Question

1. Consider the SN2 reaction shown below and answer the following questions.nobl 8 acetone no t iNaSCH SCH, + NaBr worl s1918. A. Write the rate law for the reaction. ebiler vil ar to nollurunoonoo oli nison. B. Identify the nucleophile and the electrophile in the reaction. HOCH lo noiden nog orgnismanl.d C. State how each of the following factors would affect the rate of the reaction. CHOW CHOOH glasa a. Increasing the concentration of 1-bromopropane. b. Decreasing the concentration of NaSCHs by one-half.non abiler ve orti prigion.b c. Changing 1-bromopropane to 2-bromopropane. 10007qomad- of obiter voortgninni. d. Changing 1-bromopropane to 1-iodopropane. airit mol olslot of toqxo voy bluow (a)lubong for e. Changing NaSCHə to CH3OH. wom won oy tarw navia. Snous 2. Consider the SN 1 reaction shown below and answer the following questions. Lachs Her PH + CH3OH noitoon or lo el or ons bluow privololol) to lose wodylooga bend to notando el sorol.s A. Write the rate law for the reaction.

Answer 1

SCH3 + Naso VB + NASCHE Acetone → SN2 A. Rate law: It's a bimolecular nucleophilic rate substitution reaction depends upon both the substrate and the nucleophilic concentration. Hence, rate, s = kx [ Br] [NascHz] « where k=rate constart. B. Nucleophile: SCH3 I NascHz in acetone Electrophile : BT Ito No - Free ScHz nucleophite SCH3 SCH + Br ** polar aprotic solvert line Acetone, dissolves the Cationic counterpent of the nucleophike. (a) P = [~ Br] [NA SCH3] increasing the concentration of Br, will increase the rate (2) 16 Decreasing the concentration of Nasctz by one- half slak [ B] [NascHz) = 8 Therefore, rate will also decrease by one-halt.

(c) Changing ~ Br to e (l-bromopropane) (2-bromopropane) decrease the rate of reaction. Because the nucleophilic attack from backside (180° to the c-Br bond will require more activation energy. This happens because the attacked C-atom becomes primary to secondary ; i.e. steric crowding increases at the carbon center making it difficult for the nucleophile to approach. (d) changing , ~ Br to I 1-iodo-propane). since, I is a better leaving group compare to Br, rate of reaction will be faster in this case. (e) Changing, Nascha to, CH3OH. SCH3 is a better nucleophile than ChOH. So, reaction rate will decrease up on decreasing the nucleophilicity of the nucleophite; i.e. Nasch, CHzolt & CH3OH - ph LOCH3 +HB Ph Mechanism (1) pentBr - By pro CH Cholt РЪКОСА, Unimolecular Nucleophilic Courbocatio substitution intermediate + HBY A. Rate law. The first step of the reaction is the rate rate mechasüsm determing determing step stee (eds); rate depends only on the substrate concentration; i.es rate, r = k [pi tor]

B: Nucleophile : CH3OH Electrophile : Ph (the cas bocation intermeeliate) Solvent : Sul reaction is favored by polon protic solvent; they stabilize the carbocation intermediate with their loan pairs. CH₃OH is a polar protic solvent . So, the reaction can be carried out in CH₃OH , as a solvent as well as a nucleophile. C. (a) since rate depends only on the substrate concentration (here alkyl halide, pn Br) rate will be increased upon increasing the concentration of alkyl halide. (6) Increasing the concentration of CH₂OH, will not affect or change the reaction rate, since, it is not present in the rate expression. (c) Replacing CH₂OH, by Naochz : NOOCH3 acts more like a base, than a nucleophile. In this case, reaction rate only depends on how much stable the carbocation intermediate is. Nuclepphiticity increment won't change the reaction rate. (d) Changing the halide from bromide to iodide : The rate of formation of the carbocation increases, as iodide is a better leaving group that Br. Therefore, the rate will be incoreased.

e phf Br to Bro & Bre Propyl > cation Poimary carboration highly unstable. Therefore, the rate of the reaction will be much slower. nigte Ez- + Ger substituted + Nao t o elimination (NaOt-Bu) most acidic proton (from the least carbon) is abstracted by the base. It is a bimolecular elimination reaktion. rate, p=4 Y V Naot-Bu] 33 rate depends upon both the substrate concentration. and base entration of of reaction the base, wshall (from the rate law). (a) Increasing the concentration of the b increase the rate of reaction from 16) Decoreasing the concentration of alkyl halide, Shaw decrease the rate of the reaction. S'er heat T + (9) (major) (minor) it product comes comes major from from th the abstraction most cidic acidic proton, under basic conditim. proton, under

6 (6) -Br NaOCH3 Theat → No Ez elimination should be product Obserwed Since there are no ß. protons for elimination (a) H20 th heat By For E elimination maior product comes as the most substituted alkene product; i.e. the most stable one. major) Trisubstituted (minor) Disubstituted (6.) H2SO4 OH GH ) heat + H2O (major) (minor) Trisubstituted Disubstituted The major product same reason, stated here, also above due to the
Those were a lot bunch of questions :-l