Number Of Pages: 1,275 Words Academic Level: Postgraduate

Number Of Pages 1 275 Wordsacademic Level Postgraduatedeadline

Number of pages: 1 (275 words) Academic Level: Postgraduate Deadline: 2 hours Instructions: 1. If k1 > k2, is the overall reaction SN1 or SN2? 2. Conversely, if k2 > k1 is the overall reaction SN1 or SN2? Explain why. The following molecule is chloroethane. ClCH2CH3 It is unreactive towards hydroxide and most nucleophiles under normal conditions, i.e., the chlorine-carbon bond is not a very good electrophilic center, hence 1 is not a good alkylating agent and does not transfer the CH3CH2 group easily. On the other hand, there is a class of drugs that contain a chloroethyl group which are good alkylating agents and have been used clinically to treat cancer. An example of this class of drugs is melphalan. The reason melphalan, is a good alkylating agent, and 1 is not, is due to the formation of the intermediate 3 caused by the close proximity of a basic nitrogen to the carbon-chlorine bond in 2. Reaction of 3 with a nucleophile (Nuc) gives the alkylated nucleophile, 4. In vivo, in a therapeutic setting, the Nuc is specifically nitrogen or oxygen on a purine in a DNA strand, which once alkylated cleaves from the DNA strand to create errors in reading the DNA code and ultimately death of a cancer cell. The alkylation takes place in two steps defined by the rate constants k1 and k2. 3. If the basic pKa of the nitrogen in the -N-(CH2 - CH2 - Cl)2 in molecule 2 is 1.5 and the basic pKa in the -N-(CH2- CH2 - Cl)2 in Molecule 5 is 6.5, which drug, (molecule 2 or molecule 5) is more likely to go by an SN1 reaction with the nucleophile and why? Note that the rate or ease of reaction k1 for 2 will depend on how basic is the nitrogen in -N-(CH2 - CH2 –Cl)2. The following molecule, atracurium, is a very complex drug. It is a neuromuscular blocking agent and is related to the natural product tubocurarine. Atracurium exhibits a very short half-life (20-min). The products of the fastest inactivation process are shown below. 4. Research and explain what chemical mechanism is responsible for the formation of the products. Hint: The acidic pKa of CH3(C=O)CH3 is 20. Assign2-1.gif · Follow the instructions and directions below for this lab. Disregard the outline in the manual for your LabPaq Kit. · Read this document entirely before starting your work. · Do not forget to record your measurements and partial results. · Submit a Laboratory Report through Moodle, as shown in the last section of this outline. Remember that the Laboratory Report should include the answers to the questions below. GOALS (1) To examine the concept of friction. (2) To calculate the coefficient of friction of an object. INTRODUCTION Friction is an important force, with positive and negative implications. Whenever two objects slide along each other, friction is involved; whether a skier is sliding down an icy slope, or a crate is dragged across the floor. On the positive side, friction between the road surface and the car tires is what keeps the car moving along a highway. On the negative side, friction reduces the efficiency of machines. As more work needs to be done to overcome friction, this extra work is wasted as it is dissipated in the form of heat energy. When one surface slides over another, a resisting force, friction, is encountered. Friction, and the force needed to overcome it depends on the nature of the materials in contact with each other and on the roughness or smoothness of the contact surfaces. It is also affected by the normal force (FN), but not by the contact area or on the speed of the motion. It has been determined experimentally that the force of friction (Ffr) is directly proportional to the normal force (FN). When an object is resting on a horizontal surface the normal force is just the weight of the object (mg). If an object is on an incline, FN has to be corrected via the equation FN= mg cos θ with θ being the angle of the incline. In this experiment, we will only measure friction on a horizontal surface, so we don’t need to concern ourselves with the issues affecting friction on an incline. The constant of proportionality is called the coefficient of friction, μ. The force of friction when two contacting surfaces are sliding over each other can be calculated by: where Ffr is the force of friction; FN is the normal force; and μk or μs are the coefficient of friction, which is a proportionality constant. The force of friction is parallel to the contact surfaces and opposite to the direction of motion. The term μk stands for coefficient of kinetic (or sliding) friction, which applies when the surfaces are moving with respect to each other. When an object is at rest on a surface and we attempt to push it, the