The Fundamental of Biochemistry - Case Study Example

A complete, balanced set of enzyme activities is of fundamental importance for maintaining homeostasis. An understanding of enzyme kinetic thus is important for understanding how physiologic stresses such anoxia, metabolic acidosis or alkaloids, toxicity pharmacologic agents affect that balance. For enzyme reactions,  the rate (velocity) of the reaction (v) varies with the substance concentration (ISI) as shown in the figure. A plot t of the velocity of enzymatic reaction versus substrate concentrate etically the overall enzymatic reaction is composed of two elementary reactions.

First, in which enzyme (E) and substrate (S) combine to form an unstable enzyme-substrate complex (ES,) and second, in which ES decomposes products (P) and (E). K1 is the rate constant for the formation of ES, K2 is the rate constant for t   dissociation of ES back to E and S and K3 is the rate constant for the formation of product P. The Michaelis- Menten constant, Km is defined as: From the above equation it is clear that when substrate concentration is very high, the velocity of the reaction is at its maximum and is independent of substrate concentration.

Linear-Burk plot It is also called a double reciprocal plot. It was formulated by Hans Line wear and dean Burk for determining values of Vmax and Km. The bulkiness of two methyl group attached to carbon atom develops overcrowding which results in moving apart of groups and hence the increase of each CH3-C-CH3 angle- the larger the group the greater will be that in which they are coplanar with the central carbon atom which is only possible by removal of chlorine atom (i.e.) by ionization leading to the formation of carbocation and thus expanding the bond angle to 120°.

Hence the mole (CH3)2CHCl will undergo substitution by the SN1 mechanism.Generally, primary alkyl halides react by the SN2 mechanism,tert-alkyl halides by the SN1 mechanism, and secondary alkyl halides by SN2 and SN1 mechanisms. Thus in going from left to right in the following series of alkyl halides, the mechanism changes from SN2 to SN1. This is due to the fact that the electron density on α-carbon atom goes on increasing by the inductive effect of the alkyl groups as we move from left to right. 

This increased electron density on the α-carbon atoms repels the direct attack of the nucleophile and thus retards the SN2 reaction. Moreover, the transition state of the SN2 reaction becomes more crowded as the size of the alkyl group increases. This also shows down the SN2 reaction, on the other hand, the increased electron density on the α-carbon atoms increases the tendency of ionization of the alkyl halide and thus favors the SN1 mechanism. Furthermore, the ionization relieves the steric hindrance and thus accelerates the SN1 mechanism. (b) Nucleophilic substitution rates on CH3CL are affected by the concentration of the nucleophile used.

An increase in concentration cannot alter the fraction of collisions that have sufficient energy or the fraction of collisions that the proper orientation; it can serve only to increase the total number of collisions. If more molecules are crowded into the same space, they will collide more often and the reaction will go faster. Collision frequency, and hence rate, depends in a very exact way upon concentration. ( c ). CH3CHDCL is a chiral halogenoalkane that is not racemized when undergoing nucleophilic substitution.

RCHDX undergoes nucleophilic substitution reactions through SN1 mechanisms. Slow fast CH3-CHD-CL CH3CH+D+Cl- CH3CH+D+Y→ CH3CHDY Walden inversion has been found at a primary carbon atom by the use of a chiral substrate containing a Deuterium and Hydrogen at the carbon bearing the leaving group. The carbonium ion is flat (trigonal hybridization), and hence attack by the nucleophile reagent can take place equally on either side i.e. equal amounts of (+) and (-) forms will be produced i.e. racemization results but here since the H atoms on C-1 carbon differs in density (+)& (-) forms do not occur and complete inversion or 100% optically active compound is formed. (e) The alcohol CH3CH2CH2OH cannot be synthesized by the hydration of propene.

This can be explained on the basis of the Markonikov addition rule-in which, the positive part of the unsymmetric reagent will attach to the carbon atom containing the larger member of the hydrogen atom, so instead of the formation of 1-propanol by the cleavage of UH-OH, 2-propanol will be formed. (f) Benzene undergoes substitution rather than addition reactions. Kekule’s structure of benzene is one that we would call “cyclohextriene” like the very similar compounds, cyclohexadiene or cyclohexane, to undergo readily the addition reactions characteristic of alkene structure; but this is not the case; under conditions that cause an alkene to undergo rapid addition, benzene reacts either not at all or very slowly.

In place of addition reactions, benzene readily undergoes a new set of reactions, all involving substitution. In each of these reactions, an atom or group has been substituted for one of the hydrogen atoms of benzene. The product can itself undergo further substitution of the same kind; the fact that it has retained the characteristic properties of benzene indicates that it has retained the characteristic structure of benzene. It would appear that benzene resists addition, in which the benzene ring system would be destroyed, whereas it readily undergoes substitution, in which the ring system is preserved. 2. For each of the following compounds (i), (ii), and (iii) give: (a) the number of 1H environment: (i) 3 1H environment (ii) 3 (iii) 3 (c) the integration area for all proton environments; (i) 3:2:5 (ii) 3:2:2 (iii) 6:3:1 (d) predict and explain the 1H NMR coupling patterns (i) CH3 will give 3 splittings (ii) CH3CH2CH2Cl CH3 group will give triplet CH2 group will give 12 splittings CH2Cl will give 3 splittings (iii) CH3COOCH(CH3)2 CH3 will give singlet CH will give 7 splittings (CH3)2 will give 2 splittings SECTION B 3.

Answer all parts of this question A compound is known to contain 54.5%C, 9.1%H, 36.