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In practice question 1 part a of “Localized and Delocalized Lone Pairs and Bonds”, is there a reason the delocalized lone pairs coming from the nitrogen with the arrow pointing at the single bond with carbon doesn’t turn into a double bond? Why do we just indicate the arrow push with a negative charge? Where did the electrons from the lone pair go?
Hi there,
In part a, the lone pair on the nitrogen did become a double bond (N=C) which forced the electrons of the pi bond between the two carbons to move to the right one, and as a result, there is now a lone pair on that carbon and it has a -1 formal charge.
I have a question regarding hydrolysis of nitriles. In alkaline hydrolysis, the products are carboxylic acid salt and NH3 or carboxylic acid and MeNH2 (f.e. NaNH2 if NaOH was used)? Also, which hydrolysis, acid or base, would you call superior based on yield and difficulty of purification from side products and leftover substrates?
Please check this article on nitrile hydrolysis: https://www.chemistrysteps.com/the-mechanism-of-nitrile-hydrolysis-to-carboxylic-acid/
what time are you online i have been trying to reach out
Go ahead in 5 min.
https://www.chemistrysteps.com/naming-alkenes-nomenclature-practice-problems/
According to the latest IUPAC recommendations (2013), the longest carbon chain does not need to contain C=C, hence the preferred IUPAC name for 7-methyl-3-propyloct-1-ene is 6-ethenyl-2-methylnonane
https://www.chemistrysteps.com/naming-alkanes-iupac-nomenclature-practice-problems
Similarly, according to the latest IUPAC recommendations (2013), rings always have seniority over chains, hence the preferred IUPAC name for 1-cyclopentylhexane is hexylcyclopentane
(usually locant 1 is not needed for a cycle if there’s only 1 branch)
This and the pKa of water have been on my to-do list for a while now. Thanks for bringing it up!