Question

CH3-CH=CH2 forms in the presence of B2H6 and 3H2O2/OH​

Answer 1

Answer:

Of this reagent is present follow anti markonikav rule.

Anti markonikav rule:

Negative part of the reagent goes to the carbon atom where no of hydrogen is more.

☆ H+ comes from B2H6

☆ Oh- comes from H2O2

And so the product Propanol is in attachment

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Answer 2

CH3-CH=CH2+B2H6+ 3H2O2/OH​->CH3CH2CH2OH

• This pocess is known as hydroboration oxidation reaction.

• Hydroboration-oxidation reaction  follows the Anti-Markovnikov's addition of H−OH across C=C to give alcohol.

Thus,

CH3 CH=CH2   when treated with B2H6 in presence of H2O2  will give the final product as : CH3CH2CH2OH .