Chemistry Form 6 Sem 3 Chapter 1
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CHEMISTRY FORM 6 ORGANIC CHEMISTRY CHAPTER 1 : CARBON COMPOUND
1.1 The Chemistry of Carbon Organic Chemistry ~ branch of chemistry concerning compound of carbon (except CO, CO2, CO32-) Aliphatic compounds → Open chain organic compounds. → Do not have special stability like benzene. → Compound may be unbranched or branched Alkane, alkene, alkyne, haloalkane
Alicyclic compounds → Closed ring of organic compounds. → Rings may contain single or double bonds → Compound may be branched or unbranched Cyclohexane, cyclobutene
Aromatic compounds → Contain at least one benzene ring (those with in it)
Heterocyclic compounds → Closed ring contain element other than carbon in it (like N, S, O)
Phenol, naphthalene, toluene
Pyridine
1. Hybridisation of alkane, alkene and alkyne 14element. It has the electronic Carbon is a Group ___ configuration of ______________ 1s2 2s2 2p2 The orbital diagram Ground state of C : _____ _____ _____ _____ 2s 2p Methane, CH4
Type of hybridisation :
Excited state of C : _____ 2s
Hybridised state :
sp3
_____ _____ _____ 2p
_____ _____ _____ _____ sp3
Ethene, C2H4
Type of hybridisation :
Excited state of C : _____ 2s
Hybridised state :
_____ _____ _____ 2p
_____ _____ _____
sp2 Molecular shape Trigonal planar
Angle between bond pair – bond pair 120o
sp2
_____
pz
Ethyne, C2H2
Type of hybridisation :
Excited state of C : _____ 2s
Hybridised state :
_____ _____ _____ 2p
_____ _____
sp Molecular shape Linear
Angle between bond pair – bond pair 180o
sp
_____ _____
py
pz
As a conclusion, the formation of double bond one bond (σ) and (C=C) is due to ______sigma one (π) _____pi bond While the formation of triple bond (C≡C) is one due to ______sigma bond (σ) and _____pi two (π) bond
Hybridisation in benzene Benzene (C6H6) is a flat and symmetrical molecule. All the atom (6 C atom and 6 H atom) in a benzene molecule lie in the same plane. The Carbon atoms are arranged in the form of hexagon as shown in diagram at the left. The formation of benzene can be deduced using hybridisation theory
Excited state of C : _____ 2s
_____ _____ _____ 2p
Hybridised state: _____ _____ _____ sp2
_____ pz
Since each carbon contribute an unhybridised electron, so the side touch between C–C atom will form a double bond. Hence 3 bond build between C–C in benzene ring. there are _____ double In another words, there are free delocalise electrons move about in benzene ring. ◦ The following notes are taken into consideration when drawing structure of benzene The unhybridised p orbitals do not overlap in pairs to form double bonds alternating with 3 single bonds as shown in Kekule structure.
hybrid ◦ The structure of benzene is a ……………..... of 2 forms ◦ The resonance hybrid of benzene can be expressed as ◦ Unlike ethene, the double bond in benzene has a larger volume (space) to delocalise electron. Hence the more space provided, the lower the energy in benzene. This makes benzene posses an extra stability.
