Dec 13, 2018 · What happens when alkynes react with sodium in ammonia ($\\ce{Na/NH3}$), or when they react with sodium amide ($\\ce{NaNH2}$)? Are these two sets of conditions different?

May 7, 2015 · A TA talked about this today in the context of ketones. If you stick an acetone and $\\ce{NaBH4}$ together in a protic solvent, you get 2-propanol. If you stick acetone and $\\ce{NaH}$ together, yo...

Dec 30, 2022 · We have this exercise without solutions. From a 0.2 M $\ce{NaH2PO4}$ solution and a 0.2 M $\ce{Na2HPO4}$ solution, a buffer solution with pH = 6.8 is to be prepared.

Dec 11, 2019 · I understand why compounds such as $\ce{NaH}$ or $\ce{NaNH2}$ are weak nucleophiles: as they aren't very soluble in organic solvents, they react only on the clusters' surface.

May 4, 2018 · this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.

Mar 29, 2020 · Identify the final product (B) in the following reaction: I'm not sure how $\\ce{NaNH2}$ and $\\ce{CH3I}$ will react with the different types of alcohol present here. According to my solution,ether was

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Jun 23, 2020 · Your question: Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution? I assume what you meant is your textbook is saying when two separate solutions of $\ce{NaOH}$ and $\ce{NaH2PO2}$ are combined, there should be no reaction.

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Sep 28, 2014 · I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions.