrogen and hence are basic. The basic strength of a base is related to the ease with which it

can donate its lone pair of electrons. Primary amine (R – NH2) has one alkyl group attached

to the nitrogen atom. In comparison to NH3, the electron density of nitrogen atom is more in

R – NH2 due to +I effect of the alkyl group. This facilitates the nitrogen atom to use its

unshared pair of electrons for bond formation. Thus R – NH2 is expected to be more basic

than NH3. Going by the same arguments the basic strength of amines is expected to increase

with the increase in the number of alkyl groups attached to the nitrogen atom

R3N > R2NH > RNH2 > NH3

This is indeed found to be the case when basic strength of amines is measured

in non-aqueous medium like chlorobenzene. However, this trend is not observed in aqueous

medium. The nitrogen atom in an amine shares its electron pair with H+ ion from water to

form alkyl ammonium ion,

2 2 3R NH H O R NH OH

The basic strength of the amine is expressed as equilibrium constant Kb

3b

2

[RNH ] [OH ]K

[RNH ]

The greater the value of Kb of an amine, the greater is its basicity. In fact Kb is an

index of basicity of basic strength of an amine. it gives a measure of the extent to which the

amine accepts H+ ion from water. Very often the basic strength of an amine is expressed as

pKb which is the negative logarithm of Kb. Thus, the smaller the pKb of an amine the

stronger will be the base. The pKb values of NH3 and some of the amines are given below

pKb pKb

NH3 4.75 C2H5 – NH2 3.33

CH3 – NH2 3.36 C2H5 – NH – C2H5 3.07

C2H5 – NH – CH3 3.23 C2H5 – N (C2H5) – C2H5 3.12

CH3 – N (CH3) – CH3 4.20

The basic strength of an amine in aquous medium is determined not only on the

basis of +I effect of the alkyl groups attached to the nitrogen atom but also the solvation of

alkyl ammonium ion formed by uptake of proton.

The solvation of alkyl ammonium ion is maximum as it has three H-atoms attached

to the nitrogen atom which can form hydrogen bonds with water molecules. It is less in the

case of dialkylammonium ion which has two H-atoms attached to the nitrogen atom and still

less in the case of trialkyl ammonium ion which has only one H-atom attached to the nitrogen

atom. Thus on going along the series NH3 R – NH2 R2NH R3N, the +I effect of the

increasing number of alkyl groups will tend to increase the basic strength but the gradual

decrease in the solvation of cation and hence the stabilization of cation will tend to decrease

the basic strength. The effect of alkyl dominates till secondary amines. The actual changeover

takes place on going from a secondary amine to tertiary amine. Thus, 2º amine becomes more

basic in comparison with 3º amine as far as aqueous medium is concerned.

Aromatic Amines

In aromatic amines the nitrogen atom of –NH2 group is directly attached to the

benzene ring. The unshared pair of electrons at the nitrogen is in resonance with the benzene

ring and hence not fully available for donation as in the case of aniline

R –N – H — OH+

2

H — OH2

H — OH2

> R –N – R+

H — OH2

H — OH2

> R –N – R+

H — OH2

R

If aniline is to function as a base, it has to use its unshared pair of electrons for donation at

the cost of resonance stabilization. Thus aniline is reluctant to function as a base. This is

supported by the pKb values of aniline in comparison to those of ammonia and

cyclohexylamine.

NH3 C6H5 – NH2 C6H11 – NH2

pKb 4.75 9.38 3.32

Preparation

From alkyl halides

Primary amines can be synthesized by alkylation of ammonia. A large excess of ammonia is

used if the primary amine is the desired product. Haloalkanes react with amines to give a

corresponding alkyl-substituted amine, with the release of a halogen acid.

Hoffmann Bromamide reaction

Amines (only primary) can also be prepared by Hoffmann degradation. In this method the

amine will have one carbon atom less than the amide. The reaction via formation of nitrene.

