Chemistry

• •

Freidel crafts reagent (AlCl 3 ):

CH 3 Cl + AlCl 3 ROCl + AlCl 3 + +   Cl COR + HCl + AlCl 3 + HCl + AlCl

Canizaro’s reaction:

2 aldehyde + NaOH  one is reduced & the other is oxidized.

3 2 CHO + NaOH  COONa + CH 2 OH • • • • • • •

Markonikov’s reaction:

Addition of HX to alkenes CH 3 – CH = CH 2 + HCl  CH 3 – CHCl – CH 3

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.

CH 3 – CH = CH 2 + HBr  ROOR  CH 3 – CH 2 – CH 2 Br 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 2 ry alcohol R – MgX + HCHO  R – CH 2 – OH R’ – MgX + R 2 CHO  R’R 2 C – OH 1 3 ry ry alcohol alcohol

Cyclohexane

does not react with Grignard’s reagent as it neither contains double bonds nor functional gps.

Inductive Mechanism:

– Electron donating gps: NH 2 , OH, OR, CH 3 CONH, CH 3 – Electron withdrawing gps: NO 2 , CN, SO 3 H, CHO, RC=O, X – Reactivity of halogen (X) atoms: F > Cl > Br > I – 3 ry amines are more basic > 2 ry > 1 ry > CH 3 CONH 2

Nucleophilic substitution reactions:

– Nu + R – X  R – NU + X – OH + R – X  R – OH + X – NH 3 + R – X  R – NH 4 + X

• • • • • • • • • • • • • •

Kinetics of Nucleophilic Substitution:

–

SN 2 Reaction:

Increasing conc.  increases the rate of the reaction e.g. CH 3 Cl + OH  CH 3 OH + Cl – The rate of the reaction depends on both reactants  thus Bimolecular In SN 2 reactions RX > R – CH 2 – X > R 2 – CH – X . 3 ry halides do not react by SN 2 –

SN 1 Reaction:

e.g. (CH 3 ) 3 C – Cl + OH  (CH 3 ) 3 C – OH + Cl – The rate of the reaction depends on the conc. of only one reactant.

Permanganate is an oxidizing agent:

2 MnO 4 + 6 H + + 5 NO 2  2 Mn 2+ + 3 H 2 O + 5 NO 3 This 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 NH 2 (halo amine), will offer protonated species, i.e. it will # speed of the reaction (ppt Hg).

Na barbital is H 2 O soluble

but if an acid is added  precipitate barbeturic acid (insoluble).

Oxalic acid

is the principle oxidation product of ethylene glycol.

HO–CH 2 –CH 2 –OH  HOOC–COOH

Chloroform / Trichloromethane (CHCl 3 ):

when oxidized  phosgene (a very toxic gas).

Amides are used in medicine:

as local anesthetics & anti-arrhythmics.

Sodium bisulfite:

is a mixture of NaHSO 3 and sodium meta-bisulfite Na 2 SO 3 . When dissolved in water it converts to sodium bisulfite. Bisulfites are reducing agents used to protect oxygen sensitive drugs from oxidation.

Fixed oil

upon hydrolysis

Waxes

a a High m wt fatty acid + glycerol.

Fatty acid + high mol wt alcohol

Cholesterol

is a polyalcohol a w/o emulsion.

An ampholyte:

is a substance capable of functioning both as an acid & base, e.g. NaH 2 PO 4 , water, tetracycline.

Co-precipitation:

is due to rapid cooling.

Types of isomers:

– Position (structural) isomers – Configurational isomers (single bond) – Mirror image – Mirror image & optical isomers – Not mirror image but optical isomers      Isosters Conformers (Rotomers) Stereo isomers Enantiomers Diesters (diastereomers)

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 intra molecular 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 1 st 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: Phenol gp.

HO COOH

–

CH 2

–

CH

–

NH 2 Carboxyl gp.

Amino gp. (1ry amine) – The phenol gp will solubilize – The amino gp will solubilize – The carboxyl gp will solubilize 6 – 7 carbons 6 – 7 carbons 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

–

CH 2

–

CH

–

NH 3 +

-

HO COONa +

–

CH 2

–

CH

–

NH 2 Very Soluble HO COOH

–

CH 2

–

CH

–

NH 3 + 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 N – C 6 H 5 – The molecule contains only 13 carbons.

– It is expected to be soluble in water.

