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: Atomic Nucleus!
Explanation: All atoms have a dense central core called the atomic nucleus. Forming the nucleus are two kinds of particles: protons, which have a positive electrical charge, and neutrons, which have no charge.
(Yes, it was from google.)
1.35atm(760mmHg/1atm)=1026mmHg
Indium has 49 protons
mass number (# of neutrons and # of protons combined) is 115
115 - 49 = 66
66 neutrons
Answer:
109° 27'
Explanation:
The ammonium ion is tetrahedral in shape, all the HNH bonds are exactly at the tetrahedral bond angle since there are only bond pairs in the structure and no lone pairs. Recall that lone pairs decrease the bond angke from the ideal value in a tetrahedron due to higher repulsion.
