CHEMISTRY
STRUCTURAL IDENTIFICATION & POC-1 STRUCTURAL IDENTIFICATION : (1) Calculation of Degree of Unsaturation (DU):It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE)
–2H H3C – H2C – CH3 (DU O)
–2H CH3 – C CH or CH2 = C = CH2 or
That means Deficieny of 2H is equivalent to 1 DU (i) 1DU = Presence of 1 Double Bond or Presence of 1 Ring closure (ii) 2DU = Presence of 2 Double bond or 1 Triple bond or two ring closure or 1 double bond + 1 ring closure. D.U.
(i) CxHy
y (x + 1) – 2
(ii) CxHyOz
yo (x + 1) – 2
(iii) CxHyXs
ys (x + 1) – 2
(iv) CxHyNw
y–w (x + 1) – 2
(v) CxHyOzXsNw
ys– w (x + 1) – 2
Monochlorination:When an alkane or a cycloalkane is treated with halogen (Cl2, Br2, F2, I2), a photochemical reaction takes place and a C – H bond cleaves and a C – Cl bond is formed. So, one H-atom is substituted by one halogen atom. This is known as monohalogenation reaction. Ex.1
Cl / h CH3Cl + HCl CH4 2
Cl / Sunlight Ex.2 CH3 – CH3 2 CH3 – CH2Cl + HCl
Ex.3
Ex.5
Cl / h 2
Cl / h 2
–
Cl
+ HCl Ex.4
Cl / h 2
+ HCl
Cl
–
2.
G.F.
+ HCl
Application : If a molecule has more than one type of H-atom, then on monochlorination, it forms a mixture of monochloroisomers. All these isomers are position isomers.
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CHEMISTRY Conclusion : Hence, it can be concluded that the total number of position isomers (structural) of monochloro compounds is equal to the number of different types of H-atoms present in the reactant. The different type of H-atoms are also known as non-identical Hydrogens or non-equivalent Hydrogens or chemically different Hydrogens. Ex.6
Monochlorination CH3 – CH2 – CH3 2
(structural)
Ex.7
Monochlorination CH3 – CH2 – CH2 – CH3 2
(structural)
Ex.8
Monochlorination CH3 – CH2 – CH2 – CH2 – CH3 3
(structural)
Monochlorination 4 (structural)
Ex.9
–
CH3 Monochlorination 5
Ex.10
(structural)
Remark : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated but H-atoms of benzene ring are stable. In pure benzene, no monochlorination occurs.
CH3
–
–
CH2Cl
Cl / h 2
Ex.11
3.
Catalytic Hydrogenation of C = C; C C
(i) (ii) (iii) (iv)
Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature. All C – C bonds(C = C, C C) are hydrogenated. The reaction can’t be stopped at any intermediate stage. Aromatic bonds which are stable at room temperature but can be hydrogenated at high temperature. It can be concluded that the hydrogenation product of an alkene or alkyne or any unsaturated compound is always a saturated compound. The number of moles of H2 consumed by 1 mole of compound is equal to the number of bonds presents. All positional isomers of alkenes or alkynes (due to multiple bond) always give same product on hydrogenation. During catalytic hydrogenation no rearrangement in carbon skeleton takes place. General reaction : Ni
R – CH2 – CH2 – R R – CH = CH – R + H2 Ni / Pt / Pd R – C C – R + 2H2 R – CH2 – CH2 – R
H2 R – CH CH – R R – CH2 – CH2 – R (Not isolated) H2 / Ni CH2 = CH – CH2 – CH3 CH3 – CH2 – CH2 – CH3
3H2 / Ni
H2 / Ni room temperature
CH2 – CH 3
–
CH = CH2
–
(v) (vi) (vii)
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CHEMISTRY CH2 – CH3
H / Ni 2
–
H 2/Ni (100 – 150ºC)
[Reaction cannot be stopped at any intermediate stage]
2H2 / Ni
Ex.12
CH3
(X)
Cl / h 2
+
Cl
CH3 Cl / h 2 5 Monochloro product (structural)
H / Ni 2
Ex.13
CH3
CH3 Cl / h 2 5 Monochloro product (structural)
H / Ni 2
Ex.14
4.
