In the William ether synthesis, an alkoxide reacts with an alkyl halide to give
ID: 1055869 • Letter: I
Question
In the William ether synthesis, an alkoxide reacts with an alkyl halide to give an ether. R.O. + R'-X rightarrow R.O.R' + X. You are in charge of a research group for a large company, Ethers Unlimited, and you have been assigned the task of synthesizing tert-butyl methyl ether, (CH_2)_3C-O-|CH_3. You have decided to delegate this task to two of your staff chemists. One chemist, Ima Smart, allows (CH_3)_#C-O- to react with CH_3-l and indeed obtains a good yield of the desired ether. The other chemist, Notso Bright, allows CH_3O to react with (CH_3)_3 C_Br. To his surprise, no ether was obtained. Explain why Ima succeeded but Notso Bright's reaction failed. Give the products expected when 2-bromo-2-methylhexane reacts with the following reagents. 1:1 ethanol-water sodium ethoxide in ethanolExplanation / Answer
Give the products expected when 2 bromo-2-methylhexane reacts with following reagents
Sodium ethoxide in ethanol :
This tertiary alkyl halide is subjected to a solid base in a protic dissolvable. Under these conditions, tertiary alkyl focuses are hindered to the point that they never experience SN2 substitution yet the E2 disposal can happen. While protic solvents advance SN1 and E1 responses, high centralization of base will support E2 disposal since rates of SN1 and E1 responses are autonomous of the grouping of base. The alkene items are demonstrated as follows. Take note of that the more substituted alkene D will prevail as indicated by Zaitsev's run the show.
On the off chance that the convergence of ethoxide is low, then SN1 and E1 responses will contend all the more viably with E2 disposal. E1 disposal will give an indistinguishable alkene items from above while SN1 result in the ether item
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1:1 ethanol-water:
Neither SN2 nor E2 responses can happen for tertiary alkyl halides in polar protic solvents within the sight of a poor nucleophile and a powerless base. Be that as it may, polar protic solvents advance SN1 and E1 responses, and henceforth these responses are watched. The substitution items would prevail as a result of high grouping of nucleophile, and the liquor will be the prevalent substitution item since water is the better nucleophile. The more substituted alkene will prevail as per Zaitsev's rule
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