These are the homework exercises to accompany the Textmap for McMurry's Organic Chemistry textbook.
State wither the following is para, meta, or ortho substituted.
Name the following compounds.
Draw the following structures
A – meta; B – para; C – ortho
- 2-methyl-1,3,5-trinitrobenzene. (Also known as trinitrotoluene, or TNT)
The molecule shown, p-methylpyridine, has similar properties to benzene (flat, 120° bond angles). Draw the pi-orbitals for this compound.
The nitrogen has a lone pair of electrons perpendicular to the ring.
To be aromatic, a molecule must be planar conjugated, and obey the 4n+2 rule. The following is the following molecule aromatic?
No, it is not. It does not obey the 4n+2 rule. Also it is not planar.
Draw the resonance structures for cycloheptatriene anion. Are all bonds equivalent? How many lines (signals) would you see in a H1 C13 NMR?
The following reaction occurs readily. Propose a reason why this occurs?
All protons and carbons are the same, so therefore each spectrum will only have one signal each.
The ring becomes aromatic with the addition of two electrons. Thereby obeying the 4n+2 rule.
Draw the orbitals of thiophene to show that is aromatic.
The following ring is called a thiazolium ring. Describe how it is aromatic.
This drawing shows it has 6 electrons in the pi-orbital.
Similar to the last question, the drawing shows that there is only 6 electrons in the pi-system.
This is an isomer of naphthalene. Is it aromatic? Draw a resonance structure for it.
The following molecule is adenine. It has a purine core. Of the nitrogen in the core, how many electrons are donated into the pi system?
Yes, it is aromatic. 4n+2 pi-electrons.
There is only one nitrogen of the core that contributes to the pi-system (in red). With this one lone pair the core is aromatic with 10 electrons in the pi-system.