Mechanism of above reaction has been proposed as given below

:NH2 NH2

NH2

:

NH2

NH2

:

:

R – C – NH2++ Br + 4KOH RNH + K CO + 2KBr + 2H O2 2 2 3 2

O

2Br KOH K O Br HBr

Mechanism

2 22NaOH Br NaOBr NaBr H O

(i)

(ii)

Reaction with Nitrous Acid

Primary Amines

Primary amines react with nitrous acid to produce diazonium ion as follow

2 2ArNH HNO Ar N N :

2 2R NH HNO R N N

But the diazonium ions of aliphatic amines are very unstable and produces carbocation

immediately, which can produce different products.

N 2R N N R (Carbocation )

This reaction of RNH2 has no synthetic utility, but the appearance of N2 gas signals the

presence of NH2.

Diazonium salts of aromatic amines are comparatively more stable and evolve nitrogen only

on heating. These diazonium salts can be isolated at low temperatures.

Diazonium salts of aromatic amines are very useful compound because they can undergo

nucleophilic substitution easily. Various nucleophilic substitution at benzene diazonium

chloride are given below

R – C – NH2 + OBr –

O

R – C – N – Br + OH –

O

H

N-Bromoamide

R – C – N – Br

O

R – C – N ––

Br + H O 2

O

Rearrangement

H

R – NH + CO22

H O2 R – N = C = O

CH CH CH N NCl2

–

–

3 2 N Cl + CH CH CH–

22 3 2

+ ~H (Hydride shift)–

CH CHCH33

CH CH CH OH3 2 2 CH CH CH Cl3 2 2CH CH=CH23 CH CHCH33 CH CHCH33

ClOH

1-Propanol 1-Chloropropane Propene 2-Propanol 2-Chloropropane

H O2 Cl– –H

+–H

+

H O2Cl

–

+

Coupling Reaction with Primary Amines

Diazonium ions of aromatic amines also undergo coupling reaction with aromatic rings

having a strong activating group to form diazo compounds.

Mechanism

Secondary Amines

2 2 2R NH HNO R N N O

N Nitrosoa min e

(inso lub le in amine)

Tertiary Amines

Tertiary amines except N, N-Dialkylaryl amines do not react with HNO2.

Schotten–Baumann reaction

The Schotten–Baumann reaction is a method to synthesize amides from amines and acid

chlorides: Schotten–Baumann reaction also refers to the conversion of acid chloride to esters.

H PO23 Ar – H

Ar – I

ArCl/(ArBr) (Sandmeyer reaction)

Ar – F

Ar – OH

Ar – CN

Ar – OR

CuCN

KI

CuCl(CuBr)

HBF or NaBF , 4 4

H O /H2

+

ROH

Ar – N N

Ar – N N + NH2 Ar – N = N NH2

(dye)

Ar – N N + OH Ar – N = N OH

NR2 NH SH4

HNO2

NR2

N=O

(attack by NO )+

Aromatic Nitro compounds

Reduction under acidic condition

Aromatic nitro compounds can be reduced completely to form aromatic amines under acidic

condition using metal (Sn, Fe, Zn) and conc. HCl.

Reduction under alkaline condition

Nitro benzene is reduced to azoxy, azo and hydrazobenzene under different alkaline onditions

Reduction under neutral condition

Aromatic nitro compounds can be reduced completely to form aromatic amines with Pd/C

mixture.

Aromatic nitro compounds are partially reduced to aromatic hydroxyl amines by reduction

with Zn/NH4Cl. This is the first part of Mulliken Barker reaction.

Summary of Aromatic nitro reduction

***If two nitro groups are present at meta positions to each other, one of them can be reduced

by selective reduction. For this purpose we use NH4SH or (NH4)2S or H2S in NH3.

