Enviromental Orgo HW1

1. Important Terminology (20 points) a) What is an apolar, monopolar, or bipolar compound? Give one example each. b) Look at the structure of your compound that you have chosen for the mini project. Please provide the name of your compound and classify your compound as polar or nonpolar? a) Definitions and Examples Apolar Compound (Nonpolar): Apolar basically means nonpolar. These compounds do not have significant charge separation and do not interact strongly with polar solvents like water. Lack hydrogen bond donors or acceptors. Example: Hexane (C₆H₁₄) – A hydrocarbon with only C-H bonds, making it nonpolar. Other examples: Oxygen (O₂), Benzene (C₆H₆), Methane (CH₄) (Atkins & de Paula, 2017). Monopolar Compound: A monopolar compound has a single dominant polarity but does not have distinct positive and negative poles like bipolar compounds. These compounds have either hydrogen bond donors or acceptors but not both. Example: Hydrochloric acid (HCl) –It has one highly electronegative atom (Cl), making it polar but not bipolar. It is a hydrogen bond donor but lacks an acceptor. Other examples: Ammonia (NH₃), Ethanol (C₂H₅OH), Acetone (C₃H₆O) (Silbey et al., 2021). Bipolar Compound (Dipolar or Zwitterionic): These compounds have both hydrogen bond donors and acceptors, making them highly polar. These compounds have both positive and negative charges within the same molecule but in different locations, creating distinct poles.Example: Water (H₂O) – The oxygen side is partially negative, and the hydrogen side is partially positive, making it dipolar. Other examples: Amino acids (e.g., Glycine), Phospholipids, Sulfoxides (DMSO - C₂H₆OS) (Chang & Goldsby, 2016). b) Classification of Imidacloprid Compound I picked for my mini project is Imidacloprid. It is classified as a highly Polar insecticide. Imidacloprid contains electronegative atoms like nitrogen (N), oxygen (O), and chlorine (Cl), which create uneven charge distribution. The presence of these polar groups makes the molecule interact well with water and other polar solvents. Since it has a significant dipole moment, it is classified as a polar compound (Tomizawa & Casida, 2005). 2. Persistent Organic Chemicals (20 points) a) What characteristics render an organic compound to be qualified as POP. b) “What are so-called legacy POPs”? Provide two examples. a) What characteristics make an organic compound a Persistent Organic Pollutant (POP)? Persistent Organic Pollutants (POPs) are harmful chemicals that stay in the environment for a long time, spread across the globe, and build up in living organisms. A compound is considered a POP if it has the following characteristics: 1.​ Persistence – The chemical does not break down easily in nature. It remains in soil, water, and air for years because it resists natural processes like sunlight, bacteria, and chemical reactions. 2.​ Long-Range Transport – POPs can travel long distances through air and water, reaching places where they were never used. For example, some POPs have been found in the Arctic, even though they were never used there (Stockholm Convention, 2015). 3.​ Bioaccumulation – These chemicals dissolve in fats and oils, meaning they build up in the bodies of animals over time. As smaller animals are eaten by larger ones, the concentration of POPs increases, which can harm predators like birds, fish, and humans. 4.​ Toxicity – POPs are harmful to living organisms. They can affect the immune system, disrupt hormones, damage reproductive health, and even cause cancer (Li et al., 2006). Some chemicals, like polychlorinated and polybrominated compounds, are particularly dangerous because they are hydrophobic (avoid water) and stick to fats in animals, making them more likely to accumulate in the food chain (Lohmann et al., 2007). b) What are "Legacy POPs"? Legacy POPs, also known as the "Dirty Dozen," are the first group of POPs identified and restricted under the Stockholm Convention (2001). These chemicals were widely used in farming, industry, and consumer products but later banned or controlled because of their long-term environmental and health risks. Examples of Legacy POPs: 1.​ DDT (Dichlorodiphenyltrichloroethane) – A pesticide once used to kill mosquitoes and prevent malaria but later banned in many countries because it harms wildlife and human health. 2.