Session - 2009-10
TOPIC : CONTENTS
1
GOC - I
:
IUPAC No Nomenclature
Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and and E,Z nomenclature nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections);
IUPAC NOMENCLATURE 1.1 1.1
INTR INTROD ODUC UCTI TION ON TO ORGA ORGANI NIC C COM COMPO POUN UNDS DS::Organic compounds are compounds of carbon and hydrogen and the the following elements may also be present: (Halogens, (Halogen s, N, S, P, P, O). There are large no. of organic compounds available avai lable and large no of organic compounds are synthesized every year.The year.The most important reason for large larg e no of organic compounds is the property of catenation (self –linkage) in carbon. Element
Bond Energy
C
C –C
(strongest bond)
Si
Si –Si
Ge
Ge –Ge
Sn
Sn –Sn
Pb
Pb –Pb
! Decreasing order
Bond energy depends on (i) Size of atom (Inversely proportional)
Size : C –C>Si –Si
(ii) E.N difference along period (Directly)
E.N Diff : C – H < N – H < O – H < H – F
(iii) (iii) Bond Bond order order (no. (no. of of cova covalet let bon bonds ds b/w b/w two two atoms atoms)) (Dire (Directl ctly) y)
Bond Bond order order : C – C < C = C < C " C
Catenation in carbon:-
# The element carbon has strongest tendency to show catenation or self –linkage due to – (a) its tetravalency so that it can form $ bonds with many elements elements as well as carbon itself. (b) It can form multiple bonds (C = C, C " C). Due to its small size, there is efficient colateral overlapping b/w two P –orbitals.
Does not exist (c) High bond energy of C – C bond so that it can f orm strong bonds in long chains and in cylic compounds.
1.2 1.2
CLA CLASSIF SSIFIC ICA ATION TION OF ORGAN ORGANIC IC COMP COMPOU OUNDS NDS :
IUPAC NOMENCLATURE 1.1 1.1
INTR INTROD ODUC UCTI TION ON TO ORGA ORGANI NIC C COM COMPO POUN UNDS DS::Organic compounds are compounds of carbon and hydrogen and the the following elements may also be present: (Halogens, (Halogen s, N, S, P, P, O). There are large no. of organic compounds available avai lable and large no of organic compounds are synthesized every year.The year.The most important reason for large larg e no of organic compounds is the property of catenation (self –linkage) in carbon. Element
Bond Energy
C
C –C
(strongest bond)
Si
Si –Si
Ge
Ge –Ge
Sn
Sn –Sn
Pb
Pb –Pb
! Decreasing order
Bond energy depends on (i) Size of atom (Inversely proportional)
Size : C –C>Si –Si
(ii) E.N difference along period (Directly)
E.N Diff : C – H < N – H < O – H < H – F
(iii) (iii) Bond Bond order order (no. (no. of of cova covalet let bon bonds ds b/w b/w two two atoms atoms)) (Dire (Directl ctly) y)
Bond Bond order order : C – C < C = C < C " C
Catenation in carbon:-
# The element carbon has strongest tendency to show catenation or self –linkage due to – (a) its tetravalency so that it can form $ bonds with many elements elements as well as carbon itself. (b) It can form multiple bonds (C = C, C " C). Due to its small size, there is efficient colateral overlapping b/w two P –orbitals.
Does not exist (c) High bond energy of C – C bond so that it can f orm strong bonds in long chains and in cylic compounds.
1.2 1.2
CLA CLASSIF SSIFIC ICA ATION TION OF ORGAN ORGANIC IC COMP COMPOU OUNDS NDS :
Ex.
Compounds
Classification Unsaturated
Heterocyclic , saturated Unsaturated
Saturated
Saturated, Alicylic
1.3
IMPORTANT TE TERMS: Saturated compounds: When all the valencies of an element are satisfied by $ covalent bonds. Unsaturated compound: When a compound contains one or more m ore % bonds (C = C, C = N, N = N, C = O or C " C, C " N, N " N) Molecular Formula (M.F.) : The molecular formula of a compound indicates the actual number of atoms of each element present in one molecule. Structural Formula (S.F.): It indicates the linkage due to covalent bond between different atoms in a molecule. (i) Expanded Structural Formula Form ula (E.S.F.) (ii) Condensed Structural Formula Form ula (C.S.F.) (C.S.F.) (iii) Bondline Structural Formula (B.S.F.) Ex.
M.F.
C3H8
Ex. H
Ex.
H
M.F.
C4H10
H
H
H
H
H
E.S.F. H – C – C – C – H H H H C.S.F. CH3 – CH2 – CH3
E.S.F. H – C – C – C – C – H H H H H C.S.F. CH3 – CH2 – CH2 – CH3
B.S.F.
B.S.F.
or
B.S.F.
C.S.F. M.F.
Ex. CH 3 – CH – CH 2 – CH 3 CH 3
B.S.F.
M.F.
C6H12
C5H12
Homologous series : Homologous series may be defined as a series of similarly constituted compounds in which the members possess the same functional group, have similar chemical characteristics and have a regular gradation in their physical properti properties. es. The two consecutive members differ in their molecular formula by CH 2.
Calculation of Degree of Unsaturation (DU):(a) It (a) It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE)
–2H (b) H3C – H2C – CH3 ' ' ' # (DU & O)
–2H ' ' ' # CH3 – C " CH or CH2 = C = CH 2 or
That means Deficieny of 2H is equivalent to 1 DU (c) (c)
(i) 1DU = Presence of 1 Double Bond Bond or Presence of 1 Ring closure (ii) 2DU = Presence of 2 Double bond or 1 T Triple riple bond or two ring closure or 1 double bond + 1 ring.
(d)
G.F.
D.U.
(i) CxHy
- y * (x + 1) – + ( , 2 )
(ii) CxHyOz
- y . o * ( (x + 1) – + , 2 )
(iii) CxHyXs
- y . s * ( (x + 1) – + , 2 )
(iv) CxHyNw
- y – w * ( (x + 1) – + , 2 )
(v) CxHyOzXsNw
- y . s – w * ( (x + 1) – + 2 , )
Ex
Calcul lcula ate DU DU of of fo followi owing co compoun ounds (a) C6H6O
DU = 4
(b) C6H5Cl
DU = 4
(c) C6Br6
DU = 4
(d) C5H11OCl
DU = 0
(e) C9H12N2
DU = 5
(f) C6N6
DU = 10
(g) C10H8SO5 N4Cl2
DU = 8
e. g .
M .F.
DU
S. F.
H 1.
C4H6
2
H
C –C=C=C
H H
H H C – C – C " C C – C " C – C C = C – C = C
C
CH2 C
2.
C2H2Cl2
=1
Total isomers = 3
3. C7H6O (Aromatic)
=5
4. C7H8 (Aromatic)
=4
Note : In case of the aromatic Compounds minimum DU = ‘4’. That means at least 1 Benzene ring is present.
Degree of carbon : It is defined as the number of carbon atoms attached to a carbon atom.
H | Primary (1°) carbon
:
| H
1°
Secondary (2°) carbon :
C | Tertiary (3°) carbon
:
| H
3° C |
Quaternary (4 °) carbon :
| C
Superprimary :
4°
(1 °)
Ex. 0
1 – 5 carbon
CH3 CH3 1 2 | 3| 4 5 CH3 – C – CH – CH2 – CH3 | CH3
20 – 1 carbon 30 – 1 carbon 40 – 1 carbon
Ex.
*Degree of hydrogen is same as the degree of carbon to which it is attached. Type of C – C bonds & type of replaceable H-atom s in saturated hydrocarbon. Note : Total no. of monosubstituted product does not depend upon type of C-atoms but depends upon type of replaceable H-atoms.
1.4
IUPAC NOMENCLATURE OF ORGANIC COMPOUNDS: General Scheme of Naming:Secondary Prefix + Primary Prefix _ _Word Root_ _Primary Suffix_ _Secondary suffix The organic compound is always named according to the general scheme as given by IUPAC. In every compound, two parts, viz, word root and primary suffix, always exist.
Word Root: It indicates the no. of carbon atoms present in the main chain. It is represented as Alk. Prefix : It is the first part of the name. (i) Primary Prefix : ‘Cyclo’ (ii) Secondary Prefix : Normal substitutents and junior functional groups are treated as a substituent than their name is treated as secondary prefix. The following substituent groups are always cited in the prefix. (i) R – Alkyl (ii) R – O – Alkoxy (iii) X halo (fluoro, chloro, bromo, iodo) (iv) –NO2 Nitro (v) –N = O Nitroso (vi) Junior functional group # The prefixes are always written in alphabetical order # The position of substituent group in the main carbon chain is mentioned by writing the number just before the name of substituent by writing a small dash ( –).
Suffix : (i) Primary Suffix:- It indicates saturation or unsaturation existing in the main chain. ane # single bond (saturated) ene # one = bond diene # If two double bonds Polyene # if plenty of double bond are present. yne # one " bond diyne # two " bond (ii) Secondary Suffix:- It is used for the principal functional group.
1.5
NAMING OF SATURATED HYDROCARBONS Rules :(Branched and substituted Alkanes) (1) Selection of parent chain # (a) Chain with maximum number of ‘C’ atoms (longest chain).
(b) If number of carbon atoms are same in more than one longest chain then more substituted longest chain will be the parent chain.
(c) If number of side chain are also same then that will be parent chain having its substituent at lower number.
C
C
C – C – C – C – C – C – C – C C – C – C C
(2) Numbering (a) Numbering is done from that side of the parent chain having it substituent at lower number (lowest set of locant)
(b) If position of substituent are same from both the end of the parent chain, then numbering is done from alphbetical order.
* If alphabets are also same then numbering is done from that side of the parent chain having its substituent of substituent at lower number.
Ex.
CH3 – CH3 – CH – CH3
Methylpropane
CH3 CH3 –
–
CH3 – CH – CH – CH3
2, 3-Dimethyl butane
CH3 – CH3 – C – CH3 – CH3
Dimethyl propane
2-Methylbutane
C
–
C – C – C – C – C
3-Ethyl-2-methyl hexane
–
C – C – C
C – C – C – C – C – C – C – C
4, 4-Diethyl heptane
–
C – C – C C
–
C – C – C – C – C – C – C – C
4, 4-Diethyl-2, 5-Dimethyl heptane
–
C – C – C – C 3-Ethyl hexane
C – C – C – C – C –
C – C – C CH3 – CH2 – CH2 – CH2Cl
1-Chlorobutane
C – C – C – C – C
3-Bromo-2-chloropentane
–
Cl
–
Br 2-Bromo-4-chloro pentane
C – C – C – C – C –
–
Br
Cl
Cl – C – C – C – C – C – – Br Cl
4-Bromo-2, 2-dichloro pentane
C – C – C – C – C – C
5-Bromo-2, 3-dichloro hexane
–
–
–
Cl Cl
Br
(3) Rules for writing Alkyl Radicals:Radicals:- (a)
–H ' # CnH2n + 1 CnH2n + 2 ' '
Alkyl Radical
–H ' # CnH2n CnH2n ' '
Alkenyl Radical
–
1
–H ' # CnH2n CnH2n 2 ' ' –
–
3
R (structural formula)
Name
CH3 –
Methyl
CH3 – CH2 –
Ethyl
Alkynyl Radical
CH3 | (b) CH3 – C – CH3 | H
CH3 3 2| 1 – H CH3 – C – CH2 – Isobutyl | H
– H
CH3 | CH3 – C – CH3 Tert butyl (Tertiary butyl) |
(4) Systematic Names of Radicals : Complex Radical : 5 – ( 1,2 – Dimethylpropyl) nonane
Ex
5 – (1,1 – Dimethylpropyl) nonane
5 – (1 – Ethylpropyl) nonane
5 – (Dimethylethyl) nonane
Alkene / Alkyne radicals Alkene Alkyne
Alkenyl Alkynyl
CH2 = CH2 CH3 – CH = CH2
CH2 = CH ethenyl
–
Prop – 1 – enyl Methylethenyl
Prop – 2 – enyl
CH3 – C " CH
CH3 – C " C – Prop-1-ynyl
CH3 – C " CH
–
CH2 – C " C – H Prop-2-ynyl
(5) Benzene Radical 1.
Phenyl
2.
Benzyl
3.
Benzal
4.
Benzo
5.
Tolyl
CH3 (o, m, p)
01
Methylene
7.
Alkylidene
8.
2- naphthyl
9.
3 -naphthyl
(6) Numeral Prefixes (1) Following prefixes are considered for alphabetization : (a) Iso
(b) neo
(c) Di, Tri, Tetra of complex radical are considered for alphabetization (2) Following are not considered for alphabetization (a) Di, Tri, Tetra for simple radicals. (b) Sec, Tert are not considered for alphabetization (c) Bis , Tris
(7) Retained Names of Alkanes: (1) Normal (n) : Radical or hydrocarbon which has straight chain and if it has free valency it must be present at either of ends. C – C – C – C (n butane) (2) Iso : Two methyl group at the end of linear chain (unbranched chain) (i)
(ii)
(iii) Conditions :-
C –
C – C – C
Isobutane or methyl propane
C –
C – C – C – C
Isopentane or methylbutane
C –
C – C – C – C – C
Isohexane or 2-methylpentane
4 There should be 4 to 6 carbon atoms only 4 There should not be any other alkyl group present in the chain. Q. Ans.
Iso-octane (commercial name) in petroleum industry C –
C – C – C – C – C – – C C
(3) Neo :
Ex.
IUPAC name is 2, 2, 4-trimethylpentane
4 There should be 5 to 6 carbon atoms 4 2, 2-dimethyl 4 No other substituent.
C –
C – C – C – C
C –
C – C – C – C Neohexane – C
Neopentane
(4) Secondary : It is applicable only for radical. Ex.1
(5) Tertiary :
First member
1.6
NAMING OF UNSATURATED COMPOUND (ALKENES AND ALKYNES) : (A) General formula:- CnH2n and CnH2n 2 respectively –
(B) Rules for selection of main chain:# Longest carbon chain with a multiple bond. # Longest carbon chain with maximum number of multiple bonds. # If first and second factor are common then chain with lowest locant (multiple bond) is selected as main chain. # Lowest locant Rule is followed till first point of difference. (Multiple bond prior to substituent). # Alphabetization. (C) Rules for Numbering:# Lowest locant Rule till first point of difference (irrespective of double bond or triple bond). # Then double bond is prefer over triple bond in numbering, naming and longest chain selection, If all the other factors are common Ex.
CH3 – CH2 – CH = CH2
But-1-ene
CH3 – CH = CH – CH3
But-2-ene
CH3 –
4-Methylpent-1-ene
CH3 – CH – CH2 – CH = CH2 CH3 | CH3 – CH – CH = CH – CH3
4-Methylpent-2-ene
Cl
– 5
6
4
3
2
1
CH3 – C – CH2 – CH2 – CH = CH2 –
5, 5-Dichlorohex-1-ene
Cl (a) C = C – C = C
Buta-1,3-diene
(b) C " C – C " C
Butadiyne
(c) C = C – C " C
Butenyne
C 1
"
C – C – C = C – C 2
3
5
4
2 1
4
6
Hex-4-en-1-yne
6
3 5
Hexa-1, 3, 5-triene
Ex.1
3-(2-Methylpropyl) hept-1-ene
Ex.2
4 -(1,1-Dimethylpropyl)-4-ethenylhepta-1,5-diene
Ex.3
3-Ethynylhexa-1, 5-diene
Ex.4
4-Ethenylhept-2-en-5-yne
Ex.5
4-(1, 2-Dimethylbutyl) hept-2-en-5-yne
Ex.6
2, 4-Dimethylpenta-1, 3-diene
Ex.6
2,3-Dimethylhex-1-en-4-yne
1.7
NAMING OF CYCLIC HYDROCARBON (ALICYCLIC COMPOUNDS) : (A) Main chain selection: (a) Multiple Bond > No. of carbon atoms > Maximum no. of substituents > Nearest locant > Alphabetization. (b) If all factors are similar in cyclic and acyclic part, then Cyclic > Acyclic
(B) Numbering: (a) Lowest Locant
(b) Alphabetization
(C) Naming: # Prefix ‘cyclo’ is used just before the word root if it constitutes the main chain. # If cyclic part is the main chain then the prefix ‘cyclo’ is not considered f or alphabetical order.. If cyclic part constitutes the side-chain (substituent) then prefix cyclo is considered for alphabetization:Ex:-
Cyclopropane
Cyclobutane
Cyclopentane
Cyclohexane
Cycloheptane
Cyclooctane
Methylcyclopropane
Ethylcyclopropane
Propylcyclopropane
1-Cyclopropylbutane
–
C – – C – Cl
Cl – C – – C
2-Chloroethylcyclopropane
Chloroethylcyclopropane
Cl – C – – C – C
1-Chloro-3-cyclopropylpropane
Cl 3 C – – C – C
2-Chloro-1-cyclopropylpropane
Cl – C – – C – C
1-Chloro-1-cyclopropylpropane
3
2
1
– 2
1
1
2
3
C – C – C –
–
1-Chloro-2-propylcyclopropane
Cl
Cl
– 1 – 2
4 –
Br
3 –
F
2-Bromo-1-chloro-3-fluoro-4-iodocyclohexane
I
Methylethylcyclopropane or isopropylcyclopropane
Cl
– 2 3 1 – 6 4 – 5
Br 1-Bromo-2-chloro-4-iodocyclohexane
I
6 5
1 2
3
4
3-Chlorocyclohex-1-ene
Cl Br
6 1
5
2 3
Cl
4
5-Bromo-3-chlorocyclohex-1-ene
1.8
FUNCTIONAL GROUP TABLE (Seniority order):
Class
Name
Suffix
1.
R – COOH
Alkanoic acid
–
2.
R – SO3H
Alkane sulphonic acid
3.
R – C – O – C – R || || O O
4.
Prefix
oic acid (carboxylic acid)
Carboxy
–
sulphonic acid
sulpho
Alkanoic anhydride
–
anhydride
------------
R – COOR
Alkyl alkanoate
–
alkanoate (carboxylate)
Alkoxy carbonyl
5.
R – C – X || O
Alkanoyl halide
–
oyl halide (carbonyl halide)
halo carbonyl
6.
R – C – NH2 || O
Alkanamide
–
Carbamoyl
7.
R – C " N
Alkanenitrile
–
cyano
8.
R – C – H || O
Alkanal
–
formyl / Oxo
9.
R – C – R || O
Alkanone
–
10. R – OH
Alkanol
11. R – SH
Alkanethiol
12. R – NH2
Alkanamine
1.9
amide (carboxamide)
nitrile (carbonitrile)
al (carbaldehyde)
one
Oxo / Keto
ol
hydroxy
–
–
mercapto
–
amino
thiol amine
NAMING OF FUNCTIONAL GROUP CONTAINING COMPOUNDS : (A) Selection of Main Chain: Senior F.G. > Max. no. of F.G. (Similar group) > Multiple bond > Max. no. of ‘C’ atoms > Max. no. of locants > lowest locant > alphabetization
(B) Numbering (See F.G. + Sub.): The carbon atom bearing functional group (or C – atom of terminal functional group) is given lowest possible number. (i) Lowest locant (F.G. > M.B. > substituent > Alphab)
(C) Naming:- General scheme : The senior most functional group constitutes secondary suffix. Other junior F.G. ’s are written as prefix in alphabetical order.
1.9.1
Carboxylic acid F.G.
– COOH
Prefix
Suffix
IUPAC name
Oic acid - 'C' of COOH considered in the parent chain
Alkanoic acid
Carboxylic acid - 'C' of COOH is not considerd in parent chain
Alkane carboxylic acid
Carboxy
Rule : If first alphabet of sec. suffix is begin from a, i, o, u, y then ‘e’ of primary suffix will be dropped. Ex.
(1) HCOOH
Methanoic Acid
(2) CH3 – COOH
Ethanoic Acid
(3) C – C – COOH
Propanoic Acid
(4) C – C – C – COOH
Butanoic Acid
2
3
4
5
6
(5) C – C – C – C – C – C – C – C 1 COOH
2-propylhexanoic acid
COOH
(6)
–
Cyclopropanecarboxylic acid
2
(7)
1
CH2 – COOH –
Cyclopropylethanoic acid
COOH 3' C – C – C 1'
2' –
(8)
4-(2-carboxypropyl) cyclohexane-1-carboxylic acid –
COOH
1.9.2
(9) HOOC – COOH
Ethanedioic acid
(10) HOOC – CH2 – COOH
Propanedioic acid
(11) HOOC – CH2 – CH2 – CH2 – CH2 – COOH
Hexanedioic acid
(12)
Cycloprop-1-ene-1, 2, 3-tricarboxylic acid
Sulphonic acid F.G.
Prefix
– SO3H
Sulpho
Ex.
