Question

1. Preparation of 2-chloro-2-methylbutane from 2-methyl-2-butanol involves Sw1 reaction mechanism. What does Sw1 mean? (1 pts) 2. This reaction prefers to undergo via Sw1 mechanism over Sw2. Why? (3 pts) 3. Write the structure of carbocation and nucleophile involved in this reaction. (2 pts) 4. Substrate alcohol (2-methyl-2-butanol) consists a poor leaving group (OH group). How do you covert this poor leaving group into good leaving group? (2 pts) 5. Which will be the bottom layer between aqueous and organic layer during the extraction process in this experiment? And why? (2 pts) 6. What are the roles of aqueous NaHCO, and anhydrous sodium sulfate in this experiment? (2 pts) Write a test reaction to identify 2-chloro-2-methylbutane. (2 pts) 8. Complete the following reaction by arrow pushing mechanism and identify the sequence of arrow pushing patterns. (6 pts) OH + H : - + H0

Answer 1

(1) Preparation of 2-chloro-z-metheabudane chloro-2-methelbudane from 2-methyl-2-butanol are from 2-methyl as follows: T +HCI rt +H₂O The above reaction involves a SNL mechanism. The means of SNI is nucleophilic substitution cenimolecular reaction. This reaction follows the first order kinetics. Since, the rate of reaction depends only on the concentration of substrate and not on the concentration of nudeophile So, this mechanism is called an SN1 mechanisme Rate of reaction a [2-methyl-2-butanol] (2) The reaction prefers to undergo via SNl mechanism because when the reaction proceeds the formation of a 3 - carbocation dakes place. Que to this 3 - Carbocation the intemediate becomes very stable and hence, reaction proceeds via SNl mechanism. t +H-C1 -> V tem >> 3-carbocation Due to the formation of the so-carbocation Cmost stable) reaction proceeds snt pathways. (3) The structure of the carbocation and mudeophile involved are as, or CH3 CH3 ·co or CH₂C O (structure of the carbocation) u U 7 0 CH3 RDS CH3 Fer the given reaction, chloride jen (CDacts as a nudeophile (4) oHe is a poor cleaving group so we have to convert this poor leaving group in a good leaving group. The Öt lane -pair abstract a proton from HCl and form H300 Now, Hoe becomes a good cleaving group. The whole reaction mechanism us as follows: CH CH3 HA CH3 CH3 CH₂ - & -H -4-0H CH2 6 - OH2 2 CH2 CH2 dog+ H₂O CH3 CH3 Hao H20 CH3 L OH by accepting an electo a proton from Hal I cê and becomes Ho. Since, H₂O is good leaving roup and it leaves easily now. Thus, a poor "CH₂ - 4-a leaving group (OH) changes into a good CH3 leaving group (H2O). (Product) СН3 сня сна

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