NO2

NH HS4

NH2

NH HS4

NH2

NO2 NO2NH2

Amino Acids and Carbohydrates

Amino acids

An amino acid is a bifunctional organic molecule that contains both a carboxyl

group, -COOH, as well as an amine group, 2NH . They are classified as acidic (containing

two –COOH groups), basic (containing two 2NH groups) or neutral according to number of

amine and carboxyl groups in a molecule. Neutral amino acids contain only one amine and

one carboxyl group. They are further classified according to the position of amine group in

relation to carboxyl group into and amino acids. Out of these amino acids

are most important as they are building blocks of bio-proteins.

In an - amino acid, the amine group is located on the carbon atom adjacent to the

carboxyl group (the -carbon atom). The general structure of the -amino acids is

represented as :

H

R-C-COOH

NH Amine group2

Carboxyl group

-Carbon atom

R may be alkyl, aryl or any other group.

The proteins differ in the nature of R-group bounded to carbon atom. The nature of R-

group determines the properties of proteins. There are about 20 amino acids which make up

the bio-proteins. Out of these 10 amino acids (non-essential) are synthesised by our bodies

and rest are essential in the diet (essential amino acids) and supplied to our bodies by food

which we take because they cannot be synthesised in the body.

Preparation

Gabriel phthalimide synthesis

-Halogenated acid or ester combines with potassium phthalimide. The product on

hydrolysis gives -amino acid.

CO

CONK + ClCH COOC H2 2 5

Chloro ethyl acetate

Pot.phthalimide

-KCl CO

CONCH COOC H2 2 5

2H O2

HCl

COOH

COOH+ CH NH COOH+C H OH2 2 2 5

Glycine

Phthalic acid

Strecker synthesis

An aldehyde reacts with HCN and amonia or 4NH CN and the product on hydrolysis yields

-amino acid.

3 2NH H OHCN

2 2H

H H H H| | | |

R C O R C OH R C NH R C NH

Aldehyde | | |CN CN COOH

Cyanohydrin A min onitrile A minoacid

Zwitter ion and isoelectric point

Since the - 2NH group is basic and –COOH group is acidic, in neutral solution, it exists in an

internal ionic form called a Zwitter ion where the protonof –COOH group is transferred to the

2NH -group to form inner slat, also known as dipolar ion.

In water

2 2 3

R R R| | |

H NCHCOOH H N CH COO H H N CH COOA minoacid Zwitter ion

(Dipolar ion)

The Zwitter ion is dipolar, charged but overall electrically neutral and contains both

a positive and negative charge.

Therefore, amino acids are high melting crystalline crystalline solids and

amphoteric in nature. Depending on the pH of the solution, the amino acid can donate or

accept proton.

2

H OH

3 3 2H O

H O H O H O| || | || | ||

H N C C OH H N C C O H N C C O| | |

(Pr otonre moved)R R R

Low pH(Acidicsoln.) Zwitter ion I High pH(Basicsoln.)Positive form(II) Neutralform Negativeform(III)

(Cation) (Anion)

When an ionised form of amino acid is placed in an electric field, it will migrate

towards the opposite electrode. Depending on the pH of the medium, following three things

may happen :

(i) In acidic solution (low pH), the positive ion moves towards cathode [exist as cation,

structure (II)].

(ii) In basic solution (high pH), the negative ion moves towards anode [exist as anion,

structure (III)].

(iii) The Zwitter ion does not towards any of the electrodes [neutral dipolar ion, structure

(I)].

The intermediate pH at which the amino acid shows no tendency to migrate towards

any of the electrodes and exists the equilibrium when placed in an electric field is known as

isoelectric point. This is characteristic of a given amino acid and depends on the nature of R-

linked to -carbon atom.

Carbohydrate

Definition and Classification

Historically, carbohydrates were once considered to be “hydrates of carbon” because

molecular formulas of many carbohydrates (but not all) correspond to x 2 yC H O (type i)

(a) However, certain carbohydrates do not correspond to this general formula, (type ii).

