Identification & Assay of Aspirin ,Isonicotinic acid Hydrazide (I.N.H) ,Chlorbutanol ,tetracycline,Oxytetracycline, Chlortetracycline, Nalidixic acid, Naproxen, Naproxen, Busulfan, Dipyrone

1. Done by : Marwa Abd El Kader

2. Prelab 1

3. Alkalimetry
(tab.)
Bromometry
(pd.)
Std. Alkali
Std. Br 2

4. 1) Bromometry
Aspirin
NaOH, Boil
C O O N a
O H
Salicylic
acid
Kn. Xss Std. Br2
COOH
OH
B r
O H
B rB r
Tribromo phenol
(TBP) yellow ppt
I 2
Na2S2O3
Starch
ind.
NaOHHClStd. Br2
Decarboxilative bromination
(as Br2 brominates phenolic
cpds;that’s why I changed
aspirin to salicylic acid
5 ml
CHCl3
with
shaking

5. Here we have a solution of precisely weighed potassium iodate with
added HCl that has just been served with an excess of potassium
iodide. A disproportionation has occurred in the oxidation states of
the iodine in iodate and iodide to produce elemental iodine
complexed with the excess iodide to produce the triodide ion. A
deep red-brown solution results (almost black)
Iodate is titrated with thiosulfate solution in the
process of standardizing it, the solution first turns a
dark red, then orange, then yellow.
Straw yellow color before adding
the starch indicator.

6. Straw yellow
colour

7. After adding starch
Shake vigrously , e.p 
disappearance of blue colour

8. Volatile
Toxic
(sterility to
males)
Kn. Xss std
KBrO3/KBr mix.
Sample
St. Br2
Conc. HCl

9. BrO3/Br
Br2
Na
salicylate
Salicylic
acid
KI
I2
Br2 / C. HCl
Imp.

10. Interference
Adsorption
of Br2 & I2
on the
surface
No filtration
 Bec. I2 &
Br2 are
volatile
Add CHCl3 or
CCl4
Dissolves
TBP

11. To the formation of the 1st strong
complex with I2
We can’t use bromometry in aspirin
tablets’ assay ( as tab. contains
starch as binder)

12. Calculations
:
Recovery % =
Practical wt
Theoretical wt
X 100
( Kn.xss*f – mls Na2S2O3 *f) x F x D.F
Kn xss  25 ml but bec. of
it’s 24.8 ml
Equivalent factor  1ml
std ≡ …. gm sample
N.B. Most drugs’
recovery % = 95- 105
What’s
done /
what’s
taken
Gm % W/W

13. C % =
Practical wt
Vol.
X 100 Gm % W/V
• ? Std. ≡ ? Sample ( from
balanced eq. , here in case
of aspirin we can keep that
it gave me 3 Br2 )
• Conc. of std.
F =
N.B.
• Mole = M.wt on the balance
• Molar = Mole in 1 L

14. F = Std ≡ Sample
3 Br2 ≡ 1 Aspirin
3M Br2 ≡ 1M Aspirin
1M Br2 ≡ 1/3M Aspirin
M Br2 ≡ 1M Aspirin
20 3x20
1ml M Br2 ≡ gm M.wt Aspirin
20 3x20x 1000
÷ 3
We was
working
with M/20
Any
no.s up
write it
here
down
Bel nazar ;)

15. C O O N a
O H
2) Alkalimetry:
Aspirin
Kn. Xss std NaOH
Std HCl
NaOH
+ CH3COONa
M/20
NaOH
Boil
gently
Grind 2
tab.

17. Identification Tests
Aspirin pd + 4ml 2M NaOH + H2SO4
+ FeCl3 dps 
Boil 1
min
Salicylic
Pd + 2ml Amm.molybdate 
Chloroquine + Trinitrophenol (picric) 

18. Prelab 2
1) Isonicotinic acid Hydrazide (I.N.H)  Bromometry (Redox)
I.N.H
Kn. Xss std. Br2
I 2
Na2S2O3
Starch
ind.
N/20 std Br2C. HCl
Stand
+ 2 Br2 + H3O
(oxidative cleavage)
+
+ N2 +4 HBr
Reducing agent

19. • Any redox reaction needs acid
as a catalyst .
• If the diluting solvent isn’t
mention, use H2O.

20. BrO3/Br
Br
I.N.H
Oxidative
Cleavage
KI
I2
Br2 / C. HCl
Imp.

21. Calculations
:
The same idea as before :R%, C% & F
No. of electrons gained or lost or no. of
H or OH e.g. 1M H 2 SO4 = 2 N+

22. Identification Tests
T.T 1  I.N.H pd +1ml H2O || T.T 2  2ml FehlingA+ 2ml
Fehling B Red ppt (reduction of fehling)heat
T.T 1 T.T 2

23. I.N.H pd +anh. Na2CO3  pyridine odour
I.N.H pd + H2O Vanillin + H2O
(conc. Soln.), heat
With scratching
- H2O
Schiff’s
base
N
C
O
NH N CH
OCH3
OH
Yellow ppt

24. 2) Chlorbutanol (Antimicrobial preservative):

