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Practical Manual Organic &

Medicinal Chemistry


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iii

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


iv


v

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


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


Practical Manual Organic & Medicinal Chemistry

1 | P a g e

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

*****


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



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.

*****



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



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



6 | P a g e

*****



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)



2 M NaOH = 2 * 40 g NaOH in 1000ml of water =80 g NaOH in 1000ml of water



8 | P a g e

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



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)


0.5 M H2SO4 = 0.5 * 55.5 H2SO4 in 1000ml 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



11 | P a g e

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