1. Location the alkyl halides in each group in bespeak of enhancing E2 reactivity. A. Br Br Br ns П III b. Cн, Br Нас CI CH3 П III rank the alkyl halides in each team of problem one in bespeak of boosting El reactivity. 2. I beg your pardon elimination reaction in every pair is faster? (I or II) 3. а. сн, Нас. Он Он II. B. HO DMSO он (CH),CCI (CH),CCI П. I. Н.о What alkenes are created from every alkyl halide by one E2 reaction? Make certain to incorporate all feasible stereoisomers. Label the significant product guess by Zaitsev rule. Label the major product guess by Hoffman rule. Label the significant product that would certainly be developed with the specified strong base. 4. а. \"ОСCH>) - b. CH3 Но.


You are watching: Rank the alkyl halides in the following group in order of increasing sn2 reactivity.

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Bre 1. A (11) ( I) fore 'Ej, reeactivity is 3 alkyl 2. Alkoll 1 alkyl halide ) halidé ... Halide ^ Bre X c therefore heree, et UT) ) i ( Reactivity ored C CH3 Xci l a W CH₃ . H3C together (I) (2) CO () here 'I' is much more reactive 보다 I since it is 2. Alkyl halide and 'I' is ns alkol halide top top Now, in in between 'I' and 'm', 'm oo is an ext reeactive as is weaker 보다 C-cu boad.. For this reason the reeactivita e oredere is (III) > (I) )(I)
. Halide g. Fore 'ER' reactivity is 3 alkyl halide > 2 alkol halide > i alkyl halide (a) 111 111) ns (Reactivity oreder). (b) 111 ) Illiu (Reactivity oredere ) as 'C- Bre' link is weaker 보다 ic-ci bond for this reason 'Bre' is much better leaving greoup than ar > 3.ca, a CH3 CH3 -- SES 3.(a) n CH3 ai OHS (I) hello Cici oh - CHya five then x T i ishaofastere .. Because it is halide? 2 alkol Poliole
сH3 CHI-º-c CH3 - Ho CH2= (9) (CH),cat 6H (*) \" (CHS), CC. Of CH2 = = _CH 2 DMSO уо це Both the over reaction, adhered to 'El route way. . Polare preotic solvent favoures ... The Ei reaction. For this reason 'I' is fastore 보다 I otic Solvent


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CL A.com üret öc (CH3)3 → ~ + (Hoffmann Zaitsev O product ) product - (minore (majore preoduct) preoduct) as mora bulkiere base favoures 1. Hoffmann preoduct overe Zaitsen preoduct.so below ochz. Is a bulkier base, therefore Hoffmann preoduct is majore preoduct. B. Ns CH3 inn † five - CHI ^at ^ Zaitsev polt. Hoffmano (majore) (minore) as coH is less bulkiere basic it favoures zaitsev reule.