H2 CHEMISTRY 9647

TUTORIAL 10: INTRODUCTION TO ORGANIC CHEMISTRYFORMATIVE ASSESSMENT:

Each carbon atom can form 4 covalent bonds.

Carbon needs to form equivalent orbitals of same shapes and energies by a process known as

hybridisation before bonding can occur.

Classification:

Aliphatic molecules contain open (straight or branched) chains of carbon skeletons.

Alicyclic molecules consist of closed rings of carbon atoms, which may contain single or multiple

bonds.

Aromatic molecules contain at least 1 benzene ring (ring of 6 carbon atoms where electrons in

p-orbitals are delocalized).

Functional Groups & Homologous Series:

Functional Group consists of an atom or groups of atoms or bond common to a series or family of compounds and it governs the principal chemical properties of the series.

A series of compounds containing the same functional group is called a homologous series.

Formulae:

Empirical formula – gives the simplest ratio of the atoms of each element present.

Molecular formula – gives the actual number of atoms of each element present.

Condensed structural formula – shows how the constituent atoms are joined together.

Full-structural or Displayed formula – shows both the relative placing of atoms and the number of bonds between them.

Nomenclature:

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(position — prefix substituent) Parent

Indicates the position of the substituent on the carbon chain.Number should be as small as possible.

Indicates the number of the same substituent.

Arrange in alphabetical order

Longest, continuous carbon chain (functional group)

Reactions of Organic Compounds:

(I) Types of reagent:

(a) Nucleophiles are negative ions (e.g. OH-, CN-) and compounds in which an atom has a

lone pair of electrons (e.g. NH3).

(b) Electrophiles are reagents (e.g. NO2+) which attack a region where the electron density is

high.

(II) Types of reactions:

(a) Substitution: involves replacing an atom (or a group of atoms) by another atom

(or groups of atoms).

(b) Addition : involves two molecules joining to form a single new molecule.

(c) Elimination: involves the removal of a molecule from a larger molecule.

(d) Hydrolysis: involves breaking covalent bonds by reaction with water.

(e) Oxidation / Reduction

(III) Types of bond fission:

(a) Homolytic fission: when the bond breaks, each of the bonded atoms takes one

electron. Free radicals, which are atoms or groups of atoms with unpaired electrons, are

formed. They are electrically neutral but highly reactive.

(b) Heterolytic fission: When the bond breaks, one of the atoms take both of the bonding

electrons to form an anion. The rest of the molecule becomes a cation.

Isomerism:

Isomers are compounds with the same molecular formula but different arrangement of the atoms in the molecule.

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ISOMERISM

Structuralsame molecular formula

different structural formulai.e. different arrangement of atoms

Stereoisomerismsame molecular formulasame structural formula but different spatial arrangement of atom

Chain Positional FunctionalGroup Geometric

(cis-trans)Optical

isomerism

DISCUSSION QUESTIONS:

1 Give the IUPAC name for each of the following structures.

(a) (b) (c)

(d) (e)

(f) (CH3)2CHCCl=CH2 (g) CH3C(CH3)2CHClCH3

Q1a 2,3-dimethylbutane Q1e 3,3-dimethylbut-1-ene

Functional group: alkeneQ1b 3-methylbut-1-ene

Numbering starts from the carbon with the double bond.

Q1f 2-chloro-3-methylbut-1-ene

Functional group: alkeneQ1c 2-methylpropanoic acid

Functional group: carboxylic acid

Q1g 3-chloro-2,2-dimethylbutane

2-chloro-3, 3-dimethylbutane does not give the smallest overall number assigned to the compound.

Q1d 2-bromo-3-chloro-2,3-dimethylbutane

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CH3CH CHCH3

CH3 CH3

CH3CHCH=CH2

CH3

CH3CHCO2H

CH3

CH3C CCH3

CH3 CH3

ClBr

C

CH=CH2

CH3

CH3CH3

H

* Substitutents arranged in alphabetical order

2. For each of the following compounds, i. Give the IUPAC name of the compound

ii. Predict the hybridization and geometry of the carbon atom (marked with an asterix *)iii. Indicate whether the bonds involved are σ or π bonds

(a) CH3CH2C*Br2CH3 (b) CH3CH2CH=C*HCH3

(c)

ai) 2, 2-dibromobutaneii) carbon is sp3 hybridized (bonded to 4 atoms) tetrahedraliii) All bonds involved are σ bonds.

