Name ______________________
Department of Chemistry
SUNY/Oneonta
Chem 221 - Organic Chemistry I
Examination #3 - November 11, 2002 ANSWERS
INSTRUCTIONS —
This examination has two parts. The first part is in multiple choice format;
the questions are in this Exam Booklet and the answers should be placed on the
"Test Scoring Answer Sheet" which must be turned in and will be machine
graded.
The second part requires your responding to questions in this Exam
Booklet by writing answers into the spaces provided. The Exam Booklet must be
handed in and will be returned to you with a grade.
On the Test Scoring Answer Sheet, using a soft pencil, enter the following
data (in the appropriate places): your name, instructor's name, your OSC
Student Number or Social Security number, course number (30022101) and the
test number (03); darken the appropriate bubbles under the entries (if you are
using your student number which begins with a letter, leave the bubbles under
the letter blank, but darken the bubbles under the numbers), making dark black
marks which fill the bubbles.
You may use a set of molecular models and the periodic table (last page)
provided, but no other aids, during the exam.
Answer all questions. The questions on Part I are worth 4 points each.
You have 50 minutes. Good luck!
Answers are shown highlighted.
Explanations are shown in red type.
November 11, 2002
1.
Chem 221 - Exam #3
Provide the IUPAC name for the compound shown
to the right.
H 3C
Page 2 of 8
CCl2 CCl2
CH 2
C
C
CH 3
(a) 2,2,3,3-tetrachloro-5-heptyne, (b) 2,2,3,3-tetrachloro-5-heptene,
(c) 2,2,3,3-tetrachloro-5-octyne, (d) 2,3-tetrachloro-5-heptyne,
(e) 5,5,6,6-tetrachloro-2-heptyne, (f) None of the previous answers is correct.
2.
Select the structure(s) of the intermediate(s) that is(are) formed when (Z)-2-butene reacts
with bromine.
H 3C
CH3
C
C
I
H 3C
C
H
H
H
3.
Br2
CH3
C
Br
II
H 3C
C
C
H
H
III CH
3
C
C
H
H
Br
H 3C
H
Br
CH3
(a) I only, (b) II only, (c) III only, (d) I & II, (e) I, II & III, (f) None of the previous
answers is correct.
The Br adds stereospecifically across the double bond. This means that if the methyl
groups started out cis they will be cis in the product and if they started out trans they will
be trans in the product.
Select the major organic product of the following reaction.
H 3C
H
C
C
H 3C
H
NBS, H2O
DMSO solvent
H3C (a) H
H
H 3C C C
Note: The N-bromosuccinimide
provides a low concentration of
bromine.
H3C (b) H
H
H 3C C C
OH Br
Br OH
H3C (c) H
H
H 3C C C
OH OH
H3C (d) H
H
H 3C C C
O
(e) None of the previous answers is correct
4.
Bromine attaches itself first in the bromohydrin formation, so this is a Markovnikov type
addition.
Select the major organic product of the following reaction.
H 3C
H
C
H 3C
C
H
1) BH3 / THF solvent
2) H2O2, KOH
H3C (a) H
H
H 3C C C
H3C (b) H
H
H3C C C
H OH
H3C (c) H
H
H 3C C C
OH OH
OH H
H3C (d) H
H 3C C C
H O
(e) None of the previous answers is correct
Hydroboration-oxidation gives anti-Markovnikov addition to form an alcohol.
November 11, 2002
5.
Chem 221 - Exam #3
Page 3 of 8
What is (are) the principal organic product(s) of the following reaction?
H 3C
H2C HC
CH2
(b) R-isomer only
CH3CH2CHCH3
1) Hg(OAc)2, H2O/THF
(a) CH3CH2CH2CH2OH
2) NaBH3
(c) S-isomer only
CH3CH2CHCH3
OH
(d) unequal amounts
of R & S isomers
CH3CH2CHCH3
OH
(e) racemic mixture
of R & S isomers
CH3CH2CHCH3
OH
OH
Markovnikov formation of an alcohol. When achiral reactants form chiral products
racemic mixtures form.
6.
Select the principal product of the following reaction sequence.