Suffix
Sulphonic acid
(1) CH3 – SO3H
IUPAC name
Alkanes sulphonic acid
Methanesulphonic acid
(2) C – C – C – C – C –
SO3H
Pentane-2-sulphonic acid
O = C – NH2 –
(5)
Benzene carboxamide
O = C – NH2 –
(6)
Benzene-1, 3-dicarboxamide –
C – NH2
O COOH –
(7)
3-Carbamoyl cyclohexane-1- carboxylic acid
–
C – NH2
O
O C – NH – CH3 (8) CH2
N, N’-Dimethyl propane-1, 3-diamide
C – NH – CH3 O
1.9.7
Nitrile:-
Example:-
(1) H – CN
Methanenitrile
(2) CH3 – CN
Ethanenitrile
(3) CH3 – CH – CH2 – CN – CH3
3-Methyl butanenitrile
CN – (4)
–
CN
–
Cyclopropane-1, 2, 3-tricarbonitrile
CN
CN –
(5)
(6)
(7)
Benzene carbonitrile
CH – 2 – CH2 – COOH CN
– CH CH – 2 – CH – 2 – CN
CN
CN
3-Cyano propanoic acid
Propane-1,2,3-tricarbonitrile
1.9.8
Isocyanide :-
F.Group
– N Example:-
"
C
Prefix
Suffix
Name
Isocyano
Isocyanide
Alkyl Isocyanide
(1) CH3 – N " C
Methyl isocyanide
(2) CH3 – CH2 – N " C
Ethyl isocyanide
NC –
(3)
1.9.9
Phenyl isocyanide
Aldehyde:Prefix
F.Group
Name
Formyl or oxo al or carbaldehyde Alkanal or Alkane Carbaldehyde
– CH = O
Example:-
Suffix
(1) HCHO
Methanal
(2) CH3 – CH2 – CH = O
Propanal
(3) CCl3 – CH2 – CH2 – CH2 – CH = O
5, 5, 5-Trichloro pentanal
CH = O –
(4)
Cyclohexane carbaldehyde
CH = O –
(5)
Benzaldehyde/Benzene carbaldehyde
CH = O Ph – – CH – CH – (6) Ph – CH2Cl
2-Chloromethyl-3, 3-diphenyl propanal
CH = O –
(7)
Benzene-1, 3-dicarbaldehyde –
CH = O
(8) HOOC – CH2 – CH2 – CH2 – CH = O
5-Oxopentanoic acid
(9) HOOC – CH2 – CH – CH2 – COOH
3-(2-Oxoethyl) pentane-1, 5-dioic acid
–
CH2 – CH
=O
COOC2H5 –
(10)
–
CH2 – CH2 – CH = O
`
Ethyl 3-(3-oxopropyl) cyclohexane-1- carboxylate
1.9.10 Ketone:Prefix
F.Group –
Name
one
Alkanone
Keto or oxo
C=O
–
Example:-
Suffix
(1) CH3 – C – CH3
Propanone
O
O (2)
Cyclohexanone
O (3)
Cyclohexane-1, 3, 5-trione
O
O
O
– CH
2
(4)
– C – CH3
2-(2-Oxopropyl) cyclohexanone
O COOH –
(5)
3-(3-Oxobutyl) cyclohexane -1-carboxylic acid –
CH2 – CH2 – C – CH3 O
(6) OHC – CH2 – C – CH3
3-Oxobutanal
O
1.9.11 Alcohol:F.Group
Prefix
Suffix
Name
–OH
Hydroxy
ol
Alkanol
Example:-
(1) CH3 – CH – CH2 – OH –
2-Methyl Propanol
CH3 (2) CH 3 – CH – CH 2 – CH 2 – CH – CH 3 – – CH 2 – OH CH 2 – OH
OH HO – (3)
HO
2, 5-Dimethylhexane-1, 6-diol
– –
OH Cyclohexane-1, 2, 3, 4, 5, 6-hexaol
–
–
–
OH
OH (4) CHO –
(Glyceraldehyde) 2, 3-Dihydroxy propanal
CH – – OH CH2 – OH
OH – Cl – CH2 – CH – CH2 – – CH – CH2 – CH2 – CH2 – CH2 – OH (5) CH3 – CH2 – CH2 – – OH Cl 7-Chloro-5-(3-Chloro-2-hydroxy propyl) octane-1, 6-diol
O
–
OH
HO – (6)
2, 4, 6-trihydroxy cyclohexane-1, 3, 5-trione
O
–
OH
O
(7) CH – CH – 2 – CH – 2 – OH
OH
(Glycerol) Propane-1, 2, 3-triol
OH
1.9.12 Amines:F.Group
Prefix
Suffix
Name
–NH2
amino
amine
Alkan amine
(a) R – NH2
1º = amine
Alkanamine N 5 Alkylalkan a min e (R' ) (R )
(b) R – N – R' 2º = amine H
(c) R – N – R' 3º = amine
N, N-Dialkylalkanamine
R'
(d) R – N – CH3
N-Ethyl-N-methylalkanamine
CH2 – CH 3
Ex:-
(1) CH3 – NH2
Methanamine
(2) C – C – C – C – C
Pentan-3-amine
–
NH2 (3) C – C – C – NH – C – C
N-Ethylpropan-1-amine
(4) C – C – C – N – C
N-Ethyl-N-methylpropan-1-amine
–
C C (5) C – C – C
N-Ethylpropan-2-amine
–
NH – C – C OH – (6) CH2 – CH = CH – CH – – CH3
4-(N-Ethyl amino) pent-2-en-1-ol
NH – CH2 – CH3 (7) OHC – CH 2 – CH – CH2 – CH3 –
3-(N-Phenyl amino) pentanal
NH – Ph If hetero atom ’s are count as a ‘C’ atom in the parent chain then they are written as (1) –NH – # Aza (2) –O – # Oxa (3) –S – # Thia (4) –Se # Selena
Example:-
(1) HOOC – CH2 – NH – CH2 – CH3
3-Aza pentanoic acid
H – N
(2)
Aza Cyclo pentane
H – N (3)
3-Methyl aza cyclo hexane –
CH3
1.10 ETHERS (R – O – R’):F.Group
Prefix
Suffix
Name
R – O – R'
alkoxy
–
Alkoxy alkane
Alkoxy + Alkane ! ! less no. more no. of ' C' of ' C' (i) Acyclic Ethers:(1) C – O – C
Methoxymethane
(2) C – O – C – C
Methoxyethane
(3) C – O – C – C – C
1-Methoxypropane
(4) C – C – C
2-Methoxypropane
–
O – C C
–
(5) C – C – C – O – C – C
1-Isopropoxypropane 1-(Methylethoxy) propane
–
O – C – C
(6)
Ethoxycyclohexane
– O
–
(7)
Cyclopropoxycyclohexane
Br – 3
(8)
4
2
5 6 –
– O
– C – C – C 3-Bromo-6-chloro-2-propoxycyclohexan-1-ol
1–
OH
Cl
6 In cyclic system, numbering always starts from senior most functional group. OH – 1
2
– Br 6
(9)
6-Bromo-5-methoxycyclohex-2-en-1-ol
–
3
5 4
O – CH3
OH – 2
1
(10)
Br
– 6
–
3
5 4
O – C – C
6-Bromo-5-ethoxycyclohex-3-en-1-ol
(ii) Cyclic Ether: (3-membered Ring) Hetero cyclic compounds (1) C – C O
Oxirane or Epoxyethane
(2) C – C – C O
1, 3-Epoxypropane
(3) C – C – C O
1, 2-Epoxypropane
Cl – (4) C – C – C O
1-Chloro-2, 3-Epoxypropane
Polyethers:(1) C – C – O – C – C – O – C – C 1
2
1, 2-Diethoxyethane 1
2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
6
7
8
9
10
11
(2) C – C – O – C – C – O – C – C – O – C – C 3, 6, 9-Trioxaundecane 12
13
14
15
(3) C – O – C – C – C – O – C – C – O – C – C – C – O – C – C 2, 6, 9, 13-Tetraoxapentadecane 2
3
1
4
O
O
12
5
(4) 11
O7
10O
6
8
9
1, 4, 7, 10-Tetraoxacyclododecane
1.11
AROMATIC COMPOUNDS: Classification:
Non-Benzenoid:
#
H –
– H
Cation
+
2%e
Cyclopropenium ion
2%e
Cyclobutenium dication
–
–
H H –
# –
H
– H
+ Cation +– H
–
H –
H
–
#
Anion –
Cyclopentadienyl anion
6%e
Cycloheptatrienyl cation (Tropylium cation)
–
- H – H
H
.
.
H –
#
6%e
–
H
– –
H – –
H
H
–
+ –
–
–
H
H
Benzenoid Aromatic Compounds: All organic compounds which contain atleast one benzene ring are known as benzenoid aromatic compounds.
Naming of Aromatic Hydrocarbons (Arenes): CH3 –
(1)
Methylbenzene (Toluene)
CH3
– –
CH3
(2)
(Ortho) 1, 2-Dimethylbenzene (Ortho-xylene)
CH3 –
(3)
(Meta) 1, 3-Dimethylbenzene (Meta-xylene) –
CH3
CH3 –
(4)
(Para) 1, 4-Dimethylbenzene (Para-xylene) –
CH3 Q.
Write structures of
C – C –
(i) Metadiethylbenzene –
C – C
C – C –
(ii) 1-ethyl-4-isopropylbenzene –
C – C – C (iii) Orthodiethenylbenzene –
C=C C=C –
Trialkyl Substituted Benzene: If all the three substituents are similar, then only 3 trisubstituted benzene derivative are possible. C
C
–
–
C
–
–
C 1, 2, 3-Trimethyl benzene (Vicinal Trimethyl benzene)
C
C
–
–
–
C 1, 2, 4-Trimethyl benzene (Unsymmetrical Trimethyl benzene)
–
–
C C 1, 3, 5-Trimethyl benzene (Symmetrical Trimethyl benzene)
Examples : Isopropylbenzene (Cumene)
1. –
C – C – C C
–
2.
1-Chloro-4-methyl-2-nitrobenzene
–
–
Cl
NO2
C=C
Phenylethene (Double bond > Phenyl) (Common name : Styrene)
3. – C
– C = C 3-Phenylprop-1-ene (Allylbenzene)
4. – CH
= CH – CH3 1-Phenylprop-1-ene
5. – COOH
Benzene carboxylic acid (Benzoic acid)
6.
O – C
O
– O – C–
7.
Benzene carboxylic anhydride (Benzoic anhydride)
– SO
3
H Benzene sulphonic acid
8.
O – C
– O – C2H5 Ethyl benzene carboxylate (Ethyl Benzoate)
9. – O
– C – CH3
10.
O
Phenyl ethanoate
O 11.
– C
– Cl
Benzene carbonyl chloride (Benzoyl chloride)
O 12.
– NH
– C – CH3
N-Phenylethanamide
O 13.
– C
– NH – CH3
N-Methylbenzene carboxamide
– C
"
N Benzene carbonitrile (Benzonitrile – popular)
14. – CH
=O Benzene carbaldehyde (Benzaldehyde – popular)
15.
O C–
–
16.
Diphenyl ketone (Benzophenone)
O C– CH3
–
17.
Methyl phenyl ketone (Acetophenone)
OH – Phenol
18.
OH – 19.
– –– CH3 OH –
OH
–
20.
Methylphenol Methylphenol [o-Cresol, m-cresol, m -cresol, p-cresol]
o-Hydroxyphenol (Catechol)
OH – m-Hydroxyphenol (Resorcinol)
21. –
OH
OH – p-Hydroxyphenol (Quinol)(Hydroquinone)
22. –
OH NH – 2 Benzenamine (Aniline)
23.
SH – Benzenethiol
24.
O O – C – CH3 O 25.
– C – O – H
2-Ethanoyloxy 2-Ethanoyloxy (Acetoxy) benzene -1-carboxylic acid (Aspirin)
Session - 2009-10
TOPIC : CONTENTS
1
GOC - I
:
Isomerism * Structure Isomerism * Stereo Isomerism Geomertical Optical Conformational
Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and and E,Z nomenclature nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional mono-functional and bi-functional compounds); compounds); Conformations of ethane and butane (Newman projections);
Page-29
Isomerism Isomers: Compounds with same general formula or molecular formula but different physical and chemical property. property. Ex:-
CH3 – CH2 – OH and CH3 – O – CH3
Ex:-
CH3 –COOH and HCOOCH3
Homologs: Compounds with same general formula differing by same structural unit – CH2 – or molecular weight by 14 unit. Ex:- CH3 – OH and CH3 – CH2 – OH Difference between isomers and homologs:-
Page-30
Structural isomers: When two or more number of organic compounds have same molecular formula but different structural formula these are called structural isomers.
Stereo isomers: When two or more mo re compounds have same Molecular Molecul ar Formual (M.F.) (M.F.) and same Structural Formula (S.F.) but have different stereochemical stereochemi cal formula (S.C.F.), these are called stereoisomers. stereoisome rs.
Stereo Chemical Formula (S.C.F) : It indicates different arrangements of atoms or groups in space around a stereo centre or it indicates different diff erent spatial orientations of atoms or groups around a stereo centre. M.F.
C4H8 CH 3 –
S.F. (i) CH3 – CH2 – CH = CH2 These are structural isomers. For (ii), S.C.F. are H3C
CH3
(iii) CH 3 – C = CH 2
H
C=C H CH3 Stereocentre These are stereoisomers. H
C=C
H3C
(ii) CH3 – CH = CH – CH3
H
and
Structural isomers are also known as:(i) Ske Skelletal iso isome merrs (ii) Li Linkag kage is isome merrs
(iii) Co Constitutional is isome merrs
Chain isomers : They have different size of main carbon chain and / or side alkyl chain (i) The two chain isomers should have same nature of F.G./multiple .G./multiple bonds/substituents (except –R group) (ii) The position of F.G./M.B./substituent (locants) is not. considered here. Ex:-
Alkanes:(a) C1 – C3
chain isomers not possible
(b) C4H10
H3C – CH2 – CH2 – CH3 and CH3 –
(c) C5H12
CH3 – CH2 – CH2 – CH2 – CH3 , CH3 – CH2 –
(d) C6H14
(i) CH3 – CH2 – CH2 – CH2 – CH2 – CH3, (ii) CH 3 – CH – CH2 – (iii) CH3 – CH2 –
(v) CH3 – CH2 –
–
CH3
CH2 – CH3, (iv) CH3 –
–
CH3 ,
CH3 –
–
–
–
CH3
CH3,
–
CH3,
–
CH3
–
(i) and (ii) – Chain isomers (i) and (iii) – Chain isomers (i), (iv) – Chain isomers (i), (v) – Chain isomers (ii), (v) – Chain isomers (ii), (iii) – Position isomers (iv), (v) – Position isomers
Ex.
Ans.
a nd
shows which types of isomerism ?
Chain Isomers Page-31
Q.
Write chain isomers of N-alkanamine (1 º) containing 4 carbon atoms ? C –
Ans.
