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Practical Manual Organic &
Medicinal Chemistry
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Practical Manual
Organic & Medicinal Chemistry
(As per revised PCI Syllabus-Regulation-2016)
By
Dr. A. Rama Narsimha Reddy M.Pharm., Ph.D.
Professor
Jyothishmathi Institute of Pharmaceutical Sciences,
Thimmapur, Karimnagar 505481,Telangana State, India
Email: [email protected]
Dr. L. Srividya M.Pharm., Ph.D
Associate Professor
Jyothishmathi Institute of Pharmaceutical Sciences,
Thimmapur, Karimnagar 505481,Telangana State, India
Email: [email protected]
EDUCREATION PUBLISHING (Since 2011)
www.educreation.in
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CONTENTS
S. NO. CONTENT PAGE
1. General Terminology 1
2. Preparation of Standard Solutions 2
3. Calculations of Molecular Weight and Equivalent Weight 4
4. Preparation of Molar Solutions (Molarity Calculations) 7
5. Preparation of Normal Solutions( Normality Calculations) 12
6. Titration Calculations 16
7. Synthesis of Barbitutric Acid from Diethyl Malonate 18
8. Synthesis of Phenytoin from Benzoin Or Benzil 20
9. Synthesis of Paracetamol from Para-Amino Phenol 23
10. Synthesis of 1, 4-Dihydro Pyridine from Ethyl Acetoacetate 25
11. Synthesis of Busulphan from 1, 4-Butanediol 27
12. Synthesis of Isoniazid from 4-Cyano Pyridine 29
13. Synthesis of Methyldopa from 3-hydroxy-4-methoxy phenylanine 31
14. Synthesis of Metronidazole from 2-Methyl-5-nitroimidazole 33
15. Synthesis of Benzimidazole from O-Phenylenediamine 35
16. Synthesis of Dichloramine from Para toluene sulphonyl chloride 37
17. Synthesis of Aspirin from salicylic acid 40
18. Synthesis of Quinazolinone from Anthranilic Acid (Benzoxazinone) 43
19. Synthesis of Fenofibrate from Para chloro-4-Hydroxybenzophenone 45
20. Synthesis of Isoniazid from Gamma Picoline 47
21. Synthesis of Antipyrine from Phenyl hydrazine 49
22. Synthesis of Benzocaine from Para nitro benzoic acid 51
23. Synthesis of 2, 4, 6-Tribromophenol from Phenol 54
24. Synthesis of Meta Nitro Aniline from Dinitro Benzene 56
25. Preparation of Benzoic Acid from Benzanilide 58
26. Synthesis of Para Amino Benzene Sulphonamide 60
27. Synthesis of Benzilic Acid from Benzoin 65
28. Synthesis of Dulcin from Para Phenethidine
Synthesis of Lidocaine from2,6-dimethylaniline
68
29. 70
30. Synthesis of Hexamine from formaldehyde 73
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vi
PREFACE
TThhee ppuurrppoossee ooff tthhiiss ““MMaannuuaall ooff PPrraaccttiiccaall OOrrggaanniicc aanndd MMeeddiicciinnaall CChheemmiissttrryy”” iiss iinntteennddeedd
ffoorr tthhee PPhhaarrmmaaccyy ssttuuddeennttss ooff DD..PPhhaarrmm,, BB..PPhhaarrmm aanndd PPhhaarrmm..DD ssttuuddeennttss aass ppeerr tthhee nneeww
rreegguullaattiioonnss ooff PPCCII--22001166.. TThhiiss iiss ssppeecciiffiiccaallllyy wwrriitttteenn ttoo mmeeeett tthhee pprreesseenntt nneeeeddss ooff rreevviisseedd
ccuurrrriiccuulluumm.. TThhiiss bbooookk pprroovviiddeess aa ggoooodd mmeeddiiuumm ffoorr tteeaacchhiinngg aanndd ssttuuddyyiinngg lluucciiddllyy iinn
pprraaccttiiccaallss.. BBeessiiddeess,, tthhee bbooookk wwiillll hheellppss iinn bbrriinnggiinngg tthhee uunniiffoorrmmiittyy iinn ssyyllllaabbii.. WWee aarree
ccoonnffiiddeenntt tthhaatt tthhiiss bbooookk wwiillll bbee ooff iimmmmeennssee vvaalluuee ttoo tthhee ssttuuddeennttss ooff PPhhaarrmmaaccyy..