1.2 The the (a) (b) (c) (d)
Chemical formulae of Organic Compound formula of an organic compound can be represented by empirical formula molecular formula structural formula skeletal formula
Empirical formula of a compound shows the simplest ratio of the atoms of each element in the compound. Molecular formula of a compound shows the actual number of atoms of each element in the compound
1. Derive the empirical formula of a hydrocarbon that on analysis gave the following percentage composition: C = 85.63%, H = 14.37%. given the relative molecular mass of the hydrocarbon is 84, determine the molecular formula Element
C
H
Mass
85.63
14.37
Mol
85.63 12 = 7.14 mol
14.37 1 =14.37 mol
Ratio
7.14/7.14 =1
14.37/7.14 =2
Empirical formula = CH2 (CH2)n = 84 (12 + 2(1))n = 84 n=6 Molecular formula = (CH2)6 = C6H12
2. A 1.367 g sample of an organic compound was combusted in a stream of dry oxygen to yield 3.002 g CO2 and 1.640 g H2O. If the original compound contained only C , H, and O, what is its empirical formula? CxHyOz + mO2 x CO2 + y/2 H2O Since 1 C = 1 CO2 x / 12 = 3.002 / 44 ; x = 0.8187 g Since 2 H = 1 H2O y / 2= 1.640 / 18 ; y = 0.1822 g Mass of O, z = 1.367 – (0.8187 + 0.1822) = 0.3661 g Elemen
C
H
O
Mass
0.8187
0.1822
0.3661
Mol
0.8187 12 =0.0682
0.1822 1 =0.1822
0.3661 16 =0.0229
Ratio
0.0682/0 .0229 =3
0.1822/0 .0229 =8
0.0229/ 0.0229 =1
Empirical formula = C3H8O
A 1.500 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the compound? CxHyOz + mO2 x CO2 + y/2 H2O Since 1 C = 1 CO2 x / 12 = 1.738 / 44 ; x = 0.474 g Since 2 H = 1 H2O y / 2(1) = 0.711 / 18 ; y = 0.079 g Mass of O, z = 1.367 – (0.474 + 0.079) = 0.947 g Elemen
C
H
O
Mass
0.474
0.079
0.947
Mol
0.474 12 =0.0395
0.079 1 =0.079
0.947 16 =0.0592
Ratio
0.0395/0 .0395 =1
0.079/ 0.0395 =2
0.0592/ 0.0395 = 1.5
Empirical formula = C2H4O3
Elementary analysis showed that an organic compound contained C, H, N, and O as its elementary constituents. A 1.279-g sample was burned completely, as a result of which 1.60 g of CO2 and 0.77 g of H2O were obtained. A separately weighed 1.625 g sample contained 0.216 g nitrogen. What is the empirical formula of the compound? Since 1 C = 1 CO2 x / 12 = 1.60 / 44 ; x = 0.4364g Since 2 H = 1 H2O y / 2(1) = 0.77 / 18 ; y = 0.08556 g Since 1.625 g of same sample produce 0.216 g of nitrogen Mass of N in sample, n n / 1.279 = 0.216 / 1.625 n = 0.17 g Mass of O, z = 1.279 –(0.4364 + 0.08556 + 0.17) = 0.587 g Empirical : C3H7NO3
Structural formula of an organic compound is the formula which shows how the atoms are bonded together as well as the numbers of each atom present. Structural formula can be expressed in a few ways. Example : butane, C4H10
b) Displayed formula
a) Shorthand CH3CH2CH2CH3
c) 3-D @ stereochemical
d) Skeletal formula
c) 3-D @ stereochemical
d) Skeletal formula
e) Simplified notation CH3(CH2)2CH3
Example
: 2-methylhexane
a) Shorthand CH(CH3)2CH2CH2CH2CH3 b) Displayed formula
Example :
4,4-dimethylpent-2-ene
a) Shorthand
C(CH3)3CH=CHCH3 b) Displayed formula
c) 3-D @ stereochemical
d) Skeletal formula
Example :
3-ethyl-2,4-dimethylhexane
a) Shorthand
CH(CH3)2CH(CH2CH3)CH(CH3)CH2CH3
c) 3-D @ stereochemical
d) Skeletal formula
b) Displayed formula
Example :
2,2,5-trimethylhex-3-yne
a) Shorthand
C(CH3)3CΞCCH(CH3)2 b) Displayed formula
c) 3-D @ stereochemical
d) Skeletal formula
1.3
Classification of Organic Compounds based on the Functioning Group and its General Formulae Homologous series ~ compounds with similar chemical properties in which each member differs from the previous one by addition of –CH2–. The characteristic of a homologous series are as follow. All compounds in homologous series has the same functioning group and chemical properties Each member differ from the next series by a –CH2– group, in another words, molecular mass of each compound in series 14 differ from next by ………… All the compounds in the series may be prepared by using the similar methods. Physical properties show a progressive change with increase of molecular mass. All the compounds in the series contain same elements and functioning group, thus it can be represented by same general formula.