25. Volhard's MethodMohr's MethodP.O.C
Back (acidic medium)DirectType of Titration
Std.1 (kn.xss):
AgNO3
AgNO3Titrant
Std.2 (titrant):
NH4SCN
Ferric ammonium
sulfate
(FeNH4(SO4)2)
Potassium chromate
((K2Cr2O7
Indicator
When all the chloride is
converted to AgCl, the
AgNO3 left is back
titrated against std.
NH4SCN ,the next
excess of SCN reacts
with the indicator and
gives a red color ferrous
thiocyanate (Fe(SCN)3)
complex.
When all the chloride
existing in solution is
completely precipitated
as AgCl, the next excess
drop of the titrant Ag
chromate the color of
the solution changes
from yellow to a red
ppt.
At The End Point

26. Chlorobutanol heated with NaOH
(To transform the organic aliphatic
halide to inorganic one)
3 NaCl
Chlorobutanol
Kn. Xss AgNO3
Isopropanol
Dil. HNO3
AgCl
NH4SCN
Ferric
allum
ind.
NaCl+ AgNO3 AgCl + NaNO3
AgNO3 + NH4SCN  AgSCN+ NH4NO3
NH4SCN+ FeNH4(SO4)2
[Fe(SCN)3] + 2 (NH4)2SO4

27. e.p white ppt in a buff
solution

28. 1 is Unsuccessful & always
dissociating called AgCl
( Ksp)
1 is successful,respected &
always relaxed & at ease
called AgSCN ( Ksp)

29. If the problem is with the dissociated Ag ions
of the AgCl, we must get rid of that AgCl ,
mmm boiling& filtration we’ll not work well as
coagulation of ppt will occur , thus dec. the
S.A ,so why not to coat them & cross them
over as if we’ll do that to act like a bridge in
order not to fall down & get stuck in that
tight corner?!
But How ?!!!
Ahha, I knew ,I’ll
use Isopropanol or
nitrobenzene ;)
+

30. Ksp AgCl > Ksp AgSCN
Ksp AgI,AgBr < AgSCN
To ensure no alkalinity
AgNO3 AgOH AgO2 (black ppt.)
Fe 3+ Fe(OH)3
OH
OH

31. Calculations
:
The same idea as before :R%, C% & F
AgNO3 =Chlorbutanol= 3NaCl
Cyclophosphamide &
Enteroquine have the
same principle

32. Identification Tests
Oxytetracycline Chlortetracyclinetetracycline
Pd + 1ml conc. H2SO4 [dehydration  inc. conj. color]
{tautomerism}
Deep
crimson
red (violet)
Red
(pink)
color
Deep blue
to green
color
+ FeCl3
Brown color
(phenolic
cmpd)

33. Prelab 3
Bu sulfanIbuprofenNaproxenNalidixic acid
P.O.C
Direct AFTER
HYDROLYSIS
2
Me sulfonic acid
DirectDirectBack (due to
intramolecular H-
bond & Zwitter ion
formation non polar
insol. in H2O,slightly
sol. In alc.)
Type of
Titration
NoYes, as naproxen
but till 250ml
Yes(H2O [all at
once], alc. till
100 [ shwaya b
shwaya)
NoDilution
M/20 NaOHM/20 NaOHM/40 NaOHStd. acid (we haven’t
worked it in lab)
Titrant
Acid –Base Titration

34. Calculations
:
The same idea as before :R%, C% & F
Given as “each 1 ml
0.1M NaOH is ≡ 23.03
mg naproxen” ,since we
worked with M/40 , so
we’ve to divide this
value by 4
We don’t have M/40 in
lab , so we’ll dil. M/20
to M/40 
50ml NaOH M/20 +
50ml H2O M/40
2 NaOH = Busulfan
≡ 2

35. Identification Tests
Pd = 2ml conc. HCl + 0.5ml Soln. β naphthol in alc. 
orange red ppt
Charge transfer complex
Electron
rich
Electron
poor

36. Pd. + 2ml H2O + 1ml conc. H2SO4 + 0.2 gm
vanillin yellow to orange pptBoil
- H2O
Conj.
Compound
(orange ppt.)
Dehydrating agent
is H2SO4
Active methylene
 easy to
liberate OH

37. Giriseofulvin:
Pd + 2ml conc. H2SO4 + few K2Cr2O7  Red color
Oxid. reaction
Quinoid
str. (color)

38. Prelab 4
1) Dipyrone
Dipyrone +
distilled H2O
1ml glacial
acetic acid
N/10 I2
+I2
+HI + NaHSO4
CH2
Rose red
color
(Dimer)

39. Identification Tests
Dipyrone pd +10ml dist. H2O
2ml
+2ml dil. HCl  Odour of SO2
Odour of formaldehyde
+1.5 ml dil. HCl +1ml FeCl3
bluered color disappear
+ 1ml K- pyronantimonate 
white ppt.

40. http://www.csudh.edu/oliver/demos/hh-
cubr/hh-cubr.htm
http://www.edu.xunta.es/centros/iesbreamo/
galeria/displayimage.php?album=141&pos=15
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