C – C – C – C – NH2,
C – C – C – NH2
Positional isomers: They have different position of locants Functional group (F.G.) or Multiple Bond (M.B.) or substituents in the same skeleton of C-atoms. Nature of F.G. or M.B. should not change. The skeleton of C-atom should not change. Ex.
(1) C – C – C – C – – C ,
C – C – C – – C – C
C
Q.
C
(2) C = C – C – C ,
C – C = C – C
(3) C ! C – C – C – C ,
C – C ! C – C – C
(4) C – C – C – C , C – C – C – C – OH – OH Write all Positional isomers of Dichlorobenzene ?
Cl
Cl
Cl
–
–
–
– Cl
,
Ans.
, –
–
Cl
Q.
Cl
Write all Positional isomers of Dichlorocyclopropane ? Cl
Cl– –Cl
–
Ans.
–
,
Cl
Q.
Write all Positional isomers of Dichlorocyclobutane ? Cl Cl– ,
Cl–
Cl– Cl
–
Ans.
–
, –
Cl
Q.
Write all Positional isomers of Dichlorocyclopentane ? Cl
Cl
– Cl – Cl
–
–
Cl
–
Ans.
,
, –
Cl
Ex.
(i) C – C – C – – C – C – C
(ii) C – C – C – – C – – C – C,
C – C
C
(iii) C – C – – C – C – C – – C
C
C
C
(i), (ii) – Chain (ii), (iii) – Position (i), (iii) – Chain Ex.
C6H12 (Cycloalkanes): – –
(1)
,
(2)
,
(3)
–
,
–
(4)
, –
Page-32
C – (5)
–
–
–
,
–
(6)
,
,
(8)
C – – C
,
–
C – – C (9)
C – C
–
–
–
,
(10)
,
–
Ans.
(7)
C – – C – C
(11)
–
(3) and (4) – Positional isomer (3) and (6) – Positional isomer (4) and (6) – Positional isomer (5) and (10) – Positional isomer (All are chain isomers of 1) Remaining are chain isomers
Functional isomers: They have different nature of functional group (F.G.). The chain and positional isomerism is ignored (not considered). Compound Functional isomer C – C – C – C
Nil
C – C – C = C Ex. (1)
(Ring-chain isomers are Functional isomers)
Compound C – C – C ! C
Isomer Functional (a) C = C – C = C Alkadiene
Remarks (a), (b) are not functional isomers
Alkyne
(b)
Cycloalkene
among themselve
(c)
Bicyclo
(2)
C – C – C – OH Alcohol
C – C – O – C Ethers
(3)
C – C – CH = O
C – C – C
Aldehydes
O Ketones
C – C – COOH
C – C – O – C
Carboxylic acids
O Esters
C – C – C – NH2
(a) C – C – NH – C
1º amine
2º amine
C – C – C ! N Cyanide (Propanenitrile)
C – C – NC Isocyanide (Ethane isocyanide)
(4)
.
(5)
(6)
.
.
.
(b) C – N – C – C 3º amine
1º, 2º, 3º amines are functional isomers
Important point :- Following compounds don’t exist at room temperature theref ore not consider as a structure isomer –
(i) – C = C C – OH –
–
(ii) – C ! C – OH
(iii) – C – OH –
OH –
(iv) – C – OH –
OR
–
(v) – C – O – C = C –
(vi) Any peroxy compound
OH
Page-33
Q.
Write acyclic isomers of C 5H12O C –
Ans.
(i) C – C – C – C – C – OH, C –
C – C – C – C – C, – OH C – C – C – – C – OH,
C – C – C – C – OH,
C – C – C – C – C, – OH C – C – C – C – – OH
C C – C – C – C – O – C,
C – C – C – C – OH,
(7 alcohols)
C
C – C – C – – O – C,
C – C – – C – O – C,
C –
C – C – – O – C, C
C
C
C – C – C – O – C – C,
C – C – O – C – C
(6 ethers).
G.F. CnH2n + 3N
n (1) 3
M.F. C3H9N
2º Amine (R – NH – R)
3º Amine R – N – R – R
G.F. CnH2n + 1N 2
n (1) 2
M.F. C2H5N
1º Amine (1) C = C – NH2 Ethenamine
2º Amine C – C = N Ethanimine (Unstable at room temperature)
3º Amine C – N = C N-Methylene methanamine
C –
Q.
1º Amine (R – NH2) Q
H2C– – CH2 Azine – N – H Q.
G.F. CnH2n 1N
n 2
M.F. C2H3N
Cyanides and isocyanides H3C – C ! N
CH2 = C = NH
HC ! C – NH2
–
Ans.
H3C – N
"
C
Metamers: When two isomers have same functional group (containing a hetero atom –O, N, S) but have different nature of alkyl or aryl (aromatic radical) group attached to hetero atom, then these are called metamers.
Conditions: (a) Same functional group (b) Chain or position isomerism is not considered. (c) >C = O (keto) group does not show metamerism Following functional groups show metamerism (a) R – O – R’
Ethers (All isomeric ethers are metamers)
(b) R’ – NH – R
2º Amines
(c) R’ – N – R – R'' (d) R – S – R’
3º Amines Thioethers Page-34
(e) R – C – O – R’
Esters
O
(f) R – C
–
O – C
O
R’ Anhydrides
–
O
O (g) R – S
–
O – R’
Sulphonate esters
O Q.
How many esters are possible for C 3H6O2 ?
Ans.
C # C # O # C # H and C # C # O # C are metamers || || O O
Q.
3º Amines of M.F. C5H13N (All metamers) C –
Ans.
C –
C – C – N – C – C – O
– O
– C – C and
Q.
C – C – C – N – C
–
C –
C –
C – C – N – C
– C Show which type of isomerism ?
C
Ans.
Metamerism
Q.
Acyclic compound (A) contains 18 1 º H atoms, two types of ‘C’ atoms. All H are identical. Identify (A).
C C – – C – – C – – C $ C8H18 C C
Ans.
C –
Q.
Cyclic compound (P) contains 18 1 º H atoms, two types of ‘C’ atoms. All H are identical. Identif y (P). C C –
Ans. –
C
–
or
–
– C C Compound X is an ether. It has 12 1 º H atoms. It has 2 types of H-atoms and 3 types of C-atoms. Write it Structural formula ?
C
Q.
–
CH3 –
Ans.
H3C – O – C – CH3 –
CH3
Page-35
Stereoisomers (Classification)
Stereoisomers : The stereoisomers has different orientation of groups along a stereo centre. A stereocentre can be C = C (any double bond), a ring structure, asymmetric carbon atom (*C abcd). These isomers has same general formula, structural formula and molecular formula but different stereochemical formula. E.g.
CH3
C=C
H CH3
C=C
CH3 H H
(II)
H CH3 CH3 –CH2 –CH=CH2 I, III II, III I, II
(I)
(III)
Positional Positional Stereoisomer
" " "
Configurational isomers : Configurational isomerism arises due to different orientations along a stereocentre and these isomers can be seperated and these isomers do not convert into one-another at room temperature. Therefore, they are true isomers. They can separated by physical and chemical method. E.g. Cis –2 –Butene Trans-2-Butene
Conformational isomers : When different orientations arise due to the free-rotation along a sigma covalent bond. Such isomers are called conformational isomers. E.g. eclipsed ethane and staggered ethane These isomers change into each other at room temperature and can never be isolated. So these are not considered as true isomers.
Geometrical Isomerism 1.
Cause of Geometrical isomerism : Geometrical isomerism arises due to the presence of a double bond or a ring structure C = C, C = N, N = N, or Ring structure (Stereo centres) Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecule exist in two or more orientations. This rigidity to rotation is described as restricted rotation/hindered rotation/no rotation. E.g.
a
a
C =C b b (I) The root form of geometrical isomers lie in restricted rotation.
a b
a (I)
b
Page-36
Condition (i) Restricted rotation (ii) The two groups at each end of restricted bond must be different. 1 3 C=C 2 4 Caa Caa X Caa Cbd X Cae Cbb X ! Cab Cab Cab . Cbd Cab Cde
! !
(iii) In two geometrical isomer the distance between two particular groups at the ends of the restricted bond must be changed.
Q.
Which of the following compounds show G.I. (1) Ethene
(2) Propene
(3) 2-Methylbut-2-ene CH3
(4) But-2-ene CH 3 – CH = CH – CH3
#C %
CH # CH3
| CH3 (5) Penta-1, 3-diene C = C – C = C – C
(6) 1, 2-Dideuteroethene
H2C = CH (7) Phenylethene Ans.
(8) Buta-1, 3-diene C = C – C = C
4, 5, 6
1. Geometrical isomerism across
–
Nil
and
By E / Z
2. Geometrical isomerism across (a) Imine (
)
Imine compounds are produced from carbonyl compounds on reaction with ammonia.
(Syn and anti) Imines prepared from unsymmetrical addehydes and ketones, always show geometrical isomerism. Page-37
Q.
Which of the following compounds show geometrical isomerism after reaction with NH 3. O || (a) CH3 # C # CH3 O || (d) CH3 # C # H
(g)
Ans.
O || (b) H # C # H
O || (c) H # C # D
(e) Ph # C # CH3 || O
(f) Ph # C # Ph || O
(h)
(i)
c, d, e, g, i (b) Oximes
C = N – OH :
These are prepared by reacting carbonyl compuond with hydroxyl amine (NH 2 – OH)
R C=O H
R
#H2O
+ H2 N –OH & & & & "
H
C = N –OH
(Aldoxime) R
OH
R C =N
H
(I) (syn)
and
OH
C =N H
(II) (anti)
Syn and anti in aldehyde only not for ketones. * Except formaldehyde (CH 2O) All other aldehyde form two oximes. * Unsymmetrical ketoes form two oximes. Which of the following ketones will form two oximes.
Eg.
(1) Propanone
O || CH3 # C # CH3
(2) Butanone
O || CH3 # CH2 – C # CH3
(3) 3-Pentanone
C – C – C – C – C
(4) Acetophenone
C 6H 5 – C –CH 3
O
O
O (5) Benzophenone
C 6H 5 – C –C 6H 5
(6) Cyclohexanone
O
O Me (7) Methyl Cyclohexanone Ans.
2, 4, 7
Q.
The lowest molecular weight of acyclic ketone and its next homologue are mixed with excess of NH 2 – OH to react. How many oximes are formed after the reaction ? 3
Ans.
Page-38
(c) Hydrazones
C = N – NH 2 :
R
R
#H2O C = O + H2N.NH2 & & & & " H hydrazine
C = N –NH2 H R
NH2
R +
C=N H
(I)
C=N
NH2
H
(II)
(Geo. diastereomers) Q.
Two chain isomer of a cycloalkanone which are next higher homologue of lowest molecular weight, cycloalkanone reacted with hydrazine. Identify the structure and number of isomer of hydrazones prepared ?
.. N O +
Ans.
+ H2N – NH2
NH2
+ CH3
O
CH3
..
NH2
N NH2
N ..
+ CH 3
Number of isomer = 3 (3) Geometrical isomerism across azo compounds ( – N = N – ) (i) H – N = N – H (H2N2)
(ii) Ph2 N2 (Azobenzene)
. .
.. N =N Ph
syn
Ph
..
Ph ..
N=N anti
Ph
(4) Geometrical isomerism across ring structure
(i)
Restricted rotation
(ii)
(iii)
Page-39
(5) Geometrical isomerism in cycloalkenes across double bonds : In cycloalkenes, G.I. exists across double bonds with ring size equal to or greater then 8 carbon atoms (due to ring strain)
Stereocentre:An atom or bond across which stereoisomerism exists (either G. I. or optical isomerism)
Those stereoisomers which are not mirror images of each other are called diastereomers. Those compounds which are non-superimposable mirror-images of each other are called enantiomers
are enantiomers
E/Z Nomenclature : Z (Zussamen = together) a > b and e > d
E (Entegegen = opposite) a > b and e > d Rules :(i) The group with the first atom having higher atomic number is senior. Thus – F > – OH > – NH2 > – CH3 (ii) If the first atom is identical, then second atom is observed for deciding the seniority of the group.
