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.
You might be interested in
Answer:
Explanation:
Hello,
Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.
Best regards.
Answer: All of the statements are true.
Explanation:
(a) Considering the system mentioned in the equation:-
The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B
(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.
0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,
40 to 60s is 0.14 (0.30-.16 = 0.14) mol
(0.46 > 0.14) mol leading this statement to be true as well.
(c) Average rate from t1 = 40 to t2 = 60 s is given by:
which is true as well