bi) pent-2-eneii) Carbon is sp2 hyridized (bonded to 3 atoms) trigonal planariii) double bond consists of a σ and π bond.

c) i) 3-chloromethylbenzene

(Note: Numbering of carbon atoms in ring can be clockwise or anticlockwise. Assign lowest number to substituents by alphabetical order)ii) Carbon on methyl group is sp3 hybridized (bonded to 4 atoms) tetrahedral Carbon in benzene ring is sp2 hyridized (bonded to 3 atoms) trigonal planar

iii) Carbon in methyl group only has σ bonds.Carbon in benzene ring has 3 σ bonds and 1 π electron in a delocalized π cloud.

3 Write the structural formula for each of the following compounds.

(a) 1-chloropropan-2-ol (b) 1-bromo-2-methylbutane

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*

*

Note: There are two asterixes

(c) 3-chloro-3-ethylpentane (d) 2-methylbut-2-ene

a) 1-chloropropan-2-ol

* To show all bonds and arrangement of atoms in compounds.

CH2ClCH(OH)CH3

b) 1-bromo-2-methylbutane

CH2BrCH(CH3)CH2CH3

c) 3-chloro-3-ethylpentane

CH3CH2CCl(C2H5)CH2CH3

d) 2-methylbut-2-ene

CH3CHC(CH3)2

4. (a) What do you understand by the term structural isomerism?

(b) Draw and name the full structural formulae for all possible isomers of

(i) C3H7Cl

(ii) C3H6Br2

(iii) C4H9Br

(iv) C3H8O

(v) C8H10 (Hint: this structure contains a benzene ring)

a) Structural isomerism arises due to different arrangement of atoms in organic compounds (different structural formulae) but they have the same molecular formula.

b) i) C3H7Cl

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CH3CH2CH2Cl CH3CHCl CH3

Positional isomers

1-chloropropane 2-chloropropane

(ii) C3H6Br2

CH3CH2CHBr2 CH3CBr2CH3 CH2BrCHBrCH3 CH2BrCH2CH2Br

Positional isomers

1,1-dibromopropane

2,2-dibromopropane

1,2-dibromopropane

1,3-dibromopropane

No chain isomers or functional group isomers possible for 3bi and 3bii

(iii) C4H9Br

CH3CH2CH2CH2Br CH3CHBrCH2CH3

Positional isomers

1-bromobutane 2-bromobutaneCH2BrCH(CH3)2 CH3CBr(CH3)2

Chain isomers

1-bromo-2-methylpropane 2-bromo-2-methylpropane

(iv) C3H8O

CH3CH2CH2OH (alcohol) CH3OCH2CH3 (ether)

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Functionalgroup

isomers

Propan-1-ol MethoxyethaneCH3CH(OH)CH3

Positional isomers

Propan-2-ol

(v) C8H10

Chain isomers

1,2-dimethylbenzene ethylbenzene

Positionalisomers

1,3-dimethylbenzene

1,4-dimethylbenzene

5. (a) What do you understand by the term stereoisomerism?

(b) Name the two types of stereoisomerism and explain briefly how they arise.

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(c) Which of the following structures will have cis-trans isomers?

(i) CH2=C(CH3)2 (ii) CH3BrC=CCl2

(iii) CH3CH=CHCH3 (iv) C6H5CH=CHCO2H

(v) CH2=CHCH2CH=CH2

(d) Mark the chiral carbons (if any) with an asterisk (*).

(i) (ii) (iii)

(iv) (CH3)2CHCHBrCH3 (v) H2NCH2CO2H (vi) CH3CHBrCHBrCH2CH3

(e) Draw the optical isomers for the organic compounds in parts d (i) and d (iv).

a) Stereoisomerism arises due to different arrangement of atoms in space in organic compounds with the same molecular and structural formula.

b) Geometric isomerism – arises to the restriction of rotation about the C=C double bond & when 2 different substituents are attached to

each C across the double bond. Optical isomerism – arises when compounds do not exhibit a plane of symmetry and forms mirror images which are non-superimposable.

c) (i) Does not display cis-trans as 2 similar groups (CH3) are attached to the same

carbon atom.

(ii) Does not display cis-trans as 2 similar groups (Cl) are attached to the same carbon atom.