​ Polychlorinated Biphenyls (PCBs) – Used in electrical equipment, paints, and plastics, but later found to cause cancer and immune system damage. Other legacy POPs include: ●​ Aldrin, Chlordane, Hexachlorobenzene (HCB) (pesticides) ●​ Polychlorinated dibenzodioxins (PCDDs) and dibenzofurans (PCDFs) (industrial by-products) (Secretariat of the Stockholm Convention, 2015). Even though these chemicals have been banned or restricted, they still exist in the environment today because they break down very slowly. This makes them a long-term problem for ecosystems and human health. Question 3 Acidity measurement, expressed in an acid-base reaction, is called the acidity constant, denoted as Ka. This is a measure of an acid's strength in solution and reflects the degree of dissociation of the acid into ions in water. The pKa values can be interpreted in the following manner: Part (a): Estimate the pKa values ∑σj We use the equation: p𝐾𝑎= p𝐾𝑎𝐻 -ρ 𝑗 For 4-bromo-3-ethyl-5-nitrophenol: ●​ From Table 4.9, the unsubstituted phenol has p𝐾𝑎𝐻= 9.90 and ρ= 2.25 ●​ From Table 4.10, the substituent values are: ○​ NO₂ (at ortho): σ = 1.24 ○​ Br (at ortho): σ = 0.70 ○​ C𝐻3(at ortho): σ= -0.13 ∑σj Using the equation, p𝐾𝑎= p𝐾𝑎𝐻 -ρ 𝑗 = 9.90- 2.25 • (0.70 + 1.24 -0.13) = 5.8275 Estimated pKa for 4-bromo-3-ethyl-5-nitrophenol: 5.83 For 4-bromo-3-chloro-5-methylaniline: ●​ From Table 4.9, the unsubstituted aniline has p𝐾𝑎𝐻= 4.63 and ρ= 2.90 ●​ From Table 4.10, the substituent values are: ○​ Cl (at ortho): σ = 0.67 ○​ Br (at ortho): σ = 0.71 ○​ C𝐻3(at ortho): σ= -0.13 ∑σj Using the equation, p𝐾𝑎= p𝐾𝑎𝐻 -ρ 𝑗 = 4.63- 2.90 • (0.71 + 0.67 -0.13) ​ = 1.005 Estimated pKa for 4-bromo-3-chloro-5-methylaniline: 1.005 b) 4-bromo-3-ethyl-5-nitrophenol pH α𝑖𝑎 5 0.129 7 0.9369 9 0.9993 At pH 5, 1 α= 5.8275−5 = 1+10 At pH 7, α= 1 5.8275−7 α= 1 5.8275−9 0.8275 = 1+10 1 = 1+10 At pH 9, 1 −1.1725 1 1+10 1+10 4-bromo-3-chloro-5-methylaniline pH α𝑖𝑎 5 0.9999 7 0.999999 9 1 −3.1725 = 0.129 = 1 1+0.0673 = 1 1+0.000675 1+10 = 1 1+6.73 = 0.9369 = 0.9993 At pH 5, 1 α= 1.005−5 = 1+10 At pH 7, 1 α= 1.005−7 = 1+10 At pH 5, 1 α= 1.005−9 1+10 = 1 −3.995 1+10 1 −5.995 1+10 1 −7.995 1+10 = 1 1+0.000101 = 1 1+0.00000101 = 1 −8 = 0.9999 = 0.999999 =1 1+10 Explanation For 4-bromo-3-ethyl-5-nitrophenol (phenol derivative): ●​ At pH 5, 12.94% is ionized, meaning most of the compound is in neutral form. ●​ At pH 7, 93.69% is ionized, so it's mostly in its deprotonated (phenoxide) form. ●​ At pH 9, 99.93% is ionized, meaning it's almost fully in phenoxide form. For 4-bromo-3-chloro-5-methylaniline (aniline derivative): ●​ At pH 5, 99.99% is ionized, meaning it's almost entirely in the protonated (anilinium ion) form. ●​ At pH 7, almost 100% is still protonated as the anilinium ion. ●​ At pH 9, 100% remains ionized, meaning aniline remains in its protonated (anilinium) form even at high pH. 4. Temperature Dependence of Equilibrium Constants (20 points) In table 4.2 you find the equilibrium constant of benzene distribution between air and water (Ki12 = 0.21). This value is for 25 ˚C. Table 4.2 also provides you with ∆12Hi for this phase transfer process. This allows you to calculate Ki12 at a different temperature. Calculate this Ki12 (air/water for benzene) at 10 ˚C (use equation 4-34, remember: in thermodynamics temperature must be expressed in K, also be careful with kJ vs J). Given, T1= 298 K and 𝐾𝑖12=0.21 From the table, ∆12𝐻𝑖= 29.7 KJ/mol = 29700 J/mol For, T2= 283 K we need to find 𝐾𝑖12 value Now, using the equation above, we can calculate −∆12𝐻𝑖 • 𝐾𝑖12 (𝑇2)= 𝐾𝑖12 (𝑇1)• 𝑒 1 𝑅 1 [ 𝑇2 − 1 𝑇1 ] − 𝐾𝑖12 (283 𝐾)= 0.21 • 𝑒 (29700 𝐽 1 𝑚𝑜𝑙 • 1 𝑚𝑜𝑙• 𝐾 8.314 𝑗 1 [ 283𝐾 − 1 298𝐾 ] 𝐾𝑖12 (283 𝐾)= 0.1112 5. Concentration of Benzene in Air (20 points) The benzene concentration in NYC air was reported to be as high as 2.19 µg/m3 between 2008-2013. Convert this concentration to a) mol/L and b) ppmv assuming 1 bar and 25 ˚C. a) Unit conversion from µ𝑔 to gram 3 2.19 µ𝑔/𝑚 x −6 −6 10 𝑔 1µ𝑔 = 2.19 x 10 3 g/𝑚 Molar mass of Benzene = 78.12 g/mol Gram to mole conversion Moles = 𝐺𝑖𝑣𝑒𝑛 𝑚𝑎𝑠𝑠 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = −6 3 −6 2.19𝑥10 𝑔/𝑚 78.12 𝑔/𝑚𝑜𝑙 3 Unit conversion from 𝑚 = 0.0280 x 10 3 mol/𝑚 to liter −6 Concentration (mol/L) = 0.0280 x 10 −6 ​ ​ ​ = 0.0280 x 10 ​ ​ ​ = 2.80 x 10 −11 3 mol/𝑚 • −3 • 10 3 1𝑚 1000 𝐿 mol/L mol/L b) Where P = 1 bar is pressure V is volume −6 3 n = 0.0280 × 10 mol is moles of benzene in 1 𝑚 , R = 0.08314 L.bar/mol.K is the ideal gas constant, and T = 25°C = 298 K Calculation of the volume of benzene in 1 m or 1,000 L: V= 𝑛𝑅𝑇 𝑃 Temperature in K by adding 273.15; T = (25°C + 273.15)K = 298.15 K Plug in values, V = −6 0.0280 × 10 Calculation in ppmv = 𝑚𝑜𝑙 • 0.08314 𝐿.𝑏𝑎𝑟/𝑚𝑜𝑙.𝐾 • 298 𝐾 1 𝑏𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑏𝑒𝑛𝑧𝑒𝑛𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑖𝑟 6 x 10 = −6 = 0.694 x 10 −6 0.694 𝑥 10 1000 𝐿 𝐿 6 L −4 x 10 = 6.94 x 10 ppmv