H 3C
CH2 C
1. BH3
CH
major
product
2. H2O2, KOH (pH=8)
O
(a) H3C
CH2 C
CH3
(b) (CH3)2CH
C
CH3 (c) H3C
OH
CH2 C
CH2
O
(d) H3C
CH2 CH
CH2
OH
7.
(e) H3C
CH2 CH2
CH
O
Anti-Markovnikov addition of water to the triple bond leads initially to compound (d),
but this undergoes keto-enol tautomerization to give the aldehyde.
Select the appropriate reagents to carry out the indicated reaction.
H 3C
H
(a) H2, Lindlar catalyst,
H3C C C CH3
C C
(b) H2, Pt, (c) H2O, H2SO4, HgSO4,
(d) Li, NH3,
H
CH3
(e) NaNH2 in NH3
Answer (a) would produce the cis isomer.
Which of the alkyl chlorides below is likely to be most successful in the following
reaction, where R is an alkyl group: CH3C/C:- Na+ + R-Cl 6 CH3C/C-R + Na+Cl-
?
8.
9.
(a) CH3CH2Cl, (b) (CH3)3CCl, (c) (CH3)2CHCl
(d) Bogus question, dude! Like, none of these compounds will react this way.
Primary alkyl halides work best in this reaction.
A pair of stereoisomers that are not enantiomers are
(a) constitutional isomers. (b) meso structures. (c) diastereomers. (d) conformational
isomers. (e) Another bogus question, dude! Like, all stereoisomers are enantiomers.
By definition.
November 11, 2002
10.
Chem 221 - Exam #3
Page 4 of 8
If a chiral substance that has just one chiral center is dextrorotatary the chiral center is
(a) R, (b) S, (c) E, (d) Z,
(e) Actually it is impossible to tell whether the chiral center is R or S based only on the
rotational direction of the optical activity.
11.
Select the statement that best describes the relationship
between the two structures shown to the right.
F F
F F
H
H
They are
H
H
(a) constitutional isomers. (b) enantiomers. (c)
diastereomers. (d) the same molecule and not meso structures. (e) the same molecule
and are meso structures.
The structures are the same because they are the same, i.e. superimposable. These two
structures represent a meso molecule because there are chiral centers in the molecule and
the molecule is achiral (it is superimposable on its mirror image).
12.
Select the statement that best describes the relationship
between the two structures shown to the right.
They are
(a) constitutional isomers. (b) enantiomers.
(c) diastereomers. (d) the same molecule and not meso structures.
(e) the same molecule and are meso structures.
They are non-superimposable mirror images.
14.
H F
F
H
H
Select the answer that correctly indicates the R/S configuration
around the chiral centers in the molecule to the right.
F
H
(a) 1-R, 2-R, (b) 1-R, 2-S, (c) 1-S, 2-R, (d) 1-S, 2-S.
15.
F H
H
n o CH3
CH C
N
2
H
Which of the following is not true of enantiomers?
(a) They have the same boiling point. (b) They have the same melting point.
(c) They have the same specific rotation. (d) They have the same density.
(e) They have the same chemical reactivity toward achiral reagents.
Specific rotation is the only physical property that differs between enantiomers.
16.
Which of the following molecules is chiral?
H 3C
CH3
C
C
H
H
HO
OH
(a)
C
H 3C
Br
H
Cl
H
CH3
CH3
C
(b)
CH2 C
CH
(c)
(d)
H
(d) is the only molecule in this group that is not superimposable on its mirror image.
Cl
November 11, 2002
Chem 221 - Exam #3
Page 5 of 8
17.
Which of the molecules in question #16 is a meso structure?
(a) is meso.
(a) has (two) chiral centers but is achiral (it is superimposable on its mirror image);
therefore it is meso.
18.
Which of the following is not a reactive intermediate in the free radical chlorination of
methane?
(a) H•, (b) H3C•, (c) Cl•, (d) Bogus question. All of them are intermediates.
19.
Select the two products that form in the following reaction.
NBS (N-bromosuccinimide)
UV light, CCl4 solvent
Br
I
Br
II
Br
III Br
IV
(a) I&II, (b) I&III, (c) I&IV, (d) II&III, (e) II&IV, (f) III&IV
The allylic free radical (more stable than alternatives)
C
that forms as an intermediate is shown to the right.
It can pick up a bromine at the two sites that share
C
the unpaired electron.
H
20.
