Chemistry

Limonene is found in lemon, orange, caraway and other oils. It is used to prepare terpin hydrate, used medicinally as a treatment for coughs. The commercial preparation involves addition of two moles of water to limonene in the presence of dilute sulphuric acid. Propose a structure for terpin hydrate, explaining your reasons.

TERPIN HYDRATE
    Limonene is an alkene. When reacted with water in the presence of Sulfuric acid, it is leads to the formation of an alcohol. The protonation of the double bond of the alkene occurs first to form a carbocation. The water molecule then acting as an electron lone pair donor, bonds with the carbocation to form the protonated alcohol.   Finally, a proton transfer occurs in as a water molecule accepts a proton from the protonated alcohol to leave the free alcohol product while reforming the catalytic oxonium ion H3O.  The orientation is Markonikov since the proton has added to the least highly substituted end, and the hydroxyl to the most highly substituted end (Atkinson  Hibbert, 2000).
Methylpentynol, an anxiolytic drug, has the following structure


Name the compound by the IUPAC system
Answer 3-methylpent-1-yn-3-ol
Indicate a suitable chemical test to demonstrate the presence of the alkyne group.
    When treated with Ammoniacal Silver Nitrate solution (Tollens reagent), it will form a white precipitate of Silver acetylide.
C6H10O Ag(NH3)2 OH---------------- C4H8 -CCAgH2O2NH3   
Formulate the product of, and provide a mechanism for, the reaction of Methylpentynol with dilute aqueous acidic mercuric chloride solution.
    The hydration and isomerization of 3-methylpent-1-yn-3-ol  forms (1) 3-methyl-3-hydroxy-1-pentanone and (2) 3-methyl-2pentene-1-one (McMurry, 2010).


Novocaine, a local anaesthetic, is a compound of molecular formula C13H20N2O2. It is insoluble in water and dilute NaOH, but soluble in dilute HCl. Upon treatment with sodium nitrite and HCl and then with 2-naphthol, a highly coloured solid is formed. When Novocaine is boiled with aqueous NaOH, it slowly dissolves. The alkaline solution is shaken with diethyl ether and the layers are separated. Acidification of the aqueous layer causes the precipitation of a white solid A continued addition of acid causes A to redissolve. Upon isolation A is found to have the molecular formula C7H7NO2. Evaporation of the ether layer leaves a liquid B of molecular formula C6H15NO. B dissolves in water to give a solution that turns red litmus blue. Treatment of B with acetic anhydride gives C, molecular formula C8H17NO2, which is insoluble in water and dilute NaOH but soluble in dilute HCl. Give structures for A, B, C and Novocaine, indicating the mechanisms of the reactions involved.

    In this experiment, this mixture may be readily separated by virtue of the fact that whereas aniline will react with HCl in water, the benzoic acid will react with sodium bicarbonate (NaHCO3) in water, and the chlorobenzene is simply insoluble in water.
    Thus, when the mixture is shaken with aqueous Hcl, the amine will be removed into the aqueous layer as its ammonium salt, whereas dichlorobenzene and benzoic acid will remain dissolved in the ether. The benzoic acid then can be extracted as its sodium salt by extracting the ether layer with aqueous NaHCO3
After separation of the aqueous and etheral layers, the aniline can be regenerated by basification with aqueous NaOH, while the aminobenzoic acid can be regenerated from its sodium salt by acidification with dilute acid (Cairns, 2000).
One of the reactions in the metabolism of glucose is the isomerisation of citric acid to isocitric acid. The isomerisation is catalysed by the enzyme aconitase. Propose a reasonable mechanism for this isomerisation.
    Isomerisation of citric acid to isocitric acid is the second reaction of the citric acid cycle. It is one in which the hydrogen and the hydroxyl groups change place. The enzyme aconitase requires Fe2 for this reaction. Citric acid, a symmetrical (achiral) compound, is then converted to isocitric acid an a chiral compound, a molecule that cannot be superimposed on its mirror image.    Isocitric acid has four possible isomers, but only one of the four is produced by this reaction.
    Aconitase can select one end of the citric acid molecule instead of the other. Aconitase forms an unsymmetrical three point attachment to the citric acid molecule. The first reaction is the removal of a water molecule from the citrate to produce cis-aconitate. In the second reaction, water is added back to the cis-aconitate to produce isocitrate (Campbell  Farrel, 2009).
Arrange the following compounds in order of increasing boiling point. Explain the basis for your answer.
1.CH3CH2CH3  2. CH3CH2Cl 3.CH3CH2OH      
    Ethanol has a higher boiling point due to the Hydrogen bonds between OH which require a lot of energy to break, followed by chloroethane which has the carbon-halogen which are polar and have van der Waals dipole-dipole attraction. This is because the electron pair is pulled closer to the halogen atom than the carbon. This is because chlorine is more electronegative than carbon. Finally, propane has covalent bonds which are bound together with Van Der Waal forces that are relatively strong but not as strong as the Hydrogen and polar bonds  (Clark, 2003).
5-Hydroxyhexanal readily forms a cyclic hemiacetal. Draw the structure of this hemiacetal. Would you expect this hemiacetal to be stable Explain your answer.

