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The alkene R—CH=CH2 reacts readily with B2H6 and formed the product B which on oxidation with alkaline hydrogen peroxides produces:

1. R—CH2—CHO
2. R—CH2—CH2—OH
3. R—C=O
4. R—CH—CH2 | OH

Correct answer: R—CH2—CH2—OH

Solution

The reaction involves hydroboration-oxidation of the alkene. B2H6 adds across the double bond in an anti-Markovnikov manner, and subsequent oxidation with alkaline H2O2 converts the boron intermediate into an alcohol, yielding R—CH2—CH2—OH.

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