Organic Chemistry


Professor Carl C. Wamser

Chem 334 - Fall 2003

Workshop 1

Chapter 1 Workshop Problems
Structure & Bonding: Functional Groups

1. An experimental technique called 13C Nuclear Magnetic Resonance Spectroscopy allows chemists to tell how many different kinds of carbons there are in a molecule and whether carbons are primary (1°), secondary (2°), tertiary (3°), or quaternary (4°). Give Kekule structures (i.e., use lines for electron pair bonds) for the following compounds having molecular formula C6H12. Also, on each structure, identify carbons as 1°, 2°, 3°, or 4°; tell how many different kinds of C's there are; and designate which C's are equivalent.

a. A compound having only single bonds and only secondary carbons.

b. A compound having only single bonds and primary, secondary, and tertiary carbons.

c. A compound having only single bonds and only primary, secondary, and quaternary carbons.

d. A compound having only single bonds and primary, secondary, tertiary, and quaternary carbons.

2. Draw Kekule structures (show all bonds as lines and show all non-bonding electron pairs) for constitutional isomers with molecular formula C3H6O2. Make sure that the following functional groups are included in these isomers: carboxylic acid, ester, ether, aldehyde, ketone, alcohol. Circle each functional group and indicate its appropriate name.

3. a. Draw Kekule structures for each of the compounds below. Be sure to show all bonding and non-bonding valence electrons. Also clearly indicate any formal charges on atoms.

b. Consider the molecular geometry information given for each compound below. Based on this information, specify the orbitals that each atom could use in sigma and pi bonding (sp2, sp3, p, etc.) and for holding non-bonding electron pairs. Explain how your orbital assignments are consistent with the observed geometries.

DCH=CHBr (1-bromo-2-deuterioethylene)
(two isomers)
(all atoms are coplanar)

CH3CO2- (acetate ion)
bond angle HCC = 109°
bond angle CCO = OCO = 120 °

H2CO (formaldehyde)
bond angle HCO = 120°

bond angle HCS = CSO = CSC = 109°

Materials adapted from:

Peer-Led Team Learning: Organic Chemistry, 1/e
Jack A. Kampmeier, University of Rochester
Pratibha Varma-Nelson, St. Xavier University
Donald Wedegaertner, University of the Pacific
Prentice-Hall, 2001, ISBN 0-13-028413-0