Date post: | 14-Apr-2017 |
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Practical Pharmaceutical
ChemisteryDone by : Marwa Abd El
Kader
Assay
Titrimetric
Direct (1 std)
Back (2 stds)
Spectro
Prelab 1
Alkalimetry (tab.)
Bromometry (pd.)
Std. Alkali
Std. Br 2
1 )Bromometry
AspirinNaOH, Boil HCl
COONa
OHSalicylic
acidKn. Xss Std. Br2
COOH
OH
Br
OH
BrBr
Tribromo phenol (TBP) yellow ppt
I 2
KI / HCl
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
Just For Illustration 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.
Straw yellow colour
After adding starch
Shake vigrously , e.p disappearance of blue colour
NO STD. Br 2
Volatile
Toxic (sterility to males)
Kn. Xss std KBrO3/KBr mix .
Sample
St. Br2
Conc. HCl
ROLE OF C. HCl
BrO3/Br
Br2
Na salicylate
Salicylic acid
KI
I2
Br2 / C. HCl
Imp.
TBPInterferen
ce
Adsorption of Br2 & I2
on the surface
No filtration Bec. I2 & Br2 are volatile
Add CHCl3 or CCl4
Dissolves TBP
STARCH ADDED NEAR E.P ?
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)
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 blank it’s 24.8 ml
Must be done for any vo lat i le std .
Equivalent factor 1ml std ≡ …. gm
sampleN.B. Most drugs’ recovery % = 95-
105
What’s done / what’s taken
Gm % W/W
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
F = Std ≡ Sample3 Br2 ≡ 1 Aspirin3M Br2 ≡ 1M Aspirin1M Br2 ≡ 1/3M AspirinM Br2 ≡ 1M Aspirin20 3x201ml M Br2 ≡ gm M.wt Aspirin 20 3x20x 1000
I want to reach 1ml Std ≡ … gm Sample ÷3We was working
with M/20
Any no.s up
write it here down Bel nazar);
COONa
OH
2 )Alkalimetry:
AspirinKn. Xss std NaOH
Std HCl
NaOH
+CH3COONa
M/20 NaOH
Boil gently
Cool
Grind 2 tab.
N.B. During boiling NaOH Na2CO3 :. A blank is done
Calculations: The same idea as before
F =1ml M/2 NaOH = gm M.wt Aspirin 2x2x1000
Identification Tests
Aspirin pd + 4ml 2M NaOH + H2SO4
+ FeCl3 dps Violet
Boil 1 minSalicyli
c
Chloroquine diPO4:Pd + 2ml Amm.molybdate white ppt.
Chloroquine + Trinitrophenol (picric) Yellow ppt.
Prelab 2 1 )Isonicotinic acid Hydrazide (I.N.H) Bromometry
(Redox)
I.N.HKn. Xss std.
Br2
I 2
KI / HCl
Na2S2O3
Starch ind.
N/20 std Br2C. HCl
Stand
+ 2 Br2 + H3O(oxidative cleavage)
+
+N2 +4 HBr
Reducing agent
ROLE OF STD. Br 2
Aspirin
Bromination
I.N.H
Oxidative Cleavage
STARCH
Gentle shaking
Vigrous shaking
REMEMBER THAT
• Any redox reaction needs acid as a catalyst .• If the diluting solvent isn’t mention, use H2O.
ROLE OF C. HCl
BrO3/Br
Br
I.N.H
Oxidative Cleavage
KI
I2
Br2 / C. HCl
Imp.
Calculations :
The same idea as before :R%, C% & F BUT REMEMBER:
No. of electrons gained or lost or no. of H or OH e.g. 1M H 2 SO4 = 2
N+
All std.s M=N EXCEPT: • Br2 , I2, H2SO4 1M=
2N• KMnO4 1M= 5N
Identification TestsT.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
Red ppt.
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
2 )Chlorbutanol (Antimicrobial preservative):
Any compound containing halogen(but it must be in the form
of inorganic halide)
Mohr’s Method
Volhard’s
Method
Volhard's Method Mohr's Method P.O.C
Back (acidic medium) Direct Type of TitrationStd.1 (kn.xss):
AgNO3AgNO3 Titrant
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
Chlorobutanol heated with NaOH (To transform the organic
aliphatic halide to inorganic one)
3 NaCl
Chlorobutanol
Kn. Xss AgNO3
IsopropanolDil.
HNO3
AgCl
NH4SCNFerric allum
ind.
NaCl+ AgNO3 AgCl + NaNO3
AgNO3 + NH4SCN AgSCN+ NH4NO3
NH4SCN+ FeNH4(SO4)2[Fe(SCN)3] + 2 (NH4)2SO4
e.p white ppt in a buff solution
1 is Unsuccessful & always dissociating
called AgCl ( Ksp)
1 is successful,respected & always relaxed & at
ease called AgSCN ( Ksp)
ROLE OF ISOPROPANOL
There was 2 men So Ag+ of the dissociated AgCl
tries to seek relaxing by
reacting with SCN thus inc. e.p
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 ;)
+
Ksp AgCl > Ksp AgSCN
Ksp AgI,AgBr < AgSCN
ROLE OF DIL. HNO3
To ensure no alkalinity
AgNO3 AgOH AgO2 (black ppt.)
Fe 3+ Fe(OH)3
OH
OH
Calculations :
The same idea as before :R%, C% & F BUT REMEMBER:
AgNO3 =Chlorbutanol= 3NaCl
Cyclophosphamide & Enteroquine have the same principle
Identification Tests
Oxytetracycline ChlortetracyclinetetracyclinePd + 1ml conc. H2SO4 [dehydration inc. conj. color]
{tautomerism}Deep
crimson red
(violet)
Red (pink) color
Deep blue to green color
Deep
ye
llow
color Gold
en
yello
w co
lor
+FeCl3 Brown color
(phenolic cmpd)
Prelab 3
Bu sulfan Naproxen Naproxen Nalidixic acid
P.O.C
Direct AFTER HYDROLYSIS
2
Me sulfonic acid
Direct Direct Back (due to intramolecular H- bond & Zwitter ion formation non polar insol. in H2O,slightly sol. In alc.)
Type of Titration
No Yes, as naproxen but till 250ml
Yes(H2O [all at once], alc. till 100 [ shwaya b shwaya)
No Dilution
M/20 NaOH M/20 NaOH M/40 NaOH Std. acid (we haven’t worked it in lab)
Titrant
Acid –Base Titration
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
Identification Tests
Pd = 2ml conc. HCl + 0.5ml Soln. β naphthol in alc. orange red ppt
Charge transfer complexElectron rich
Electron poor
Imp
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
Giriseofulvin:Pd + 2ml conc. H2SO4 + few K2Cr2O7 Red
color
Oxid. reaction
Quinoid str. (color)
Prelab 41 )
Dipyrone
Dipyrone + distilled H2O
1ml glacial acetic acid
N/10 I2
+I2
+HI + NaHSO4
CH2Rose red color
(Dimer)
Identification Tests
Dipyrone pd +10ml dist. H2O2ml
+2ml dil. HCl Odour of SO2 Odour of formaldehyde
+1.5 ml dil. HCl +1ml FeCl3 bluered color disappear
+1ml K- pyronantimonate white ppt.
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