(a)
(b)
<
(c)
<
(d)
> Page-40
Ex.
Z
Ex.
E (iii) If the first atom has same atomic number but different atomic mass, that is isotopes. Then heavier isotope has higher seniority. (iv) If the group has unsaturation, then a hypothetical hypothetical equivalent in drawn for it and it is compared with other group for seniority.
(1) (2) – CH = CH2 < – C ! CH
(Hypothetical)
(v) Bond pair is always senior to lone pair. CH2 = CH
Ex.
(1)
HC
C
C == C
C(CH3)3
E CH2 – CH = CH 2
(2)
Z
N (3)
C
H2C = C = CH
C == C
CH2 – C(CH3 )3 E C
CH
(4)
Z
(5)
E
(6)
E
Page-41
Number of geometrical isomers Case I : Compounds having dissimilar ends No. of G.I. = 2n where n = number of stereocentre Ex. Ans.
Stereocentre = 2 Geometrical isomerism = 4
(a)
(b)
(c)
(d)
Case II : Compounds with similar ends & even no. of stereocentre. n –1
Ex.1 Ans.
No. of Geometrical isomerism = 2 + CH3 – CH = CH – CH = CH – CH3 Stereocentre = 2 Geometrical isomerism = 2 1 + 20 = 3
n #1 22
Case III : Compounds with similar ends but odd number of stereocentre. No. of Geometrical isomerism = 2 Ex.1 Ans.
n –1
+
n#1 2 2
CH3 – CH = CH – CH = CH – CH = CH – CH3 Stereocentre = 3 Geometrical isomerism = 6
Physical properties of Geometrical isomers : The physical properties of organic compounds can be compared through the knowledge of their molecular formula, structural formula and stereochemical formula (S.C.F.) More Polar geometrical isomer is more soluble in water.
(i)
>
(ii)
Cis > Trans
<
COOH
COOH C=C
(iii) H
Trans > Cis
C=C
> H
COOH
H COOH
Cis > Trans H
Page-42
Page-43
Optical Isomerism Some organic compounds can rotate the plane of plane-polarised light. Such compounds are called optically active compounds. The optically active compound can show optical isomerism. Measurement of optical activity It is measured by an instrument called polarimeter.
= (1)
(5)
(3)
(6)
Recorder (8)
(1) " Source of light (2) " Polychromatic Non-polarised light (3) " Slit (Monochromator) (4) " Monochromatic Non-polarised light (5) " Polariser (Prism-setting) (6) " Monochromatic plane-polarised light (7) " Sample tube ( ! = path length) (8) " Recorder for measurement of optical rotation Observations:- When plane polarised light is passed through sample tube, then following changes can be observed in the recorder:Angle of rotation ( ) (1) ( = 0º (2) ( = +xº (clockwise rotation) (3) ( = –xº (anticlockwise rotation)
Inference Compound is optically inactive. Optically active, dextrorotatory or d or (+). Optically active, laevorotatory or l or ( –).
Specific Rotation:- The specific rotation of a compound indicates the optical rotation of unit concentration (1 g/mL) present in a sample tube of 1 dm of path-length at given temperature and given wavelength of light. It is represented as follows:-
t % 25 º C
[* ] + %580 nm % where
(
c)!
( = observed angle of rotation
c = concentration in g/mL (or density) ! = Path length of Sample tube in dm Q.
Compound ‘X’ has ( = +70º for 2 g/mL solution in sample tube of
Ans.
* =
(1) (2)
! =
1 dm. Calculate * ?
70 = +35º 2 Its concentration is made twice, the observed rotation( () will be 140º but * = 35º If ( = +70º, it will be a (i) d compound or (ii) ! compound of ( = –(360º – 70º) = –290º It can be decided by changing the concentration or by changing the length of the tube (c or ! ). If concentration is reduced to half, d will have +35 º and ! will have –145º (not distinguish) still halfed, it will give +17.5º and –72.5º to distinguish.
(I) (II) With symmetrical With asymmetric molecule molecule [*] = 0 [*] 0 Optically inactive compounds:For symmetrical molecule optical rotation observed after interaction of light is zero. Page-44
But in case of optically active compounds. The molecules are asymmetric in nature and show non-zero optical rotation.
Symmetry of elements :(1) Centre of symmetry (C.S.) (2) Plane of symmetry (P.S.) (3) Axis of symmetry (4) Alternating axis of symmetry Plane of symmetry : - The imaginary plane which divides a molecule into two equal halves which are related as mirror image is known as plane of symmetry
Centre of symmetry : - A centre of symmetry in a molecule is said to exist if a line is drawn from any atom or group to this points and then extended to an equal distance beyond this point, meets the identical atom or group. Cl H
R
H
C=C H
R
H
Cl Centre of symmetry
Centre of symmetry
Axis of symmetry : It is defined as Cn axis of symmetry that means of the molecule is rotated by
360º n
angle, then its original or identical or superimposable. The molecule is said to have n-fold axis of symmetry
Ex.
Ex.
C2 & & & " (180 ºrot )
(C4 axis)
(axis)
Ex.
There is no relation whatsoever with chirality and axis of symmetry. Meso molecule does not have axis of symmetry
,&& no
axis of symmetry
.
Page-45
Alternating Axis of Symmetry : When a molecule is rotated by
360º n
angle and its mirror image is taken in the perpendicular plane of
rotation, then its original molecule is obtained. It is represented by S n
Ex.
H H
Me
Me
H
Me
180 º rotation
(I)
H
Me Me
H
H
Reflextion
(II)
Me
(III)
I and III are equivatent. Dissymmetry:- The molecules or objects in which minimum two elements of symmetry (Centre of symmetry and Plane of symmetry) are absent are called dissymetric molecules/objects. Asymmetry:- When all the elements of symmetry (23 including Centre of symmetry and Plane of symmetry are absent) the molecule is said to be asymmetry. So, all asymmetric molecules are always dissymmetric, but the reverse is not true. Dissymmetry and chirality:- All dissymmetric molecules are chiral and are optically active. Dissymmetry, Chirality and Optical Activity:- Dissymmetry/Chirality is the minimum and sufficient condition for a molecule to be optically active. So, if in a molecule, a C.S. and P.S. are absent, the molecule will be optically active. All asymmetric molecules are also optically active. Chirality (Hand-like property or Handedness):- The human hand does not have Centre of symmetry and Plane of symmetry , so it is dissymmetry. It does not have any element of symmetry, so it can be called asymmetry. Any molecule which has this property is called chiral. Note : All dissymmetric compounds are chiral.
Optical Isomers:Enantiomers:- The mirror-image stereoisomers are called enantiomers. Two types:(a) Dextrorotatory (b) Laevorotatory -
Enantiomers are always mirror image isomers. They have same molecular formula, structural formula but have different orientation in space. They have dissymmetry/chirality. Every enantiomer is optically active. The two enantiomers can rotate the plane-polarised light with equal magnitude and opposite signs. They have similar physical properties except the sign of optical rotation. These are the isomers which have maximum resemblance with each other. These can be distinguished only by polarimeter.
Chirality (Dissymmetry) and Optical Isomers:(a) (b) (c) (d) (e)
The dissymmetric molecules have two orientations in space. These two orientations are called stereoisomers, optical isomers, enantiomers. These are mirror images (enantiomers). Non-superimposability of enantiomers:- The dissymmetric molecules are always non-superimposable on their mirror-image orientations. The non-superimposable orientations are non-identical orientations, so are called isomers. Because of dissymmetry, these two isomers are capable of rotating plane-polarised light. So, these are called optical isomers. Page-46
Optically Active Carbon Compounds:If a carbon-atom is attached with four different group, then if does not have any element of symmetry. It is known as asymmetric carbon atom, which is represented as *C abde. a C* e d If a molecule contains only one asymmetric carbon atom, then the molecule as a whole becomes chiral and optically active and show optical isomers. b
Centre of symmetry
Plane of symmetry
Optical active
H C
(1)
H
H
H
Absent
Yes
No
Absent
Yes
No
Absent
Yes
No
Absent
No
Yes
H C
(2)
H
H
Cl
H C
(3)
Br
H
Cl
H C
(4)
Br
Cl
F Ex.1:- Mark the chiral objects:(i) Cup (ii) Plate (vii) Shoe (viii) Glove Ans.
(iii) Letter A
(iv) Letter G
(v) Fan
(vi) Door
IV, VII VIII
Projection Formula of Chiral Molecules:(i) Wedge-Dash Projection formulae down up
Ex:-
(i) Butan-2-ol
CH3 C H
C2H5 OH
(ii) Fisher Projection formula Ex:-
(i) Butan-2-ol
(CH3 – CH2 – CH – CH ) – 3
OH Rules of writing Fisher Projection formula :(i) It is represented by a cross (+). (ii) Groups at Vertical line are away from observer. (iii) Groups at Horizontal line are towards the observer. (iv) Central ‘C’ atom of the cross is chiral. (v) High priority group lies at the top of vertical line (IUPAC Numbering starts from top). CH3 H
OH CH2CH3
/ / / / / / / / / / / / / / / / / / / / / / /
Page-47
Ex:-
Draw Fisher Projection formula of following molecules:-
(1) 2-chlorobutane
(2) Pentan-2-ol
(3) CH2 – CH – CH –
–
OH
OH O Glyceraldehyde
1
2
*
3
4
*
5
(4) CH3 – CH – CH – CH2 – CH3 – – OH OH
,
,
,
(I, II), (III, IV) " Enantiomers. All other diastereomers. They are true isomers (not superimposable) (5) Glucose
CHO – CHOH – CHOH – CHOH – CHOH – CH2OH
Mark:- (i) No. of C* " 4 (ii) Draw one Fisher Projection Formula CH = O H
OH
H
OH
H
OH
H
OH
Page-48
Configuration nomenclature in Optical Isomers :Relative configuration : The experimentally determined relationship between the configurations of two molecules, even though we may not know the absolute configuration of either. Relative configuration is expressed by D-L system. Absolute configuration : The detailed stereochemical picture of a molecule, including how the atoms are arranged in space. Alternatively the (R) or (S) configuration at each chirality centre. (I) D - L System (Relative configuration) : Application on correct Fisher Projection Formula This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceraldehyde. The configuration of (+)-glyceraldenyde has been assigned as D and the compounds with the same relative configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L.
Examples :
Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde ( – OH group on right side) is designated as a Dsugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is designated as an L-sugar.
e.g.
(I) R/S Configuration : R " Rectus
S " Sinister
Examples :
a b
c
d
(1) Seniority order a > b > c > d (2) Put junior most group at dotted line b $
(3)
Q.
c a d Now go from a to b, b to c. (i) If it follows clockwise route, it is (R). (ii) If it follows anticlockwise route, its configuration is (S). Br Br
Cl
I
F (R)
Cl F (S)
I
Page-49
R/S Configuration in Fisher Projection Formula :(c)
Me(c) Ex:-
H(d)
Et(b)
!
(d)
Pr (a)
"
(b)
a"b"c & & & & & & " Clock wise
!
R
(a)
"
If the junior most group is at horizontal line clockwise " S & anticlockwise " R If juniormost at vertical line, then clockwise " R & anticlockwise " S
Q.
(1)
Ans. R
(2)
Ans. R
(3)
Ans. S
Properties of enantiomers :One chiral carbon : (i) Number of optical isomer = 2 (d or !) (ii) Number of Racemic Mixture " Equimolar mixture of d and optical rotation.
!