CCuurrrriiccuulluumm ddeevveellooppmmeenntt iiss aa ddyynnaammiicc aanndd ccoonnttiinnuuoouuss pprroocceessss.. NNoo oonnee ccaann ccllaaiimm ttoo hhaavvee
ddeevveellooppeedd aa ppeerrffeecctt ccuurrrriiccuulluumm mmaatteerriiaall.. TThheerree wwiillll aallwwaayyss bbee ssoommee ssccooppee ffoorr ffuurrtthheerr
iimmpprroovveemmeenntt,, ssoo II rreeqquueesstt aallll tthhoossee wwhhoo wwiillll bbee uussiinngg tthhiiss tteexxtt bbooookk ttoo eevvaalluuaattee tthhee
mmaatteerriiaall wwiitthh aann ooppeenn mmiinndd aanndd ooffffeerr tthheeiirr vvaalluuaabbllee ssuuggggeessttiioonnss ffoorr iimmpprroovviinngg tthhee qquuaalliittyy
iinn ffuuttuurree..
DDrr.. AA.. RRAAMMAA NNAARRSSIIMMHHAA RREEDDDDYY
DDrr.. LL SSRRIIVVIIDDYYAA
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Practical Manual Organic & Medicinal Chemistry
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1.0 GENERAL TERMINOLOGY
________________________________________________________________________
Atomic weight: Atomic weight is the sum of the mass of all the atomic particles
(neutrons, electrons and protons) in an atom.
Molecular weight: Molecular weight is the sum of the atomic weights of all the atoms in
a molecule. Also called formula weight.
Mole: A mole is also called gram-molecular weight and defined as number of particles
whose total mass in grams was numerically equivalent to the molecular weight or it is the
amount of substance containing the Avogadro number (6.022 × 1023
) of particles.
Molarity: Molarity or molar concentration denotes the number of moles of a given
substance per liter of solution.
Molar solution: It is a solution that contains 1 mole of solute in each liter of solution.
The units for molar concentration are mol/L denoted by ‘M’.
Equivalent weight: An equivalent weight is equal to the molecular weight divided by the
valence (replaceable H ions).
Normal: A normal is one gram equivalent of a solute per liter of solution.
Normality: Normality is the total no of gram equivalents of the solute present per liter of
the solution. The units for normal concentration are Eq/L denoted by ‘N’
Normal solutions: The definition of a normal solution is a solution that contains 1 gram
equivalent weight (gEW) per liter solution.
Examples:
Preparation of 1N NaOH solution- Dissolve 40 g of NaOH in a 1000 ml (1 liter) of
water
Preparation of 1N Ca(OH)2-Dissolve 37 g of Ca (OH)2 in a 1000 ml (1 liter) of water
Preparation of 1N KCL solution- Dissolve 74.55 g of KCl in a 1000 ml (1 liter) of
water
*****
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Dr A Rama Narsimha Reddy
2 | P a g e
2.0 PREPARATION OF
STANDARD SOLUTIONS
________________________________________________________________________
For Solid Substances
Place a weighing boat or weighing paper on a balance and add substance (for eg-
NaOH) until the balance reads desired grams. If you're using weighing paper, first
fold it diagonally both ways and unfold leaving a creased "x". This makes it easier to
empty the solid into a flask.
Pour the solid into a 1000 ml volumetric flask using a funnel and bending the
weighing boat or paper diagonally. The flask will have a line on the neck that
represents 1000 ml.
Fill it up to about 2 cm below the line, using distilled or dionized water. Put the cap
on the flask, invert it and swirl gently.
Add more water, drop by drop, until the meniscus (low point) of the water is at the
line.
Repeat the mixing procedure and check the volume again. Add more water if needed
to get 1000 ml of solution.
The reason for not filling to the line initially is that the combined volume of the solute
and solvent can be less or more than the volume before mixing. (NaOH will tend to
have a lower volume after mixing).
For Liquid Substances
If the substance is a liquid (for eg- HCl) measure accurate liquid by pipette or
measuring cylinder and transfer into volumetric flask then make up the volume.
First of all, fill the volumetric flask about halfway with deionized water to avoid
violent reactions. NEVER add water to concentrated acid.
Choose a clean pipette of suitable size and transfer the liquid to the volumetric flask.
When the whole solution has been drained, touch the tip of the pipette to the side of
the volumetric flask to allow the last of the liquid to drain out. DO NOT blow out the
remaining solution.
Allow the solution to reach room temperature because a volumetric flask is only
accurate at the temperature at which it has been calibrated (usually 20 °C).Very
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Practical Manual Organic & Medicinal Chemistry
3 | P a g e
carefully fill the flask to the mark on the neck of the flask, using a dropping pipette to
add the last few milliliters of liquid.
Mix your solution thoroughly, by inverting the flask and shaking. NEVER hold large
volumetric flasks by the neck alone provide support at the bottom.
Transfer the prepared solution to a clean, dry storage bottle and label it.
NOTE
When making chemical solutions, always use the appropriate safety equipment.
As a general rule, always add the more concentrated solution to the less concentrated
solution.
All chemicals that you are unfamiliar with should be treated with extreme care and
assumed to be highly flammable and toxic.
NEVER store solutions in a volumetric flask.
*****
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Dr A Rama Narsimha Reddy
4 | P a g e
3.0 CALCULATIONS OF MOLECULAR
WEIGHT AND EQUIVALENT WEIGHT
________________________________________________________________________
Calculation of the Molecular Weight
Molecular weight of Sodium hydroxide NaOH
To do this, look up the atomic masses of the elements on the periodic table (provided at
the end of the experiment).
The atomic mass is the mass in grams of 1 mole of atoms.
mass of Na . g/mol
mass of O . g/mol
mass of H = 1.00 g/mol
So, the mass of one mole of NaOH is:
mass of NaOH = mass of Na + mass of O + mass of H
mass of NaOH = 23 g + 16 g + 1 g
mass of NaOH = 40g
Molecular weight of sulphuric acid H2SO4:
mass of 2 hydrogen atoms 2 X 1 g = 2 g
mass of 1 sulfur atom 1 X 32.06 g = 32.06 g
mass of 4 oxygen atoms 4 X 16 g = 64 g
So, the mass of one mole of H2SO4 is:
mass of H2SO4 = mass of 2 H + mass of S + mass of 4 O
mass of H2SO4 = 2 + 32.06 + 64
mass of H2SO4 = 98.06 g
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Practical Manual Organic & Medicinal Chemistry
5 | P a g e
Calculation of Equivalent Weight
Examples:
HCL the MWt= 36.5 the EWt = 36.5/1 = 36.5
H2SO4 the MWt = 98 the EWt = 98/2 = 49
H3PO4 the MWt = 98 the EWt = 98/3 = 32.7
Examples:
NaOH the MWt = 40 the EWt = 40
Ca (OH)2 the MWt = 74 the EWt = 74/2 = 37
Examples:
KCL the MWt= 74.55 the EWt = (74.55 / 1 X 1) = 74.55
CaCl2 the MWt= 110.98 the EWt = (110.98 / 1 X 2) or (110.98 / 2 X 1) = 55.49
Na2CO3 the MWt= 105.98 the EWt = (105.98 / 1 X 2) or (105.98 / 2 X 1) = 52.99
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6 | P a g e
*****
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Practical Manual Organic & Medicinal Chemistry
7 | P a g e
4.0 PREPARATION OF MOLAR SOLUTIONS
(MOLARITY CALCULATIONS)
________________________________________________________________________
Preparation of 1 Molar (1M) Sodium Hydroxide Solution
First calculate molecular weight of Sodium hydroxide
molecular weight of Sodium hydroxide = 40g
40 g mass is 1 mole of Sodium hydroxide
1M NaOH (Molarity) = 40 g NaOH in 1000ml of water
Preparations of Sodium Hydroxide
0.1 M NaOH =0.1 * 40 g NaOH in 1000ml of water
=1 g NaOH in 1000ml of water
= 0.5g in g NaOH in 500ml of water
=0.25 g NaOH in 250ml of water
=0.01 g NaOH in 100ml of water
0.5 M NaOH = 0.5 * 40 g NaOH in 1000ml of water (or)
=20 g NaOH in 1000ml of water (or)
= 10g in g NaOH in 500ml of water (or)
=5 g NaOH in 250ml of water (or)
=1 g NaOH in 100ml of water (or)
2 M NaOH = 2 * 40 g NaOH in 1000ml of water =80 g NaOH in 1000ml of water
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3 M NaOH = 3 * 40 g NaOH in 1000ml of water= 120 g NaOH in 1000ml of water
10 M NaOH = 10 * 40 g NaOH in 1000ml of water =400 g NaOH in 1000ml of water
2/3 M NaOH = 2/3 * 40 g NaOH in 1000ml of water = 26.67 g NaOH in 1000ml of
water
Preparation of 1 Molar (1M) Hydrochloric Acid Solution
First calculate molecular weight, specific gravity of hydrochloric acid
Molecular weight of hydrochloric acid = 36.46 g
Specific gravity of hydrochloric acid = 1.19
1ml of HCl = 1.19 g HCl
1000ml of HCl = 1190 g HCl
Molarity of HCl = Amount / MWt × %purity
= 1190/36.46 × 36/100
= 12M
Concentrated HCl means 12M HCl
Now make 1:12 dilutions to arrive at 1M HCl
M 1 V 1 = M 2 V2
1 × 1000 = 12 × ?
