Chemistry 241
Clark College
Homework 7
Please put your answers on a separate sheet of paper.
Due Tuesday, December 2nd, 2008, IN CLASS.
1. Recopy the following chart, and complete the reactions. Be sure to indicate stereochemistry, where
appropriate.
**I indicated stereochemistry for every stereospecific reaction. Note that in some instances, a racemic
mixture would form, so technically stereochem would not need to be indicated! I am more concerned
that there is recognition of which reactions are stereospecific and which are not!
Hydrogenation,
Syn addition of two H's
H
H
Markovnikov addn of HCl
Cl
OH
Anti-Markovnikov, syn
addition of an alcohol
OH
H
1) BH3
2) H2O2,
NaOH
H2
Pt
HCl
O
1) Hg(OAc)2, H2O
2) NaBH4
OH
OH
Syn formation of a 1,2-diol
H
O
(+ DMSO)
1) O 3 ) S
H 2
2) (C 3
H
Non-rearranging, Markovnikov,
anti addition of an alcohol
Oxidative
cleavage
"green" anti addition of Br2
HBr, H2O2
Br
OsO4
HOO
Cl2
H2O
OH
Br
H+
O
Cl
OH
Markovnikov addition of an alcohol
Halohydrin reaction.
Markovnikov, Anti addition of Cl, OH
Homework 7
Fall 2008
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Chemistry 241
Clark College
2. Give the reagents needed to affect the following transformations.
Br2
H2O
H
1) BH3
2) NaOH, H2O2
OH
OH
Br
1) OsO4
2) NaHSO4
or OsO4,
HOO
1) Hg(OAc)2, H2O
2) NaBH4
OH
H2
Pt, Pd or Ni
H2O2
Na2WO4
OH
HO
O
+ CO2
OH + H O
2
3. a) Complete the following two-step synthesis, providing the structures for compounds A and B.
Compound A would be formed as a racemic mixture, so I did not indicate stereochemistry for that
product. However, Whatever stereochemistry was assigned for the alcohol would be retained in the
product.
OH
1) BH3
2) H2O2, NaOH
H2O
O
H+
A
B
b) What prominent feature in the IR would you expect to see for compound A?
Compound A is an alcohol! You should see a large, broad peak round 3400 cm-1.
c) Provide three peaks that you should expect to see in the IR for compound B – give the
approximate frequency and what part of the molecule it corresponds to (e.g. 3000-3100 cm-1, sp2
C-H stretch). You should refer to the larger IR reference charts in your textbook for this.
You will see a C-O single bond stretch, between 1000-1150 cm-1. You will also see sp3 C-H peaks: the
stretching frequency just below 3000 cm-1 and the bending frequency around 1450 cm-1. Most
importantly, the alcohol peak will be gone!
Homework 7
Fall 2008
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Chemistry 241
Clark College
4. Give the products expected when the following compounds are treated hydrogen peroxide in the
presence of a sodium tungstate catalyst.
Don’t forget, when using hydrogen peroxide with the sodium tungstate catalyst, the products of doublebond cleavage are ketones and carboxylic acids. Aldehydes only form during ozonolysis.
4
3
5
O
2
6
1
O
HO
OH
O
7
HO
OH
H2O + CO2
O
Numbering can help get the product!
O
O
OH
COOH
O
OH
O OH
1
2
34 5
6
COOH
O
OH
O
Homework 7
Fall 2008
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Chemistry 241
Clark College
5. Use the following experimental data to determine the structure of an unknown alkene.
a. The molecular formula for this alkene in C10H14. What is the index of hydrogen deficiency?
(See p.184, or your lecture notes from Ch. 5)
If this molecule had no double bonds, it should have 2n+2 hydrogens, or (2x10) + 2 = 22 H’s. It only
has 14, so it is missing 8 H’s, which means that the Index of Hydrogen Deficiency = 4.
b. One mole of the molecule reacts with three moles of H2 gas in the presence of a platinum
catalyst. The product of this hydrogenation/reduction is 1-isopropyl-4-methylcyclohexane.
? + 3 H2
this part gives a skeleton structure to work from, and
indicates that there are 3 double bonds. 3 double
bonds + 1 ring = 4 units of hydrogen deficiency.
Pt
c. When the molecule is treated first with ozone, then with dimethylsulfide, the following
products are formed:
O
O
O
O
O
H
H
H
H
O
If you had troubles visualizing this, make some models! Two carbonyl (C=O) groups would link
together to form an alkene in the original molecule.
O
1) O3
2) (CH3)2S
H
O
H
O
O
H
O
H
O
H
H
O
O
H
O
O
O
O
H
Homework 7
Fall 2008
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Chemistry 241
Clark College
Extra Credit (10 points)
Propose a mechanism for the following reaction. (hint: count carbons, and watch your lone pairs on N)
You may work with each other on this one, but you may not get help from any other instructors!
N
N
H2C N N
If we consider the starting alkene as the Lewis base, we need to identify the "best" Lewis acid on the
dinitrogen compound. You might think that the middle nitrogen is the Lewis acid, since it has a positive
formal charge. However, consider the following resonance structures:
H2C N N
H2C N N
Although this second structure
does not have an octet, the
positive charge is now on the
less electronegative element.
In the second resonance structure, the carbon has a full positive charge, which makes it a good
candidate for the Lewis acid. Also, in the first resonance structure, the positive nitrogen inductively
withdraws a lot of electron density from the carbon, so the carbon has a large !+ charge.
H2C N N
Homework 7
N
Fall 2008
N
N
N
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