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

PTT 102: Organic Chemistry

Sem I 2011/2012Khadijah Hanim Abdul Rahman

School of Bioprocess EngUniMAP

Week 1: 22/9/2011

Page 2: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 3: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 4: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Alkanes:hydrocarbons containing only single bonds

Straight-chain alkanes: the carbons are in a continuous chain with no branches.

General formula: CnH2n+2

Page 5: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Nomenclature of Alkyl Substituents

Removing a hydrogen from an alkane results in an alkyl substituent

Page 6: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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”.

Page 7: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.
Page 8: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

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

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

Page 10: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 11: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 12: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 13: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 14: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 15: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Some constituents have only a systematic name.

Page 16: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Examples

Draw a structure for:- 2,3-dimethylhexane

Give systematic name:

- Answer: 3-ethyl-6-methyl-5-propylnonane

Page 17: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Provide an acceptable name for the alkane shown below.

 Answer: 2, 5-dimethylheptane

Page 18: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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: CnH2n

Cycloalkanes are named by adding prefix ‘cyclo’ to the alkane name.

Page 19: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 20: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

2. Name the two substituents in alphabetical order

- The no 1 position is given to the substituent listed 1st.

Page 21: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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 2nd substituent getting as low no as possible.

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

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:

Page 23: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 24: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Attractive forces between alkanes- relatively weak- van der waals forces.

van der Waals force

Dipole–dipole interaction

Hydrogen bonds

Page 25: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 26: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 27: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

•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.

Page 28: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 29: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Nonpolar compounds- no net charge.Dissolves in nonpolar solvents- van der

waals interactions between solvent molecules and solutes molecules

Page 30: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Exercises

41) Which of the following has the lowest boiling point?A) CH3CH2CH2CH2CH2CH3B)

C)

D)

E) CH3CH2CH2CH2CH2CH2CH3

Page 31: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 32: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 33: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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

Page 34: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

Different Conformations of Ethane

Figure shows the potential energies of all conformers of ethane obtained in one complete 3600 rotation about the C-C bond. Noticed that staggered conformers are at energy minima, whereas eclipsed conformers are at energy maxima.

Page 35: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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:

Page 36: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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:

Page 37: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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)

Conformers CH3 substituents Energy

Anti conformer (D) Opposite of each other

Lower energy

Gauche conformer (B&F)

Adjacent to each other

Higher energy

Page 39: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

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.

Page 40: PTT 102: Organic Chemistry Sem I 2011/2012 Khadijah Hanim Abdul Rahman School of Bioprocess Eng UniMAP Week 1: 22/9/2011.

•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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