Answers to Practice Exam 1 Question 1 20 points The following questions refer to the compound propyne shown below sp3 H3C sp C sp CH propyne a On the structure above identify the hybridization state of all carbon atoms b Draw a picture below that clearly shows the interacting orbitals of all of the C C single bonds in propyne H3C C C H C C bonding in propyne small lobes of the orbitals are not shown c Draw a picture below that clearly shows the interacting orbitals of all of the multiple bonds in propyne H3C C C bonding in the triple bond H Question 2 20 points The compound guanidine has the chemical formula CH5N3 a Complete the Lewis structure of guanidine on the template below by adding all the hydrogen atoms missing bonds and missing nonbonded electrons Make sure to identify any non zero formal charges on the atoms H H N C H N N H H b Guanidine is a very good bronstead base Addition of H to guanidine leads to the guanidinium polyatomic ion CH6N3 Complete the Lewis structure of the guanidinium ion on the template below by adding all the hydrogen atoms missing bonds and missing nonbonded electrons Make sure to identify any non zero formal charges on the atoms H H N H 1 N C H N H H c Does it matter which pair of electrons accepts the H Why or why not Clearly drawn pictures are worth a thousand words Yes because only addition of H to the lone pair shown above leads to a highly resonance stabilized structure H H H N H H H N C H H H N C H H C H N N N N N N H H H H H H Each nitrogen bears 1 3 of the positive charge It is the stability of the conjugate acid that makes guanidine a good base Question 3 20 points Label each of the following pairs of molecules as identical structural isomers conformational isomers enantiomers or diastereomers Answer a 2 chloropentane 3 chloropentane structural isomers Answer CH3 b identical CH3 CH3 CH3 Answer c enantiomers CH3 CH3 CH3 CH3 Answer d identical H3C H3C CH3 CH3 H e CH3 H3C H Cl H Answer diastereomers H Cl Cl Cl Question 5 25 points The following questions refer to the compound shown below Cl E H3C S CH3 CH3 a Label the double bond as E Z or not applicable Use the space below to show your work 2 H Cl 1 H3C 1 CH3 CH3 2 b Assign R or S to all stereocenters Use the space below to show your work 4 H3C H 2 1 Cl CH3 3 CH3 c Provide the complete IUPAC name for the compound S E 4 chloro 3 methyl 2 pentene Note no points were deducted for where the E and S are placed c Draw a diasteromer of the compound CH3 Cl CH3 CH3 e The double bond in the compound can be removed by reaction with hydrogen gas and a metal catalyst shown below How many possible stereoisomers will result from this reaction Make sure to briefly explain how you arrived at your answer Cl Cl H2 H3C CH3 metal catalyst H3C CH3 CH3 CH3 One stereocenter created in the reaction Remains S unchanged during reaction Cl H3C CH3 CH3 New stereocenter is either R or S Therefore there are two possible stereoisomers that can result from the reaction R S and S S