Jul 29, 2011 · “NaNH2 is usually made by dissolving sodium metal in liquid ammonia, so this is the most common reagent/solvent combination.” If so, how will we solve 1-bromo-1-phenylethene + Na/liq.NH3 =? Will Na/liq.NH3 work as a reducing agent and form 1 …

First, the nucleophilic NH 2− group adds to the δ+ ring carbon atom, pushing electrons onto the ring nitrogen atom and forming the anionic σ-adduct, which is stabilized by sodium. Electrons from the N atom are then pushed towards the ring to regain aromaticity, forming a C=N bond and ejecting a hydride ion in the process.

Consequently, other reagents of the type Y–NH 2 have been studied, and found to give stable products (R 2 C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed below, together with the structures and names of their carbonyl reaction products.

Sodium amide, commonly called sodamide (systematic name sodium azanide), is the inorganic compound with the formula NaNH2. It is a salt composed of the sodium cation and the azanide anion.

Jun 11, 2013 · You treated alkane alkenes with NaNH2. Here this acts NaNH2 acts as an base, which deprotonates the substrate giving elimination product. Also, using NaOH(OH^-) when treated with substrate (mostly alkane derivatives) gives substitution reactions, whereas the same NaOH, acts as a base for the substrate, like in above bromoalkene did (when ...

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?

H2O = 2 carboxylic acids (cleavage rxn) Terminal Alkynes 1. KMnO4 or O3/2. H2O. be CH3, CH2CH3, CH2 CH2CH3, etc. New carbon-carbon bond formed.

Why is NaNH2 used as a base to deprotonate and alkyne? It's a strong enough base to deprotonate the alkyne. If something like NaOH or NaOMe is used, the negatively charged oxygen is more stable than the alkane, making the base a weaker base, so the reaction won't favor deprotonation.

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Dec 29, 2024 · React the terminal alkyne with sodium amide (NaNH₂) or sodium hydride (NaH). The reaction is: R-C≡C-H + NaNH₂/NaH → R-C≡C⁻Na⁺. The terminal alkyne (R-C≡C-H) reacts with sodium (Na) to form R-C≡C⁻Na⁺ and hydrogen gas (H₂). Similarly, it reacts with sodium amide (NaNH₂) or sodium hydride (NaH) to form R-C≡C⁻Na⁺.