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Unformatted text preview: CEM 351
1st EXAM/Version A
Friday, September 19, 2003
1:50 – 2:40 p.m.
Room 138, Chemistry
Student # .
. Section Number (2 pts extra credit) Grade? TOTAL 1.(20 pts.)
100 pts.) Score Note: Answer any 5 questions for a total score of 100 pts. Be sure to look over
all the questions first before beginning the exam, and indicate which five
questions are to be graded by checking the corresponding box.
M Day and Time
11:30 - 12:20
4:10-5:00 p.m. Instructors:
Yu Zhang Room
136 (new) 1. (20 pts) Consider the following eight representations of brominated hydrocarbons: 2-Bromo-2,2-dimethylpropane CH3CHBrCH=C(CH3)2
A B Br (CH3)2CHCH=CHCH2Br
C D 2-Bromo-4-methyl-4-pentene 1-Bromo-4-methylcyclopentene E F CH3
H Br H3C H
Br H G H In this collection, identify…
a) Two compounds that could have cis and trans isomers: C and D . b) A chemically meaningful but IUPAC incorrect name: __E__; write the correct name
here: _____ 4-Bromo-2-methyl-1-pentene_____________.
c) A chemically impossible structure (violates octet rule): __ B___.
d) Two representations of the same compound: __A__ and __H__.
e) All but one of the chemically reasonable structures [i.e. not the answer to question
(c)] have the same formula, which is: ___C6H11Br___. Which is the one with a
different formula? (hint: think of degrees of unsaturation): __F__. 2 2. (20 pts)
(a) (3 pts each) Form acceptable Lewis structures for the following uncharged
compounds by adding the necessary nonbonding electrons and multiple bonds:
H C C H H C O
Acetic Acid Acetylene (ethyne) H (b) (3 pts each) Assign formal charges as necessary to the atoms in the following Lewis
H C H N H O
Nitromethane H N O C O
Glycine (an amino acid) (c) (4 pts) A full description of nitromethane (above) requires two main resonance
structures. The one you completed was one; draw the other one:
O (d) (4 pts) Shown below is another commonly published representation of Glycine. Is this
a valid resonance structure of the version in (b)? Please circle your answer and explain.
Yes No H
H N C H O
C H O H Explanation:
The H (circled) has moved from N to O. Resonance structures
of a given species cannot have atoms in different positions. 3 3. (20 pts)
(a) (12 pts) Ions H2+ and He2+ have similar bond strengths (ca. 60 kcal/mol) but
neutralization of their positive charges by adding an electron results in bond
strengthening for one and dissociation for the other. Label the simple MO interaction
pictures below, and use them to help explain the similarity of the cations and the
difference between the neutralized systems.
Add 1e– Antibonding 1sA 1sB Add 1e– 1sA 1sB 1sA HB HeA 1sB 1sA HeB He A 1sB Bonding H2+ HA HB+ HA H2 He2+ "He2" HeB Explanation:
Both H2+ and He2+ have “half bonds” (more formally, bond order = 1/2 for
each)—i.e. there is one electron in the bonding orbital not cancelled by
one in the antibonding orbital, so their similar bond strengths make sense.
But the electron added to H2+ goes into the bonding orbital, strengthening
the H-H bond (from BDE = 63 to 104 kcal/mol), whereas that added to
He2+ goes into the antibonding orbital canceling the second bonding
electron and rupturing the bond (from BDE = 57 to 0); hence the quotation
marks around “He2”, a nonexistent entity except in He-He collisions.
