MOis the least basic due to its transition element having the highest oxidation state, which in turn, makes the molecule have more covalent bonds. The covalent bonds prevent the formation of OH ions when the molecule is dissociated in water.

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CHM 262


Name ______________________

Test 3, Winter 2011


1. If M represents a transition element, which of its oxides should be the least basic and

why? MO, M
2
O
3
, MO
2
, and MO
3



MO
3

is the least basic due to its transition element having the highest
oxidation state,
which in turn, makes the molecule have more covalent bonds. The covalent bonds
prevent the formation of OH
-

ions when the molecule is dissociated in water.



2. Several years ago people who were taking supplements of tryptophan (an essenti
al

amino acid that we must take in with our food for normal growth and maintenance of

a healthy body; without it we could not live) started becoming very ill. Some

individuals suggested that tryptophan was what was causing the disease. For most

amino acids
, the alpha carbon (i.e., the carbon to which a hydrogen, the amino group,

and the carboxyl group are bound) is asymmetric, which means that the four groups

on this carbon can be arranged in two different manners. This causes amino acids to

have two stereo
isomers. One stereoisomer is called “
D
” and the other is called “
L
”.

Only “
L
” amino acids are incorporated into proteins. Suppose the problem with the

tryptophan was that it was contaminated with the “
D
” isomer and this is what caused

the problem. How woul
d you go about determining that the tryptophan is

contaminated with the
D
isomer?



Do determine if the D isomer is contaminated, a polarimeter

would be used to analyze
the rotation of light on each isomer. If the D isomer is not contaminated, it should have
the same rotation as the L isomer. If the D isomer is contaminated, the rotation of light
will be intermediate of the D and L isomers.



3.
The
S
values for the following organic compounds, CH
4
, C
2
H
6
, C
3
H
8
, and C
4
H
10
, is, in

J/mol

K, 186.1, 229.5, 269.9, and 310.0, respectively. What would you estimate the
S
0

value for n
-
pentane (C
5
H
12
) to be? Briefly explain your answer.



The
S
0

value would approximately be 350
-
352 J/mol. As each alkane increases its
amount of atoms, the

S
0

value goes up by about 40 J/mol for each new molecule. Using
an average of the J/mol amounts added to each alkane, and adding it to the
S
0

value of
C
4
H
10
, giv
es the estimated amount of 351.3 J/mol.


4. You can make an ester from a carboxylic acid and an alcohol or hydrolyze an ester to

the corresponding alcohol and carboxylic acid under similar conditions but at

identical temperatures. Does this negate the seco
nd law of thermodynamics? In

which direction is the reaction spontaneous and what must you do to make the
nonspontaneous

reaction spontaneous, without changing temperature?



This reaction does not negate the second law of thermodynamics. When the water is

removed from the right side of the equation, it leaves the temperature identical, an
d
allows the ester to hydrolyze into carboxylic acid and an alcohol. The reaction is
spontaneous on
the left side of the equation. To make the non
-
spontaneous side
spontan
eous, water can be added to the other side as well as a catalyst.



5. Two isomers of [Pt(NH
3
)
2
Cl
2
] exist, cis
-
platin and trans
-
platin. First, show the

structures of these two compounds. Then explain why cis
-
platin is an excellent anticancer

agent and tran
s
-
platin is not.











Cis
-
platin





Trans
-
platin


Cis
-
platin is an excellent anti
-
cancer agent due to its structure in which it can intercalate
between DNA bases and prevent unwinding. The Cl

atoms placement allows them to
react with nitrogen bases. Trans
-
platin does not have the same structure because of the
placement of the Cl atoms. The trans
-
platin Cl atoms end up too far from the nitrogen
bases to react with them.



6. Using crystal field

theory, explain why the type of ligand used in a coordinate

covalent complex determines the color of the complex.


Differences in s
plitting energies, means that the ligands will absorb various wavelengths
of visible light. Wavelengths that are not absorbe
d are shown as their respective color
which corresponds to specific wavelength.





7. A transition metal complex (complex 1) with a particular ligand is orange in color,

which means it absorbs blue light at 460.0 nm. A second complex of the metal

(complex

2) with a different ligand is purple in color, which means it absorbs yellow

light at 570.0 nm. Which ligand (i.e., ligand 1 or ligand 2) is a strong ligand, also,

what are the splitting energies (

’s) in kJ/mol for the two complexes, and finally,

what wa
velength of light is equivalent to the difference in the splitting energies?

Fully explain your answers. h = 6.62606931 × 10
-
34
J

s, c = 2.99792458 × 10
8

m/s.


Ligand 1 is a strong ligand because it has a larger splitting energy.

Ligand 1 has a splitting e
nergy of 260.05 kJ/mol

Ligand 2 has a splitting energy of 209.87 kJ/mol

The difference between the splitting energies of these two ligands is 50.18 kJ/mol,
which
is equivalent to a wavelength of 2383.34 nm.



8. The atomic radius of Sn is 140 pm. If Sn

forms a closest packed crystal structure,

what is the diameter of the octahedral hole formed in the Sn crystal and what atom(s)

would fit in them?









9. Amino acids are composed of an amine group (i.e.,
-
NH
2
) and a carboxylic acid group

(i.e., COOH).

However, for the amino acids found in protein, the amino acid group

and carboxylic acid group are both on the
α
-
carbon that also contains some other

group that defines the amino acid. For example, if the fourth group is an H, the

amino acid is glycine, if

it is a methyl group (CH
3
-
), the amino acid is alanine. For the

amino acid alanine, the
K
a
for the carboxylic acid is 4.467 × 10
-
3

and the
K
b
of the

amino group is 7.413 × 10
-
5
. If you dissolve 0.1250 moles of alanine in 1.000 L of

water, what will be the

pH of the solution?










10. If you take
d
-
2
-
butanol, which is optically active because there are four different

groups on the second carbon, and treat it with HI, you produce 2
-
iodo
-
butane, which

is

optically inactive even though there are still four different groups on the second

carbon. Fully explain this observation.


In the d
-
2
-
butanol molecule, the second carbon has is a tetrahedral, asymmetrical
shape. When HI is added to the compound, the carb
on’s hybridization changes from sp
3
to

sp
2
. This causes the molecular geometry to change to a flat, symmetrical, trigonal
planar, shape.

Because the shape is now flat and symmetrical, this gives two sides.
making the 2
-
iodo
-
butane molecule a racemic

mixture
. U
sing a polarimeter to measure
the light rotations, it is determinable that the D and L sides are rotating in opposite
directions,
cancelling each other out, therefore making the molecule optically inactive.