practical pharmaceutical chemistery

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 Ibuprofen 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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