These are the homework exercises to accompany the Textmap for McMurry's Organic Chemistry textbook.
Label the following orbitals:
1= 3px; 2= 3s ; 3= 2pz
Give the electron configurations for Al, Br, Fe.
Al = 1s22s22p63s23p1
Br = 1s22s22p63s23p64s23d104p5
Fe = 1s22s22p63s23p64s23d6
List the bond angles for each of the following compounds: BH3, CF4, H2O.
Why is sulfur dioxide a bent molecule (bond angle less than 180°)?
HBH = 120°
FCF = 109.5°
OHO = 104°
This deviation is due to the lone pairs on the sulfur. These force the molecule to exhibit a “bent” geometry and therefore a deviation from the 180°.
Draw an energy diagram for energy vs. intermolecular distance for a fluorine molecule (F2) and describe the regions of the graph.
A - Repulsive Forces are present, p-orbitals are too close together
B - Optimal distance between the two p-orbitals to have a bond (the bond length)
C - Cannot form a bond, orbitals are too far away
Draw pentane, CH3CH2CH2CH2CH3, predict the bond angles within this molecule.
All the bond angles will be the same size.
Consider the following molecule:
At each atom, what is the hybridization and the bond angle? At atom A draw the molecular orbital.
A - sp2, 120°
B - sp3, 109°
C - sp2, 120° (with the lone pairs present)
D - sp3, 109°
1-Cyclohexyne is a very strained molecule. By looking at the molecule explain why there is such a intermolecular strain using the knowledge of hybridization and bond angles.
The alkyne is a sp hybridized orbital. By looking at a sp orbital, we can see that the bond angle is 180°, but in cyclohexane the regular angles would be 109.5°. Therefore the molecule would be strained to force the 180° to be a 109°.
Identify geometry and lone pairs on each heteroatom of the molecules given.
Diethyl ether would have two lone pairs of electrons and would have a bent geometry around the oxygen.
Dimethyl amine would have one lone pair and would show a pyramidal geometry around the nitrogen.
Below is the molecule for caffeine. Give the molecular formula for it.