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Summary Ieb Intermolecular Forces With Organic Chemistry

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Intermolecular Forces in Organic Chemistry Notes

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  • June 15, 2022
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  • 2021/2022
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Intermolecular Forces
Affect the physical properties (melting point and boiling point) of organic compounds

Alkanes and Alkenes
CH4 – methane C8H18 – octane C23H48 – candle wax
∆𝐸𝑛 = 2.5 − 2.1 = 0.4
Polar bonds, but due to shape, non-polar molecules → London Forces
Bigger molecules have more electrons and this means stronger London forces, because temporary dipoles are easier
to set up and hence more energy required to break the IMFs during melting
London forces increases with
Intermolecular Bonds and Intermolecular Forces an increase in molecular
surface area
Bonds (Interatomic forces (more places for temporary
dipoles to be set up)

or

number of electrons
Coulomb Forces
(temporary dipoles are easier
Electrostatic Forces
to set up)




Covalent Bonding
Metallic Bonding
Nucleus---shared e- pairs---nucleus
Positive atomic residues---mobile electrons


Ionic Bonding

Positive ions---negative ions

Intermolecular forces only



Van Der Waals Forces

weak
non-polar ---- non-polar
LONDON FORCES (Dispersion) strong strong
polar ---- polar dipole ---- dipole
H2, Cl2, CO2, CH4 KEESOM FORCES (Dipole-dipole) HYDROGEN BONDING FORCES

1. Alkanes HCl and CO NH3, H2O, HF
2. Alkenes
3. Alkyl Halides 1. Alkyl halides 1. Alcohols
4. Alcohols (except for C-I) 2. Carboxylic Acids
5. Carboxylic acids 2. Esters
Molecules containing the
6. Esters
-OH group

, Explain how London forces Originate:

1. As a result of the random movement of electrons, a temporary (instantaneous) dipole is set up in one
molecule
2. Dispersing the electrons in the other (neighbouring) molecules (i.e. polarizing the molecules around it)
3. Resulting in an induced dipole forming (the induced dipole is also temporary)
4. The two temporary dipoles are then able to attract each other

Alkyl Halides → London and Keesom Forces




Cl, I, Br, F




∆𝐸𝑁(𝐶 − 𝐹) = 4 − 2.5 = 1.5 → Keesom

∆𝐸𝑁(𝐶 − 𝐶𝑙) = 3 − 2.5 = 0.5 → Keesom

∆𝐸𝑁(𝐶 − 𝐵𝑟) = 2.8 − 2.5 = 0.3 → Keesom

∆𝐸𝑁(𝐶 − 𝐼) = 2.5 − 2.5 = 0 → London

Esters → London and Keesom Forces

not very polar

∆𝐸𝑁(𝐶 − 𝑂) = 3.5 − 2.5 = 1



Hydrogen bonding forces exist between polar molecules in which hydrogen is bonded to:

• a small atom
• of high electronegativity
• with at least one lone pair of electrons
• N, O, F

Alcohols → Hydrogen Bonding Forces




Carboxylic Acids → Hydrogen Bonding Forces (2 sites per molecule)

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