(b) Moreover, several compounds are although not carbohydrates, their molecular formula

correspond to the above general formula, (type iii).

(i) 6 12 6

Glu cose and Fructose

C H O 12 22 11

Sucross

C H O 6 10 5

Cellulose and starch

C H O

(ii) 6 12 5

Rhamnose

C H O 7 14 6

Rhamnohexose

C H O

(iii) 2

Formaldehyde

CH O 2 4 2 3

Acetic acid

C H O or CH COOH 3 6 3

Lactic acid

C H O 6 12 6

Inositol

C H O

Simple carbohydrates are also known as sugars or saccharides, and name of most

sugars end in –ose.

Carbohydrates are polyhydroxyaldehydes, polyhydroxyketones or compounds that

can be hydrolyzed to them. A carbohydrate that cannot be hydrolyzed to simpler compounds

is called a monosaccharide. A carbohydrate that can be hydrolyzed to simpler and indefinite

monosaccharide molecules is called a disaccharide, a trisaccharide, and a polysaccharide

respectively.

Aronosaccharide may be further classified; if it contains an aldehydic group it is

called an aldose, and if it contains a ketonic group it is called a ketose. Alternatively, a

monosaccharide may be classified on the basis of the number of carbon atoms present it it,

viz. triose, tetrose, pentose, hexose, eptose, and so on. Most naturally occurring

monosaccharides are pentoses or hexoses. These two classifications are frequently combined.

A 4C aldose, for example, is called on aldotetrose, a 5C ketose is called a ketopentose.

n

2

An aldose

CHO|

CHOH

|CH OH

2

n

2

A ketose

CH OH|CO|

CHOH

|CH OH

2

An aldotriose

CHO|CHOH|CH OH

2

2

A ketotetrose

CH OH|CO|CHOH|CH OH

Tlyceraldehyde (an aldoteiose) is considered to be the simplest chiral carbohydrate.

Carbohydrates that reduce Fehling’s or benedict’s or Tollen’s reagent are known as

reducing sugars. All monosaccharides, whether aldose or ketose, and most disaccharides are

reducing sugars. However, the most common disaccharide, sucrose is a non-reducing sugar.

Oxidation-Reductions

Oxidation

When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called

an aldonic acid. Because of the 2º hydroxyl functions that are also present in these

compounds, a mild oxidizing agent such as Br2, H2O must be used for this conversion. If both

ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.

A stronger oxidising agent like warm HNO3 is used for this purpose. By converting an aldose

to its corresponding aldaric acid derivative, the ends of the chain become identical.

Reduction

Sodium borohydride or H2/Pt reduction of an aldose makes the ends of the

resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same

configurational change produced by oxidation to an aldaric acid.

Mutarotation

Mutarotation is the change in the optical rotation because of the change in the equilibrium

between two anomers, when the corresponding stereocenters interconvert. Cyclic sugars

show mutarotation as α and β anomeric forms interconvert

Osazone formation

Reaction of carbohydrates with phenylhydrazine requires special attention. At first stage,

carbohydrates react with phenylhydrazine to form phenylhydrazones. However, if an excess

of phenylhydrazine is used, the reaction proceeds further to yield products known as osazone,

which contain two phenylhydrazine residues per molecule.

6 5 2 6 5 2

6 5 6 5

C H NHNH C H NHNH

6 5 2 32nd mole1st mole

CHO CH NNHC H CH NNHC H| | |

CHOH CHOH CO C H NH NH

6 5 2

6 5 2

6 5 6 5

C H NHNH H

3rd mole 2C H NHNH

6 5

CH NNHC H CH NNHC H CHO

|| |

CO C NNHC H CO

sazone Osone

O

Both of the phenylhydrazine residues of osazone can be removed to form

dicarbonyl compounds, known as osones.