V2 = 1 × 1000/12
= 83.3ml
Subtract the volume of solute from the total solution volume
1000-83.3 = 916.7 ml of water is required to make 1N solution
1M HCl is 83.3ml of concentrated HCl in 1000ml water
Preparations of Hydrochloric Acid
0.1 M HCl =0.1 * 83.3 ml HCl in 1000ml of water (or)
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=8.33ml HCl in 1000ml of water(or)
= 4.165 ml in g HCl in 500ml of water (or)
=2.0825 ml HCl in 250ml of water (or)
= 0.833 ml HCl in 100ml of water (or)
0.5 M HCl = 0.5 * 83.3 HCl in 1000ml of water (or)
=41.65 ml HCl in 1000ml of water (or)
= 20.0825 ml HCl in 500ml of water (or)
=10.4125 ml in 250ml of water (or)
=4.165 ml HCl in 100ml of water
2 M HCl = 2 * 83.3 ml HCl in 1000ml of water =166.6 ml HCl in 1000ml of water
3 M HCl = 3 * 83.3 ml HCl in 1000ml of water= 249.9 ml HCl in 1000ml of water
10 M HCl = 10 * 83.3 ml HCl in 1000ml of water =833 ml HCl in 1000ml of water
2/3 M HCl = 2/3 * 83.3 ml HCl in 1000ml of water = 55.53 ml HCl in 1000ml of water
Preparation of 1 Molar (1M) Sulphuric Acid Solution
First calculate molecular weight, specific gravity of hydrochloric acid
Molecular weight of sulphuric acid = 98.08 g
Specific gravity of sulphuric acid = 1.84
1ml of H2SO4 = 1.84 g H2SO4
1000ml of H2SO4 = 1840 g H2SO4
Molarity of H2SO4 = Amount / MWt × %purity
= 1840/98.08 × 98/100
= 18.38M
Concentrated H2SO4 means 18M H2SO4
Now make 1:18 dilutions to arrive at 1M H2SO4
M 1 V 1 = M 2 V2
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10 | P a g e
1 × 1000 = 18 × ?
V2 = 1 × 1000/18
= 55.5ml
Subtract the volume of solute from the total solution volume
1000-55.5 = 944.5 ml of water is required to make 1N solution
1M H2SO4 is 55.5ml of concentrated H2SO4 in 1000ml water
Preparations of Sulphuric Acid
0.1 M H2SO4 =0.1 * 55.5 ml H2SO4 in 1000ml of water (or)
=5.5ml H2SO4 in 1000ml of water(or)
=2.7 ml in g H2SO4 in 500ml of water (or)
=1.38 ml H2SO4 in 250ml of water (or)
=0.55 ml H2SO4 in 100ml of water (or)
0.5 M H2SO4 = 0.5 * 55.5 H2SO4 in 1000ml of water (or)
=27.75 ml H2SO4 in 1000ml of water (or)
=13.8 ml H2SO4 in 500ml of water (or)
=6.93 ml in 250ml of water (or)
=2.775 ml H2SO4 in 100ml of water
2 M H2SO4 = 2 * 55.5 ml H2SO4 in 1000ml of water =111 ml H2SO4 in 1000ml of
water
3 M H2SO4 = 3 * 55.5 ml H2SO4 in 1000ml of water= 166.5 ml H2SO4 in 1000ml of
water
10 M H2SO4 = 10 * 55.5 ml H2SO4 in 1000ml of water =555 ml H2SO4 in 1000ml of
water
2/3 M H2SO4 = 2/3 * 55.5 ml H2SO4 in 1000ml of water = 37 ml H2SO4 in 1000ml of
water
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