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-(M/S) A Solvent 1 [alkly chloride] .100 .100 .200 .100 0.0520 0.104 OSIA OUG melder Isicoge-3+01 MARS de] [iodide]) rate (M/s) is an Hairw ni eno ai nodosen nouitedua A .100 moitos 0.0518 edua s to elqmsxe elqmia s yuzimerinis .200 woled nworia as 1ertions yd sbilsd A 10+ 1-SHOЯ+ 1+ 10-HO-Я 916 mainsrk m 10 ebio brigos sluoslomid bris (nebio fat) elucelominus rito8 Solvent 2 Jnsvloe bns [alkly chloride] [iodide] .100 .100 0.0381 .100 .200 0.0763 HO-Я -SHO-A stni srit lua to ior erit no gribneqeb noitose air hot eldizzoq rate (M/s) 200 gate 0.0760 .100 (M2) sluɔslominu B-C HO-Я (S Evaluate the values of x and y in the rate law, r=k[alkyl chloride][iodide], for solvent 1 and solvent 2. toubong bns in 91 sill to bris (312 162 State the mechanism, unimolecular (SN1) or bimolecular (SN2), favored by solvent 1 and solvent 2. Justify your statements. If the reaction is done with a single enantiomer of the reactant the outcome is as follows. Explain the stereochemical outcomes in terms of the mechanism followed. Why is a single enantiomer produced in Solvent 2 (with inversion of chirality relative to the starting material) but two enantiomers in equal amounts are produced in solvent 1. 8) CH3 CoH 10 SHO 1 + 10-sh vino ser es esibermain of al CH3 sribem 2 CH3 CH3 CHE art smil solvent I C\"H en wold ba CH₂ CH₂ CH3 - 50% I I CH3 C !!!! I H 50% solvent 2 I- CH₂ CHE Owly H product

-(M/S) A Solvent 1 [alkly chloride] .100 .100 .200 .100 0.0520 0.104 OSIA OUG melder Isicoge-3+01 MARS de] [iodide]) rate (M/s) is an Hairw ni eno ai nodosen nouitedua A .100 moitos 0.0518 edua s to elqmsxe elqmia s yuzimerinis .200 woled nworia as 1ertions yd sbilsd A 10+ 1-SHOЯ+ 1+ 10-HO-Я 916 mainsrk m 10 ebio brigos sluoslomid bris (nebio fat) elucelominus rito8 Solvent 2 Jnsvloe bns [alkly chloride] [iodide] .100 .100 0.0381 .100 .200 0.0763 HO-Я -SHO-A stni srit lua to ior erit no gribneqeb noitose air hot eldizzoq rate (M/s) 200 gate 0.0760 .100 (M2) sluɔslominu B-C HO-Я (S Evaluate the values of x and y in the rate law, r=k[alkyl chloride][iodide], for solvent 1 and solvent 2. toubong bns in 91 sill to bris (312 162 State the mechanism, unimolecular (SN1) or bimolecular (SN2), favored by solvent 1 and solvent 2. Justify your statements. If the reaction is done with a single enantiomer of the reactant the outcome is as follows. Explain the stereochemical outcomes in terms of the mechanism followed. Why is a single enantiomer produced in Solvent 2 (with inversion of chirality relative to the starting material) but two enantiomers in equal amounts are produced in solvent 1. 8) CH3 CoH 10 SHO 1 + 10-sh vino ser es esibermain of al CH3 sribem 2 CH3 CH3 CHE art smil solvent I C\"H en wold ba CH₂ CH₂ CH3 - 50% I I CH3 C !!!! I H 50% solvent 2 I- CH₂ CHE Owly H product

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section11.1: Reaction Rate
Problem 11.2CE: Instantaneous rates for the reaction of hydroxide ion with Cv+ can be determined from the slope of...
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I've already calculated the x and y values for Solvent 1 (x=1 y=0) and Solvent 2 (x=1 y=1) and determined Solvent 1 to be unimolecular and Solvent 2 to be bimolecular but can you please help with the third part to the question. 

Thank you! 

-(M/S) A Solvent 1 [alkly chloride] .100 .100 .200 .100 0.0520 0.104 OSIA OUG melder Isicoge-3+01 MARS de] [iodide]) rate (M/s) is an Hairw ni eno ai nodosen nouitedua A .100 moitos 0.0518 edua s to elqmsxe elqmia s yuzimerinis .200 woled nworia as 1ertions yd sbilsd A 10+ 1-SHOЯ+ 1+ 10-HO-Я 916 mainsrk m 10 ebio brigos sluoslomid bris (nebio fat) elucelominus rito8 Solvent 2 Jnsvloe bns [alkly chloride] [iodide] .100 .100 0.0381 .100 .200 0.0763 HO-Я -SHO-A stni srit lua to ior erit no gribneqeb noitose air hot eldizzoq rate (M/s) 200 gate 0.0760 .100 (M2) sluɔslominu B-C HO-Я (S Evaluate the values of x and y in the rate law, r=k[alkyl chloride][iodide], for solvent 1 and solvent 2. toubong bns in 91 sill to bris (312 162 State the mechanism, unimolecular (SN1) or bimolecular (SN2), favored by solvent 1 and solvent 2. Justify your statements. If the reaction is done with a single enantiomer of the reactant the outcome is as follows. Explain the stereochemical outcomes in terms of the mechanism followed. Why is a single enantiomer produced in Solvent 2 (with inversion of chirality relative to the starting material) but two enantiomers in equal amounts are produced in solvent 1. 8) CH3 CoH 10 SHO 1 + 10-sh vino ser es esibermain of al CH3 sribem 2 CH3 CH3 CHE art smil solvent I C\"H en wold ba CH₂ CH₂ CH3 - 50% I I CH3 C !!!! I H 50% solvent 2 I- CH₂ CHE Owly H product
Transcribed Image Text:-(M/S) A Solvent 1 [alkly chloride] .100 .100 .200 .100 0.0520 0.104 OSIA OUG melder Isicoge-3+01 MARS de] [iodide]) rate (M/s) is an Hairw ni eno ai nodosen nouitedua A .100 moitos 0.0518 edua s to elqmsxe elqmia s yuzimerinis .200 woled nworia as 1ertions yd sbilsd A 10+ 1-SHOЯ+ 1+ 10-HO-Я 916 mainsrk m 10 ebio brigos sluoslomid bris (nebio fat) elucelominus rito8 Solvent 2 Jnsvloe bns [alkly chloride] [iodide] .100 .100 0.0381 .100 .200 0.0763 HO-Я -SHO-A stni srit lua to ior erit no gribneqeb noitose air hot eldizzoq rate (M/s) 200 gate 0.0760 .100 (M2) sluɔslominu B-C HO-Я (S Evaluate the values of x and y in the rate law, r=k[alkyl chloride][iodide], for solvent 1 and solvent 2. toubong bns in 91 sill to bris (312 162 State the mechanism, unimolecular (SN1) or bimolecular (SN2), favored by solvent 1 and solvent 2. Justify your statements. If the reaction is done with a single enantiomer of the reactant the outcome is as follows. Explain the stereochemical outcomes in terms of the mechanism followed. Why is a single enantiomer produced in Solvent 2 (with inversion of chirality relative to the starting material) but two enantiomers in equal amounts are produced in solvent 1. 8) CH3 CoH 10 SHO 1 + 10-sh vino ser es esibermain of al CH3 sribem 2 CH3 CH3 CHE art smil solvent I C\"H en wold ba CH₂ CH₂ CH3 - 50% I I CH3 C !!!! I H 50% solvent 2 I- CH₂ CHE Owly H product
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