Ozonolysis:
(i)
It tells about position of unsaturation.
(ii)
Alkene and polyalkene on ozonolysis undergo oxidative cleavage.
(iii)
(a) The reagent of reductive ozonolysis is (i) O3 (ozone) (ii) Zn and H2O or Zn and CH3COOH or (CH3)2 S (b) The reagent of oxidative ozonolysis is O3 and H2O2.
(iv)
The products are carbonyl compounds (aldehydes or ketones). This type of ozonolysis is known as reductive ozonolysis.
(v)
Ozonolysis does not interfere with other functional groups. General Reaction : (1) O
R – CH = CH – R 3 R – CH = O + O = CH – R + ZnO + H2O ( 2) Zn / H2O
(1) O
Ex.15
CH2 = CH2 3 CH2 = O + CH2 = O
Ex.16
CH3 – CH2 – CH = CH2 3 CH3 – CH2 – CH = O + O = CH2
Ex.17
CH2 = CH – CH2 – CH = CH – CH3 3 CH2 = O + O = CH – CH2 – CH = O + O = CH – CH3
Ex.18
( 2) Zn / H2O
(1) O
( 2) Zn / H2O
(1) O
( 2) Zn / H2O
(1) O
3 OHC – CH – CHO (Propandial) 2 ( 2) Zn / H2O
Applications: The process is used to determine the position of C = C in a molecule. If the products are rejoined, the position of C = C can be determined in the reactant molecule. All C = C (except aromatic ones) undergo oxidative cleavage under normal conditions. At higher temperature, the aromatic double bonds can also undergo ozonolysis.
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CHEMISTRY O3 O = CH – CH2 – C – CH2 – HC = O + O = CH2
Ex.19
Zn
O
O () 3
Ex.20
Zn
– Ex.21
Ex.22
low temperatur e
O3 / Zn
– CH = CH –
CH = O
–
CH = CH – CH 3
+ O = CH – CH3
O3
Zn/H2O C6H5 – CH = O
POC-I : Introduction The main objective of an organic chemist is the determination of the structure of a new organic compound which has been obtained in pure state either from a natural source or synthesised in the laboratory.
In order to establish the correct structure of an organic compound, it is necessary to detect element and functional group present in the organic compound.
Detection of elements (Qualitative Analysis) :Most of the organic compounds contain 2 to 5 different elements
The principal elements present are carbon, hydrogen and oxygen.
Less commonly present elements are nitrogen, sulphur and halogens.
In few organic compounds , phosphorus and metal may also be present
The order of abundance of these elements in organic compounds is given below: Always present Carbon Hydrogen Nearly always present Oxygen Generally present Less commonly present Nitrogen, halogen, sulphur Rarely present Phosphorus and metal Detection of nitrogen, sulphur and halogens are tested in an organic compounds by lassaigne’s test.
The organic compound (N,S or halogen) is fused with sodium metal as to convert these elements in ionisable inorganic substance i.e. nitrogen into sodium cyanide, sulphur into sodium sulphide and halogens into sodium halides.
These cyanide, sulphide or halide ions can be confirmed in the aqueous solution by usual test. The aqueous solution (fused sodium extract) is called lassaigne’s filterate.
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CHEMISTRY
4.1
Identification of Elements in Organic Compounds Element
1. Nitrogen
Test / Reaction
Remark
Lassaigne’s test
The appearance of green or
Na + C + N NaCN
prussian blue colour confirms
FeSO4 + 6NaCN Na4 [Fe(CN)6] + Na2SO4
the persence of nitrogen.
3Na4[Fe(CN)6] + 4FeCl3 Fe4[Fe(CN)6]3 + 12NaCl
2. Sulphur
Formation of a white ppt.