Osazones are highly coloured, crystalline compounds and can be readily identified,

isolated and purified. Hence formation of osazone is used in the identification of

carbohydrates. Osazone formation is not limited to carbohydrates, but in general it is a typical

reaction of α-hydroxyaldehydes and α-hydroxyketones (e.g. benzoin, although not a

carbohydrate, 6 5 6 5C H CHOHCOC H , it forms osazone). Further osazone formation involves

only the first two carbon atoms, i.e. CHOCHOH- in aldoses and 2CH OHCO in ketoses, of a

compound without affecting the configuration of the rest of the molecule. Thus hexoses

having similar configuration on 3 4C ,C and 5C will form same osazone, e.g. glucose, mannose

and fructose.

CHO

H OH

HO H

H OH

H OH

CH OH2

Glucose

;

CHO

HO H

HO H

H OH

H OH

CH OH2

Mannose

;

CH OH2

HO H

H OH

H OH

CH OH2

Fructose

CO

3C H NHNH6 5 2 Same osazone

Structural relation between glucose, mannose and fructose. A pair of

diastereomeric aldoses that differ only in the configuration about one of its chiral carbon are

called epimers, e.g. glucose and mannose are examples of C-2 epimers; while glucose and

fructose, similarly mannose and fructose are simply isomers.

Carbon chain elongation and degradation in carbohydrates

Chain elongation

Chain degradation

Crystal Field Theory

This theory advanced by Brethe and Van Vleck was originally applied mainly to ionic

crystals and is, therefore, called Crystal Field Theory (CFT). It was not until 1952 that Orgel

popularised its use for inorganic chemists.

Important features of CFT

(i) The central metal cation is surrounded by ligands which contain one or more lone pairs of

electrons.

(ii) The ionic ligands (e.g. F–, Cl–, CN– etc.). are regarded as negative point charges (also

called point charges and the neutral ligands (e.g. H2O, NH3 etc.) are regarded as point

dipoles or simply dipoles, i.e. according to this theory neutral ligands are dipolar. If the

ligands is neutral, the negative end of this ligand dipole is oriented towards the metal cation.

(iii) The CFT does not provide for electrons to enter the metal orbitals. Thus the metal ion

and the ligands do not mix their orbitals or share electrons, i.e. it does not consider any orbital

overlap.

(iv) According to CFT, the bonding between the metal cation and ligand is not covalent but it

is regarded as purely electrostatic or coulombic attraction between positively-charged (i.e.

cation) and negatively-charged (i.e. anions or diploe molecules which act as ligands) species.

Complexes are thus presumed to form when centrally situated cations electrically attract

ligands which may be either anions or dipole molecules. The attraction between the cations

and the ligands is because the cations are positively charged and the anions are negatively

charged and the dipole molecules, as well, can offer their negatively incremented ends for

such electrostatic attractions.

Grouping of five d-orbitals into t2g and eg sets of orbitals

On the basis of the orientation of the lobes of the five-d-orbitals with respect to

coordinates these have been grouped into the following two sets.

1. eg set of orbitals : dz2 and dx2-y2 orbitals. This set consists of the orbitals which have

their lobes along the axes and hence are called axial orbitals. Quite obviously these are dz2

and dx2-y2 orbitals. Group theory calls these eg orbitals in which e refers to doubly degenerate

set.

2. t2g set of orbitals : dxy, dyz, dzx orbitals : This set includes the orbitals whose lobes

lie between the axes and are called non-axial orbitals. Group theory calls these t2g orbitals,

‘t’ refers to triply degenerate set.

Splitting of five d-orbitals in an octahedral complex.

(a) Five degenerate d-orbital on the central metal cation which are free from any ligand

field. (b) Hypothetical degenerate d-orbitals at a higher energy level (c) Splitting of d-orbitals

into t2g

and eg orbitals under the influence of six ligands in octahedral complex.

