Chemistry
•
•
Freidel crafts reagent (AlCl3):
CH3Cl + AlCl3 +

ROCl + AlCl3 +

Cl
+ HCl + AlCl3
COR
+ HCl + AlCl3
Canizaro’s reaction:
2 aldehyde + NaOH  one is reduced & the other is oxidized.
2
•
CHO
COONa
+ NaOH 
CH2OH
+
Markonikov’s reaction: Addition of HX to alkenes
CH3 – CH = CH2 + HCl
 CH3 – CHCl – CH3
•
Markonikov’s rule: Hydrogen is added to the more hydrogenated carbon (with HCl or HI).
•
Anti-markonikov’s reaction: due to free radicle reaction to alkenes.
CH3 – CH = CH2 + HBr  ROOR  CH3 – CH2 – CH2Br
HCl & HI cannot give anti-markonikov’s reaction. Only HBr can because of free radicles.
•
Grignard’s reagent:
Mg + RX  R – MgX.
Mg + ArX  Ar – MgX
R – MgX + R’CHO  R R’ – CH – OH
2ry alcohol
R – MgX + HCHO  R – CH2 – OH
1ry alcohol
R’ – MgX + R2CHO  R’R2 C – OH
3ry alcohol
•
Cyclohexane does not react with Grignard’s reagent as it neither contains double bonds nor
functional gps.
•
Inductive Mechanism:
•
–
Electron donating gps:
NH2, OH, OR, CH3CONH, CH3
–
Electron withdrawing gps:
NO2, CN, SO3H, CHO, RC=O, X
–
Reactivity of halogen (X) atoms: F > Cl > Br > I
–
3ry amines are more basic > 2ry > 1ry > CH3CONH2
Nucleophilic substitution reactions:
–
Nu + R – X  R – NU + X
–
OH + R – X  R – OH + X
–
NH3 + R – X  R – NH4 + X
•
Kinetics of Nucleophilic Substitution:
– SN 2 Reaction: Increasing conc.  increases the rate of the reaction
e.g. CH3Cl + OH  CH3OH + Cl –
The rate of the reaction depends on both reactants  thus Bimolecular
In SN 2 reactions RX > R – CH2 – X > R2 – CH – X . 3ry halides do not react by SN 2
–
SN 1 Reaction:
e.g. (CH3)3C – Cl + OH  (CH3)3C – OH + Cl –
The rate of the reaction depends on the conc. of only one reactant.
•
Permanganate is an oxidizing agent:
2 MnO4- + 6 H+ + 5 NO2-  2 Mn2+ + 3 H2O + 5 NO3This is an example of oxidation-reduction reaction in which nitrite is oxidized to nitrate
in acidic medium.
•
Hg acetate (mercuric acetate): in a non-aqueous titration of NH2 (halo amine), will offer
protonated species, i.e. it will # speed of the reaction (ppt Hg).
•
Na barbital is H2O soluble but if an acid is added  precipitate barbeturic acid (insoluble).
•
Oxalic acid is the principle oxidation product of ethylene glycol.
HO–CH2–CH2–OH  HOOC–COOH
•
Chloroform / Trichloromethane (CHCl3): when oxidized  phosgene (a very toxic gas).
•
Amides are used in medicine: as local anesthetics & anti-arrhythmics.
•
Sodium bisulfite: is a mixture of NaHSO3 and sodium meta-bisulfite Na2SO3. When
dissolved in water it converts to sodium bisulfite. Bisulfites are reducing agents used to
protect oxygen sensitive drugs from oxidation.
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Fixed oil upon hydrolysis
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Waxes
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Cholesterol is a polyalcohol
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An ampholyte: is a substance capable of functioning both as an acid & base, e.g. NaH2PO4,
water, tetracycline.
•
Co-precipitation: is due to rapid cooling.
•
Types of isomers:
a
a
High m wt fatty acid + glycerol.
Fatty acid + high mol wt alcohol
a w/o emulsion.
–
Position (structural) isomers

Isosters
–
Configurational isomers (single bond)

Conformers (Rotomers)
–
Mirror image

Stereo isomers
–
Mirror image & optical isomers

Enantiomers

Diesters (diastereomers)
– Not mirror image but optical isomers
Predicting Water Solubility
I - The Empiric Method: The chemical properties of a functional gp. are not usually affected by
the presence of another chemical gp. within the molecule.
•
The presence of a single functional gp. (Liberal Method)  there is no liability of intramolecular bonding; however there is a probability of inter-molecular bonding; e.g. hexanol
binds to another molecule of hexanol (each having only 1 functional gp., -OH) through
dipole-dipole bonding. In order to dissolve hexanol in water, one must 1st overcome (break)
this dipole-dipole bond so that water molecules can bind to the functional gps.
•
The presence of a Poly-functional gps. (Conservative Method)  there is probability of
intra-molecular bonding as well as of inter-molecular bonding. See the example of Tyrosine:
Carboxyl gp.
COOH
– CH2 – CH – NH2
HO
Phenol gp.
Amino gp.
(1ry amine)
–
The phenol gp will solubilize
6 – 7 carbons
–
The amino gp will solubilize
6 – 7 carbons
–
The carboxyl gp will solubilize
5 – 6 carbons
–
The total solubilization potential
17 – 20 carbons.
–
The molecule contains only 9 carbons.
–
It is expected to be soluble in water.
–
However, in reality it is insoluble (0.5 gm / L)
– This is mainly due to intra-molecular bonding. Amino-acids undergo intra-molecular
ion-ion bonding (Zwitterion effect).
–
As a result this intra-molecular bonding destroys the ability of these 2 functional gps.
(amino & carboxyl) to bond with water. Meanwhile, the phenol gp. is unable by itself to
dissolve the molecule.
–
If the intra-molecular bonding is destroyed by the addition of either HCl or NaOH, the
resulting compound becomes water soluble.
COO - Zwitterion Effect
HO
– CH2 – CH – NH3+
COONa+
-
HO
– CH2 – CH – NH2
Very Soluble
COOH
HO
– CH2 – CH – NH3+ Cl -
Very Soluble
Example 2:
•
Consider the shown compound which contains two 3ry amine functional gps.
•
Using the liberal method: (mono-functional gp.)
•
•
–
Each 3ry amino gp will solubilize
7 carbons
–
The 2 amino gps will solubilize
14 carbons
–
The molecule contains only 13 carbons.
–
It is expected to be soluble in water.
N – C6 H5
Using the conservative method: (poly-functional gp.)
–
Each 3ry amino gp will solubilize only
3 carbons
–
The 2 amino gps will solubilize
6 carbons
–
The molecule contains 13 carbons.
–
It is expected to be insoluble in water.
N
CH3
CH3
In fact: the compound is soluble because it contains similar functional gps (no liability for
intra- nor inter-molecular bonding).
Example 3:
•
•
Using the liberal method: (mono-functional gp.)
–
The aldehyde gp will solubilize
5 carbons
–
The 3ry amino gp will solubilize
7 carbons
–
The molecule contains only 9 carbons.
–
It is expected to be soluble in water.
CHO
Using the conservative method: (poly-functional gp.)
–
The aldehyde gp will solubilize
1 carbon
–
The 3ry amino gp will solubilize
3 carbons
–
The molecule contains 13 carbons.
–
It is expected to be insoluble in water, which is the case.
–
The conservative method is more appropriate for this example.
N
CH3
CH3
II - The Analytical Method: This method is based on the partitioning of a drug between octanol
(a standard lipophilic medium) & water (a standard hydrophilic medium).
Log P
•
=
Conc. of drug in octanol / Conc. of drug in water
–
Where P is the partition coefficient of the drug.
–
The P value measures the solubility characteristics of the molecule as whole.
How to measure Log P: This is calculated as the summation of the hydrophilic-lipohilic
values (p value) of different fragments of the molecule
Log P
=
Sp
–
In order to use this method you must fragment the molecule into basic units.
–
Hence for each fragment, assign appropriate p values depending on the atoms & groups
of atoms present (in this specific fragment).
–
Positive p values: means that the fragment, relative to hydrogen, is lipophilic, or favors
solubility in octanol.
–
Negative p values: means that the fragment, relative to hydrogen, is hydrophilic, or
favors solubility in water.
•
Normal log P value = + 0.5
•
Log P value > + 0.5  compound is water insoluble ( solubility in water < 3.3%).
•
Log P value < + 0.5  compound is water soluble.
Example 1: Calculating water solubility of salicylic acid + intra-molecular H-bonding (IMHB).
Log P – IMHB
Log P + IMHB
Phenyl gp
+2
+2
Hydroxyl
-1
-1
Carboxylic
- 0.7
- 0.7
IMHB
-----
+ 0.6
Sum
+ 0.3
+ 0.9
Result
•
Soluble
O
C – OH
OH
Insoluble
In fact: the compound is insoluble (IMHB is present & is expected to $ water solubility).
Example 2: Predicting the water solubility of procaine.
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6 C atoms X
0.5 each
+ 3.0
1 Phenyl
2.0
+ 2.0
X
2 Nitrogen X - 1.0 each
- 2.0
1 C=O
- 0.7
X - 0.7
NH2
Sum = 2.3
CO–CH2–CH2 – N (C2H5)2
 Water insoluble.
N.B: There is no need to investigate the possibility of IMHB, since the result is insoluble,
adding the value for IMHB will only make the result more water insoluble.
1.
The addition of a catalyst, the reaction will:
a. Give more products
b. Give less products
c. Get products rapidly (proceed rapidly)
2.
What is the type of reaction of HCl + H2O:
a. It is an ionization reaction (as HCl is a proton donor).
3.
Perchloroacetic acid can protonate acetic acid because:
a. Acetic acid is stronger than perchloroacetic acid
b. Perchloroacetic acid is stronger than acetic acid
c. Acetic acid accepts protons easily
An acid capable of
protonating another acid
is the stronger acid
d. Acetic acid is a strong oxidizing agent
e. Perchlorate is more basic than acetate
f. Acetic acid is more basic than perchloroacetic acid.
4.
In 1,4 dimethyl cyclohexane, the best confirmation of the
2 methyl gps. Is when they are:
Axial
a. Equatorial - equatorial
b. Axial - axial
Equatorial
c. Equatorial – axial
5.
A drug in its state of 1S2 2S1 P2 is in:
a. Its ground state.
b. Its excited state.
c. Hybrid.
6.
Phenols are acidic because they have:
a. Mesomeric interaction.
c. Strong induction anion.
b. Strong induction cation.
7.
The titrable result of Codeine phosphate + acetic acid is:
a. [Codeine] PO43-.
b. Codeine H2PO4. (codeine hydrogen phosphate)
c. Codeine HPO4
8.
To increase the rate of chemical reaction:
a. Remove the resultant as it is formed
d. Increase temp
b. Increase stirring force
e. Addition of catalyst
c. Decrease pH
9.
In gravimetric analysis the pH is adjusted to:
a. To complete pptn (facilitate pptn).
b. To avoid unwanted pptn (avoid co-pptn).
c. To obtain fine filterable particles.
d. Favorable pH for pptn.
10. Why is mercuric acetate used in non-aqueous titration:
a. To produce protonation species.
11. For the given structures, which is soluble in NaOH:
a. The one with the COOH gp.
12. For the given structures, which is soluble in HCl:
a. The one with the NH2 gp.
13. Which of the following is a differentiating solution in non-aqueous titration:
a. Acetic anhydride.
c. Glacial acetic acid.
b. Quinine / Quinidine
d. Acitonitrile.
14. In sulphonamide titration, what is the role of dimethyl formaldehyde:
a. It acts as a basic solvent.
15. From the given structures which is an azodye:
a. Ph – N = N – naphthyl with ortho hydroxyl.
b. Ph – NH – naphthyl.
c. Ph – NH – NH – naphthyl
16. What is the base for titration of perchloric acid:
b. Cl-
a. Acetic acid
c. SO4--
CH3COOH + HClO4  CH3CH(OH)2 + ClO4
17. The structure shown is:
a. A zwitterion.
c. A resonant form.
b. An ion pair
18. In the titration of codeine phosphate with glacial acetic acid, color change is due to the
formation of:
a. R3 – NH+
19. In the structure shown, which carbon is least acidic:
a. C 1
b. C 2
c. C 3
20. In the structure shown, which carbon is least basic:
a. C 5
b. C 2
21. NO2 + O2 (KMnO4) in acidic medium:
a. NO
c. C 3
22. Which of the following is more basic in water:
23. Soap is:
a. Na+ salt of fatty acid.
24. Which can give cis & trans isomers:
