1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.

hex-1-ene with a hydroboration-oxidation reaction

Explanation:

In this case, we can start with the structure of 1-hexanol. In this alcohol, the "OH" is on the first carbon. Thus, the alkene that this alcohol can produce must have a double bond between carbons 1 and 2, that is, hex-1-ene.

Now, it is important to know which of the reactions we can use, hydroboration-oxidation, or oxymercuration-reduction. To decide which reaction to using, we must look again at the alcohol we want to produce. In this alcohol, the "OH" is attached to a primary carbon. Therefore, the addition of the "OH" must be made on the least substituted carbon, that is, carbon number 1.

Because of this, an anti-Markovnikov reaction (a reaction in which the "OH" is added on the less substituted carbon) must be used. The reaction, which serves us then is: hydroboration-oxidation since this reaction is anti-markovnikov.

See figure 1

I hope it helps!

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-