Using the conservative method:

(poly-functional gp.) – Each 3ry amino gp will solubilize only 3 carbons – The 2 amino gps will solubilize – The molecule contains 13 carbons.

– It is expected to be insoluble in water.

6 carbons CH 3 N CH 3

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.

Using the conservative method:

(poly-functional gp.) – The aldehyde gp will solubilize – The 3ry amino gp will solubilize – The molecule contains 13 carbons.

1 carbon 3 carbons – It is expected to be insoluble in water, which is the case.

– The conservative method is more appropriate for this example.

CH 3 CHO N CH 3

•

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 = S p – 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

O Phenyl gp + 2 + 2 C – OH Hydroxyl - 1 - 1 Carboxylic - 0.7

- 0.7

OH IMHB ---- + 0.6

Sum + 0.3

+ 0.9

Result Soluble Insoluble

In fact:

the compound is insoluble (IMHB is present & is expected to $ water solubility).

•

Example 2:

Predicting the water solubility of procaine.

6 C atoms X 0.5 each + 3.0

NH 2 1 Phenyl X 2.0

+ 2.0

CO–CH 2 –CH 2 – N (C 2 H 5 ) 2 2 Nitrogen X - 1.0 each - 2.0

1 C=O X - 0.7

- 0.7

Sum = 2.3

 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 + H 2 O:

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 d. Acetic acid is a strong oxidizing agent e. Perchlorate is more basic than acetate f.

Acetic acid is more basic than perchloroacetic acid.

An acid capable of protonating another acid is the stronger acid

4.

5.

6.

In 1,4 dimethyl cyclohexane, the best confirmation of the 2 methyl gps. Is when they are:

a. Equatorial - equatorial

b. Axial - axial c. Equatorial – axial A drug in its state of 1S 2 2S 1 P 2 is in: a. Its ground state.

b. Its excited state.

c. Hybrid.

Phenols are acidic because they have: a. Mesomeric interaction.

b. Strong induction cation.

Axial c. Strong induction anion.

7.

8.

The titrable result of Codeine phosphate + acetic acid is: a. [Codeine] PO 4 3 .

b. Codeine H 2 PO 4 . (codeine hydrogen phosphate)

c. Codeine HPO 4 To increase the rate of chemical reaction: a. Remove the resultant as it is formed b. Increase stirring force c. Decrease pH

d. Increase temp e. Addition of catalyst Equatorial

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 NH 2 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: a. Acetic acid b. Cl CH 3 COOH + HClO 4  CH 3 CH(OH) 2 c. SO + ClO 4 4 - 17. The structure shown is:

a. A zwitterion.

b. An ion pair c. A resonant form.

18. In the titration of codeine phosphate with glacial acetic acid, color change is due to the formation of:

a. R 3 – 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. NO 2 + O 2 (KMnO 4 ) in acidic medium:

a. NO

3 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 > OCH 3 > CH 3

26. Isoxazole & oxazole contain:

a. Elecron withdrawing gps.

27. RCH 2 NH 2 + A  RCH 2 COCH 3 , A stands for:

a. CH

3

COCl

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. H 2 O .

c. HO .

31. How can we differentiate by a simple reaction between 1 ry , 2 ry , & 3 ry alcohol:

a. By using CrO 4 / SO 4 which can oxidize 1

ry

alcohol to aldehyde, 2

ry

alcohol to ketone while 3

ry

alcohol does not react.

32. NH 4 + + OH –  NH 3 + H 2 O; 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 b. Rotate plain polarized light c. Are optical isomers

41. How many isomers are there for 2 butene:

a. Four

c. Two b. None 42. E – Z isomers are :

a. Geometric isomers

b. Diasteriomers b. Enantiomers 43. Which gp forms salt with:

a. NaHCO 3



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

b. Bases 47. SN 2 reactivity is more with:

a. CH 3 X

48. SN 1 reactivity is more with: a. (CH 3 ) 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  CH

3

COOR, A is :

a. CH 3 COOH / pyridine b. Br 2 / UV

53. Nicotinic acid nucleus is:

a. Pyridine

c. KMnO

4

/ H

2

O 54. Folic acid nucleus is : a. Ptyredine 55. Zn differs from Ca in their reactions because:

a. Zn is a Lewis base

56.

+ KMnO 4 + Br 2 / CCl 4   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 NaHCO 3 :

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 Br 2 / CCl 4 :

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 Na 2 CO 3 )

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. RN 2 ???

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