Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
Answer:
I don't see any following statements
Explanation:
could you give me some and I'll be happy to help : )
Sodium (Na) has a +1 charge and Iodine ( I ) has a -1 charge. To create a molecule of sodium iodide the charges will need to balance.
Because the charges on anion and cation are the same; the molecular formula will be NaI
A net ionic equation simply means to cancel out any ions which appear on both sides of the chemical equation that are not involved in the reaction - they're called spectator ions.
We'll first write out the full ionic equation, showing all ions and compounds formed, then rewrite and not include spectator ions.
2FeBr3(aq) + 3Na2S(Aq) --> Fe2S3(s) + 6NaBr(aq) [original eqation]
2Fe3+(aq) + 6Br-(aq) + 3Na+(aq) + 3S2-(aq)--> Fe2S3(s)+6Na+(aq) + 6Br-(aq)
[full ionic equation]
2Fe3+(aq) + 3S2-(aq)--> Fe2S3(s) [net ionic equation]
notice that Br- and Na+ appear unreacted on both sides of the full ionic equation, so they cancel out and do not appear in the net ionic.
*Please give me a 'brainliest' if you can! Thanks!
