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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.