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PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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Presentation on theme: "PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011."— Presentation transcript:

1 PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011

2 Learning Outcomes Able to explain and differentiate the chemical and physical properties of each functional groups carry out theoretical reaction mechanism at molecular level Able to explain and differentiate the chemical, physical properties and reactions of alcohol, ether, aldehyde, ketone and carboxylic acids. Able to apply the knowledge of organic chemical process in biotechnology industry

3 CHAPTER 1 ALKANES - Structure, nomenclature and physical properties of alkanes and cycloalkanes: Define and identify the principle in naming and physical properties of alkanes. - Conformational analysis: Discuss and interpret the conformational analysis or rotation about carbon-carbon single bonds in molecules such as ethane and butane.

4 Alkanes:hydrocarbons containing only single bonds Straight-chain alkanes: the carbons are in a continuous chain with no branches. General formula: C n H 2n+2

5 Nomenclature of Alkyl Substituents Removing a hydrogen from an alkane results in an alkyl substituent

6 Nomenclature of Alkanes The systematic name of an alkane is obtained using the following rules: 1. Determine the number of carbons in the longest continuous chain. - This chain is called the parent hydrocarbon. - The no of carbons in the parent hydrocarbon becomes the alkane’s “last name”.


8 2. Number the chain so that the substituent gets the lowest number - Name of any alkyl substituent that hangs off the parent hydrocarbon is placed in front + the no to designate the carbon to which the alkyl substituent is attached.

9 3. Number the substituents to yield the lowest possible number in the number of the compound. - If more than 1 substituent is attached to parent hydrocarbon, the chain is numbered in direction that will produce the lowest possible numbers. - Substituents listed in alphabetical order

10 If 2 or more substituents are same: prefixes di, tri and tetra are used- indicating the no. of substituents of the compound. Note: Prefixes are ignored in alphabetizing substituent groups except for iso and cyclo- not ignored

11 4. Assign the lowest possible numbers to all of the substituents

12 5. If the same substituent numbers are obtained in both directions, the first group cited receives the lower number

13 6. If a compound has two or more chains of the same length, the parent hydrocarbon is the chain with the greatest number of substituents

14 7. Certain common nomenclatures are used in the IUPAC system- systematic substituent names are preferred. - Systematic substituent names obtained by numbering the alkyl substituent starting at the carbon attached to the parent hydrocarbon. Carbon attached to parent hydrocarbon is always no 1 carbon. Note: if prefix such as di is part of a branch name, it is included in the alphabetization

15 Some constituents have only a systematic name.

16 Examples Draw a structure for: - 2,3-dimethylhexane Give systematic name: - Answer: 3-ethyl-6-methyl-5- propylnonane

17 Provide an acceptable name for the alkane shown below. Answer: 2, 5-dimethylheptane

18 Nomenclature of Cycloalkanes- skeletal structures Cycloalkanes: alkanes with their carbon atoms arranged in a ring Has 2 fewer hydrogens than acyclic alkane General molecular formula: C n H 2n Cycloalkanes are named by adding prefix ‘cyclo’ to the alkane name.

19 Rules for naming cycloalkanes Resemble the rules for naming acyclic alkanes. 1. No number is needed for a single substituent on a ring - The ring is parent hydrocarbon unless the substituent has more carbon than the ring

20 2. Name the two substituents in alphabetical order - The no 1 position is given to the substituent listed 1 st.

21 3. If there are more than two substituents, they are cited in alphabetical order - Substituent given the no 1 position is the one results in a 2 nd substituent getting as low no as possible.

22 Exercises: Provide the systematic name of the compound shown. 4-butyl-1-ethyl-2- methylcycloheptane Give the systematic name: Answer: 4-butyl-1,2- dimethylcyclohexane Answer:

23 The physical properties of alkanes Boiling points: - Boiling point (bp) of a compound is the temperature at which the liquid form becomes a gas (vaporizes). - To vaporize- must overcome the forces that hold the individual molecules close to each other in the liquid - If the molecules are held by strong forces- a lot of energy will be needed to pull the molecules away from each other- high boiling point. - In contrast for molecules that being held by weak forces- low boiling point.

24 Attractive forces between alkanes- relatively weak- van der waals forces. van der Waals force Dipole–dipole interaction Hydrogen bonds

25 van der Waals Forces The boiling point of a compound increases with the increase in van der Waals force magnitude of the van der waals forces that hold alkane together depends on- area of contact between molecules. the greater the area of contact- the stronger the van der waals forces- more energy to overcome.

26 Melting points Melting point (mp) of a compound is the temperature at which its solid form is converted into liquid. mp of alkanes- as molecular weight increase, mp increases. In addition to intermolecular interactions mentioned, mp is influenced by the type of packing- arrangement of the molecules.

27 Alkanes with odd no of carbons- pack less tightly The end molecules are facing and repelling the methyl group on the end of the other- increasing the average distance between the chains- weaker interactions.

28 Solubility Polar compounds dissolve in polar solvents and vice versa. “like dissolves like”. Reason ‘polar dissolves polar’- polar solvent, ie water has partial charges that can interact with partial charges on polar compound. Clustering of the solvent molecules around the polar molecules separates them from each other- dissolve The interaction between solvent and solute molecules- solvation

29 Nonpolar compounds- no net charge. Dissolves in nonpolar solvents- van der waals interactions between solvent molecules and solutes molecules

30 Exercises 41) Which of the following has the lowest boiling point? A) CH3CH2CH2CH2CH2CH3 B) C) D) E) CH3CH2CH2CH2CH2CH2CH3

31 Explain why has a lower boiling point than CH3CH2CH2CH3. Answer: CH3CH2CH2CH3 has greater van der Waals forces because it has a greater contact area than isobutane. Therefore, the boiling point of CH3CH2CH2CH3 is higher.

32 Conformations of Alkanes: Rotation about Carbon–Carbon Bonds Conformations: different spatial arrangements of the atoms that result from rotation about a single bond. Conformational isomers/conformers- molecules with different conformations. The conformers produced by rotation about carbon-carbon bond ethane: staggered conformer & eclipsed conformer.

33 Staggered conformer- more stable and lower in energy than eclipse conformer. Lower energy- low intramolecular repulsions between hydrogens due to large distances.

34 Different Conformations of Ethane Figure shows the potential energies of all conformers of ethane obtained in one complete 360 0 rotation about the C-C bond. Noticed that staggered conformers are at energy minima, whereas eclipsed conformers are at energy maxima.

35 Conformations of n-butane Butane has 3 C-C single bonds Rotation can occur about each of them. The Newman projections below show the staggered and eclipsed conformers that result from rotation about the C1- C2 bond:

36 The staggered conformers resulting from rotation about the C1-C2 bond in butane- have the same energy The staggered conformers resulting from rotation about the C2-C3 bond- do not have the same energy. The staggered and eclipsed conformers from rotation about the C2-C3 bond in butane are:

37 Of the staggered conformers- D is more stable than B and F- the 2 methyl groups are far apart from each other. Most stable conformers- anti conformer (anti= opposite) The other 2 conformers, B and F- gauche conformers (gauche=left) ConformersCH 3 substituentsEnergy Anti conformer (D) Opposite of each other Lower energy Gauche conformer (B&F) Adjacent to each other Higher energy


39 Anti and gauche conformers do not have the same energy  Steric strain Steric strain: strain experienced by a molecule when atoms or groups are close enough to one another for their electron clouds to repel each other. Thus possesses additional energy.

40 Eclipsed conformers resulting from rotation about the C2-C3 bond in butane- different energies. The eclipsed conformer A is less stable than C and E.

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