CFSE

The difference of energy between the two sets of d-orbitals is called crystal field

splitting energy or crystal field startilization energy. It is denoted by . 10 q ,0

4t

g

(a) Five degenerate d-orbitals on the central metal cation which are free from any

ligand field

(b) Hypothetical degenerate d-orbitals at a higher energy level (c) Splitting of d-orbitals

into e– orbitals and t2 orbitals under the influence of four ligands in tetrahedral complex.

Crystal field splitting of d-orbitals in tetrahedral complex

In case of a free metal ion (Mn+) all the d-orbitals are degenerate, i.e. these have the

same energy. Now let us consider a tetrahedral complex ion, [ML4]n+ in which the central

metal ion (Mn+) is surrounded tetrahedraly by four ligands. A tetrahedron may be supposed

to have been formed from a cube. The centre of the cube is the central of the tetrahedron at

which is placed the central metal ion (Mn+). Four alternate corners of the cube are the four

t2geg

xy yz zx z2

x –y2 2

t2g

0.4

= + 4Dq

Dq

= + 0.6

= + 6 Dq

No splitting

State

eg

(b)Ener

gy

incr

easi

ng

t2geg

xy yz zx z2

x –y2 2

eg

4 Dq

No splitting

State

t2g

(b)

Ener

gy i

ncr

easi

ng

xy yz zx

t

z2

x –z2 2

(a)

(c)

6 Dq

corners of the tetrahedron at which the four ligands, L, are placed. It may be seen from the

figure that the four ligands are lying between the three axes viz, x, y and z axes which pass

through the centres of the six faces of the cube and thus go through the centre of the cube.

Now since the lobes of t2g orbitals (dxy, dyz and dzx) are lying between the axis, i.e. are lying

directly in the path of the ligands, these orbitals will experience greater force of repulsion

from the ligands than those of eg orbitals (dz2 and dx2-y2 ) whose lobes are lying along the

axes, i.e. are lying in space between the ligands. Thus the energy of t2g orbitals will be

increased while that of eg orbitals will be decreased. Consequently the d orbitals are again

split into two sets as shown in fig from which it may be seen that the order of energy of t2g

and eg sets is the reverse of that seen for t2g and eg sets in octahedral complexes.

The energy difference between t2g and eg sets for tetrahedral complex is represented as

t . It has been shown that 0t the cause of which is that t2g -orbitals, although now

closest to the ligands, do not point directly at the ligands, i.e. in an octahedral complex there

is a ligand along each axis and in a tetrahedral complex no ligand lies directly along any axis.

For this reason and also because there are only four ligands in the tetrahedral complex, while

in an octahedral complex there are six ligands, the tetrahedral orbital splitting, t is less than

0 for the same metal and ligands and same internuclear distances. It has also been shown

that t = 0.45 0 . Thus the energy level of the t2g set is raised by 0.4 t = 0.18 0 while that

of eg set is lowered by 0.6 t = 0.27 0 . The relation namely t = 0.45 0 also shows that,

other things being equal, the crystal field splitting in a tetrahedral complex will be about half

the magnitude of that in an octahedral complex.

In case of tetrahedral complex, since t is generally less than P ( t <P), the electrons

tend to remain unpaired and hence only high spin tetrahedral complexes are known, i.e. low

spin tetrahedral complexes are not known.

Since t < 0 crystal field spliting of d-orbitals favours the formation of octahedral

complexes.

Factors Influence the Magnitude of 0Δ .

A mass of experimental data show that the magnitude of depends on the following factors.

A. Nature of the metal cation

The influence of this factor can be studied under the following four heading :

1. Different charges on the cation of the same metal : The cations from atoms of the same

transition series and having the same oxidation state have almost the same value of 0

but the cation with a higher oxidation state has a larger value of 0 than that with lower

oxidation state, e.g.

(a) 0 for [Fe2+ (H2O)6]2+ = 10,400 cm–1 .........3d6

0 for [Fe3+ (H2O)6]2+ = 13,700 cm–1 ..........3d5

(b) 0 for [Co3+ (H2O)6]2+ = 9,300 cm–1 ..........3d7

0 for [Co3+ (H2O)6]3+ = 18,200 cm–1 ..........3d6

This effect is probably due to the fact that the central ion with higher oxidation state

(i.e. with higher charge) will polarise the ligands more effectively and thus the ligands

would approach such a cation more closely than they can do the cation of lower

oxidation state, resulting in larger splitting.

2. Different charges on the cation of different metals : Two different cations having the

same number of d-electrons and the same geometry of the complex but with different

charge can also be compared. The cation with a higher oxidation state has a larger value

of 0 than that with a lower oxidation state. For example, the behaviours towards the

same ligand of V(II) and Cr(III), which are both d3 ions can be compared. It is observed

that the value of 0 in [V2+ (H2O)6]2+ is less than that in [Cr3+(H2O)6]3+ as in shown

below :

0 for [V2+ (H2O)6]2+ = 12,400 cm–1 ....... 3d3

and 0 for [Cr3+ (H2O)5]3+ = 17,400 cm–1 ....... 3d3

This fact can be explained in terms of the charge on the cation. The Cr3+ ion, which

has greater positive charge than V2+ ion, exerts a greater attraction for water molecules

(ligands) than does the V2+ ion. Hence the water molecules approach the Cr3+ ion more

closely than they approach the V2+ ion and so exert a stronger crystal field effect on the

d-electrons of Cr3+ ion.

3. Same charges on the cation but the number of d-electrons is different : In case of

complexes having the cations with the same charges but with different number of d-

electrons in the central metal cation the magnitude of 0 decreases with the increase of

the number of d-electrons, e.g.

0 for [Co2+(H2O)6]2+ = 9300 cm–1

0 for [Ni2+(H2O)6]2+ = 8500 cm–1

From the combination of 1, 2 and 3 mentioned above it can be concluded that :

(a) For the complexes having the same geometry and the same ligands but having different

number of d-electrons, the magnitude of 0 decreases with the increase of the number of

d-electrons in the central metal cation (No. of d-electrons 0

1

)

(b) In case of complexes having the same number of d-electrons the magnitude of 0

increases with the increase of the charges (i.e. oxidation state) on the central metal cation

(oxidation state 0 )

4. Quantum number (n) of the d-orbitals of the central metal ion : 0 increases about 30%

to 50% from 3dn to 4dn and by about the same amount again from 4dn to 5dn

complexes, e.g.

0 for [Co3+ (NH3)6]3+ = 23,000 cm–1 .......3d6

0 for [Rh3+ (NH3)6]3+ = 34,000 cm–1 .......4d6

0 for [Ir3+(NH3)6]3+ = 41,000 cm–1 .......5d6

Presumably the 5d and 4d valence orbitals of the central ion are better than the 3d-

orbitals in -bonding with the ligands.

B. Strong(er) and weak(er) ligands and spectrochemical series

Ligands which cause only a small degree of crystal field splitting are termed weak field

ligands. Ligands which cause a large splitting are called strong field ligands. Most

values are in the range 7000 cm–1 to 30000 cm–1. The common ligands can be

arranged in ascending order of crystal field splitting . The order remains practically

constant for different metals, and this series is called the spectrochemical series.

Spectrochemical series

weak field ligands : I– < Br– < S2– < Cl– < NO3– < F– < OH– < EtOH < oxalate < H2O

< EDTA < (NH3 and pyridine) < ethylenediamine < dipyridyl

< o-phenanthroline < NO2– < CN– < CO strong field ligands

Limitations of Crystal Field Theory

(i) CFT considers only the metal ion d-orbitals and gives no consideration at all to other

metal orbitals such as s, px, py and pz orbitals and the ligand -orbitals. Therefore, to

explain all the properties of the complexes dependent on the π-orbitals. Therefore to

explain all the properties of the complexes dependent on the π-ligands orbitals will be

outside the scope of CFT. CFT does not consider the formation of π-bonding in

complexes.

