Chem2O6 - 1996/97

MidTerm Test #1 ANSWERS November 22, 1996

Please see your Tutorial Leader before January 15, 1997 if there is an addition error in your paper, or if you believe that you were robbed of a significant (i.e. >5) marks.

1. Lysergic acid N,N-diethylamide ("LSD") has the general structure shown below:

Structures MT1 #1a

(a) The condensed formula of LSD is C20H25N3O.

(b) The hybridization of the three nitrogen atoms are N1: sp2; N2: sp3; N3: sp2.

(c) The most basic nitrogen atom in the molecule is N2.

(d) How many stereocenters does LSD possess? - two. How many configurational isomers are there for a molecule of this general structure? - four. How many of them are chiral ? - all of them

(e) Draw the structure of one of the configurational isomers of LSD. Label the stereocenters in your structure with its absolute configuration, determined using the Cahn-Ingold-Prelog rules.

Structures MT1 #1e

(f) LSD reacts with one equivalent of bromine in carbon tetrachloride. In the structure you have drawn for part (e), identify the site in the molecule which is expected to be most reactive towards Br2. Draw the structure of the major product(s) expected from the reaction of the particular stereoisomer you have drawn in (e) with this reagent:

Structures MT1 #1f

All the other double bonds are aromatic!

2. As you may know, insects communicate with each other by means of species-specific organic compounds known as pheromones. Mammals, including humans, do so as well. One approach to controlling insects without the use of pesticides is through the use of simple traps, which are smeared liberally with the synthetic sex pheromone of the bug of interest in order to lure it to its untimely doom. At least one perfume on the market today employs an analogous ingredient. It is not directed at insects, however.

A laboratory synthesis of the sex pheromone of the codling moth involved the following reactions:

Structures MT1 #2

See Ege, Problem 8.46

(a) Name the alkene (1) used as starting material in the synthesis, including its stereochemistry:


(b) Supply the reagents and solvents necessary to carry out the two steps of the transformation shown above.

A: 4-toluenesulfonyl chloride (TsCl) in pyridine.

B: NaBr in acetone, methanol, ethanol, acetone, or THF.

(c) Alcohols can also be converted to alkyl halides by reaction with the corresponding hydrohalic acid. For example,

Structures MT1 #2c

(i) Write the names of these two compounds (see above).

(ii) Write a complete mechanism for this reaction.

Structures MT1 #2cii

(iii) Why can't this procedure be used for the direct conversion of 1 to 3 in the pheromone synthesis on the previous page?

Because the C=C bond would react.

Draw the products you would expect, and the mechanisms by which they are formed:

Structures MT1 #2ciii

3. cis-1-Bromo-2-phenylcyclohexane reacts with potassium tert-butoxide (KOC(CH3)3) in tert-butyl alcohol at 50 oC to give exclusively 1-phenylcyclohexene. trans-1-Bromo-2-phenylcyclohexane does not react under the same conditions, but gives 1-phenylcyclohexene when the reaction mixture is refluxed (82 oC) for longer periods of time. Draw structures for the two compounds, and briefly explain these observations.

See: Ege, Problem 7.40

This is E2 elimination. In cis-1-bromo-2-phenylcyclohexane, the H and the leaving group can assume the anti-periplanar arrangement which is required for fast elimination. In the trans-compound, this is not possible. (Syn-) elimination to give the same alkene is much slower than from the cis-isomer. However, it is favoured over anti-elimination on the other side because the nonconjugated alkene that would result (3-phenylcyclohexene) is much less stable than 1-phenylcyclohexene.

4. For the following equilibrium, Go = -2.85 kJ/mol at 25 oC.

(a) Given that for methanol, pKa = 15.5, the pKa of cyclopentadiene is (circle the correct answer; R = 8.3 J mol-1 K-1):

Rough work:

Go = -RTln Keq = -2.303RTlog Keq
log Keq = (-2850)/[(-2.303)(8.3)(298)] = 0.5

Keq = KaCP / KaMeOH      (CP = cyclopentadiene; MeOH = methanol)
log Keq = log KaCP - log KaMeOH
             = pKaMeOH - pKaCP
 pKaCP = pKaMeOH - log Keq
             = 15.0

(b) In tetrahydrofuran (C4H8O) solution, which is present in higher concentration, C5H6 or C5H5Na? C5H5Na.

(c) In methanol solution, which is present in higher concentration, C5H6 or C5H5Na? C5H6.

(d) Draw the structure of C5H5Na:

(e) Cyclopentadiene is unusually acidic for a hydrocarbon. Draw all the resonance structures for its conjugate base. Why is C5H6 so acidic?

The five carbons in the cyclopentadienyl anion are all sp2 hybridized. Thus, the -system consists of a cyclic array of p-orbitals containing six electrons - the molecule is AROMATIC!


6. See Ege, Problem 7.38


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