Chemistry Survey - Lab Report Example

Chemistry Q1. a) The required reaction is Mg(s) + Br2(l)MgBr2(s) Given, Therefore, b) The Born-Haber cycle for formation of MgBr is presented below Mg+(g) + Br(g) Mg+(g) + Br(g) EA +IE Mg+(g) + Br+(g) Lattice Energy Mg(g) + Br(g) Hf Energy Mg(g) + Br2(l) +Hatm c) kJmol-1 = -78 kJmol-1 d) Value of is negative, besides, value of will be very small in magnitude, therefore, value of will be negative and its value will be vary close to -78 kJmol-1. Because, value of is negative, therefore, MgBr(s) would be expected to be stable with respect to the elements. e) From a) = -504.12 kJmol-1 and from d) -78.0 kJmol-1 Because is much more than and is negative in magnitude, therefore, MgBr2(s) is much more stable than MgBr(s). Hence MgBr(s) does not exist. Q2. a) From equation (1), a possible rate equation for the given reaction can be written as b) Because, and also looking at the data provided in table (1). The reaction is pseudo first order reaction and a simplified rate equation for equation (1) is c) (i) The kinetic reaction profile for is presented in figure 1, below: (ii) The value of rate constant is almost constant at 150 s and 250s. This implies that the reaction is pseudo first order reaction. (iii) From figure 1, three successive half lives of the reaction are 125s, 125s and 125s. These values also imply that the reaction is pseudo first order reaction. (iv) A plot of rate of reaction (J) vs. concentration of AsO33- i.e.[A] was made (figure 2, below). A linear trendline was drawn using MS excel. Slope of this plot gives the rate constant. The value of the rate constant is 5.3x10-3s. d) From data given in table (2) a plot of lnkR' vs. 1/T was made (figure 3, below). Slope of this curve equals -E/R, where E is activation energy of the reaction and R is the Universal gas constant, which equals 8.38 Jmol-1K-1. Therefore, -E/R = 8122.6 K Or, E = 8122.6*8.38 Jmol-1= 68067.4 Jmol-168.7 k Jmol-1 Q3. a) The mechanism in this case is SN1 i.e. Nucleophilic Substitution of First Order. In this mechanism the carbonium ion forms first, to which the nucleofile OH- gets attached in no time. Because formation of carbocation is the slow or rate controlling step therefore this is a first order reaction. Therefore, rate of this reaction is proportional to concentration of hexan-3-ol. Because, carbocation is planer therefore, the nucleofile (OH-) has equal probability of attaching to it from both sides (above as well as below) of the plane. Thus, a mixture of both the optical isomers (R-isomer and S-isomer) forms in equal proportions. The mechanism is shown below: b) (i) In this case S-isomer will form. (ii) The mechanism in this case is SN2 i.e. Nucleophilic Substitution of Second Order. In this case both molecules R-3-bromohexane and NaOH are involved in the reaction mechanism. The nucleofile OH- attacks the positively polarized carbon atom from the other side of the leaving group i.e. from other side of the C-Br bond, because of steric hindrance. This backside attack by the nucleofile leads to inversion of the stereo configuration of the molecule. The reaction mechanism is shown below: c) This is because, in the beginning, when the reaction is being carried out in pure water, the mechanism is SN1 according to the mechanism described in section a) above and this leads to formation of the two stereo isomers in 50:50 ratio. This has been discussed in detail in section a) of this question. In this case the rate of reaction will depend on the concentration of 3-bromohexane only. When the concentration of hydroxide ion is increased in the solution, the reaction mechanism is a mixture of SN1 and SN2. While SN1 mechanism continues to produce and equal mixture of the two stereo isomers, R and S, the SN2 mechanism will produce the S isomer preferentially. As the operating mechanism is a mixture of the two mechanisms SN1 and SN2, therefore, the product will have more of S isomer, than R isomer with increasing concentration of the hydroxide ion. The rate of reaction will have some dependence of concentration of the hydroxide ion as well, besides, that of the 3-bromohexane. The order of reaction will be fractional and its value will be between 1 and 2, depending on the concentration of the hydroxide ion. When concentration of the hydroxide ion reaches 1.0 mol dm-1, then the mechanism is fully SN2 i.e. rate of the reaction is equally dependent on the concentration of 3-bromohexane as well as that of the hydroxide ion. The order of the reaction is 2 and the product is S isomer only according to the reaction mechanism discussed in b) (ii). Q4. a) (i) For this reaction the major product is formed by Elimination, the mechanism is E2. This is because potassium tert-butoxide anion is very bulky nucleofile, so it cannot bond with carbon. In other words, it is a poor electrofile. However, it is strong enough base to extract a H+ and therefore capable of carrying out elimination by E2 mechanism. The major products are (CH3)3N, C2H4 and tertiary butanol. The energy profile for the reaction is drawn below: (ii) The reaction mechanism here is SN2, i.e. Nucleophilic Substitution Reaction of Second Order. This is because CN- is a very strong nucleophile and therefore, it makes bond with carbon atom. The product is CH3CH2CH2CN. The reaction profile is shown below: (b) (i) The reaction mechanism is E2. The likely product is trans-CH3CH==CHCH2CH3. This compound forms according to Saytzev's rule, which predicts, most substituted or most stable alkene, because the substrate is neutral and -hydrogen are on different carbon atoms. Besides, the Elimination will occur preferentially from an antiperiplanar conformation, therefore, the product will be a trans and not a cis stereoisomer. (ii) The likely mechanism is E1. The products are C5H10C==CHCH3 and (CH3)2S. More substituted alkene forms in this reaction as the operating mechanism is E1. (iii) The operating mechanism is E2. The products are (CH3)3N and (C6H5)(CH3)CHCH==CH2. In this case because the substrate is charged, therefore, least substituted alkene forms according to Hoffman rule. The product is shown below: Read More