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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