Borane (BH3) adds to alkenes to form an alkylborane. In the first box draw the mechanism arrows, and in the second box draw the
correct product. Be sure to add lone pairs of electrons and nonzero formal charges to all species.
1 answer:
Answer:
The reaction when the Borane (BH3) is add to an alkene and form an alkylborane is shown below.
Explanation:
The boron of the borane does not have extra electron pairs, in this way the double bond of the alkene attacks the boron and the hydrogen belonging to the borane adheres to the carbon that is more substituted, thus forming an alkyl borane.
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Solutions are made up of two non reacting species called solute and solvent. The amount of solute in solvent is known as concentration of that solute. Concentration is often measured in Molarity. Molarity is the amount of solute dissolved in 1 dm3 of solution. Answer to your question is as follow;
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
And![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
</span>
For triehtylamine,
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4And
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7