2020年8月18日 · As @MaxW already pointed out, 1:4 means that you have to use four times as much water as t-BuOH. But I also want to explain why this is, because you'll probably see this more often. In fact, 1:4 is a ratio, so what it represents is the ratio between t-BuOH and H 2 O:

2019年8月13日 · Several reasons: t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.

2018年2月19日 · Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d 1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating, 3 favoring protonation rather than deuteration (isotope effect).

2015年5月1日 · The short answer is that the steric hindrance of the tert butyl alcohol makes it hard to add an protonate the hydroxy group.

2019年9月28日 · 1-chloro-2-methylpropane --> 2-chloro-2-methylpropane. My understanding is that an E2 reaction produces two types of alkenes, the major product for non-sterically hindered bases is the more subst...

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2017年4月27日 · Can Tert-butoxide displace bromide via SN2 reaction? I read that Tert-butoxide is bulky and has steric hindrance, even with primary substrates. How is this reaction (picture below) possible? Sour...

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$\begingroup$ You are assuming that the phenol and alcohol are in water. In a less polar solvent I suspect no or very little dissociation of phenol occurs, thus the polar nature of the solvent has a real role to play in stabilising the ions and driving the reaction what ever the …

2018年2月18日 · Why not E2? The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product.