frictional force is opposing the pushing force. As long as the pushing force is less than the friction force, the object will not move. There is a threshold value of the pushing force beyond which a larger pushing force will cause the body to start sliding. It is this threshold value which is related to the coefficient of static friction, μs. However, when an object is already in movement, the friction that it experiences is related to its coefficient of kinetic friction, μk. When comparing published μk and μs values for identical materials, we see that μs is slightly larger than μk. This indicates that it takes more force to start moving a material than keeping it moving. PROCEDURE For purposes of data recording, we will use Newtons (N) as this is the unit of Force in the International Systems of Units. For the purposes of this experiment, we can convert any mass to force using the following formula: Measurement setup For each measurement setup described in this section, we will measure two forces. First, we will measure the force that we need to exert on a wooden block so it starts moving from a resting position. This force is related to the coefficient of static friction. The second force is the force that we need to exert on the wooden block to keep it moving. This force is related to the coefficient of dynamic friction. Because the wooden blocks in your lab kit are too light to produce accurate results, we will augment their mass by adding an object of constant weight, for example a can of soda. The experimental setup can be seen in figure 1 below. You can use any other object that you may have around instead of the soda can. Figure 1: Experimental Setup We will perform two types of measurements: · Force of Static Friction (Fs): Starting from the block at rest, pull from the spring scale until the block starts moving. Record this force. · Force of Kinetic Friction (Fk): Starting from the block at rest, pull from the spring scale until the block starts moving and continue pulling until it moves at a constant speed. Record this force. You should observe that Fk is lower than Fs. We will repeat these measurements using several materials as well as in different configurations for the wooden blocks. Question 1 What is the mass of the system made of the wooden block and the soda can (or the other object that you are using)? Question 2 Convert the mass measured in Question 1 to its weight in Newtons. This is the value of FN that will be used in the calculations for the tables. SURFACES Wood / Wood (larger surface) For this case, we will use the larger surface of the wooden block as was shown in Figure 1. Measure Fs and Fk as indicated in Section 3.1. Run 5 trials, completing Table 1 below. Wood / Wood (smaller surface) For this section, turn the wooden block on its side as shown in Figure 2 and repeat the experiment, completing Table 2. Figure 2: Wooden block on its side (courtesy of Chad Saunders, TESU student) Wood / Glass For this section, use the glass surface of your wooden block to repeat the measurements and complete Table 3. Other surfaces Repeat the measurements using other surfaces (for example, Wood / Sandpaper, Wood / Carpet, Glass / Carpet, etc.) When completing Table 4 below, make sure that you indicate the surfaces you used. ANALYSIS OF RESULTS QUESTION 3 Study the results from Table 1 and Table 2. What can you conclude about these results? QUESTION 4 Using the data from the same tables, do you think that the ratio of μk to μs is constant in both cases? If so, what do you think this indicates? QUESTION 5 Studying the standard deviation data from all the 4 tables, which experiment do you think is the most reproducible? Why? QUESTION 6 In general, how does the coefficient of static friction compare to the coefficient of dynamic friction? QUESTION 7 In designing machinery, would we prefer to use materials with larger or smaller coefficient of friction? Explain your reasoning. QUESTION 8 In driving a vehicle, would you prefer to use materials for the contact between the wheels and the road with larger or smaller coefficient of friction? Explain your reasoning. LABORATORY REPORT Create a laboratory report using Word or another word processing software that contains at least these elements: · Introduction: what is the purpose of this laboratory experiment? · Description of how you performed the different parts of this exercise. At the very least, this part should contain the answers to questions 1-8 above. You should also include procedures, etc. Adding pictures to your lab report showing your work as needed always increases the value of the report. · Conclusion: What area(s) you had difficulties with in the lab; what you learned in this experiment; how it applies to your coursework and any other comments.

