Inorganic-Chemical Practical Course
Spring Semester 2014
[I(py)
2
][NO
3
], K[ICl
4
] and Ph
3
P
.
BH
3
26.05.2014
Yann Baumgartner
A1

Yann Baumgartner
1 Aim of the experiment
Exp #6
The goal of this experiment was the synthesis of dipyridineiodine nitrate [I(py)
2
][NO
3
], potassium tetrachloroiodate K[ICl
4
] and triphenylphosphane borane Ph
3
P
.
BH
3
. Some exercise was provided in shapes, isomerism and mass spectroscopy in the question part.
2 Introduction
Three different p-block compounds had to be formed in this experiment, showing unexpected oxidations states of halogens and some coordination chemistry. P-block and coordination chemistry is a versatile field of chemistry covering a large section of chemistry. P-block chemistry is, as the name says, the chemistry of the p-orbitals. All elements of the group 13 –
18 are part of this group, where the electron configurations of the noble gases are their aim.
The noble gases are often referred to as inert gases, which is not totally true. The noble gases can still undergo some reactions leading to unexpected compounds. Some research has been made in the field of oxidation states of halogens to determine if they can form stable compounds being oxidized, as the halogens are close to reach the goal of the noble gas configuration. The findings lead to surprising results as e.g. I
+ is not stable but forms clusters to stabilize the oxidation such as I
3
+
, I
5
+
and IO
+
. I
+
opened new possibilities in organic chemistry as it can be used to replace a proton in an organic compound, inserting a new reactive group. Other elements of the p-block chemistry such as boron also form interesting compounds as it binds with four rests even though it only has 3 valence electrons, defying the rules. The fourth bond is a so called coordination bond. The only difference between a coordination bond and a normal bond is that both electrons needed for the coordination bond come from one compound and not one from each as usual. This means that the fourth compound needs to be a lone pair donor. This attracted several research groups into investigating phosphine-boranes for a long time, creating a large amount of different phosphine-boranes which can be used as activating and protecting groups at the same time in stereospecific synthesis.
[1-5]
2

Yann Baumgartner
3 Experimental part
3.1 Synthesis of [I(py)
2
][NO
3
]
[1]
Exp #6
AgNO
3
170 g/mol
1.69 g
10.0 mmol
1.0 eq.
I
2
127 g/mol
2.61 g
20.5 mmol
2.0 eq.
C
10
H
10
IN
3
O
3
347 g/mol
2.06 g
5.94 mmol
59 %
Silver nitrate ( 1 , 1.69 g, 10.0 mmol, 1.0 eq.) was added to pyridine (10 mL) and stirred at
50°C until complete dissolution. The solution was cooled in an ice bath and chloroform (10 mL) was added to the reaction mixture. Iodine ( 2 , 2.61 g, 20.5 mmol, 2.0 eq.) was added and the reaction mixture stirred for 2.0 hours. Silver iodide was filtered off, and the filtrate was slowly poured into stirring diethyl ether (20 mL). The solid was collected by filtration and washed with ether containing a little pyridine (18 mL/ 2 mL) to obtain [I(py)
2
][NO
3
] ( 3 , 2.06 g, 5.94 mmol, 59 %) as a yellow solid.
IR (

/cm
-1
):3095 (w,

C-H arom.
), 1600 (m,

C=C arom
), 1453 (m,

C=C arom
), 1331 (s,

C=N arom
),
1052 (m,
 δ
C-H
), 1009 (w, δ
C-H
), 756 (st,
 δ oop, C-H
), 687 (st,
 δ oop, C-H
), 635 (st,
 δ oop, C-H
).
1
H-NMR (250 MHz, 295 K, CDCl
3
,

/ppm) : 9.18 – 8.48 (m, 4H, 1 ), 8.48 – 7.87 (m, 2H, 3 ),
7.87 – 7.38 (m, 4H, 2 ), ac_prakt_.2898
3

