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Assignment 3 - 4 Questions of Organic Chemistry I - Fall 2001 | CHEM 231A, Assignments of Organic Chemistry

Material Type: Assignment; Class: Organic Chemistry I; Subject: Chemistry; University: Whittier College; Term: Fall 2001;

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Organic Chemistry - CHEM 231A
Problem Set #3
Due November 26, 2001
1a. Iodide ion is a good nucleophile and sodium iodide is quite soluble in acetone. On the other hand,
sodium chloride and sodium bromide have low solubilities in acetone. As a result, the reaction of
alkyl bromides and alkyl chlorides with NaI/acetone can serve as a simple test reaction as
indicated below.
R Br + NaI (soluble) Acetone R I + NaBr (insoluble)
The reactivity order of the following alkyl bromides with NaI/acetone is:
Br
Br Br
Write a mechanism for this reaction. Explain how your mechanism accounts for the observed
reactivity order. (It will help to build molecular models.)
1b. The optical rotation of a solution of (+)–2–bromobutane in diethyl ether does not change with
time. However, when tetrabutylammonium bromide is dissolved in this solution, the rotation
decreases slowly with time to zero. The NMR spectrum of the solution, however, does not change.
Clearly explain what is happening. Why does the rotation decrease to zero? Why doesn't it become
negative?
1c. Explain the following observation: Reaction of either diastereomeric pair of 1-bromo-3-
ethylcyclohexane in the presence of tetrabutyl ammonium bromide yields the same diastereomeric
ratio, with the trans diastereomer predominating.
Br Br
Diastero merica lly Pure Diasterom erically Pure
Bu4N+ Br-
Et2O
Bu4N+ Br-
Et2O
Br Br
+
Minor Major
1d. Discuss how these three experiments above are related to our general understanding of the
mechanism of bimolecular nucleophilic substitution reactions.
2a. When the optically active tosylate (A) shown below was reacted with CH3SNa in CH3OH, the
reaction was observed to be second order and the substitution product shown was formed almost
exclusively. Show the mechanism for this reaction and predict the stereochemistry of the product.
OTos + CH3S Na CH3OH SCH3
A
2b. If the solvent in part (a) was changed from methanol to dimethylformamide (DMF) the same
product was formed but at a much faster rate. Explain.
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Organic Chemistry - CHEM 231A

Problem Set # Due November 26, 2001

1a. Iodide ion is a good nucleophile and sodium iodide is quite soluble in acetone. On the other hand, sodium chloride and sodium bromide have low solubilities in acetone. As a result, the reaction of alkyl bromides and alkyl chlorides with NaI/acetone can serve as a simple test reaction as indicated below.

R Br + NaI (soluble)

Acetone R I + NaBr (insoluble)

The reactivity order of the following alkyl bromides with NaI/acetone is:

Br Br Br Write a mechanism for this reaction. Explain how your mechanism accounts for the observed reactivity order. (It will help to build molecular models.)

1b. The optical rotation of a solution of (+)–2–bromobutane in diethyl ether does not change with time. However, when tetrabutylammonium bromide is dissolved in this solution, the rotation decreases slowly with time to zero. The NMR spectrum of the solution, however, does not change. Clearly explain what is happening. Why does the rotation decrease to zero? Why doesn't it become negative?

1c. Explain the following observation: Reaction of either diastereomeric pair of 1-bromo-3- ethylcyclohexane in the presence of tetrabutyl ammonium bromide yields the same diastereomeric ratio, with the trans diastereomer predominating.

Br Br

Diasteromerically Pure (^) Diasteromerically Pure

Bu 4 N+^ Br- Et 2 O

Bu 4 N+^ Br- Et 2 O

Br Br

Minor Major

1d. Discuss how these three experiments above are related to our general understanding of the mechanism of bimolecular nucleophilic substitution reactions.

2a. When the optically active tosylate ( A) shown below was reacted with CH 3 SNa in CH 3 OH, the reaction was observed to be second order and the substitution product shown was formed almost exclusively. Show the mechanism for this reaction and predict the stereochemistry of the product.

OTos (^) + CH 3 S Na CH^3 OH^ SCH 3

A 2b. If the solvent in part (a) was changed from methanol to dimethylformamide (DMF) the same product was formed but at a much faster rate. Explain.

2c. When A was reacted with NaOCH 3 in CH 3 OH, the reaction was still kinetically second order. However, the major product was the alkene shown below. Account for the shift from predominant substitution for the reaction in part (a) to predominant elimination for the reaction with CH 3 ONa/CH 3 OH. Also, show the stereochemistry of the methyl ether substitution product.

OTos (^) + CH 3 O Na CH 3 OH^ OCH (^3)

A

Minor Product Major Product

2d. How would you change reagents or reaction conditions so that A would give almost exclusively the alkene (C 6 H 5 CH=CH 2 )?

2e. When A is refluxed in methanol (no added CH 3 ONa), both substitution and elimination products are formed as in part (c). However, substitution predominates and the stereochemical result is different then that observed for part (c). Predict the stereochemistry of the substitution product and explain why these results are observed from the reaction of A in methanol.

OTos OCH^3

A

Major Product Minor Product

CH 3 OH Reflux

  1. Consider the following reaction conditions and NMR spectra:

The starting material, compound A, C 7 H 15 BrO, has the following NMR spectrum:

Doublet 1.2 ppm int 5.6 cm Quintet 1.23 ppm int 3.8 cm Quartet 1.51 ppm int 3.7 cm Doublet 1.79 ppm int 5.7 cm Quartet 1.87 ppm int 3.6 cm Sextet 3.83 ppm int 1.8 cm Sextet 4.13 ppm int 1.9 cm

When A is reacted with NaH in Et 2 O at 0 C, one product, compound B, C 7 H 14 O, is formed with the following NMR spectrum:

Doublet 1.21 ppm int 4.2 cm Quintet 1.25 ppm int 1.3 cm Quartet 1.49 ppm int 3.0 cm Sextet 3.85 ppm int 1.2 cm

3a. What is the structure of compound A? 3b. What is the structure of compound B? 3c. Write a mechanism that accounts for the formation of compound B. 3d. In a different reaction, when compound A is heated in dimethyl formamide, not only is compound B formed, but a new product, compound C, C 7 H 14 O, is formed with the following NMR spectrum:

Doublet 1.22 ppm int 4.4 cm Quartet 1.53 ppm int 3.1 cm Doublet 1.61 ppm int 4.5 cm Quartet 2.05 ppm int 3.0 cm Sextet 3.79 ppm int 1.5 cm Quintet 5.35 ppm int 1.4 cm Quartet 5.39 ppm int 1.5 cm