Homologous series Alkanes
R–H
Examples + Structural formula
General formula pentane CnH2n+2
Cycloalkane
CH3CH2CH2CH2CH3
heptane CH3CH2CH2CH2CH2CH2CH3
cyclobutane
Cyclohexane
hex-1-ene
but-2-ene
CnH2n
Alkenes
RCH=CHR
CnH2n
CH2=CHCH2CH2CH2CH3
cyclohexene
Cycloalkene CnH2n–2
CH3CH=CHCH3
cyclobutene
Alkynes
RCΞCH
propan-1-ol
pentan-3-ol
Butanal
hexanal
Propanone
heptan-3-one
CnH2nO
Ketone
R–CO–R
2-bromopentane
CnH2n+1OH / CnH2n+2O
Aldehyde
R–COH
1-chlorobutane CnH2n+1X (X = Cl, Br, I)
Alcohol
R–OH
pent-1-yne
CnH2n–2
Haloalkane
R–X
Propyne
CnH2nO
Carboxylic acid
RCOOH
propyl butanoate
ethylamine
butylamine
propylamide ;
pentylamide
CnH2n+1NH2
Amide
RCONH2
ethyl ethanoate ; CnH2n+1COO– CmH2m+1 / CnH2nO2
Primary amine
R–NH2
pentanoic acid
CnH2n+1COOH CnH2nO2
Ester
RCOOR
butanoic acid ;
CnH2n+1CONH2
1.4 Alkyl and Type of Alkyl groups Alkyl group ~ obtained by removing a hydrogen atom from an alkane. Symbol of alkyl is R, where –R has the general formula of CnH2n+1.
Alkane
Alkyl
Alkane
Alkyl
Methane
Methyl
–CH3
Ethane
Ethyl
–C2H5
Propane
Propyl
–C3H7
Butane
Butyl
–C4H9
–C5H11
Hexane
Hexyl
–C6H13
Pentane
Pentyl
Alkyl can be categorise into 3 groups Type of alkyl group
Example
Comment
Primary
Only one alkyl group attached to carbon atom
Secondary
Two alkyl group attached to carbon atom
Tertiary
Three alkyl group attached to carbon atom
1.5 Isomerism in Organic Compound Isomers ~ substances which have the same molecular formula but different molecular structure Isomerism
Structural isomerism
Chain isomerism
Position isomerism
Stereoisomerism
Functional isomerism
Geometrical isomerism
Optical isomerism
1.5.1 Structural Isomerism ~ are isomers with same molecular formula but different structural formula (link differently) As mentioned above, structural formula can be separate into 3 different categories ◦ Chain isomerism ◦ Positional isomerism ◦ Functional group isomerism
1. Chain isomerism ~ isomers which have different carbon chain (straight or branched chain) Pentane (C5H12)
Butan-1-ol (C4H9OH)
CH3CH2CH2CH2CH3
CH3CH2CH2CH2OH C(CH3)4 CH(CH3)2CH2OH
CH3CH2CH(CH3)2
1-chloropentane (C5H11Cl)
Hexanal (C6H12O)
CH(CH3)2CH2CH2CHO
CH3CH2CH2CH2CH2CHO
CH3CH2CH2CH2Cl CH3CH2CH(CH3)CH2Cl
C(CH3)3OH
C(CH3)3CH2CHO C(CH3)3CH2Cl CH(CH2CH3)2CHO
2. Position isomerism ~ isomers which the position of functioning group is different Hexene (C6H12)
Bromohexane (C6H13Br) CH3CHBrCH2CH2CH2CH3
CH2=CHCH2CH2CH2CH3 CH3CH=CHCH2CH2CH3
CH3CH2CH=CHCH2CH3 Pentanol (C5H11OH)
CH3CH2CH2CH2CH2OH
CH3CH2CH2CH(OH)CH3
CH3CH2CH(OH)CH2CH3
CH2BrCH2CH2CH2CH2CH3
CH3CH2CHBrCH2CH2CH3 Dichlorobenzene (C6H4Cl2)
3. Functional isomerism ~ isomers which has the same molecular formula but different molecule with different functioning group. Alkene and cycloalkane – C5H10
Alcohol and ether – C4H10O
Aldehyde and ketone – C6H12O
Carboxylic and ester – C7H14O2
1.5.2 Stereoisomerism Geometrical Isomerism ~ same structural formula but different spatial arrangement. ~ also known as cis-trans isomer ~The essential requirement for the existence of geometrical isomerism in organic compound must contains a carbon– carbon double bond (C=C) ~ A ring structure which hinders the rotation of a C–C single bond in a ring. However, cis-trans isomers cannot occur if one of the carbon atoms in the double bond has 2 identical atoms / groups. 1,2-dichloroethene
1,1-dichloroethene
pent-2-ene
1,2-dimethylcyclobutene
But-2-ene-1,4-dioic acid
1,4-dichlorocyclohexane
Physical Properties of Geometric Isomers Cis-isomer usually has a lower melting point as the structure of cis-isomer is less symmetrical. Therefore, cisisomer cannot be closely packed in the crystal lattice resulting the intermolecular forces to become weaker than in trans-isomer. On the other hand, cis-isomer has a higher boiling point because the space arrangement in cis isomer caused the compound to become a polar molecule. As a result, the intermolecular forces fo cis-isomer in liquid is stronger, causing the temperature required to boil the substance become higher. Trans-isomer on the other hand, has 0 dipole moment.
