A to Z Directory – Virginia Commonwealth University
Chapter 2
Types of bonds
• Sigma – the overlap is directly between the two nuclei
o Every bond contains one (and only one) sigma bond
• Pi – the overlap is above and below the sigma bond
o These are formed by the overlap of unhybridized p orbitals
o There is one sigma bond and one pi bond in a double bond.
o There is one sigma bond and two pi bonds in a triple bond.
• How many sigma and pi bonds in the following molecule?
[pic]
o You need to draw out all the bonds (including hydrogens which were left off) in order to answer this.
[pic]
o There are 14 sigma bonds and 3 pi bonds.
Hybridization
▪ When orbitals hybridize, the s and some or all of the p orbitals of the outer shell combine to form hybrid orbitals
o The p orbitals that don’t get involved remain p orbitals
[pic]
[pic]
[pic]
▪ What’s the hybridization?
o Count the charge clouds (sigma bonds and lone pairs)
▪ Call it #
o Hybridization is sp#-1
▪ So if there are three charge clouds, then the hybridization is sp3-1=sp2
▪ What’s the electronic geometry?
o sp → linear
[pic]
o sp2→ trigonal planar
[pic]
o sp3 → tetrahedral
[pic]
▪ What’s the bond angle?
o sp→ 180°
o sp2 → 120°
o sp3 → 109.5°
▪ What’s the molecular geometry?
o If all the charge clouds are bonds, then same as electronic geometry
o If 2 bonds and one lone pair, then bent
▪ [pic]
o If 3 bonds and one lone pair, then trigonal pyramidal
[pic]
o If 2 bonds and 2 lone pairs, then bent
[pic]
Big picture!
▪ When you identify the hybridization of an atom in a molecule you are identifying what type of orbitals that atom has.
▪ Bonds form from the overlap of these hybrid orbitals.
▪ A frequently missed type of question in this class is something like “the carbon-oxygen sigma bond of acetone is formed from the overlap of what orbitals?”
o Identify the hybridization of both atoms involved.
[pic]
o Look to see if you are asked for the sigma or pi bond.
▪ If you were asked for the pi bond, then it’s always the overlap of two p orbitals.
▪ If you were asked for the sigma bond, then it’s just the two types of hybrid orbitals of each atom.
• In this case because both the oxygen and carbon are sp2 hybridized the sigma bond is formed from the overlap of two sp2 orbitals.
Bond rotation
▪ Single bonds rotate
[pic]
▪ Double and triple bonds don’t
[pic]
o You would have to break the pi bonds in order to rotate around that bond.
▪ It’s really that simple!
o If this isn’t making sense, play with your models.
Drawing in 3D
• Anything coming out of the page should be drawn on a wedge and anything drawn going into the page should be drawn on a dash.
• If an atom is sp3 hybridized, make sure you draw the dashed and wedged pieces outside the angle formed by the two flat pieces.
[pic]
o This doesn’t make that much of a difference now, but from quiz 2 and on, you will miss points (a lot of them) if you do this incorrectly.
Isomerism
▪ Structural isomers and constitutional isomers are the same thing
o They have the same molecular formula, but the atoms are connected in a different order.
o If you can’t tell whether two molecules are identical or isomers, see if you would name them differently
▪ Ex. 2,4-dimethylhexane vs 2,3-dimethylhexane
▪ Stereoisomers
o The atoms are connected to each other in the same order, but they differ in their arrangement in space
▪ Ex. R/S, E/Z, cis/trans
▪ At this point in the course, you are only responsible for cis/trans isomerism
• There are two types of cis/trans isomers
• cis/trans double bonds
o cis double bonds have both “pieces” on the same side of the double bond
[pic]
o trans double bonds have both “pieces” on opposite sides of the double bond
[pic]
o When do I have cis/trans isomerism possible?
[pic]
▪ You only have cis/trans isomerism possible when A≠B and C≠D
[pic]
▪ Be careful to make sure the double bond is in the same position when you are comparing two similar structures.
▪ The two compounds on the right above are stereoisomers of each other and the compound on the left is a structural isomer of the right two.
• cis/trans on rings
o When both substituents on the ring are facing the same way (both on wedges or both on dashes), then you have a cis isomer.
[pic]
o When the two substituents on the ring or facing different ways, then you have a trans isomer.
[pic]
o Be careful when comparing two compounds to make sure that the two substituents are still on the same carbons of the ring.
[pic]
▪ These two are structural isomers, not stereoisomers because the chlorines are 1,3 to each other on the first ring and 1,2 to each other on the second ring.
Intermolecular Forces
• Boiling point is a good measure of intermolecular forces.
[pic]
• Solubility is also an expression of intermolecular forces.
o Like dissolves like.
▪ Like in polarity.
o What will be soluble in water?
▪ Salts
▪ Polar organic molecules where the polar part is not overcome by huge nonpolar R-groups.
▪ Ex. As the carbon chains of alcohols get longer, they become less soluble in water and more soluble in nonpolar solvents.
[pic]
▪ With larger carbon pieces, the more branched isomer will be more soluble in water than the less branched isomer.
[pic]
o What will be soluble in hexane?
▪ Most organic compounds, as long as they’re not too polar.
• London dispersion
o Present in all molecules
o The weakest of the attractions
o Which of the following has the lowest boiling point?
[pic]
▪ Branching lowers boiling point, so the third molecule has the lowest.
▪ This is because more branched isomers are more compact, so the London dispersion forces are smaller.
▪ The unbranched chain, then, has the highest boiling point.
[pic]
• Dipole-dipole
o The interaction between two polar molecules
o Stronger than London dispersion
▪ Remember that polar molecules still have London dispersion forces.
• Hydrogen-bonding
o Only happens in molecules where hydrogen is bonded to oxygen, nitrogen, and fluorine.
o Remember that hydrogen-bonding compounds still experience dipole-dipole attractions and London dispersion forces.
Classes of compounds
|Classification |Functional group |
|Alkanes |Just carbon and hydrogen |
| |No multiple bonds |
|Alkenes |At least one carbon-carbon double bond |
|Alkynes |At least one carbon-carbon triple bond |
|Aromatics |For now, benzene and compounds that contain benzene rings |
| |[pic] |
|Alcohols |-OH (hydroxyl group) |
|Ethers |R-O-R’ |
|Aldehydes |[pic] (carbonyl group) |
|Ketones |[pic](carbonyl group) |
|Carboxylic Acids |[pic](carboxyl group) |
|Acid chlorides |[pic] |
|Esters |[pic] |
|Amides |[pic] |
|Amines |[pic](amino group) |
|Nitriles |R-CN |
You’re not responsible for naming these compounds until we cover them later.
• For Test 1 of 301, we’ll have nomenclature of alkanes.
• For Test 3 of 301, we’ll likely have nomenclature of alkenes.
• For the Final of 301, we’ll likely have nomenclature of alkynes and alcohols.
A molecule can belong to more than one class of compound!
I can guarantee that at some point you will be asked to circle the functional groups on a large molecule and state what class of compound each functional group makes the molecule.
• Understand the difference between functional group and class of compound!
• Ex. –OH is a functional group called a hydroxyl. If a molecule has that functional group, then it is an alcohol.
• You don’t need to circle the boring bits of the molecule and say “alkane.” That’s just the default.
• “Cyclic” is not a class of compound or functional group.
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