Paper For Above instruction

Analysis of Nucleophilic Substitution Mechanisms and Friction Coefficients

This paper investigates the mechanistic pathways of SN1 and SN2 reactions often encountered in organic chemistry, especially focusing on how rate constants (k1 and k2) influence the reaction pathway. Additionally, the comprehensive study of friction on various materials aims to establish the static and kinetic coefficients of friction, their implications for machinery design and vehicle safety, and how molecular properties influence chemical reactivity.

Understanding SN1 and SN2 Reactions: Role of Rate Constants

Nucleophilic substitution reactions are fundamental in organic synthesis, with SN1 and SN2 pathways distinguished primarily by their mechanisms and kinetic signatures. In SN1 (substitution nucleophilic unimolecular), the rate-determining step is the formation of a carbocation intermediate. This pathway favors substrates that stabilize carbocations, such as tertiary alkyl halides, and involves a single rate-limiting step characterized by the rate constant k1. Conversely, SN2 (substitution nucleophilic bimolecular) involves a one-step concerted mechanism where the nucleophile attacks the electrophilic carbon simultaneously as the leaving group departs, favoring primary alkyl halides and characterized by the rate constant k2.

Impact of Rate Constants on Reaction Pathways

The comparison of k1 and k2 provides insight into the dominant mechanism. When k1 > k2, the formation of the carbocation intermediate is faster than the backside attack, indicating an SN1 pathway. However, if k2 > k1, the reaction proceeds predominantly via SN2, as the nucleophile directly displaces the leaving group in a single step.

Mechanistic Insights in Chloroethane and Alkylating Agents

Chloroethane (ClCH2CH3) is relatively unreactive under normal conditions because its carbon-chlorine bond does not form a stable carbocation intermediate readily, favoring an SN2 mechanism when it reacts with strong nucleophiles. The reaction is characterized by a single step with a rate influenced by the nucleophile's strength, indicative of SN2 kinetics.

In contrast, drugs such as melphalan, which contain a chloroethyl group, can form a more stable intermediate due to the proximity of a basic nitrogen atom that facilitates the formation of a carbocation-like transition state. This stabilization makes the pathway more consistent with SN1, where the rate-limiting step involves carbocation formation.

Influence of Basicity on Reaction Mechanism (k1 and k2)

The basicity of the nitrogen atom in molecules influences the rate constant k1 substantially because a more basic nitrogen (higher pKa) can assist in stabilizing the transition state or intermediate, thus decreasing the energy barrier for carbocation formation. For molecule 2, with a pKa of 1.5, the nitrogen is less basic, making carbocation formation less favorable, and reactions tend to follow the SN2 pathway if k1 is lower than k2. Conversely, for molecule 5, with a pKa of 6.5, the higher basicity enhances carbocation stabilization and favors an SN1 mechanism when k1 exceeds k2.

Role of Chemical Mechanisms in Drug Inactivation and Formation of Products

The formation of products from atracurium involves mechanisms typical of hydrolysis and esterification reactions, often facilitated by the environment's pH and the presence of water molecules. The ester bonds, with a high pKa (around 20), are susceptible to nucleophilic attack by water or other nucleophiles, leading to hydrolysis and the formation of inactivation products. These mechanisms are primarily nucleophilic acyl substitution reactions, where water acts as the nucleophile attacking the electrophilic carbonyl carbon.

Friction and Its Practical Implications in Engineering

Friction, a resistive force opposing relative motion between surfaces, depends on the nature of the contact materials, surface roughness, and normal force. The coefficients of static and kinetic friction (μs and μk) vary with material combinations, affecting energy efficiency and safety. Static friction must be overcome to initiate motion, while kinetic friction resists ongoing movement. The proportionality between normal force and frictional force is well established through empirical studies.

Experimental Measurement of Friction Coefficients

In laboratory settings, the force required to initiate movement (static friction) and to maintain constant motion (kinetic friction) is measured using spring scales. The mass and normal force are calculated, and multiple trials are performed across various surface combinations to determine the coefficients of friction. Data analysis involves calculating mean values and standard deviations to assess reproducibility and material performance.

Implications for Machinery and Vehicle Design

Materials with higher coefficients of friction are advantageous where grip and braking are essential, such as in tires, whereas lower coefficients are preferred for minimizing wear and energy loss in machinery. In vehicle design, selecting materials with appropriate frictional properties enhances safety and efficiency by balancing grip and wear resistance.

Conclusion

Understanding the mechanistic pathways of nucleophilic substitution and the factors influencing reaction rates provides insight into drug design and toxicology. Similarly, quantifying friction coefficients enables the optimization of mechanical systems and transportation safety. The principles learned underscore the interconnectedness of chemical reactivity and physical properties, vital for advancements in pharmaceuticals and engineering applications.

References

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