(a) Oxidation test
indicates presence of sulphur
3KNO3 3KNO2 + 3[O] Na2CO3 + S + 3[O] Na2SO4 + CO2 BaCl2(aq) + Na2SO4(aq) BaSO4 + 2NaCl(aq)
Appearance
(b) Lassaigne’s test
purple
colouration confirms the
2Na + S Na2S Na2S + Na2[Fe(CN)5NO] Na4[Fe(CN)5NO.S]
3. Halogens
of
Lassaigne’s test
presence of sulphur
A white ppt. soluble in NH4OH
X + Na NaX
solution indicates chlorine.
NaX + AgNO3 NaNO3 + Ag X
A dull yellow ppt. partly soluble in NH4OH solution indicates bromine. A yellow ppt. completely in-
,
soluble in NH4OH solution indicates iodine
A white ppt. of magnesium
4. Phosphorus
pyrophosphate indicates phosphorus H3PO4 + Magnesia mixture MgP2O7 + H2O 2MgNH4PO4 Mg2P2O7 + 2NH3 + H2O
5. Nitrogen and Sulphur
Blood red colouration confirms
Lassaigne’s test FeCl
Na + C + N + S NaSCN 3 Fe(SCN)3
presence of both nitrogen & sulphur
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Bubbles of H2 come out (3)° Cloudiness appears immidiately (2°) Cloudiness appears within 5 min. (1°) Cloudiness appear after 30 min.
Lucas Reagent [Conc. HCl + anhyd. ZnCl2]
ROH 3° 2° 1°
White ppt.
"manishkumarphysics.in" R – OH + HCl
white ppt
+ H2 O cloudiness
R – C C Ag (white)
R – C C Cu (red)
RCOOH + RCOOH
2HCHO
2ROH + Na 2RONa + H2
R – C CH + Ag+
R – C CH + CuCl
Red ppt.
Na
(b) AgNO3 + NH4OH
(a) Cuprous chloride + NH4OH
R – C C – R
Acid formed.
= O Compounds
O3(ozone)
O3
Red colour decolourises
H2C = CH2 + O3
--------------
NR NR NR NR Pink colour Disappears
Reaction
Observation
Br2 / H2O
[Bayer’s reagent] alk. dil. cold KMNO4
conc. H2SO4 conc. NaOH KMnO4 LiAlH4
Reagent
(R – OH)
R – C CH
CC
CC
C=C/ CC
C–C
Functional Groups
4.2 Identification of Functional Groups by Laboratory Tests
Lucas Test I. ter.alcohol II. sec. alcohol III. pri.alcohol
Presence of active ‘H’
Ozonolysis
Ozonolysis
Bromination
Hydroxylation
Inert paraffins
Remarks
CHEMISTRY
6
"manishkumarphysics.in"
Amides
Ester
CH3CHO
or ArCOCH3 or
R – COCH3
R – CHO
Ar – OH Enols
Functional Groups
(yellow orange ppt.)
Sodium bicarbonate test
Effervescence evolve.
Pink colour disappear on heating.
Smell of NH3
Conc. NaHCO3 solution
NaOH, phenophthalein.
Conc. NaOH,
RCOONa + NH3
Schiff’s reagent : p-Rosiniline hydrochloride saturated with SO2 so it is colourless. The pink colour is resumed by RCHO.
RCONH2 + NaOH
R COOH + R’ OH (Colourless solution)
(pink)
R COOR’ + NaOH + Phenophthalein
Litmus test.
Iodoform reaction
Tollen’s test
Fehling’s test
DNP-test
Test of enols / phenols
Remarks
Litmus change to red.
H2O + CO2
RCHO + Ag+ RCOOH + 2Ag (Silver mirror)
RCHO + Cu+2 RCOOH + Cu2O + 2H2O Fehling soln. Red
H
+ H2
Reaction
Blue litmus
Yellow ppt of CHI3 (iodoform)
Pink colour resume
Schift’s Reagent *
I2 / NaOH
Black ppt. or silver mirror
Tollen’s reagent
Red ppt.
Yellow orange ppt.
2, 4-Dinitrophenyl hydrazine (2, 4-DNP) solution
Fehling solution A&B
Coloured ppt. (violet, blue, green buff)
Observation
FeCl3 (Neutral)
Reagent
CHEMISTRY
7
Violet colour Blue colour
Ninhydrin reagent (0.2 % sol.n)
Amino acids
Reddish violet colour.
red colouration Liebermann test
Molisch’s reagent (10% -naphthol in alcohol).
