Answer:
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1)
HI(aq) → H⁺(aq) + I⁻(aq)
So this is an Arrhenius acid because it releases H⁺.
2)
LiOH(s) → Li⁺ + OH⁻
So this is an Arrhenius base because it releases OH⁻
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
Answer:
tetrahedral geometry
<h3>CHCH2O- CH2CH3</h3>
Explanation:
There are several centers of interest. Each carbon with all single bonds is the center of a tetrahedral geometry.
