Why does $\ce{CH2NH2}$ have lower priority than $\ce{CN}$? The carbon on the first molecule is bonded to (N,H,H) while the carbon on the second molecule is just bonded to (N). I can't seem to find this scenario anywhere online.

What is CH2NH2 called Ethylamine or something else Thanks. Disclaimer: Information in questions, answers, and other posts on this site ("Posts") comes from individual users, not JustAnswer; JustAnswer is not responsible for Posts.

The image above depicts three mesomeric structures of diazomethane ($\ce{CH2N2}$). When you count the valence electrons for every atom, you will find that only the left and middle structure satisfy the octet / duet (for hydrogen) rule, i.e. that every $\ce{C}$ and $\ce{N}$ atom has 8 valence electrons, and every $\ce{H}$ has two valence electrons.

2015年2月21日 · $\ce{R-CH=O}$ has higher priority over $\ce{R-CH_2OCH_3}$. Because, if $\ce{R}$ group is the same, then we have to compare between the next substituents: For $\ce{R-CH=O}$ you have one carbon atom connected to 2 oxygen atoms (a double bond is equivalent to two simple bonds) and one hydrogen atom.

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2023年6月6日 · Karan: Structure 1 not only has the issue that you have addressed, the priority of -CHO vs. - $\ce{CH2OH}$, but also deals with double bond geometry and the configuration at $\ce{C2}$.

2020年5月7日 · As OP suggested there are three resonance structures of diazomethane contributing to the structure: If you want to neutralize the formal charges of last resonance structure, you can put a double bond on negatively charged carbon and adjacent positively charged nitrogen to get the fourt resonance structure.

2015年10月2日 · When the atome of nitrogen participates through its free electron pair in resonance, the amine group plays the role of a donating group by conjugation.

2022年10月21日 · Which of the following groups has the highest priority in the Cahn-Ingold-Prelog sequence rules? a) CH2CH3 b) CH=CH2 c) C≡CH d) C(CH3)3 I narrowed it down to either C or D.

2013年1月17日 · $$\ce{R-CH2OH -> R-CH2NH2}$$ direct alkylation of ammonia normally is the last thing you want to do in the lab. Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent further alkylation of the initial product.