(iii)

cis-but-2-ene trans-but-2-ene

(iv)

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C

CO2H

H

NH2CH3

C

CH3

N OHH

C

OH

CH3

OH

OH

cis-isomer trans-isomer

(v) Does not display cis-trans as 2 similar groups (H) are attached to the same carbon atom.

d)(i)

Mirror

(ii) No chiral carbon

(iii)

Mirror

(iv)

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*

*

CO2H

CO2H

Mirror(v) No chiral carbon

(vi) 2 chiral carbons

6. Give the structural formulae for each of the following compounds (including geometrical isomers, if any).

(a) hex-1-ene (b) hex-2-ene

(c) 2-methylbut-1-ene (d) 1-chlorobut-2-ene

(a) No geometric isomers1

(b)

cis-hex-2-ene trans-hex-2-ene

1 To express C2H5 as CH2CH3, C3H7 as CH2CH2CH3 and C4H9 as CH2CH2CH2CH3 when representing solutions to students.

*

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(CH3)2CHCHBrCH3

CH3CHBrCHBrCH2CH3*

CH3CHBrCHBrCH2CH3*

(c) No geometric isomers

(d)

cis-1-chlorobut-2-ene trans-1-chlorobut-2-ene

7. A monocarboxylic acid P occurs in goat’s milk. P has the following composition by mass: C, 62.1%; H, 10.3%; O, 27.6%. Neutralisation of a sample of 0.10 g of P requires 8.6 cm 3 of 0.10 mol dm-3 sodium hydroxide.

Calculate the empirical and the molecular formulae of P. (Ans: C3H6O; C6H12O2)

C H O% by mass 62.1 10.3 27.6

Ar 12.0 1.0 16.0No of moles 5.18 10.3 1.73

Simplest mole ratio 3 6 1

Empirical formula: C3H6O

Let molecular formula be (C3H6O)n

Mr of carboxylic acid = (36.0+6.0+16.0)n = 58.0n

No. of moles of NaOH

Since the acid is a monocarboxylic acid, NaOH acid No. of moles of acid = 8.60 x 10-4

Molecular formula of acid = C6H12O2

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ASSIGNMENT QUESTIONS

8. (a) A carboxylic acid of the formula C3H5O2Br is optically active. What is its structure?

(b) Give the structure of the lowest-molecular-mass alcohol (containing only C, H and O) that is chiral.

(c) A solution of 2-hydroxypropanoic acid isolated from a natural source rotated polarised light to the right, whereas a solution of 2-hydroxypropanoic acid synthesised from ethanol did not rotate polarised light. Why is this so?

a) A carboxylic acid with formula C3H5O2Br has two possible structural formulae as shown below:

3-bromopropanoic acid 2-bromopropanoic acid However only 2-bromopropanoic acid is optically active

(b)*

* Note: Ans to 8(b) should be

O─H │ H─C≡C─C─CH3

│ H

Lowest molecular mass alcohol (containing only C, H and O) that is chiral.(c) A racemic mixture containing an equal amount of both 2-hydroxypropanoic acid

enantiomers is formed when the acid is synthesised from ethanol. When 2-hydroxypropanoic acid is isolated from the natural source, only one of the pair of enantiomers is obtained, thus the acid is able to rotate polarised light.

9. A hydrocarbon (containing 14.3% by mass of hydrogen) is a gas of density 2.50g dm-3 at s.t.p.

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*

(a) Calculate its empirical formula.

(b) Calculate its molecular formula.

(c) Draw three possible structural formulae for the compound.

(a)C H

% by mass 85.7 14.3Ar 12.0 1.0

No of moles 7.14 14.3Simplest mole ratio 1 2

Empirical formula: CH2

(b) Molar gas volume at s.t.p = 22400 cm3

At s.t.p, 1 mole of the hydrocarbon has a volume of 22.4 dm3

Mass of 1 mole of hydrocarbon =

Let molecular formula of hydrocarbon be (CH2)n

Mr of hydrocarbon = 14.0n

Molecular formula of hydrocarbon = C4H8

(c)

CH2CH2CH2CH2 CH3CH2CHCH2 CH3CHCHCH3 CH2C(CH3)2

Structural isomers

cyclobutane But-1-ene But-2-ene 2-methylpropene

ADDITIONAL QUESTIONS (ON ISOMERISM)

1) Nov 2004 P3 Q6 (a) 2) Nov 2002 P2 Q5 (a)3) Nov 2000 P2 Q4 (a) 4) Jun 2000 P1 Q6 (a)

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