(ii) CFT is unable to account satisfactorily for the relative strengths of ligands, e.g. it gives

no explanation as to why H2O appears in the spectrochemical series as a stronger ligand

than OH–.

(iii) According to CFT the bond between the metal and ligand is purely ionic. It gives no

account of the partly covalent nature of the metal-ligand bonds. Thus the effects directly

dependent on covalency cannot be explained by CFT.

CHEMISTRY PRACTICAL QUESTIONS

1. Describe the Lassaigne’s test

A small piece of metallic sodium was taken in a dry fusion tube. Then the tube was heated in

non luminous flame to melt the sodium into a shiny silver coloured droplet. A small pinch of

organic sample was added to the tube and it was heated to red hot condition. The lower

portion of the red hot tube was then plunged quickly into 10 mL distilled water taken in a

mortar. The cracked lower part of the tube containing the fused mixture was ground with a

pestle. The resulting solution was filtered and divided into three portions to perform the

following tests.

Experiment Observation Inference

i) To one portion of the fusion filtrate, few

crystals of FeSO4 or a pinch of Mohr salt

was added. The mixture was boiled for a

few minutes, cooled and acidified with

dilute H2SO4.

Prussian blue colouration. Nitrogen present.

ii) To another portion of the fusion filtrate,

two drops of sodium nitroprusside solution

was added.

Violet colouration. Sulphur present.

iii) Last portion of fusion filtrate was

acidified with dil. HNO3, and boiled to

reduce the volume in half. The solution was

cooled and few drops of AgNO3 solution

was added to it.

Curdy white precipitation.

Soluble in NH4OH but

insoluble in dilute HNO3.

Chlorine present.

2. Write down the formula of Prussian blue product produced during the detection of nitrogen

by Lassaigne’s test

3. Explain the observations of chlorine detection test in Lassaigne’s method

During Lassaigne’s test the chlorine in the organic sample reacts with metallic Na to form

NaCl. NaCl in reaction with AgNO3 produces curdy white precipitate of AgCl. This AgCl is

soluble in NH4OH due to formation of soluble salt but insoluble in HNO3.

Na + Cl → NaCl

NaCl + AgNO3 → AgCl↓ + NaNO3

4. Explain the observations of sulfur detection test in Lassaigne’s method

During Lassaigne’s test the sulfur in the organic sample reacts with metallic Na to form

NaSCN and Na2S. Sodium nitroprusside (Na2[Fe(CN)5NO]) solution reacts with Na2S to

form the violet coloured complex Na4[Fe(CN)5NOS]

Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS]

5. Describe the Dye test for aromatic primary amine sample with equation.

To a cold 5% HCl solution of the organic sample, chilled aqueous NaNO2 solution was added

at -5ºC. The resulting mixture was quickly added to alkaline was quickly added to alkaline β-

Naphthol soluntion kept on ice salt bath. Bright orange or red coloured dye was formed

confirming the presence of aromatic primary amine group.

6. Describe Mulliken Barker Test

Aromatic nitro groups can be confirmed in presence of primary amine group by Mulliken

Barker Test. Nitro compounds are partially reduced to aromatic hydroxyl amines by

reduction with Zn dust and NH4Cl in ethanol. The resulting mixture is filtered upon freshly

prepared Tollen’s reagent. Formation of silver mirror or grey-black precipitate proves the

presence of nitro group, as hydroxyl amine is oxidised to nitroso group and tollen’s mixture is

reduced to metallic silver (Ag).

7. Write down the reaction of FeCl3 test of phenolic –OH group detection

Two drops of neutral ferric chloride solution was added to ethanolic solution of the sample

compound. Red, green or violet colouration due to chelate formation indicates presence of

phenolic –OH group.