Yann Baumgartner
3.2 Synthesis of K[ICl
4
]
Exp #6
K[IO
3
]
214 g/mol
3.00 g
HCl
36.5 g/mol
5.51 g
10 mL (37%)
1.49 g/mL (37%)
K[ICl
4
]
308 g/mol
1.02 g
14.0 mmol
1.0 eq.
151 mmol
11 eq.
3.31 mmol
24 %
Conc. HCl ( 5 , 10 mL, 5.51 g, 151 mmol, 11 eq.) was added in 1 mL portions to K[IO
3
] ( 4 ,
3.00 g, 14.0 mmol, 1.0 eq.) and stirred for 2 hours at 60°C. The deeply orange solution was slowly cooled to room temperature and then put in the fridge for 45 minutes. The solid was collected by filtration to obtain K[ICl
4
] ( 6 , 1.02 g, 3.31 mmol, 24 %) as yellow crystals. The product was then destroyed with NaOH (1M).
3.3 Synthesis of BH
3
.
PPh
3
[3]
NaBH
4
37.8 g/mol
0.681 g
C
4
H
10
BF
3
O C
18
H
15
P
142 g/mol 262 g/mol
3.01 g
2.67 mL
4.72 g
1.13 g/mL
18.0 mmol 21.3 mmol 18.0 mmol
1.0 eq. 1.2 eq. 1.0 eq.
C
18
H
18
BP
276 g/mol
4.12 g
14.9 mmol
83 %
This reaction was carried out under inert conditions. A suspension of NaBH
4
( 7 , 0.681 g, 18.0 mmol, 1.0 eq.) in THF (25 mL) was cooled in an ice bath and stirred. A solution of BF
3
.
Et
2
O
( 8 , 2.67 mL, 3.01 g, 21.3 mmol, 1.2 eq.) in THF (10 mL) was added drop wise and the reaction mixture stirred for 45 minutes. A solution of PPh
3
( 9 , 4.72 g, 18.0 mmol, 1.0 eq.) in
4

Yann Baumgartner Exp #6
THF (20 mL) was added drop wise and the reaction mixture stirred for 1.0 hour. The reaction mixture was allowed to warm to room temperature and filtered. The solvent of the filtrate was evaporated to obtain BH
3
.
PPh
3
( 10 , 4.12 g, 14.9 mmol, 83 %, Lit
[3]
: 100 %) as a white solid.
Mp.: 185°C (Lit
[3]
: 184.5 - 187°C)
1
H-NMR (250 MHz, 295 K, CDCl
3
,

/ppm) : 7.63 – 7.39 (m, 15H, 1-3 ), 1.52 (q, 3H,
1
J
HH
=
45 Hz, 4 ), ac_prakt_.2889
IR (

/cm
-1
):2981 (w,

C-H arom.
) , 2376 (m,

B-H
), 1434 (m,

C=C arom
), 1103 (m, δ
C-H
), 1055 (m,
δ
C-H
), 733 (st,
 δ oop, C-H
), 689 (st,
 δ oop, C-H
).
4 Discussion
4.1 Synthesis of [I(py)
2
][NO
3
]
The synthesis of dipyridineiodine nitrate ( 3 ) worked well. Small adjustments to the synthesis were made to allow better handling as follows. The amount of pyridine was raised to allow
AgNO
3
( 1 ) to dissolve completely, therefore the crude product did not crystallise after cooling the solution and the chloroform was not used to dissolve the crude product but was just added to the reaction mixture. The reaction times were also longer than prescribed due to simultaneous work and the lunch break. The analyses were conclusive with the product. In the first step the pyridine forms a dipyridine complex with silver, which then is replaced by iodine forming the final product.
5

Yann Baumgartner
4.2 Synthesis of K[ICl
4
]
Exp #6
The synthesis of potassium tetrachloroiodate ( 6 ) went well and special attention was paid to avoid any contact with metal spatula as it is corrosive. There were no adjustments made to the procedure and the final product 6 was destroyed to avoid decompositions and reaction with anything else which may get in contact with. No analyses were performed for this experiment.
In this experiment the iodine is reduced by losing its 3 oxygen atoms and gaining 4 chloride atoms. As iodine would much prefer to have a full valence shell, the final product decomposes by losing two chlorine, resulting in further reduction of the iodine.
4.3 Synthesis of BH
3
.
PPh
3
The synthese of triphenylphosphane borane ( 10 ) was conducted under inert conditions using
Schlenk line techniques. The synthesis went well and no adjustments had to be done. In the first step, NaBH
4
( 7 ) and BF
3
.
Et
2
O ( 8 ) reacted together to form BH
3
which then coordinates with PPh
3
( 10 ) to form the desired product. The characterisation of the protons in the
1
H-
NMR spectrum could not be conducted in much detail because the signals of the protons are overlapping each other in a higher order spectrum. The integral also could not be used as the only reference being the signal of the protons coupling with
11
B are overlapping with the water peak of the residual water in the solvent.
5 Conclusion
The synthesis of [I(py)
2
][NO
3
], K[ICl
4
] and Ph
3
P
.
BH
3 were successfully achieved. We had again the opportunity to practice working under inert conditions, and we had the opportunity to learn a little bit about p-block and coordination chemistry. The questions which had to be answered gave practice in isomers, shapes and electrochemistry as well as in mass spectroscopic analysis.
6