. [*] = 0 due to external compensation of
Properties:(i) Dipole moment (ii) Boiling Point (iii) Melting Point (iv) Solubility (v) Specific rotation [ *]
same same same same different
Molecules with more than one chiral carbons:(I) Calculation of No. of optical isomer :Case I:- When all Chiral carbon atoms are differently substituted (all are dissimilar C*) optical isomer = (2) n Case II:- When all the Chiral carbon atoms are similar at ends. (i) n = even
$x=2
n – 1
+
n –1 22
(ii) n = odd
$ x = 2n
–
1
Page-50
Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:CH 3 – CH – CH – CH2 – CH3 – – Cl Cl (ii) Optical isomer = 2 n = 22 = 4 (iii) Stereochemical Formula:(i) Structural formula
CH3
CH3
CH3
CH3
H
Cl
Cl
H H
Cl
Cl
H
H
Cl
Cl
H; Cl
H
H
Cl
C2H5
C2H5
C2H5
I [*] = +xº
II –xº
C2H5
III +yº
Analysis:(a) I, II – Enantiomers I, III – Diasteromers II, III – Diasteromers
IV –yº
III, IV – Enantiomers I, IV – Diasteromers II, IV – Diasteromers
(b) No. of Racemic Mixture – Two (I + II, III + IV). Q.
CH3 – CH – CH – CH – CH2 – CH3 – – – Cl Cl Cl Calculate total number of optical isomer.
Ans.
n = 3, x = 23 = 8 (4 d- ! pairs)
(i)
(ii)
(ii)
(iii)
(iv)
and their mirror images. 4 Racemic Mixtures (4 different fractions on fractional distillation)
Compounds with 2 Asymmetric Carbons of Similar Nature:Ex:-
CH3 – CH – CH – CH3 – – OH OH (i) n = 2 n – 1
(ii) x = 2
+
n –1 2 2 =
3 (Optical stereoisomers).
(iii) Stereo chemical formula :Me H
OH
H
OH
Me (I) (II) " Plane of symmetry present. In (I and II) " Superimposable on its mirror image. " Thus, I and II are identical. " [*] = 0, optically inactive. " Meso isomer " In (III) Specific rotation [*] = +xº " In (IV) Specific rotation [ *] = –xº
(III)
(IV)
Page-51
No. of d –
!
pairs = 1 (III + IV) = Racemic mixtures
In (I) and (II) : [ *] = 0 due to internal compensation and Non-Resolvable In (III) and (IV) : Resolvable [can be separated into two isomers (enantiomers)]
Meso isomers:"
The optical stereoisomers which have more than one asymmetric carbon atoms but have a plane of symmetry are called meso compound. " They are achiral (optical rotation = 0). " They have [*] = 0 due to internal compensation of optical rotation. " They are diastereomer of d – ! pair. So, it has different physical properties than d – ! -pair.. " Presence of more than one asymmetric ‘C’ atoms. " They are non resolvable. Q.
Mark meso compound among following H
COOH OH
H (1)
(2)
OH COOH
H Ans.
Cl Me
Cl
(3)
(4)
Me
OH H
(5)
*
OH
(1) (2) (4) and (5)
Molecules with three similar chiral carbon:Ex:-
.
.
.
CH3 – CH – CH – CH – CH3 ! M.F.. –
Cl
–
Cl
–
Cl
n=3 x = 2n 1 = 4 stereoisomers. –
Stereo Chemical Formula : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
(I) 2S 3S 4R
(II) 2S 3R 4R
(III) 2S 3 " achiral 4S
(IV) 2R 3 " achiral 4R Optically active
(i) & (ii) Achiral due to Plane of symmetry. (Optically inactive).
Properties of Optical Diastereomers:The optical isomer which are neither mirror image nor superimpossible to each other are called optical diastereomers.
Conclusion:- Optical Diastereomers have different physical properties, so these can be separated by normal physical methods of separation. (i) By fractional distillation (different B.P.) (ii) By fractional crystallization (different solubility) (iii) Chromatography (different solubility) (iv) Differential Melting (M.P. diff.) Page-52
(III) By salt formation (R .COOH + R’.NH2)
+
+
Diastereomers (Separable)
Optical Purity or Enantiomeric Excess (O.P. or E.E.):Ex:Compound X has [*] = / 70º We have mixture of enantiomers in different percentage. %E.E. %
((obs)
[*]specific
) 100%
Page-54
Optically active compounds without a chiral carbon:Rapid R – R Inversion – 0 0 - Asymmetric N :R' – – N N – – R' R'' R'' " A nitrogen atom attached with 3 different groups (sp 3) is chiral, asymmetric. " In case of acyclic compounds, the Nitrogen, having trigonal pyramidal shape undergoes rapid inversion of shape at room temperature. So at room temperature, everytime a racemic mixture of ‘d’ and ‘ ! ’ forms exists. This racemic mixture can never be resolved at room temperature. " Nitrogen converts into its enantiomer..
So, acyclic molecules with chiral nitrogen are chiral but optically inactive ([ *] = 0) and are non-resolvable. If asymmetric Nitrogen atom is present in ring structure, then inversion does not take place and such molecules are optically active 1[*] 3 02 . (otherwise ring will break)
Ex:-
. Me N
[* ] 3 0
C % C % C – C – C (Cumulated double bond) 4 One after another
C % C – C % C – C (Conjugated double bond) 4 Alternate
C % C – C – C % C (Isolated double bond) 4 Separated
(I) Case of allene : (a) Allenes with even 5 bonds :
e.g.
the orbital diagram of this structure will be Since the groups at the end of allene are in perpendicular plane, it will not show geometrical isomerism. The molecule lacks centre of symmetry as well as plane of symmetry. Overall the structure has molecular dissymmetry which is the sufficient condition for optical activity. The molecule will exist in two enantiomeric forms.
(b) Allenes with odd
5 bonds
:
e.g.
the orbital diagram of this structure will be
-
The groups at the end of allene structure lie in same plane (ZX plane). Therefore it will have a plane of symmetry (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence optical isomerism. But the compound will exist in two geometrical diastereomeric forms.
Page-55
(II) Case of spiranes : A similar case like allenes is observed in spiranes. The spiranes with even rings and different groups at terminal carbons show optical activity & optical isomerism, while the spiranes with odd rings shows geometrical isomerism. (a) spiranes with even rings : shows optical isomerism.
(b) spiranes with odd rings : shows geometrical isomerism. (III) Case of cycloalkylidene :
(IV) Case of ortho-ortho-tetrasubstituted biphenyls :
becomes non-planar at room temperature in order to have minimum electronic repulsion among the substituent. In this orientation (phenyl planes perpendicular to each other) the free rotation of C – C single bond is restricted and molecule shows optical activity due to m olecular disymmetry. e.g.
Page-56
CONFORMATIONAL ISOMERISM : 1.
Free Rotation : A sigma covalent bond undergoes free rotation at room temperature. (C –C, C –O, C –N, N –N, O –O)
2.
Conformers / Rotamers or conformations : The infinite number of spatial oreintation of molecule arises due to free rotation along a sigma covalent bond
3.
Conformational Isomers : Those conformations which are most stable and have minimum P.E. are defined as conformational isomers. These are not true isomers and can never be isolated.
4.
Conformational Energy : The rotational energy barrier is known as conformational energy. It is the P.E. difference between conformational at potential energy minima and maxima.
5.
Angle of rotation/Dihedral Angle (D.H.A.)/Torsion Angle : The interfacial angle between the groups attach at two 6-bonded atoms is defined as dihedral angle. Projection formula (a) Saw horse Projection formula Ex.
CH3 – CH3 H(a) H(c) rotation
& & & & & " H (b)
Eclipsed
H (c)
H(b)
H(a)
Staggered
(b) Newmann Projection formula
& & & " 1 #12.5 kJ / mol#
Energy - Level Diagram of ethane
Ex.
CH3 –CH2 –CH3
Page-57
Saw horse Projection formula
Eclipsed
Staggered
Newman Projection Formula CH3 H
CH3 H
H
H
H
& " 60º
H
120 º & & & "
HH
II
Eclipsed
CH3 H
H H
H
&"
H
III
Staggered
H
H
H
IV
Eclipsed
Staggered
Propane :
Butane : Ex.
Ethyl-hydrogen repulsion is less than methyl-methyl repulsion. So draw the newman between C 2 - C3 CH3
CH3 CH3
60 º & & & "
H H
H (I)
H
CH3 CH3
H
120 º & & & "
H
H (II)
180 º & & & "
H
H
H
H
CH3
(IV )
300º
& & & & "
CH3 H
H
240 º & & & "
H CH3 H H ( ) V
360º
& & & & "
Page-58
I/VII = Fully eclipsed II/IV = Gauche form III/V = Partially eclipsed IV = Anti form
Stability Order : IV > II > III > I
Anti > Gauche > Partially eclipsed > Fully eclipsed
P.E. Order : IV < II < III < I Potential Energy Diagram of butane :
Conformational Isomers : Due to free rotation from 0 º to 360º those conformations which are most stable and have minimum P.E. are defined as conformational isomers.If these conformers have same energy then there isomers are called degenerate isomers or equienergic isomers. The conformational isomers with different energies are called non-degenerate isomers. The conformational isomes lie at the P.E. minima in the P.E. diagram with respect to rotational angle.
Strains : (i) Torsional Stran (eclipsing strain) : It is defined as the electronic repulsion between the bond-pairs electrons of two adjacent eclipsed bonds. It is active at torsional angles 0 º, 120º, 240º in which the molecule has eclipsed conformation. It is considered almost zero in the staggered conformation. (D.H.A. = 60 º, 180º, 300º) (ii) Vander Waal strain : It is the repulsion between the group attached at adjacent bonds. The vander waal strain is maxim um in eclipsed conformation, and minimum in anti conformation, while intermediate in gauche conformations. It is almost zero for H-atom since the sum of vander waal radii is less than internuclear distance between two H atom. (iii) Angle Strain : It arises due to distortion in normal bond-angles. In acyclic compounds, there is no distortion of bond-angle due to free rotation. So, all conformations have zero angle strain in acyclic compound
Page-59
Ques. Draw conformation isomers of following compound (a) CH3 – CH – CH3
(b) CH3 – (CH2)3 – CH3
CH3 Ques. Draw conformation isomers of following compound CH3 – CH –CH – CH3 with respect to C 2 and C3 carbon CH3 CH3 atoms. H H
H 60 º
CH CH3 CH3 3 Eclipsed
180 º
& & & "
CH3
CH3 CH3
CH3
CH3
120 º
& & & "
H3C
H3C
H
CH3
H
H & & & "
H3CCH
3
H
CH3
H3C
CH3
H Staggered
Page-60
In case of G – CH2 – CH2 – OH, where G = – OH, – NH2 , – F, – NR2, – NO2, – COOH, – CHO the Gauche form is more stable than the anti form due to intramolecular hydrogen bonding i.e.
stability : Gauche form > anti form. CH2 –OH (Glycol).
Example 1.
CH2OH OH H
OH
& &"
& &"
H
& &"
H H (II) Gauche
Hydrogen Bonding : ( II)
Gauche form most stable due to hydrogen bonding. Stability Order : II > IV > III > I Example 2.
O2N – CH2 – CH2 – OH
Interconversion of projection formulae : Example
Tartaric Acid : COOH – CHOH – CHOH – COOH
no. of Sterioisomers : 3 COOH
1. Meso
H
OH
H
OH COOH
Page-61
2. d/
#
COOH
COOH H
Meso
& &"
OH
H
OH COOH
&" COOH &
H
& &"
OH
OH
H H
COOH OH
COOH
COOH COOH
H H
COOH OH
H
OH
COOH , &&
OH
H
OH
H
OH
d/ COOH
COOH COOH OH H
OH
COOH
& &"
H OH
& &"
H
H
OH
COOH
COOH OH
HH
&" COOH & H OH
COOH
H , &&
H OH
OH
OH COOH
COOH
H
COOH
OH
H
OH H COOH OH
Conformation isomers of cyclohexane : Chair form and Boat form
Page-62
Session - 2009-10
TOPIC : CONTENTS
1
GOC - I
:
Structure Identification * Monochlorination * Catalytic Hydrogenation * Ozonolysis * Elements detection * Identification of Functional Group by Lab. Test
Refer sheet GOC- I JEE Syllabus [2009] Practical organic chemistry: Detection of elements (N, S, halogens); Detection and identification of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro; Chemical methods of separation of mono-functional organic compounds from binary mixtures.
