2017年2月27日 · There are no resonance structures. There is only one valid Lewis structure for the bromomethyl radical. The Bromomethyl radical has only one possible Lewis structure. This molecule as written in the question has an odd number of electrons Carbon: 4 Hydrogen: 1 * 2=2 Bromine: 7 4+2+7=13 Resonance structures are multiple valid Lewis structures for a molecule that differ only in the electron ...

2017年5月4日 · This type of formulae can be confusing. Might be helpful to redraw it as $$\ce{CH3(CH2)2CH(CH2CH3)CH2Br}$$ or $$\ce{CH3CH2CH2CH(CH2Br)CH2CH3}$$ There is only one carbon atom in methyl, so 2-bromomethyl is nonsense. The locant means on which carbon atom is the functional group attached.

2020年7月27日 · The question asks for the rate of reaction in $\beta$-elimination using alcoholic $\ce{KOH}$.. Arrange the following alkyl halides in decreasing order of rate of $\beta$-elimination reaction with alcoholic $\ce{KOH}$:

2018年2月15日 · The IUPAC name is 3-(bromomethyl)-1-hydroxypentane-2,4-dione. > I think this is the structure you are trying to draw. Here's how you generate the IUPAC name. Step 1. Find the longest continuous chain of carbon atoms The longest continuous chain contains five carbon atoms, so the parent hydrocarbon is pentane. Step2. Identify the functional group of highest priority The molecule contains "OH ...

2014年9月5日 · Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible.

2018年10月20日 · 1-CH(CH3)CH2F 2-CHCl2 3-CH2Br 4-C(CH3)3. this is the order i did using the system but i am having a hard time understanding it and was wondering if anyone could tell me if my answer is correct, and if it is not, explain to me please

2016年4月18日 · We have the reaction of the alkyl halide $\ce{CH3-CHBr-CH2Br}$ with 1) $\ce{OH-}$ 2) $\ce{NaNH2}$ The answer is propyne. I am unable to understand why we have performed an elimination reaction here. Why can't it be a substitution reaction instead? Also, how do we decide which two carbon atoms would lose their hydrogen atoms?

2017年11月1日 · According to my textbook (A Level Chemistry a for OCR Student Book, Pages 84-5) $\\ce{CH2Br2}$ is more polar than $\\ce{CH3I}$. The molecules are both polar, they both contain different bonded atoms,...

2016年1月19日 · Huh, I can't think of more than one good path. There are others, but they won't give as good of a yield because the alkyl halide is a secondary alkyl halide, which is not great for "S"_N2, ruling that out in the first step for the best possible synthesis. Here's what I came up with: The first step is a second order elimination (E2) via a …

2019年12月14日 · $$\ce{CH3CH=CH2 ->[Br2] CH3-CHBr-CH2Br}$$ In the reaction mechanism, why does the nucleophile attack at the 2nd carbon atom? Shouldn't it attack at the 1st one, so that in the transition state, the carbanion is relatively more stable?