C
C
H
Which of the following alcohols will be most reactive toward HBr in terms of converting
the alcohol to an alkyl bromide?
(a)
CH3CHCH2CH3
OH
(b)
HOCH2CH2CH2CH3
(c)
H 3C
CH3
C
CH3
OH
(d) These alcohols would be essentially equally reactive.
Order of reactivity: tertiary>secondary>primary. Reason: tertiary alcohol forms tertiary
carbocation, etc. and tertiary carbocation is more stable than secondary, etc.
Directions for Part II --- Answer the questions in the space provided. If there is
insufficient space continue your answer on the back of the sheet but clearly indicate on
the front of the sheet that you have done this.
November 11, 2002
Chem 221 - Exam #3
Page 6 of 8
Mechanism.
(a) Show the mechanism for the following reaction.
>
Be certain to clearly show the configurations around the chiral carbon(s) in the
product(s).
>
Be certain to show all intermediates and their stereo (3-dimensional) structure(s).
>
Be certain to show direction(s) of approach of reacting species if it has
stereochemical consequences. For example, if an approach from one direction
leads to one stereoisomer and approach from a different direction leads to a
different stereoisomer, show this.
H
H
CH3
C
C
H 3C
+
H C
H 3C
HCl
H
CH3
H
C
H 3C
C
CH3
C Cl
H
H
+
HCl
H
CH3
C Cl
H
H C
H 3C
(a)
H
H3C C
H
CH3 + Cl
C H
(b)
from
path
(a)
H
H 3C C
H
from
path
(b)
H
H 3C C
H
Cl
C
C
CH3
H
H
CH3
Cl
The carbon that is the site of positive charge in the intermediate carbocation is
2
trigonal/planar (sp ) and has a vacant 2p orbital, as shown. This orbital can join to a
chloride ion from the top [path (a)] or from the bottom [path (b)]. These paths are
equally accessible so the products (enantiomers) will be formed in equal amount.
Now, it is true that I’ve shown a conformation of the intermediate carbocation in which
the methyl group on the left is pointed toward the front, not toward the top or bottom. If
this methyl group is oriented up toward the top, instead of front, it will be easier for
chloride to approach via path (b). However, there will be an equal number of
carbocations in which the methyl group is oriented down (mirror image of the
carbocation in which it is up). For these carbocations approach of chloride via path (a)
will be easier. These mirror-imaged conformations of the carbocation will have the same
energy and hence will be present in equal amount, thus leading to equal amounts of the
product enantiomers.
November 11, 2002
Chem 221 - Exam #3
Page 7 of 8
(b) Is the product, or mixture of products taken as a whole, of this reaction optically
active?
No.
A racemic mixture of enantiomers results in this reaction as explained above. However,
one need not know the mechanism of this reaction to know the answer to this question
because optically inactive reactants (the case here where neither reactant is chiral) give
optically inactive products.
2.
Synthesis.
Outline syntheses which would produce each of the following compounds in good yield.
[Note: In outlining a synthesis you should show explicitly what compounds you are
using and any special conditions. You need not balance equations or show mechanisms;
doing so correctly will gain you no additional credit, doing so incorrectly will cost you.]
(a) You must start each synthesis with cyclopentene, and may use any other materials
you need to carry it out. More than one step may be required.
(i) 3-bromocyclopentene
C
H
H
NBS
UV light or heat
C Br
H
H
C Br
3-Bromocyclopentene is a chiral molecule. Will just one enantiomer form, or will
unequal amounts of the two enantiomers form, or will the product be racemic?
Racemic.
Achiral reactants give racemic mixtures of enantiomers when chiral products are formed.
November 11, 2002
Chem 221 - Exam #3
(ii) 3-(3-cyclopentenyl)cyclopentene
H
C Br
from
part (i)
Li
C
H
H
C
C
H
Li
CuI
Page 8 of 8
C
H H
Cu C
Li
H
C Br
C
H H
C
It should be apparent from the question that the idea here is to join two cyclopentenyl
units together. Currently, you know (or should know) only two synthetic methods that
will join two smaller units together to give a larger one. One of these involves using
terminal alkynes; that method will not work here. The other method is to employ a
Gilman reagent. That is the method employed here.
Part I (80)________
PartII
1. (10)________
2. (15)________
Total(105)________