STRUCTURE INDENTIFICATION Monochlorination:(a)
Cl 2 / h# ! ! " CH Cl + HCl (i) CH 4 ! ! ! 3 Cl 2 / Sunlight ! ! ! ! " CH – CH Cl + HCl (ii) CH3 – CH3 ! ! ! 3 2 Cl
–
Cl 2 / h#
(iii)
+ HCl
! ! ! ! ! "
(iv) C –
C C – – Cl2 / h# ! " C – C C – – C ! ! ! ! – – C – Cl + HCl C C Cl
Cl 2 / h#
(v)
–
! ! ! ! ! "
+ HCl
Remarks:- When an alkane or a cycloalkane is treated with halogen (Cl 2, Br2, F2, I2), a photochemical reaction takes place and a C – H bond cleaves and a C – Cl bond is formed. If one H-atom is substituted by one halogen atom. This is known as monohalogenation reaction. 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. Conclusion:- Hence, it can be concluded that the total no. 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.
Monochlorination
(a) C – C – C
! ! ! ! ! ! ! ! ! "
(b) C – C – C – C
! ! ! ! ! ! ! ! ! "
(c) C – C – C – C – C
! ! ! ! ! ! ! ! ! "
(d) C – C – – C – C
! ! ! ! ! ! ! ! ! "
Monochlori nation Monochlori nation
Monochlorination
2 Products (structure isomers) 2 Products (structure isomers) 3 Products (structure isomers) 4 Products (structure isomers)
C CH3 –
Monochlori nation
(e)
! ! ! ! ! ! ! ! ! "
Q.
Cl 2 / h# ! ! " only one monochloro isomer. X(C5H12) ! ! ! .
Ans.
X = Neopentane
Q.
Cl2 / h# ! ! " Two monochloro P(C6H14) ! ! !
5 Products (structure isomers)
How many isomers of P will give two monochloro compounds ? C –
Ans.
C –
C – C – C – C only one isomers
Remark : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated, but H-atoms of Benzene ring are stable.
CH3
CH2Cl
–
Ex.
–
Cl2 / h#
! ! ! ! ! "
Q.
Cl2 / h# ! ! " Two mohochloro X(C8H10) (Aromatic) ! ! ! Cl 2 / h# ! ! " One monochloro Y(C8H10) (Aromatic) ! ! !
Ans.
(X)
(Y)
Catalytic Hydrogenation of C = C; C $ C General reaction:Ni ! " R – CH – CH – R (a) R – CH = CH – R + H2 ! ! 2 2 Ni / Pt / Pd ! " R – CH – CH – R (b) R – C $ C – R + 2H2 ! ! ! ! ! 2 2 H2
R – CH % CH – R (Not isolated)
H2
! ! ! "
R – CH2 – CH2 – R
2H2 / Ni ! " CH – CH – CH – CH (c) CH2 = CH – CH = CH2 ! ! ! ! 3 2 2 3 3H2 / Ni
(d)
! ! ! ! ! "
CH = CH 2
(e)
–
H2 / Ni ! ! ! ! ! ! ! ! " room temperatur e CH 2 – CH 3
–
H 2/Ni (100 – 150ºC)
(f)
H / Ni, & ! !2 ! ! ! "
CH2 – CH3
–
[Reaction cannot be stopped at any intermediate stage]
Remarks:(a) Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature. (b) All C – C ' bonds(C = C, C $ C) are hydrogenated. The reaction can ’t be stopped at any intermediate stage.
Exceptions: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 no. of moles of H 2 consumed by 1 mole of compounds is equal to the no. of ' bonds presents.
"
All positional isomers of alkenes or alkynes (due to multiple bond) always give same product on hydrogenation.
"
During catalytic hydrogenation and monochlorination, carbon skeleton remain unchanged.
Ex:- C = C – C = C + Cl 2 X
Ex.
C – C – C – C – Cl
2H2 / Ni
(1)
! ! ! ! ! "
Cl
Y =
Cl / h
# 2 ! ! ! ! ! "
(X)
Y+Z
Z =
Cl CH3
CH3 H2 / Ni
(2)
CH3
5 Monochloro product
CH3 H2 / Ni
(3)
Q.
Cl / h
# 2 ! ! ! ! ! "
! ! ! ! "
5 Monochloro product
! ! ! ! "
H / Ni Cl 2 / h# ! ! " Z(only one monochloro product) X(C4H6) ! !2 ! ! " Y ! ! !
Identify X, Y, Z Ans.
DU = 2 Cl
X"
,
Y"
,
Z"
–
Q.
Identify the lowest molecular weight alkane which gives four structural isomeric monochloro products ?
Ans.
C – C – – C – C C
C5H12 Q. Ans.
= 72g
Identify the structure of hexane which gives 3 monochloro products ? C –
C – C – – C – C,
C – C – C – C – C – C
C
Q.
Find the no. of monochloro products of a fully saturated isomer of C 4H6.
Ans.
DU = 2 (
Q.
2 monochloro product
Find the structural isomers of product?
fully saturated cycloalkane of M.F. C 6H12 which gives two monochloro
–
Ans. –
Q.
–
Cl 2 / h# ! ! ! ! ! " A C 8H17 Cl (C8H18 ) (Only one type )
Identify A ?
C C – – C – – C – – C C C
Ans.
C –
Q.
Write all isomeric alkynes which produce an isomer of heptane which on further monochlorination gives (a) three monochloro products. C – C –
Ans.
(i) C $ C – C – C – C
(ii) C $ C –
C – C – – C – C C
(b) Ans.
two monochloro Nil
Q.
Find the structure of lowest molecular weight hydrocarbon and maximum unsaturation which on hydrogenation produce such an alkane which gives two monochloro products ?
Ans.
C = C = C or C – C $ C
Q.
Determine the M.W. of maximum unsaturated hydrocarbon which on hydrogenation gives C 6H12 which on further chlorination gives two monochloro. CH2
Ans. CH2
CH2
C6H6 = 78g
Ozonolysis: It tells about position of unsaturation. Remarks:(1)
Alkene and polyalkene on ozonolysis undergo oxidative cleavage.
(2)
(a) The reagent of reductive ozonolysis is (i) O3 (ozone) (ii) Zn and H 2O or Zn and CH 3COOH or (CH3)2 S (b) The reagent of oxidative ozonolysis is O 3 and H2O2.
(3)
The products are carbonyl compounds (aldehydes or ketones). This type of ozonolysis is known as reductive ozonolysis.
(4)
Ozonolysis does not interfere with other F.G.s.
General Reaction:- R – CH = CH – R
Ex:-
(1) CH2 = CH2
(1) O3 R – CH = O + O = CH – R + ZnO + H2O (2) Zn/H2O
(1) O3 CH2 = O + CH 2 = O (2) Zn/H2O
(2) CH3 – CH2 – CH = CH2
(1) O3 CH3 – CH2 – CH = O + O = CH 2 (2) Zn/H2O
(3) CH2 = CH – CH2 – CH = CH – CH3
O3 / Zn
CH2 = O + O = CH – CH2 – CH = O + O = CH – CH3
O3 / Zn
(4)
! ! ! ! "
+ OHC – CH2 – CHO (Propandial)
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.
C
O
3 ! ! ! " O
(1)
Zn
= C – C – C O
C
O3 ( & ) ! ! ! ! "
(2)
Zn
CH = CH – CH 3
(3)
(4)
Q.
C – C = O + O = CH2
–
–
CH = O
–
low temperatur e
! ! ! ! ! ! ! ! "
+ O = CH – CH3
O3 / Zn
O3
– CH = CH –
! ! ! " C
6
Zn
(P) CnH2n – 2 Single Compound (no isomer )
H 2 / Ni
! ! ! ! "
CnH2n (Q )
H5 – CH = O
Cl / h
)
2
# 2 ! ! ! ! ! "
CnH2n 1Cl (m products ) m%3 )
CH 3 –
HCOOH + CH3 – C – COOH –
CH 3 Identify P ?
Ans.
P =
Q.
An unsaturated hydrocarbon on ozonolysis produces 1 mole of
, 1 mol CO 2, 1 mol
Find the structure of the hydrocarbon and the no. of monochloro products formed followed by hydrogenation. Ans.
,
5 monochloro product
–
X Q.
Cl2
H2 / Ni
(Unsat. hydrocarbon)
! ! ! "
! ! ! ! "
–
h#
m products (m % 3)
O 3 (Zn/H2O)
O identify structure of X ?
O
Ans.
Q.
X =
X (Unsat. H.C.)
H 2 / Ni
! ! ! ! "
Cl / h
# 2 ! ! ! ! ! "
O3 (Zn/H2O)
HCHO + 4(1-oxoethyl) Cyclohexan-1-one. Identify X ?
C Ans.
X" –
C = C – C
Q.
Identify structure of X ?
Ans.
X is
Q.
H2 ! " C – C – C – C – C – C X ! ! O3 (Zn)
CH3CHO + CHO – CHO Identify structure of X ? Ans.
X is C – C = C – C = C – C
m products (m % 7 )
Q. CH3
O3
–
Sol.
2
! ! ! "
+
Zn, H2O
–
CH 3
CH3
–
CH3
O3
! ! ! "
–
CH 3
Zn, H2O
CH3 – C – C – +2 O
O
Q.
A
Methyl glyoxal + Formaldehyde
Ex.1
Identify A,B & C with the help of following reactions. Cl2 / h*
(A) (C 9H18 )
! ! ! ! ! "
Single monochloro produc
( Saturated Hydrocarbon )
t (B) (C8H18)
Cl2 / h*
! ! ! ! ! "
(C) (C7H14 )
Cl2 / h*
! ! ! "
Single monochloro product
Two monochloro productss
(Saturated Hydrocarbon)
Sol.
A=
or
C C | | B = C+C+C+C | | C C
C =
Identify ,-Terpinene and P-Menthane.
Ex.2
Sol.
,-Terpinene
Ex.3
Identify A & B
Sol.
A = CH3 + C % CH + CH % CH2 | CH3
B = CH3 + CH + CH2 + CH2 – CH3 | CH3
Identify A & m
Ex.4
A = Ph + C % C + Ph | | CH3CH3
Sol.
m=3
Ex.5
H 2 / Ni A ! ! ! ! "
Cl / h
# 2 ! ! ! ! ! "
O3 Zn/H 2O
+
n-products
H – C – H +
H – C – C – CH2 – CH2 – C – C – H
Sol.
