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Determination of melting point of organic compounds

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melting point of organic compounds and a brief explaination of theeffect of intramolecular forces on it
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Determination of melting point of organic compounds Dotsha J. Raheem College of Science- university of salahaddin Department of Chemistry
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Page 1: Determination of melting point of organic compounds

Determination of melting point of organic compounds

Dotsha J. RaheemCollege of Science- university of salahaddin

Department of Chemistry

Page 2: Determination of melting point of organic compounds

Melting Point

1. Is the temperature at which a solid is converted to liquid at normal atmospheric pressure.

2. A more specific definition of m.p. (or f.p.):

Is the temperature at which the solid and the liquid phases are at equilibrium at a specific pressure (normally taken atmospheric unless stated otherwise)

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Page 3: Determination of melting point of organic compounds

3. Is the temperature at which the forces which unite the crystals of a solid are ruptured. This results in a change from the crystalline to the amorphous state. This change involves the absorption of a characteristic amount of heat called the heat of melting or fusion.

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Page 4: Determination of melting point of organic compounds

Importance of m.p.

1. It is a physical property used for identification.

2. It is an important indicator for purity.

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Page 5: Determination of melting point of organic compounds

Factors affecting m.p.

M.wt. or size of the molecule

Branching

Intermolecular forces

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Page 6: Determination of melting point of organic compounds

CompoundMolecular

formula Molecular

weightMelting

point

methane CH4 16 -182

ethane C2H6 30 -183

propane C3H8 44 -188

butane C4H10 58 -138

pentane C5H12 72 -130

M.wt. or size of the molecule

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Page 7: Determination of melting point of organic compounds

Branching

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Page 8: Determination of melting point of organic compounds

1. Ionic, Dipole - Dipole attractions

Intermolecular forces

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Page 9: Determination of melting point of organic compounds

2. Hydrogen Bonding

• a special kind of dipole-dipole

force that occurs when a H atom

is bonded to one of the very

electronegative atoms, F, O, or N.

• Electronegative atom must have

at least a lone pair of electrons

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Page 10: Determination of melting point of organic compounds

- 82 oC0 oC

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Page 11: Determination of melting point of organic compounds

The H-F, H-O, and H-N bonds are very polar, because the electronegative atom draws the bonding electron pair strongly to itself. This leaves the hydrogen nucleus exposed

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Page 12: Determination of melting point of organic compounds

Intermolecular forces

Non-polar molecules do not have dipoles like polar molecules, but they can solidify. This is due mainly to the very important van derwaals forces that exist in these molecules

3. Van der Waals forces

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Page 13: Determination of melting point of organic compounds

dispersion forces < dipole-dipole interactions < hydrogen bonds 13

Page 14: Determination of melting point of organic compounds

Types of m.p. and effect of impurities

In the presence of impurities

m.p. will be lower

needs less energy to destroy the

crystal lattice

Melting range will be broader

Due to the presence of different

types of interactions

Pure compound,melting range 1-2 oC

impure compound,melting range >2 oC 14

Page 15: Determination of melting point of organic compounds

Ionic vs. covalent compounds

NaCl m.p. 801 oC

In ionic compounds the structural units are ions, and the strong intermolecular interactions are overcome only at very high temperatures

Na+

Na+

Na+

Na+ Na+

Na+

Na+

Na+ Na+

Na+

Na+

Na+

Na+Na+

Cl-

Cl-Cl- Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl-

Cl- Cl-

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Page 16: Determination of melting point of organic compounds

Methane m.p. -182 oC

In covalent compounds, the intermolecular forces are very week if compared to the forces in ionic compounds (and m.p. represents only the amount if heat needed to overcome the intermolecular interactions but not the covalent bonds)

CH4CH4

CH4 CH4

CH4

CH4

CH4 CH4

CH4

CH4

CH4

CH4CH4CH4CH4

CH4 CH4CH4

CH4 CH4 CH4CH4

CH4CH4

CH4

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