a. 2 butene
b. 1 butene
c. 3 butene
25. Arrange the following gps in decreasing order of reactivity:
a. COO– > COOH > OCH3 > CH3
26. Isoxazole & oxazole contain:
a. Elecron withdrawing gps.
27. RCH2NH2 + A  RCH2COCH3 , A stands for:
a. CH3COCl
28. A hydroxyl containing compound undergoes a reaction to give COOH, The OH gp is:
a. Terminal hydroxyl
c. 3ry hydroxyl
b. Secondary hydroxyl
29. Which is formed by alcohol metabolism:
a. Aldehyde
b. Alkene
c. Alkane
30. Which is free oxygen radicle:
a. O.
b. H2O.
c. HO.
31. How can we differentiate by a simple reaction between 1ry, 2ry, & 3ry alcohol:
a. By using CrO4 / SO4 which can oxidize 1ry alcohol to aldehyde, 2ry alcohol to ketone
while 3ry alcohol does not react.
32. NH4+ + OH–  NH3 + H2O; this is a :
a. Neutralization reaction
b. Esterification reaction
33. In the assay of ephedrine HCl we use:
a. NaOH
b. Acetic acid
34. Which compounds are affected by hydrolysis:
a. Esters
b. Amides
35. Which compounds react with schiffs base but not with fehling:
a. Ketones
b. Aldehydes
36. Ortho dinitro phenol is a stronger acid than phenol because of:
a. The electron withdrawing effect of the nitro gps.
37. Compared to formic acid, acetic acid is a weaker acid because:
a. The methyl gp in acetic acid is electron donating.
38. Hinsberg reaction is used for differentiation of amines since:
a. 1ry amines react with the reagent & dissolve in NaOH
b. 2ry amines react with the reagent but do not dissolve in NaOH
c. 3ry amines do not react with the reagent
39. How many chiral carbon atoms are in the shown compound:
a. Zero
c. One
b. Two
40. Enantiomers:
a. Have a chiral carbon
c. Are optical isomers
b. Rotate plain polarized light
41. How many isomers are there for 2 butene:
a. Four
c. Two
b. None
42. E – Z isomers are :
a. Geometric isomers
b. Enantiomers
b. Diasteriomers
43. Which gp forms salt with:
a. NaHCO3  4
b. NaOH
 2&4
c. HCl
 1
44. Hg acetate is used in the assay of histamine to:
a. Provide a protonated species
45. Acidity of carboxylic acid (COOH) but not phenol (OH) is due to:
a. Anaionic resonance.
46. Amphoteric agents can form salts with:
a. Acids
47. SN 2 reactivity is more with:
a. CH3X
b. Bases
48. SN 1 reactivity is more with:
a. (CH3)3 – X
49. How can we obtain a di-carbonyl gp:
a. By Claisen’s condensation.
50. Grignard’s reagent is:
a. R – Mg – X
51. Which compound does not react with Grignard’s reagent:
a. Cyclohexane (as it contains no functional gps nor double bonds)
52. ROH + A  CH3COOR, A is :
a. CH3COOH / pyridine
c. KMnO4 / H2O
b. Br2 / UV
53. Nicotinic acid nucleus is:
a. Pyridine
54. Folic acid nucleus is :
a. Ptyredine
55. Zn differs from Ca in their reactions because:
a. Zn is a Lewis base
+ KMnO4 
56.
+ Br2 / CCl4 
57. In acid base titration, which affects the reaction:
a. The change in pH of the solution
58. Boron 1 S , 2 S 2 , 2 P 2; this is in th:
a. Excited state
b. Hybrid
c. Ground state
59. Which compound reacts with schiffs base but does not reduce fehling’s solution:
a. Buterophanones
60. Which gp forms salt with NaHCO3:
a. Strong acid
61. Which compound reacts by SN mechanism:
a. Butene (C = C)
62. Amphetamine (weak base) is extracted by:
a. ether
63. Which of the following does not discolour Br2 / CCl4:
a. ????
64. Phenol is weakly acidic because:
a. It weakly dissociates
65. In the assay of ephedrine HCl, we dissolve in water, add a reagent X, filter then extract with
methylene chloride, what is the reagent X:
a. 0.1 N NaOH (or Na2CO3)
66. Perchloric acid is stronger than acetic acid because:
a. Acetic acid is more basic ???
67. Codeine phosphate is dissolved in acetic acid & titrated with perchloric acid,:
a. RN2 ???
68. Several organic compounds are given, which compound has water solubility characteristics:
a. The structure showing a phosphate salt
69. Given the chemical structures of quinine & quinidine, these are:
a. Optical isomers
70. In the shown structure:
a. Which carbon is the most acidic  C 6
b. Which carbon is the most basic  C 2