Question

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1. Write a reaction equation for the synthesis of nylon 4,6. (Hint: the reaction equation for making nylon 6,10 is shown at the top of page 2. Make appropriate changes to the reactants to make nylon 4,6.) Но - + H₂N. wit 2. Which side product is formed when sebacyl chloride and hexamethylenediamine react to form nylon? (Hint: In the reaction equation at the top of page 1, which atoms are removed when one carboxylic acid molecule reacts with one amine molecule? What is the side product of that reaction? For the reaction of sebacyl chloride and hexamethylenediamine, different atoms are involved, and the side product is different.) NH (CH2), NH - 2 (CH₃)4 E-R + HCl 3. Complete the table below: grams of mL of hexamethylene diamine used density of hexamethylene hexamethylene- diamine=0.93 g/ml diamine used molar mass of moles of hexamethylene hexamethylene- diamine 116 g/mol diamine used 1 mL mLx (g/ml) = 8 8x (mol/g) = mol mL of sebacyl chloride used density of sebacyl chloride = 1.12 g/mL grams of sebacyl chloride used molar mass of sebacyl chloride =239 g/mol moles of sebacyl chloride used 1 ml ml x (g/mL) - 8 gx (mol/g) = mol Name of the limiting reactant → Hexamethyldiamine or sebacyl chloride? (What is there fewer moles of; which runs out first?) 4. Complete the table below: % yield of nylon moles of limiting reactant from the table in #3 molar mass of nylon = 282 g/mol Theoretical yield of nylon, in grams Actual yield of nylon obtained in the experiment. molx (g/mol) = 8 0.958 *You didn't run the experiment, but assume that you obtained 0.95 g. **% yield = 100 x (actual yield in grams/theoretical yield in grams)

Answer 1

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1.) Hcxamethylene-damine used 2 Using Stoichiometry with data guan: used 0.939 (1 ml) - 10.939 ml 0.93g ( mal 8.02% 10° 3 rrales K 116 g 2.) Sebacyl chloride a 1.12 g (1 ml) limiting reactant - 1.129 used M Sebacyl chloride 1.129 imal 4.69x10 males 239g 4.) 4.69x10 "males Sebacyl 1 mal Nylon 2829 =) 1.329 Nylon Imal sebacyl Almal Nylon % geeld 100 0.959 1. 329 71.97%