(i) NaNO2 + H2SO4 (ii) Phenol
R2NH Sec. Amines
Orange red dye is formed
Effervescence of N2
Nauseating odour (Offensive smell) (Carbylamine)
black ppt
Observation
Carbohydrate
HNO2 (NaNO2 + HCl) + -Naphthol
HNO2 (NaNO2 + HCl)
CHCl3, KOH
Mulliken’s test
Reagent
Ar. amines. ArNH2
Amines (pri.) RNH2
Nitro Compounds (RCH2NO2) or ArNO2
Functional Groups
ArNHOH
"manishkumarphysics.in" C
C=N – C (Blue colour)
CO
CO
C OH
CO
+ H2N.CHR.COOH (Amino acid) OH
OH
(Ninhydrin)
CO
CO
Benzenediazonium chloride – Naphthol
N=N-Cl +
OH N=N
OH
+ CO2+ RCHO + H2O
orange-red dye
+ 2H2O
N2Cl
NH2.HCl
ROH + N2 + H2O NaCl + HNO2
+ HNO2
Ag Ag
RNC + 3KCl + 3H2O
NaNO2 + HCl
R NH2 + HONO
R NH2 + CHCl3 + 3KOH
Reaction
Ninhydrin test
Dye test
Carbylamine Reaction
Remarks
CHEMISTRY
8
CHEMISTRY
MISCELLANEOUS SOLVED PROBLEMS (MSPS) 1.
Identify X, Y & Z with the help of following reactions. Cl2 / h
Single monochloro structural product
(X) (C 9H18 ) ( Saturated Hydrocarbon)
Cl2 / h Single monochloro structural product (Y) (C8H18) Cl / h
(Z) (C7H14 ) 2 s Two monochloro structural products ( Saturated Hydrocarbon)
Ans.
C, D. CH3 CH3 | | Y = CH3 – C – C – CH3 Z = | | CH3 CH3
Sol.
X=
or
2.
Cl / h Two monochloro X(C8H10) (Aromatic) 2
(structural)
Cl / h One monochloro Y(C8H10) (Aromatic) 2
(structural)
,
Ans.
(X) 3.
An organic hydrocarbon on oxidative ozonolysis produces oxalic acid and butanedioic acid. Its structure is (A)
Ans.
(Y)
(B)
(C)
(D*)
(D)
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CHEMISTRY O3 Oxidation
Sol. 4.
An optically active acyclic compounds X (molecular formula C5H9Br) give following reactions
Ans.
The incorrect statements about 'X' are: (A*) It has two stereoisomers (C*) It has two asymmetric carbon atoms (A,B,C)
(B*) It gives an achiral product (Y) on ozonolysis (D) It has four stereoisomers and all are optically active
O3
Sol.
H2 (X)
5.
A set of reagents (1 to 8) are successively reacted with the following compound
1. NaHCO3
2. 2, 4, DNP
3. Na metal
4. AgNO3 + NH4OH
5. Fehling’s solution
6. Cu2Cl2 + NH4OH
7. Br2 / H2O
8. NaNO2 + HCl
The reagents which give positive test with the given compound are : (A) 1, 2, 3, 4, 5 (B) 3, 4, 5, 6, 8 (C) 1, 2, 3, 4, 8
(D*) All reagents except 1 and 8
Ans. Sol.
(D) There is no COOH group or NH2 group. Regents 1 and 8 will fail to give test.
6.
Compounds I and II can be distinguished by using reagent. () (II) 4-Amino-2-methylbut-3-en-2-ol 4–Amino–2,2-dimethylbut- 3-yn-1-ol.
Ans.
(A) NaNO2 / HCl
(B) Br2 / H2O
(C*) HCl / ZnCl2 (anhydrous) (C)
(D) Cu2Cl2 / NH4OH
Sol.
(I) gives immediately turbidity by Lucas reagent and (II) does not give turbidity appriciably.
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