Chemistry Department

 

University of Alberta

           

CHEM 261

 

Exam II

 

March 13, 2008

                                                                       

 

1. Using acceptable chemical nomenclature, provide an unambiguous name for the following compounds:

 

a.     

 

                       

 

                        meso-3,4-dimethylhexane

                       

 

b.     

 

                       

 

                                (R)-2-methoxypentane

 

c.     

 

                       

 

                                cis-1-bromo-3-chlorocyclohexane

                       

 

2.  Complete the following partial structure of (R)-2-methyl-3-bromoheptane:

 

 

 

3. Complete the following partial Fischer structure of (2R,3S)-2,3-dibromopentane:

 

           

 

4. Consider the diastereoisomers of 1-bromo-4-chlorocyclohexane.

 

a.    Explain the term "diastereoisomer".

 

Diastereoisomers are stereoisomers that are not enantiomers.

 

b.    Draw the most stable conformation of cis-1-bromo-4-chlorocyclohexane.

 

 

c.    Draw the most stable conformation of trans-1-bromo-4-chlorocyclohexane.

 

 

d.    Use the principles of conformational analysis to predict which isomer is the more stable. Explain your answer.

 

Both isomers are free of angle strain and torsional strain. The trans isomer has no steric strain whereas the cis isomer has two 1,3-diaxial interactions between Cl and the two axial hydrogens:

 

 

e.    Is the trans isomer chiral?

 

No. there is a plane of symmetry.

 

5. Propose a synthesis of 2,2-dimethylpentane starting from (CH₃)₃CH.  You can use any other organic compounds having a maximum of three carbons and any inorganic reagents that you wish!

 

 

6. Bromotrichloromethane will brominate alkanes in the presence of light. The relative reactivity of the hydrogens in this reaction (tertiary : secondary : primary) is 1600 : 82 :1. The overall reaction is shown below:

 

a.    Draw the structure of the major product of bromotrichloromethane monobromination of (CH₃)₃CH.

 

(CH₃)₃CBr

 

b.    Propose a mechanism for the formation of the product drawn in “a”. Do not worry about termination steps.

 

 

c.     Propose a mechanism to explain the light induced bromination of (CH₃)₃CH using Br₂. Do not worry about termination steps.

 

 

d.    Calculate DH for the propagation steps in your mechanism from part “c”.

 

DH = 390 – 366 = +24kJ

 

DH = 193 – 263 = -70kJ

 

e.    Calculate DH for the analogous steps for the chlorination of (CH₃)₃CH.

 

DH = 390 – 432 = -42kJ

 

DH = 243 – 330 = -87kJ

 

f.     The bromination of (CH₃)₃CH using Br₂ is slowed down by the addition of fairly large amounts of HBr. Explain why this occurs.

 

As the concentration of HBr increases, it will compete with the Br₂ to react with the tert-butyl radical as both processes will be exothermic (-24 kJ for HBr and -70 kJ for Br₂).  Thus in both cases the activation energy will be low.

 

g.    Why does HCl not react in this way?

 

HCl will not react in this fashion because its reaction with the tert-butyl radical is endothermic (+ 87 kJ), a process that will have a far higher activation energy than the reaction with chlorine

 

7. What reagents would you use to effect the following conversions?

 

a.     

                       

 

b.     

                           

           

 

8. Give the major product(s) of the following reaction: