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CHE
311
Answers
in
BOLD
RED
EXAM
3
Answer
Key
(Ch.
11­15)
Multiple
Choice
(60%;
2%
each)
Please
mark
the
letter
of
the
BEST
answer
to
each
question
clearly
on
your
answer
sheet.
Thank
you!
1.
The
MO
pi
bonding
model
of
benzene
consists
of
______
molecular
orbitals
in
total,
______
of
which
are
bonding.
a.
four
;
two
b.
eight
;
four
c.
six
;
three
d.
three
;
all
2.
The
common
name
xylene
is
used
to
refer
to
a
benzene
derivative
having
______________
on
the
ring.
a.
two
methyls
b.
a
hydroxyl
c.
an
amino
(‐NH2)
d.
a
carboxyl
3.
Which
of
these
benzylic
groups
will
NOT
oxidize
to
a
carboxyl
(‐COOH)
when
heated
with
a
strong
oxidant?
a.
–CH3
b.
–CH2CH3
c.
–C(CH3)3
d.
–CH2OH
4.
Which
of
these
annulenes
is
aromatic
according
to
Hückel’s
Rule?
only
C
(10
pi
electrons)
a.
b.
c.
d.
5.
The
most
common
alkenylbenzene
compound,
used
in
making
a
variety
of
polymers,
is:
a.
styrene
b.
anthracene
c.
butadiene
d.
naphthalene
6.
In
order
to
be
useful
for
substitution
reactions
of
aromatic
rings,
a
reagent
must
be
classified
as
a(an)
a.
carbocation
b.
nucleophile
c.
electrophile
d.
Lewis
base
7.
Sulfonation
of
benzene
is
unusual
in
two
ways;
which
two?
I.
The
active
reagent
is
NOT
an
ion
II.
Yields
are
quite
low
III.
It
is
very
reversible
IV.
Products
may
explode
a.
I
and
II
b.
II
and
III
c.
II
and
IV
d.
I
and
III
8.
In
Friedel‐Crafts
Alkylation,
one
of
the
main
complications
is
NOT
that
a.
alkyl
groups
may
rearrange
b.
alkyl
groups
may
come
off
easily
c.
products
have
increased
reactivity
d.
alkyl
halides
are
expensive
9.
An
example
of
an
activating
group
which
is
also
ortho,
para
directing
would
be
a.
–NO2
b.
–Cl
c.
–COOH
d.
–CH3
10.
The
normal
reagent
sequence
used
to
place
an
n­butyl
group
on
a
benzene
ring
would
be
a.
(CH3)3CHCl
+
AlCl3
b.
CH3CH2CH2CH2Br
+
AlBr3
c.
CH3CH2CH2COCl
+
AlCl3
d.
step
c,
then
HCl
+
Zn(Hg)
11.
If
you
treat
1‐bromo‐2‐nitrobenzene
with
Cl2
+
AlCl3,
where
will
the
Cl
most
likely
be
placed?
a.
C‐3
and
C‐4
b.
C­4
and
C­6
c.
C‐4
and
C‐5
d.
C‐3
and
C‐6
12.
Which
analytical
method
does
NOT
rely
on
the
presence
of
energy
levels
of
some
type
within
a
molecule?
a.
UV
b.
FTIR
c.
NMR
d.
MS
13.
The
two
common
forms
of
‘hyphenated
MS’
combine
Mass
Spectroscopy
with
a.
extraction
b.
filtration
c.
distillation
14.
The
hydrogens
on
the
second
carbon
atom
of
1‐chlorobutane
should
have
a
H‐NMR
signal
which
is
a.
a
triplet
b.
a
triplet
of
triplets
c.
a
quarter
d.
a
quintet
15.
Which
is
a
limitation
of
13C‐NMR
as
it
is
usually
performed?
a.
peaks
fall
over
a
narrow
range
of
ppm
values
b.
not
every
13C
atom
will
give
an
NMR
signal
c.
there
is
a
poor
correlation
of
shifts
with
structure
d.
peak
area
does
not
correspond
to
number
of
C
16.
In
Infrared
Spectroscopy,
the
peaks
that
are
largest
are
produced
by:
a.
the
most
frequent
functional
group
b.
nonpolar
bonds
c.
highly
polar
bonds
d.
symmetrical
bonds
17.
Which
of
the
following
should
have
a
significant
UV
spectrum
(produce
absorbances
in
the
UV
range)?
a.
CH2=CHCH2CH2COOH
b.
(CH3)2C=C=O
c.
Cyclopentene
d.
2­Phenylpropane
18.
Organometallic
reagents
like
Grignard
and
Lithium
reagents
are
usually
made
by
reacting
a.
metal
halides
+
alkenes
b.
metals
+
alkanes
c.
alkyl
halides
+
metals
d.
alcohols
+
metal
oxides
19.
When
a
Grignard
reagent
is
reacted
with
formaldehyde
(H2C=O)
and
then
worked
up
normally,
the
product
is
a.
a
primary
alcohol
b.
a
secondary
alcohol
c.
a
tertiary
alcohol
d.
a
hydrocarbon
+
CO
gas
d.
chromatography
CHE
311
Exam
3
Page
2
20.
If
we
react
excess
CH3CH2MgBr
with
the
ester
CH3CH2COOCH3,
the
major
product
formed
after
workup
is
a.
(CH3CH2)3COH
b.
CH3CH2C(=O)CH2CH3
c.
CH3CH2CH(OH)CH2CH3
d.
C(CH3CH2)4
21.
The
Simmons­Smith
reagent
(CH2I2
+
Zn(Cu))
is
used
to
transform
alkenes
into
a.
epoxides
b.
vicinal
di‐iodides
c.
cyclopropyls
d.
geminal
di‐iodides
22.
When
organometallic
chemists
study
the
bonding
&
stability
of
complexes,
they
see
if
the
electron
count
on
the
metal
is
______,
which
indicates
greater
stability.
a.
10
b.
8
c.
18
d.
32
23.
Ethanol
is
most
often
made
commercially
(for
use
as
a
solvent,
not
beverage)
by:
a.
reduction
of
CH3COOH
b.
hydrolysis
of
ethylene
oxide
c.
oxidation
of
CH3CH3
d.
hydration
of
CH2=CH2
24.
Reduction
of
aldehydes
and
ketones
by
NaBH4
has
several
advantages.
Which
is
NOT
one
of
them?
a.
NaBH4
is
a
mild
and
safe
reagent
to
use
b.
NaBH4
is
an
easier
&
safer
to
use
reagent
than
H2
gas
c.
NaBH4
is
stable
in
acidic
environments
d.
NaBH4will
not
reduce
C=C
bonds
like
H2
would
25.
To
reduce
carboxylic
acids,
esters,
or
their
salts
to
alcohols
the
preferred
reagent
is
usually
a.
NaBH4
b.
H2
+
Pt
c.
LiAlH4
d.
Na
in
liquid
NH3
26.
Epoxides
are
useful
starting
points
for
a
variety
of
products;
treatment
with
____________
gives
diols
while
their
reaction
with
Grignard
reagents
or
lithium
reagents
will
give
_______________.
a.
H3O+
;
alcohols
b.
HCl
;
alkenes
c.
H2O2
;
aldehydes
d.
OsO4
;
ethers
27.
Simple
symmetrical
ethers
can
be
produced
by
a.
heating
2°
alcohols
with
strong
acids
c.
warming
1°
alcohols
with
a
trace
of
acid
28.
Esters
are
made
via
Fischer
esterification
by
reacting
carboxylic
acids…
a.
with
alkenes
and
strong
acids
b.
with
alkyl
halides
and
AgNO3
c.
with
warm
alcohols
with
a
trace
of
acid
d.
with
epoxides
in
dilute
aqueous
acid
29.
To
convert
a
primary
alcohol
into
an
aldehyde,
the
best
of
these
reagents
to
use
would
be
a.
H2CrO4
b.
KMnO4
c.
OsO4
d.
PCC
or
PDC
30.
The
biochemical
reagents
that
are
most
often
used
to
oxidize
alcohols
to
aldehydes
are
called
a.
oxidases
b.
aldases
c.
dehydrogenases
d.
lipases
b.
treating
epoxides
with
alkyl
halides
and
base
d.
treating
epoxides
with
dilute
aqueous
acid
Problems
(40%;
15
&
25%
respectively)
I.
Bromination
of
nitrobenzene
produces
only
one
major
product.
Begin
by
drawing
the
proper
structure
for
nitrobenzene.
Then
show
how
bromination
occurs,
including
how
the
reactive
form
of
bromine
is
made.
Draw
cyclohexadienyl
cation
intermediates
(&
resonance
forms)
and
explain
why
only
one
product
is
formed.
:Br—Br:
+
AlBr3

Br+
+
AlBr4–
generates
the
positive
Br
ion
(or
at
least
polarizes
the
Br2
like
it)
O
N+
O–
O–
O
N
O
N
etc.,
puts
partial
+
at
the
ortho
and
para
O– positions,
making
them
less
reactive.
[can
omit
this
structural
detail
but
need
to
at
least
explain
meta]
Br
only
product
–O
N+
–
H+
O
O
N+
O–
O
Br+
resonance
contributors
keep
+
from
being
on
C
of
­NO2
group
O
O–
O
O–
O–
+
+
+
N
N
N
Br
Br
Br
H
H
H
CHE
311
Exam
3
Page
3
II.
A
clumsy
stockroom
worker
(totally
unlike
ours
at
TU)
has
forgotten
to
label
three
bottles
of
similar
organic
substances.
One
is
2‐phenylethanol.
Another
is
acetophenone
(C6H5C(=O)CH3)
and
the
third
is
phenylacetic
acid
(C6H5CH2COOH).
Smell
could
help
with
the
identity,
but
the
worker
has
a
cold,
so
samples
of
each
of
the
three
bottles
are
taken
and
H‐NMR,
C‐NMR,
and
FTIR
spectra
of
each
bottle
are
made.
Based
on
the
spectra
and
the
structures,
which
substance
is
in
each
bottle
(A,
B,
and
C).
You
must
explain
your
answer
with
at
least
TWO
spectroscopic
features
from
each
sample
(each
of
the
two
coming
from
a
different
method
–
for
example
you
could
use
the
IR
and
H‐NMR
or
samples
A
&
B
and
perhaps
the
H‐NMR
and
C‐NMR
of
C
to
determine
the
correct
identity
of
each,
which
ever
two
are
most
helpful).
Identities:
For
purposes
of
the
key,
spectra
are
shrunk
and
put
with
the
relevant
answer
Substance
A
is____Acetophenone__________
I
know
this
because:
There
are
two
key
features
in
the
spectra
that
make
this
ID
very
reasonable:
1. The
C­NMR
has
a
peak
at
198
ppm,
indicating
a
carbonyl
2. The
H­NMR
has
no
peak
below
8
ppm,
so
it
cannot
have
a
carboxyl
group
(which
would
occur
at
10­13
ppm).
3. There
is
no
sign
of
an
–OH
in
the
H­NMR
or
in
the
FTIR,
which
is
almost
featureless
at
the
area
that
should
occur.
4. Finally,
the
FTIR
shows
a
very
strong
carbonyl
signal
at
1686
WN
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!
!
!
CHE
311
Exam
3
Page
4
Substance
B
is
___Phenylacetic
acid____
I
know
this
because:
1. The
FTIR
has
BOTH
a
carbonyl
at
c.
1720
WN
and
a
broad
–OH
signal
at
c.
3100
WN,
confirming
the
carboxylic
acid
group.
2. The
H­NMR
has
the
intense
methylene
signal
at
3.7
ppm,
not
found
in
the
other
substance
still
unidentified
(phenylethanol);
more
importantly,
the
carboxyl
H
at
c.
12
ppm
is
a
strong
proof.
3. The
C­NMR
is
not
very
helpful
in
this
case.
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CDCl3
QE-300
240
220
200
180
160
140
120
100
80
60
40
20
0
12
11
10
9
8
7
6
5
4
3
2
1
0
CHE
311
Exam
3
Page
5
Substance
C
is
___2­Phenylethanol____
I
know
this
because:
1. The
FTIR
has
a
broad
–OH
signal
at
3348
WN,
confirming
the
alcohol
group,
but
no
strong
carbonyl
signal
(the
weak
one
at
c.
1722
may
be
a
trace
of
oxidized
material).
2. The
H­NMR
is
a
very
strong
confirmation
because
it
has
the
proper
shifts
and
splitting
for
the
compound
in
question
(note
the
alcohol
OH
is
at
the
unusual
shift
of
2.0
ppm).
The
aromatic
H
are
at
c.
7.2;
the
–CH2­O
is
at
3.8
while
the
other
–CH2–
is
found
at
2.8
ppm.
These
methylenes
couple
and
split
each
other
into
the
triplets
that
are
observed.
3. The
C­NMR
is
most
helpful
in
a
negative
way:
it
has
no
peak
in
the
carbonyl
region
(170­210).
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!
!
!