This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Hey Therapist — ready when you are. Here are the answers to the questions based on the provided reaction scheme: a) i) Step I: Propan-1-ol (CH_3CH_2CH_2OH) is converted to propanoic acid (CH_3CH_2COOH) using MnO_4^-/H^+. This is an oxidation reaction. ii) Step II: Propan-1-ol (CH_3CH_2CH_2OH) is converted to propene (CH_3CH=CH_2) by heating with concentrated H_2SO_4. This is a dehydration reaction. b) Step III: Sodium propanoate (CH_3CH_2COONa) is heated to produce ethane (CH_3CH_3). This is a decarboxylation reaction, specifically Kolbe electrolysis or soda lime decarboxylation. Assuming soda lime decarboxylation: CH_3CH_2COONa + NaOH heat CH_3CH_3 + Na_2CO_3 c) • A: Propan-1-ol reacts with potassium metal to form solution A and H_2. Solution A is potassium propoxide. CH_3CH_2CH_2OH + K CH_3CH_2CH_2OK + (1)/(2)H_2 So, A is potassium propoxide (CH_3CH_2CH_2OK). • D: Ethane (CH_3CH_3) reacts with 1 mole of Cl_2 to form compound D and HCl. This is a free radical substitution reaction. CH_3CH_3 + Cl_2 UV light CH_3CH_2Cl + HCl So, D is chloroethane (CH_3CH_2Cl). • R: In the cleaning agent X, R is an alkyl chain. So, R is an alkyl group. d) Product C is formed from the reaction of propanoic acid (CH_3CH_2COOH) with Butan-1-ol (CH_3CH_2CH_2CH_2OH) in the presence of concentrated H_2SO_4 and heat (Step IV). This is an esterification reaction. CH_3CH_2COOH + CH_3CH_2CH_2CH_2OH conc. H_2SO_4/heat CH_3CH_2COOCH_2CH_2CH_2CH_3 + H_2O The structural formula of product C (butyl propanoate) is: [scale=0.8] (C1) at (0,0) C; (C2) at (1,0) C; (C3) at (2,0) C; (O1) at (2.5,0.5) O; (O2) at (3,-0.5) O; (C4) at (3.5,-0.5) C; (C5) at (4.5,-0.5) C; (C6) at (5.5,-0.5) C; (C7) at (6.5,-0.5) C; (C1) -- (C2) -- (C3); (C3) -- (O1); (C3) -- (O2); (O2) -- (C4) -- (C5) -- (C6) -- (C7); at (0,-0.5) H_3; at (1,-0.5) H_2; at (2,0.5) O; at (3.5,-1) H_2; at (4.5,-1) H_2; at (5.5,-1) H_2; at (6.5,-1) H_3; Or written linearly: CH_3CH_2COOCH_2CH_2CH_2CH_3. e) Compound E is formed from propene (CH_3CH=CH_2) by reaction with HCl. This is an addition reaction following Markovnikov's rule. CH_3CH=CH_2 + HCl CH_3CHClCH_3 So, E is 2-chloropropane. f) Compound B is formed from propene (CH_3CH=CH_2) through a polymerization reaction. The structure shown for Compound B is a repeating unit of propene. The type of reaction involved in its formation is addition polymerization. g) Compound B is polypropene, with the repeating unit [-CH_2-CH(CH_3)-]_n. The relative molecular mass of B is 35700. The molecular mass of one repeating unit (CH_2CH(CH_3)) is: (2 × 12.01) + (3 × 1.008) + (12.01) + (3 × 1.008) = 24.02 + 3.024 + 12.01 + 3.024 = 42.078 g/mol Using approximate atomic masses: C=12, H=1. Mass of one repeating unit = (3 × 12) + (6 × 1) = 36 + 6 = 42. The relative molecular mass of B is n × (mass of one repeating unit). 35700 = n × 42 n = (35700)/(42) n = 850 The value of n is 850. h) i) Identify the cleaning agent suitable to be used with water containing magnesium chloride. Magnesium chloride (MgCl_2) causes hard water. Cleaning agent X is a soap (R-COONa). Soaps form scum with hard water ions like Mg^2+. Cleaning agent Y is a synthetic detergent (R-O-SO_3^-Na^+). Synthetic detergents do not form scum with hard water ions. Therefore, cleaning agent Y is suitable. ii) State one advantage of using cleaning agent Y. One advantage of using cleaning agent Y (a synthetic detergent) is that it does not form scum with hard water, making it effective in both hard and soft water. Send me the next one 📸