(2) Optical isomerism Optical activity is the ability of certain crystal or solution of certain substances to rotate the plane of plane-polarised light. Such substances are said to be optically active (sometimes are known as chiral molecule) Optical isomers are optically active substances which possess the same molecular structure but different in their effect on plane polarised light. For an optically active isomer, the object and mirrored image cannot be superimposable to one and another. Example, in 1bromo-1-chloro-1fluoromethane (CHClBrF), it shows optical active when brought to plane-polarised light.
*
An organic molecule will exhibit optical isomerism if it contains one chiral carbon atom. Chiral carbon atom is a carbon atom which is attached by 4 different atoms / groups. Also known as asymmetrical carbon often shown as C* Example : State whether the organic compound below exhibit optical isomerism. If yes, mark * at chiral carbon atom.
CH3CH2CH(Cl)CH3
C(CH3)2(Br)(OH)
X
CH3CBr=CH2
X
HOCH2CH(NH2)COOH
CH3CH2CH(OH)CH3
1.6 Lewis Acid & Lewis Base : Nucleophiles and Electrophiles Substance which receive lone pair electron Lewis Acid : ………………………………………………………………………… Substance which donate lone pair electron Lewis Base : …………………………………………………………………………
Ammonia react with hydrogen ion to form ammonium ion
Boron trifluoride react with ammonia to form a complex
Aluminium trichloride react with chloride ion to form tetrachloroaluminate ion
Water molecule attached to cobalt ion to form hexaaquacobalt (III) ion.
Methylamine react with water to form a basic solution
Ethanoic acid react with water to form ethanoate ion and hydroxonium ion
In organic reaction, such reaction can be classified as either nucleophile or electrophile, depending on whether they attack regions of high electron density (δ –) or region with low electron density (δ +) Nucleophile
Definition
Examples
Electrophile
Nucleo = nucleus ; phile = love Nucleophile mean love nucleus. In terms of Lewis acid-base theory, nucleophiles are often Lewis base, which donate lone pair electron.
Electro = electron ; phile = love Electrophile mean love electron. In terms of lewis acid-base theory, electrophiles are often Lewis acid, which accept lone pair electron.
:OH–
RO:–
:Cl–
H+ (H3O+)
NO2+
Cl2 ; Cl+
:Br–
:I–
CN:–
Br2 ; Br+
I2 ; I+
carbonium
:C– carbanion
R–OH
–C=C–
RN2+
R3C+
BF3
:NH2–
R–NH2
H2O
AlCl3
FeBr3
ZnCl2
C⊕
1.7 Inductive Effect and Mesomeric Effect Inductive effect is defines as the shift in electron density from one atom to another to form a polar bond. For example, in the bonding of –C–Cl. In the presentation of inductive effect, the arrow shows that carbon atom repel electrons δ+
δ–
Chlorine atom attract the electrons as it has a higher elctronegativity 1.7.1 Electron–withdrawing group Most atoms and groups of atoms are more electronegative than carbon atom, thus withdraw electron from carbon. When this occurs, atoms/molecules are said to exert a negative inductive effect (–I effect). Example of electron withdrawing group
–Cl ; –Br ;–I
–NO2
–CN
–COOH
–COOR
–C=O
–SO3H
–C6H5
1.7.2 Electron donating group Some atoms/groups are less electronegative than carbon thus donate electrons to carbon atom. These atoms/groups are also known as electron donating group and they exert a positive inductive effect (+I effect). Alkyl group are known to have a +I effect. The effect increase with the increase of alkyl group attached to the C atom. The inductive effect produces a permanent dipole and this will influence its physical properties such as boiling / melting point, acid and basic strength, reaction of functional group and rate of reaction. Inductive effect may also occur when there is a permanent shift of electron due to the polarisation of sigma bond (σ-bond). The shift of electron can also occur in a pi (π) bond. The shift of π electron in multiple bonds toward a more electronegative atom is known as resonance effect (mesomeric [M] effect)
Example in methanal, oxygen atom is more electronegative than carbon atom. Consequently, the π electron shifted to oxygen atom. The structure can be represented as a resonance hybrid between structure I and II.
Same phenomenon can occur in carboxylate anion. M-effect play an important role in the structure and stability of many compounds. The π electrons are spread and delocalised over the carbon and oxygen atoms. We can represent the delocalisation of electrons in carboxylate ion as follow
1.8
Type of chemical reaction in Organic Chemistry
Type of Organic Reaction Substitution
Addition
Elimination
Definition : one or more atom / group is substitute by another atom / group of atoms
Definition : a small Definition : 2 reactants atom / group is react together and removed (eliminated) form 1 product from the molecule involve
Example : CH3CH2Cl + OH– CH3CH2OH + Cl–
Example : CH2=CH2 + Br2 CH2(Br) CH2(Br)
Example CH3CH2OH CH2=CH2 + H2O
1.9 Bond Fission ~ Ways of Breaking Chemical Bond In a covalent bond, a pair electron is shared between 2 atoms. When the covalent breaks, these 2 electrons are redistribute between 2 atoms. There are 2 ways how these electrons are redistribute. Bond fission Homolytic fission • The 2 shared electrons distribute evenly between the 2 atoms • Free radicals are formed after homolytic fission and it is usually represented by •X • Since it is a single electron, free radicals tend to become unstable, thus are reactive. • Example : Cl – Cl 2 Cl• (chlorine radical) H3C–H CH3• + H• methyl radical hydrogen radical
Heterolytic fission o Shared electron in the covalent bond only goes to one of the atom. The pair of electron usually goes to the atom with higher electronegative o Eventually, one will become negatively charged (as a result of receiving extra electron) while the other become positively charged (lose electron to the other party) o Example : Cl – Cl :Cl– + Cl⊕ (CH3)3C–Br :Br– + (CH3)3C⊕
Practice 1.Base on the molecular structure, draw the hybridisation diagram of the following molecules, state the type of hybridisation and its shape, and state the angle of each hybridised structure a) CH3CH=CHOH
H
H
H
σ
σ σ
σ
σ
π σ σ
H
σ
H
H
b) CHΞCCH2NH2 H
σ π π H
σ
σ
σ
σ
σ σ H
σ H
H
Displayed formula
Short hand / notation
CH(CH3)2C(CH3)2CH2CH2CH3
C(CH3)3CH(CH2CH3)2
CH(CH3)2CH(CH3)CH=C(CH3)2
CH(CH3)2CH(OH)CH(CH2CH3)2
Skeletal formula
CH2(NH2)CH(C6H5)CONH2
COHCH(Br)C(CH3)2CH2COOH
COHC(CH3)2CH(OH)CΞCCH3 COOHCH(CH3)CH2C(CH3)(CH2CH3)CH2CH(OH)(CH3)
Name the functioning group presence Alkene, alcohol, aldehyde
CH2ClCOCH2CHNH2CH2CH2CONH2
Haloalkane, Ketone ; Amine , Amide Aldehyde ; Alcohol ; Alkyne Carboxylic acid
Molecules
Type of isomerism
geometrical
geometrical
Optical
Diagram of isomers
CH2ClCH=C(CH3)CH2COOH
geometrical geometrical
optical
optical
Particles
Particles
Particles
Br2
electrophile
Br+
electrophile
Br-
nucleophile
NH3
nucleophile
CH3NH2
nucleophile
NH2+
electrophile
CH3-
nucleophile
CH3COH
electrophile
CH3+
electrophile
nucleophile
nucleophile
electrophile
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