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Examination 1 - Organic Chemistry | CHEM 345, Exams of Organic Chemistry

Material Type: Exam; Class: Organic Chemistry; Subject: Chemistry; University: University of Wisconsin - Milwaukee; Term: Fall 1998;

Typology: Exams

Pre 2010

Uploaded on 09/02/2009

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Last Name _____________________________ First Name _______________________________
Please Print. Please put your name on the back of the test also.
EXAM 1 ORGANIC CHEMISTRY 345
September 28, 1998 Schwabacher
page 1. (12 points) Quiz scores so far
page 2. (25 points) Q1
page 3. (18 points) Q2
page 4. (35 points)
page 5. (10 points)
TOTAL (100 points)
I. (12 points) Nomenclature: Provide a name for the given structure, or a structure for the
given name. Don't forget stereochemistry!
A) O
B) 3
(E)-1,4-hexadiene
D) 1,2-epoxycyclohexane
pf3
pf4
pf5

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Download Examination 1 - Organic Chemistry | CHEM 345 and more Exams Organic Chemistry in PDF only on Docsity!

Last Name (^) _____________________________ First Name (^) _______________________________

Please Print. Please put your name on the back of the test also.

EXAM 1 ORGANIC CHEMISTRY 345

September 28, 1998 Schwabacher

page 1. (12 points) Quiz scores so far

page 2. (25 points) Q

page 3. (18 points) Q

page 4. (35 points)

page 5. (10 points)

TOTAL (100 points)

I. (12 points) Nomenclature: Provide a name for the given structure, or a structure for the given name. Don't forget stereochemistry!

A)

O

B) 3

(E)-1,4-hexadiene

D)

1,2-epoxycyclohexane

II. (25 points) Identify the compounds from the spectral data given.

A. Compounds X and Y each have the formula C 2 H 6 O. Compound X has a strong broad absorption at 3500-3600 cm-1^ in the IR, while Y has no absorption in that region.

X = Y =

B. A compound of formula C 2 H 4 O shows no strong absorption at 3500-3600 cm-1^ or at 1600-1800 cm-1. Propose a structure.

C. One compound of formula C 6 H 10 shows 3 signals in the 13 C NMR, and has a UV absorption at 227 nm. The 1 H NMR shows a doublet at δ 1.6 that integrates to 6 hydrogens, and a multiplet at δ 5 that integrates to 4 hydrogens. Propose a structure.

D. Another compound, also of formula C 6 H 10 , shows 6 signals in the 13 C NMR, has no measurable absorption in the UV above 200 nm, and in the 1 H NMR shows (among other things) a doublet at δ 1.6 that integrates to 3 hydrogens.

E. A molecule shows m/e = 58 as the largest mass peak in the mass spectrum. The IR shows a strong band at 1715 cm-1, and no band 3300-3650 cm-1. The 1 H NMR is a single peak at δ 2.1.

IV. (35 points) Provide products, showing stereochemistry where relevant.

A.

O

  • CH 3 OH (solvent)

H 2 SO 4

(catalytic amount)

B.

O +^ CH 3 SH

NaOH

(catalytic amount)

C.

Cl +

O O OH

D. Show stereochemistry of both products

O

CH 3

H CH 3

  • HBr (1 equivalent)

E.

NO 2

O 2 N

F.

O (^) Na Br

NOTE: Choose one of these, and cross off the other. Do either the synthesis (V.a) or the mechanism (V.b) for 10 points.

V.a (10 points) Synthesis: Show a sequence of reactions that would produce the molecule shown. You may begin with any alkenes of 4 carbons or fewer. You may use any other reagents, as long as all the carbon atoms for the product come from alkenes of 4 carbons or fewer.

O OH

V.b (10 points) Provide a mechanism for this reaction, showing all intermediates and all steps. Notice the stereochemistry of the starting material and the product! This requires two proton transfer steps, and two substitution steps.

Cl OH

+Na - O ethanol

heat O

  • HO