It is a strong base because it is highly soluble in water forming the NaOH and NH3 which further produces NH4OH .
an alcohol reacts with a strong base to from an alkoxide ion
it's ionic !
A terminal alkyne with a proton missing; reacts as strong base and nucleophile RC(triplebond)C(-) The C- has a lone pair and a formal negative charge You can make one by reacting a terminal alkyne with a strong base such as NaNH2
Pentyne (C5H8)
sodium amide, then ethyl bromide --> 1-butyne. Add Hg/H2SO4 --> 2butanone
Yes
Well ammonia is a weak base and and NaNH2 is a strong base, so overall, you've got a pretty strong base.
Well ammonia is a weak base and and NaNH2 is a strong base, so overall, you've got a pretty strong base.
an alcohol reacts with a strong base to from an alkoxide ion
it's ionic !
A terminal alkyne with a proton missing; reacts as strong base and nucleophile RC(triplebond)C(-) The C- has a lone pair and a formal negative charge You can make one by reacting a terminal alkyne with a strong base such as NaNH2
Pentyne (C5H8)
sodium amide, then ethyl bromide --> 1-butyne. Add Hg/H2SO4 --> 2butanone
It is possible to make KOH work for the double dehydrohalogenation of an alkane if the base is in a high concentration. If a solvent like triethylene glycol is used, KOH will be less solvated than in solutions of say water, and therefore act as a stronger base.
According to wikipedia, the formula is: 2 Na + 2 NH3 → 2 NaNH2 + H2 I hope I helped! ;)
Methanol reacts with sodamide (sodium amide) to produce sdoium methoxide and ammonia. CH3OH + NaNH2 --> NaOCH3 + NH3
Sodium is one of the elements in the compound sodamide, which is scientifically called sodium amide (NaNH2). It is formed by reacting sodium metal (Na) with ammonia (NH3).