Yann Baumgartner
6 References
[1] Garret and Gillespie, Inorg. Chem., 1965 , 4, 563.
[2] Tsuneo Imamoto, Pure & Appl. Chem. 1993, 65 , 655 – 660
[3] R.C. Moore et al. Inorg. Synth.
1970, 12 , 110.
[4] McNulty et al , Tetrahedron Letters , 2004, 45 , p. 407
[5] C.E. Housecroft, AC1 lecture skript, Uni. Basel
7 Answers to the questions
7.1.1 What is the oxidation state of I in [I(py)
2
]
+
?
The oxidation state of I is +1
7.1.2 Use VSEPR rule to work out the structures of [I
3
]
+
and [I
3
]

.
Exp #6
7.1.3 Draw the structure of [I(py)
2
]
+ and explain how you decided what the structure is.
Why is the structure of [I(py)
2
]
+ similar to that of [I
3
]

and not [I
3
]
+
?
The structure of [I(py)
2
]
+
is as shown because the iodine in the middle possess 6 electrons, and the binding electrons come from the nitrogen. This means that iodine has 3 lone pairs. As the lone pair – lone pair repulsion is greater than bond – bond repulsion, the lone pairs are placed trigonal planar around the iodine, and the pyridines linear towards the iodine. The structure of [I(py)
2
]
+
is similar to that of [I
3
]

because in [I
3
]

the iodine in the middle also possess three lone pairs which have to be placed with the least interaction between each other.
7

Yann Baumgartner Exp #6
7.1.4 The mass spectrum (positive mode) of [I(py)
2
)]
+ is shown below. Work out from the spectrum how many isotopes naturally occurring iodine has. Explain how you reach your answer.
The peak pattern arises from the [I(py)
2
]
+
ion with its basic peak at 285 m/z. The peak at 285 m/z has an intensity of 100, and the peak at 286 m/z has an intensity of 11. As there are 10 carbon atoms in [I(py)
2
]
+
and the natural occurrence of
13
C is 1.1 % it gives the intensity of 11 in the mass spectrum. The isotopes of nitrogen can be neglected because there are 2 nitrogen atoms for 10 carbon atoms, and the natural occurrence of
14
N is 99.64%. The peak with the very low intensity at 287 m/z belongs to a [I(py)
2
]
+ cation containing 2
13
C atoms. This all points to the fact, that iodine does not have a stable isotope occurring at an observable level.
7.2.1 Using the half equations below, explain what happened in the reaction, including careful balancing of the oxidation state changes. Write a balanced equation for the overall reaction. What role does the K
+
ion play?
Cl
2
+ 2e

2Cl

[IO
3
]

+ 6H
+
+ 2e

+ 4Cl

[ICl
4
]

+ 3H
2
O
Balanced equation: K
I
[I
V
O

II
3
] + 6H
I
Cl

I
Cl
0
2
+ K
I
[I
III
Cl

I
4
] + 3H
I
2
O

II
8

Yann Baumgartner Exp #6
The iodine ion is reduced from ox. state V to III, the four chlorine ions that end up in [ICl
4
]
 do not change their ox. state, but the two chlorine ions that end up in Cl
2
are oxidized from ox. state

I to 0, balancing the ox. state changes from the overall reaction. The potassium ion plays the role of the counter-ion, but is not directly involved in the reaction.
7.2.2 Predict the structures of the following using VSEPR rules: a) [ICl
2
]
 b) [BrF
2
]
+
linear
bent c) [ClF
4
]
+ d) I
2
Cl
6 bi square planar bisphenoidal e) [ICl
4
]
 f) [BrF
6
]
+
square planar octaeder g) IF
5 h) IF
7
square pyramidal pentagonal bipyramidal
7.2.3 In the mass spectrum (negative mode), the highest mass peak of K[ICl
4
] appears as follows. Explain how the isotope pattern arises.
9

