In the hydroboration-oxidation reaction of propene with diborane, H$$_2$$O$$_2$$ and NaOH, the organic compound formed is:

The hydroboration-oxidation reaction of propene involves two main steps: hydroboration with diborane (B2H6) followed by oxidation with hydrogen peroxide (H2O2) and sodium hydroxide (NaOH). Propene has the structure CH3CH=CH2, which is an unsymmetrical alkene. The double bond is between the second carbon (CH group) and the third carbon (CH2 group), with the third carbon being less substituted.

In the hydroboration step, diborane dissociates into BH3, which acts as an electrophile. Boron attaches to the less substituted carbon (the terminal CH2 group) due to its higher electron density, while hydrogen attaches to the more substituted carbon (the CH group). This addition follows anti-Markovnikov orientation. The reaction proceeds as follows:

CH3CH=CH2 + BH3 $$\rightarrow$$ CH3CH2CH2BH2

However, diborane typically forms trialkylboranes. With excess propene, the reaction yields tripropylborane:

3 CH3CH=CH2 + (1/2) B2H6 $$\rightarrow$$ (CH3CH2CH2)3B

In the oxidation step, the alkylborane is treated with H2O2 and NaOH. This replaces the boron group with a hydroxyl group (OH), attaching it to the same carbon that was bonded to boron. The reaction for the trialkylborane is:

(CH3CH2CH2)3B + 3 H2O2 + NaOH $$\rightarrow$$ 3 CH3CH2CH2OH + sodium borate

The organic product is propan-1-ol, CH3CH2CH2OH, where the OH group is attached to the terminal carbon.

Now, comparing with the options:

• A. CH3CH2CH2OH (propan-1-ol)
• B. CH3CH2OH (ethanol)
• C. CH3CHOHCH3 (propan-2-ol)
• D. (CH3)3COH (tert-butanol)

The product matches option A. This is consistent with anti-Markovnikov addition, where the OH group attaches to the less substituted carbon.

Hence, the correct answer is Option A.