A=
n=7
1.1
Identification of Elements in Organic Compounds Element
1. Nitrogen
Test / Reaction
Lassaigne’s test
Remark
The appearance of green or
Na + C + N " NaCN
prussian blue colour confirms
FeSO4 + 6NaCN " Na4 [Fe(CN)6] + Na 2SO4
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 NH 4OH
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 Na + C + N + S " NaSCN
FeCl3
! ! "
presence of both nitrogen & Fe(SCN)3
sulphur
’
H ‘
s k r a m e R
n o i t a l y x o r d y H
s n i f f a r a p t r e n I
n o i t a n i m o r B
t p p e t i h w
s t s e T y r o t a r o b a L y b s p u o r G l a n o i t c n u F f o n o i t a c i f i t n e d I 2 . 1
-
n o i t c a e R
n o i t a v r e s b O
R R R R N N N N
s i s y l o n o z O
O H C H 2
3
O + 2 H C = C 2 H
r s u r o a l o e c p p k a s n i i P D
s e s i r u o l o c e d r u o l o c d e R
s d n u o p m o C O =
s i s y l o n o z O
H O O C / R + H O O C R
/
R –
C $
t n e g a e R
l a n o s i t p C – c u n o C u r F G
O 2 H / 2 r B
/ C C = $ C C
) e n o z o ( 3 O
) d e r ( u C C $ C –
R
) e t i h w ( g A C $ C –
R l C u C + H C $
+
g A + H C $
C
C
C
–
–
R
R
R
–
. d e m r o f d i c A
4
O ] t N n M e K g 4 a l d O H e O r o S 2 a s c . H N 4 4 ’ r l i . . O e d c c n H y . n n M l a k A B l o o i c c K L [ a
e v i t c a f o e c n e s e r P
3
O
. t p p d e R
. t p p e t i h W
H O 4 H N + e d i r o l h c s u o r p u C ) a (
H O 4 H N + 3 O N g A ) b ( H C $
C % C
C $ C
C –
R
t l s l o l o e o h o h T h c l o s o c c a l a l . a c a . . c i u r r L t e e p s . . I . I I I I I
O 2 s H s e n + i d u o l c 0 2 H + a N O R 2 " a N + H O R 2
t u o e m o c 2 H f o s e l b b u B
a N ) H O –
R (
l C H + H O –
R n i h r t i e t w f a s s r r r a a a e e e p p p p p p a a a y s s s l s e s s e t e e . a n n n n i i . i i i d d d i d n u i u m u m o m o l o l l 0 m C C i 5 C 3 ) ) ° ) ° °
3 ( 2 ( 1 (
] 2 l C n Z . d y h t n n a e + g l a C e R H . s c a n c o u L C [
H O ° ° ° R 3 2 1
s l o f n e o h s t p k s / r e s a T l m o e n R e
t s e t s ’
t s e t P N D
g n i l h e F
) . t p p e g n a r o w o l l e y ( -
2
H +
n o i t c a e R
H
n o i t c a e r m r o f o d o I
t s e t s ’
n e l l o T
e t a n o b r . a t c s i e b t s m u u i t m d t o s i e L S t
) r o r r i m O 2 r e H v 2 l i + S ( g d O 2 e A u R 2 C + + H H O O . O O n C l R o C s R " g n " 2 i + l u h + g e A C F + + O H C R
0 2 O C + O 2 H
) n o i t u l o s s s e l r u o l o C (
n i e l a h t ) h k n p i o ( p n e h P H + O H ’ O R a + N + H ’
R O O C R
O H C R
O O C R
0 3 H N + a N O O C R
H O a N + 2 H N O C R
. O H C R y b d e m u s e r s i r u o l o c k n i p e h T . s s e l r u o l o c s i t i o s 2
) f f u b n e e r n . g t o , i p e t p a u d l v b r e , r e u t s e l o o b l O o i v C (
t n e g a e R
) l a r t u e N ( l 3 C e F
l H a n O s o l s i – o t p r n c u n o A E r u F G
. t p p e g n a r o w o l l e Y
e n i z a r d y h n l o y i t n u l e h o s p ) o r P t i N n i D D - 4 4 , , 2 2 (
. t p p d e R
n o i t u l o s g B n & i l h A e F
r o r r i m r e v l i s r o . t p p k c a l B
t n e g a e r s ’
n e l l o T
O H C –
R
) m r o f o d o i (
e m u s e r r u o l o c k n i P
3 I
H C f o t p p w o l l e Y
* t n e g a e R s ’
H O a N /
t f i h c S
2 I
. e v l o v e e c n e c s e v r e f f E
n o i t u l o s 3 O s C u H m a t i l N . e c u n l o B C
. g n i t r a u e o h l o n c - o k r n a i e P p p a s i d
H N f o l l e m S
. n i e l a h t h p o n e h p , H O a N
& , H O a N . c n o C
r e t s E
s e d i m A
3
3
H H C O C O H O C r C C r o 3 – A H r C R 3
. d e r o t e g n a h c s u m t i L
o
O S h t i w d e t a r u t a s e d i r o l h c o r d y h e n i l i n i s o R p : t n e g a e r s ’
f f i h c S
e n i m n a i o l y t c b r a a e C R
s k r a m e R
g A
H O H N r A n o i t c a e R
n o i t a v r e s b O
t n e g a e R
t p p k c a l b
t s e t s ’
n e k i l l u M r
s o l d ) a n 2 n u O o s o N i 2 t p o 2 r p H O c u t i m n o N o C N r u r C R F G ( A
O 2 H 3 + l C K 3 + C N R
H O K 3 + l 3 C H C + 2 H N R
r u o d ) o e n g i n i t m a a l y e b s r u a a N C (
H O K , l 3 C H C
t s e t n i r d y h n i N
t s e t e y D
O 2 H + 2 N + H O R
O N O H + 2 H N R
2
N f o e c n e c s e v r e f f E
) l C H + 2 O N a N ( 2 O N H
) . i r p ( s e 2 n H i m N A R
H O N = N 2
O N H + l l C C a 2 N N
l C H + l 2 C O H N . 2 a H N N
O 2 H 2 +
e y d d e r e g n a r o
O 2 H + O H C R + 2 O C +
l o h t h p a N
H O ) O i d C . c a R o n H i m C . ( A N 2 H +
H – O + 1 l m 2 u C - i O N n N = o e z d H N i a i r + d l o e h n c e z n e B
d e m r o f s i e y d d e r e g n a r O
H O
H O )
n i r d y h n i O N (
C O C
. r u o l t o c s t n t e e o i n l t o i a r n v a u h o l m s i o r e d c b d d i e e e r L R
) l C H + 2 O l o N h a t h N ( p 2 a O N N 1 H +
O S 2 H + l 2 o O n e N h a P N ) ) i i i ( (
. s e n i 2 m H a . N r r A A
s e n i m A H . c N 2 e R S
4
r u o l o c t e l o i V
r u o l o c e u l B
) n . l . ) l o o s h % o c 2 l . a 0 ( n t t i n n l e o e h g g t a h a e e r p a r n s ’ n i - r h 1 d c y s % h i l n o 0 i M 1 ( N e t a r d y h o b r a C
s d i c a o n i m A
C
O C
H C O
C –
N = C O C
) r u o l o O c C e u l B (
Examples : Structure Determination 1.
Identification of Organic Compounds on the bacis of Physical Properties (a) Physical state (b) Odour (c) Water solubility (d) m.p. / b.p. (e) relative
Ex. 1
Ex. 2
Ex. 3
Ex. 4.
2.
Identification of Organic Compounds on the bacis of Chemical Properties
Ex.5
Which of the following will not give (+ve) L.S. test for Nitrogen. (A) CH3 –CH2 –NH2 (B) (C) (D) NH4NO3 (2 Here is no carbon for the formation of CN ) –
Instant turbidity with Lucas reagent (
H2C = CH – CH2 – OH
CH2 = CH – CH+ " CH2+ – CH = CH2 Allyl C+ , 1°
(
Benzyl C+ , 1°
" Angle strain / less stable / unstable C +
Although 3° , but has angle strain , slow r×n with HCl/ZnCl2
Other Examples
Ex.1
Ans.
Ex.2
Ans.
Ex.3
Ans.
Ex.4
-
Ans.
CH3 – CH2 – C $ C – CH2 – CH3
Ex.5
Ans.
CH3 – CH2 – CHO
Ex.6
C4H8
Ans.
CH3 – CH = CH – CH3
Ex.7
(A)
Ans.
A=
Ans.
CH3 – CH2 – COOH
Ans.
O || CH3 + C + O + CH3
Ans.
A=
Ex.8
Ex.9
Ex.10
A (C7H8O)
Na metal –ve FeCl3 (neutral)
–ve
Lucas –ve Reagent Ex.11
R gives following tests.
Ans.
1.
How will you distinguish the following pair of com pounds. Compounds Isomers
Reagents
(I) (a) CH3 – CH2 – COOH (b) CH3 – C – O – CH3
1
2
3
NaHCO3 (+) NaHCO3 ( –)
Acidic odour Fruity/Sweet odour
Na metal (+) Na metal ( –)
O (II) (a) Ph – CH2 – C – OH Same
O (b) Ph – C – O – CH3
Same
Same
O (III) (a) Ph – CH2 – CH = CH2
Br2/H2O (+)
Dil KMnO4 (+)
(b) Ph –
Br2/H2O ( –)
Dil KMnO4 ( –)
(IV) (a) CH3 – CH2 – CH2 – OH
Na metal (+)
(b) CH3 – CH2 – O – CH3
Na metal ( –)
(V)
Ceric Ammonium Nitrate (+) ,, ( –)
CH3 –
Lucas Reagent instant turbidity
Na metal (+)
Ceric Ammonium (+)
(b) CH3 – CH – CH2 – CH3
In 5 min.
Na metal (+)
(+)
OH (c) CH3 – CH2 – CH2 – CH2 – OH
In 30 min.
Na metal (+)
(+)
(VI) (a) CH3 – CH2 – CH2 – CH2 – NH2
NaNO2/HCl (N2 gas )
Na metal (+)
(b) CH3 – CH2 – CH2 – NH – CH3
N2 gas not liberated
Na metal (+)
(a) CH3 – C – OH – CH3 –
2.
Tick mark the reagents which will give positive response with the following compound. (A) Na metal O O H2N
O
Ans. A C D F G H I J
(B) NaHCO3 (C) 2, 4-DNP (D) AgNO3 + NH4OH
OH
(E) Fehling solution (F) NaNO2 /HCl (cold) (G) ZnCl2 (Anhyahous)/HCl (H) Cu2Cl2 + NH4OH (I) Br2 /H2O
3.
(J) Dil KMnO4 (cold) T (C7H6O2) is an aromatic white solid which liberates a colourless odourless gas on heating with NaHCO 3. Write S.F. of X and its all possible functional isomers (all aromatic)
Sol.
T =
;
,
POC - II Separation of binary mixture of organic compounds Theory of separation : Organic compounds have different solubilities in different solvents. So they can be separated by use of appropriate solvents.
Purification of organic compounds : 3
The organic compounds derived from natural sources or prepared in the laboratory are seldom pure. They are usually contaminated with other substances.
3
Purification means the removal of undesirable impurities associated with a particular organic compound, i.e to obtain the organic compound in pure state.
3
Various methods have been developed to purify organic compounds
1.
Physical methods : (i) Crystallisation (ii) Sublimation (iii) Distillation (iv) Solvent extraction (v) Chromatography
2. 3
Chemical methods:Chemical methods of separation depend upon the nature of the functional group present in the component. Hence these can be applied to solid as well as liquid compounds.
3
A chemical method can be applied only when one of the components of the mixture is soluble in a particular solvent while the other is insoluble in the same solvent .
3
Separation is the first step during the actual analysis of organic mixture. It is the most important step in the sense that if separation is incomplete the result will not be correct because the impure compound will give tests of different functional group and its melting point will also be very much different from that of the pure compound obtained from complete separation.
(i) (ii) (iii) (iv)
Separation of Binary mixtures of organic compounds. The usual systematic scheme for separating a solid binary mixture is discussed below. Separation with water Separation with sodium bicarbonate Separation with sodium hydroxide Separation with hydrochloric acid Solubility of two components. Separation Scheme Solvent H2O
! ! ! ! ! ! "
(1)
Some important point :3
The mixture of organic compounds can be separated by using appropriate solvent.
3
Most of the aromatic compounds are water insoluble due to large hydrophobic group of six carbon atom
3
Aromatic acids are insoluble in water but soluble in aqueous NaHCO 3 solution or NaOH solution, due to salt formation.
3
Aromatic hydroxy compounds are water insoluble but are soluble in aqueous NaOH solution due to salt formation.
3
Aromatic amine (Aniline 1 º, 2º, 3º) are organic base and water insoluble but are soluble in aqueous HCl solution due to salt formation.
3
Aliphatic compoud with atleast two functional group ( which can form H-bonding) are water soluble. Ex. Diacids, diols. diamines, hydroxy acids ( – OH, –COOH), Amino acids ( – NH2 , –COOH) . oxalic acid , malonic acid, maleic acid, fumaric acid, glycol, glycerol, sucrose, glucose.
Ex.1
Binary mixtures - (Two components)
Compound A
Compound B
Appropriate Solvent
(1)
+
H2O
(2)
+
H2O
(3)
Fructose
+
H2O
(4)
+
aq. NaHCO3
(5)
+
aq. NaOH
(6)
+
aq. HCl
+
Ex.2
Sol.
Identify I & II. I ( A, II ( B
+
Ex.3
Identify I & II Sol.
4 " Q ,
44 " P
Ex.4
+
+
P+Q+R
Identify P, Q, R Sol.
P"I
Q " II
R " III