    It is not very stable as stability of hemiacetal depends on the number and strength of radicals attached to the quinone ring. It has only one quinone C(O) group (Remsen  Rouillu, 1895).
A solution of -D-glucose has a specific rotation of 112 a solution of -D-glucose has specific rotation of 19. On mutarotation, the specific rotation of each solution changes to an equilibrium value of 52. Calculate the percentage of -D-glucose in the equilibrium mixture.
19y112(100-y)40100.
19y(11200-112y)4000
19y-112y4000-11200
-93y-7200
Y77.41
    The specific rotations for -D-glucose and -D-glucose are about 112 and 18, respectively. It will be apparent that the product of the present invention, at 40, is mainly -D-glucose and that the commercially available products are mainly -D-glucose. An approximate calculation suggests that the present product is, in fact, 75 to 80 -D-glucose (Crystalline glucose and process for its production United States Patent 4342603 ) (Fuson  Snyder, 1942).
Raffinose is the most abundant trisaccharide in nature and has the structure below.

Name the three monosaccharide units in raffinose.
Answer Galactose, fructose and glucose.
There are two glycosidic links in raffinose. What type are they
    Raffinose consists of D-galactose, D-glucose, and D-fructose with the galactose and glucose linked by an -1,6 glycosidic bond, and the fructose linked to the glucose by an , -1,2-glycosidic bond. 
Would you expect raffinose to be a reducing sugar
    It is not a reducing sugar since there are no open chain forms possible. A sugar is only a reducing sugar if it has an open chain with an aldehyde or a ketone group (Chow  Halver, 1980).

Examine the structures of vitamins A, D3, E and K2. From their structural formulae would you expect them to be more soluble in water or olive oil
    Solubility is affected by polarity of the solute and solvent molecules. Polar solute molecules will dissolve in polar solvents and non-polar solute molecules will dissolve in non-polar solvents.  There is a definite connection between structure and solubility.  The fat soluble vitamins A, D, E, and K are composed mostly of carbon and hydrogen atoms that have similar electronegativities, therefore non polar. This causes them to be soluble in non polar materials such as olive oil which is also largely composed of Carbon and Hydrogen but not soluble in polar solvents such as water (Zumdahl, 2004, p. 854).
b) Would you expect them to be soluble in blood plasma. Explain your answers.
    The vitamins A, D, E, and K are not soluble in plasma. The vitamins that are soluble in blood plasma must either be soluble in water (the main component of blood plasma), or be solubilised by some other particles (e.g., proteins) that are carried in the blood (Casidy  Frey, 1999).

Calculate the saponification value of pure stearin, molecular weight 890.
If MMean molecular weight and KSaponification value then, M56100K
Therefore for stearin 5610089063.033 (Benedikt, 2007).
Glutathione is one of the most common small peptides in animals, plants and bacteria.
Name the three amino acids in this tripeptide
Answer Cysteine, Glutamic acid and Glycine

What is unusual about the peptide bond formed between the first two amino acids of this tripeptide
    The amide bond in the peptides should be made in the order that the amino acids are written. The amine end (N terminal) of an amino acid is always on the left, while the end (C terminal) is on the right. In the case of Glutathione, the oxidising agents react with the SH group of Cysteine of the glutathione (Tsen  Tappel, 1958)
c) Write an equation for the reaction of two molecules of glutathione with oxygen.
4GSHSeO-3             GSSGGS-Se-SG2OH-H2O (1)
2OH-GS-Se-SGO2                         GSSGSeO3H2O (2)
Which of the following amino acids will migrate towards the () electrode and which towards the (-) electrode on electrophoresis firstly at pH 6.0 and secondly at pH 8.6
Tyrosine       arginine       cysteine       aspartic acid       asparagine       histidine
    At any pH above the isoelectric point, the molecule will have a net negative charge and move towards the anode. Similarly, at pHs below the isoelectric point, the molecule will have a net positive charge and move towards the cathode (Davies, 2002).
    At PH 6.O Tyrosine Cysteine, aspartic acid, and asparagines will move towards the anode (-) while Arginine and Histidine will move towards the cathode () (Davies, 2002).
    At ph 8.6, Tyrosine Cysteine, aspartic acid Histidine and asparagine will move to anode (-) while Arginine will move towards the cathode () (Davies, 2002).
At high temperatures, nucleic acids become denatured, that is they unwind into disordered single strands. Account for the fact that the higher the content of G-C base pairs, the higher the temperature required to denature a given molecule of DNA.
    G (guanine) and C (cytosine) undergo a specific hydrogen bonding whereas A (adenine) bonds specifically with T (thymine). The GC base pair has three hydrogen bonds, whereas the AT base pair has only two as a consequence, the GC pair is more stable. This stability, however, is not caused solely by Hydrogen bonds but is mainly due to stacking interaction. DNA with high GC content undergo autolysis, thereby reducing the longevity of the cell. Denaturing the stable GC pair will therefore require high temperatures (Garrett  Grisham, 2008).


An important drug in the chemotherapy of leukaemia is 6-mercatopurine, a sulphur analogue of adenine. Draw a structural formula for 6-mercaptopurine. How do you think it exerts its anti-cancer action

    6- mercaptopurine forms yellow prisms of a monohydrate form when removed from an aqueous solution, becoming anhydrous at 140 degrees centigrade and decomposing at 313-314 degrees centigrade.
    6- mercaptopurine is a DNA antimetabolite. This means that it mimics a substance necessary in DNA synthesis necessary.
    6- mercaptopurine is converted to the corresponding ribonucleotide in the body,. 6- mercaptopurine ribonucleotide is a inhibits  the conversion of a compound called inosinic acid to adenylic acid. Without which DNA cannot be synthesized. Each component of DNA  A (adenine), T (thymine), G (guanine), and C (cytidine), is linked in turn to deoxyribose, a sugar.
    6- mercaptopurine also works by being incorporated into nucleic acids as thioguanosine, rendering the resulting nucleic acids (DNA, RNA) unable to direct proper protein synthesis (Feist, n.d.).