Yann Baumgartner Exp #6
The peak pattern is due to the four chlorine atoms present in [ICl
4
]

. Chlorine has two isotopes which occur in an observable amount. The
35
Cl has an occurrence of 75.78 % and the
37
Cl has an occurrence of 24.22 %. As there are four chlorine atoms in the anion, there are 5 possible combinations for the [ICl
4
]
 anion. The peak at 266.78 m/z has four
35
Cl and no
37
Cl and the peak at 274.78 has no
35
Cl and four
37
Cl. The intensity of the peaks can be explained by statistics using the given percentages.
7.2.4 K[ICl
4
] slowly decomposes. One of the products has a strong smell. No other reagents (e.q. H
2
O or O
2
) are involved in the decomposition. Write an equation for the decomposition reaction. Which product gives the smell?
The product which gives the strong smell is chlorine.
7.3.1 What does the {
1
H} notation mean? The spectrum below shows the
31
P{
1
H} NMR spectrum of another R
3
PBH
3
compound, structurally analogous to your product.
The {
1
H} notation means that the spectrum is proton decoupled. That way the coupling towards protons are not taken in account. The observable signal splits to a non-binomial quartet because
11
B has a spin of 3/2.
10

Yann Baumgartner Exp #6
7.3.2 The spectrum below is the 96 MHz
11
B NMR spectrum of PhMe
2
PBH
3
. Explain the appearance of the spectrum. What coupling constants (in Hz) can you measure from the spectrum? Copy the spectrum into your protocol, and mark on where to measure these coupling constants.
1
J
BH
= 73 Hz
1
J
BP
= 44 Hz
7.3.3 How does a Lewis base differ from a Br ø nsted base?
A Br ø nsted base is a proton acceptor and a Lewis base is a lone pair donor.
7.3.4 Give two examples of each of a Lewis base, a Lewis acid, a Br ø nsted base and a
Br ø nsted acid.
Lewis base: Et
2
O, THF
Lewis acid: AlCl
3
, BF
3
Br ø nsted base: NH
3
, NaOH
Br ø nsted acid: H
2
SO
4
, HCl
11

Yann Baumgartner Exp #6
7.3.5 How does the
11
B{
1
H} NMR spectrum of a solution of THF
.
BH
3
differ from the
11
B
NMR spectrum of the same solution?
In the
11
B{
1
H} NMR there would be a singlet, in the
11
B NMR there would be a binomial quartet.
7.3.6 When PMe
3
is added to a solution of THF
.
BH
3
, the
11
B{
1
H} NMR spectrum changes from a singlet to a doublet. Why?
PMe
3
forms a more stable compound with BH
3
than THF. This means, that if PMe
3
is added, a ligand exchange will happen and the obtained compound is PMe
3
.
BH
3
. Because
31
P has a nuclear spin of ½, the signal of the B will split to a doublet.
7.3.7 The mass spectrum of Et
3
N
.
BH
3 is shown below. Assign as many peaks as you can in the spectrum.
The peak at 115 m/z belongs to the [Et
3
N
.
BH
3
]
+ ion, the peak at 101 m/z belongs to [Et
3
N]
+
, the peak at 87 m/z belongs to [Et
2
MeN]
+
, the peak at 72 m/z belongs to [Et
2
N]
+
, the peak at
12

Yann Baumgartner Exp #6
58 m/z belongs to [MeEtN]
+
, the peak at 43 m/z belongs to [NEt]
+
and the peak at 29 m/z belongs to [Et]
+
.
8. Spectra
13

Yann Baumgartner i)
1
H-NMR spectrum of dipyridineiodine nitrate ( 3 )
Exp #6
14

Yann Baumgartner ii) IR spectrum of dipyridineiodine nitrate ( 3 )
Exp #6
15

Yann Baumgartner iii)
1
H-NMR spectrum of triphenylphosphane borane ( 10 )
Exp #6
16

Yann Baumgartner iv) IR spectrum of triphenylphosphane borane ( 10 )
Exp #6
17