(b) (8 pts) Tetraalkyl hydrazines, R2N-NR2 (where R is some alkyl substituent), have sp3
hybridized N centers and a single N-N bond essentially similar to the C-C bond in ethane,
with low barriers to rotation like those in the analogous ethanes. However, oxidation of
such hydrazines by one electron appears to (i) flatten out the N centers to sp2
hybridization, (ii) create a much larger rotation barrier for N-N rotation; and (iii) actually
strengthen the N-N bonding. Use orbital reasoning and diagrams like those above,
together with the thinking you went through above, to rationalize this result. (Hint: think
about the π system).
neutral π2 π*2 This story is like He2/He2+, but now it is π
orbitals that are filled (4e–; 2 in πbond, 2 in
π*antibond). They’re held together by a s
bond, but there’s no net π bonding, so
rotation is easy. Losing 1e– leaves only 1e–
in π* while π still has 2e– (i.e. π2 π*1), a net
π bond of order 1/2. Then, (i) to optimize π
overlap the N’s rehybridize to sp2 (flat);
(ii) the π bond creates a barrier to rotation;
(iii) it strengthens/shortens the N-N bond. Lose
cation π2 2pB
π π*1 R2NA R2N-NR2 NBR2 4 4. (20 pts) Consider the five Newman projections below.
H3C H3C H
CH3 H3C H H3C CH3 H CH3 H3C H CH3 H CH3 A H3C H3C H3C H H3C H3C H3C H H3C H3C H H CH3 B C D E (a) (2 pts) All of these diagrams depict alkanes of formula ___C6H14___.
(b) (4 pts) Structure __C__ represents a compound different from the other four. This
compound’s IUPAC name is ______2,2-Dimethylbutane______.
(c) (2 pts) The other four depict (IUPAC name):__ 2,3-Dimethylbutane______.
Of the four in (c),…
(d) (2 pts) …which two structures represent staggered conformations? __B__ and __D__.
(e) (2 pts) …which two structures represent energy maxima (high points) on the torsional
potential energy surface? __A__ and __E__.
(f) (4 pts) For two of the structures there are additional conformations of equal energy.
Which ones are they? __A__ and __D__. Draw their same-energy conformations by
completing the Newman projections below.
H3C H3C CH3
H3C H H3C H3C H
CH3 CH3 H
H F G (g) (4 pts) In the box below, take structure E to be at 0˚ and sketch the rotational potential
energy diagram, placing structures A-G (except for the misfit) in your picture.
Y F A
B -180˚ B -120˚ -60˚ 0˚ 5 60˚ 120˚ 180˚ 5. (20 pts)
(a) (12 pts) Draw structures to represent the following IUPAC names
Cl F (ii) 2,4-dimethyl-2-pentene (iii) 3,3-diethylpentane (b) (2 pts) Which of the compounds in (b) has no dipole? Please circle your answer (i) (ii) (iii) (c) (3 pts) Why don’t we have to give a number to show the double bond position in
compound (i), a substituted alkene?
In propene, there is only one place the double bond could go (C1-C2)
because its position is what defined the numbering direction, and the
chain is too short for there to be a non-terminal location for the alkene. (d) (3 pts) Why don’t we have to indicate (E) or (Z) in the name of (ii), an internal
One end of the double bond has two substituents (methyl groups) the
same, so switching them would not make a different compound. There is
no E/Z isomerism in this compound.
6 6. (20 pts)
(a) (6 pts) Fill in > or < symbols to rank the following pairs of groups according to the
Cahn-Ingold-Prelog priority rules:
Example: Br > C(CH3)3
> (b) (8 pts) Provide IUPAC names for the following two compounds: (i) (ii)
Cl (i) ______3,3-Dichlorocyclohexene_________________
(c) (6 pts) Imagine moving the internal double bond among the carbons inside the dotted
oval in the compound from (b)(ii). Don’t change the framework or move the vinyl
group around; it’s outside the oval. Seven possible alkenes could be generated that
(i) Draw the one you would expect to be lowest in energy (hint: double bonds like to
be next to double bonds, so 2,4-heptadiene is lower in energy than 2,5-heptadiene).
next to vinyl group (ii) How about highest (there are two essentially similar candidates). Double bond here
is also OK (though
not inside oval) Terminal
alkene 7 ...
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- Fall '08