Hydroboration-oxidation converts an alkene into an alcohol with anti-Markovnikov orientation and overall syn addition of water. In the first step diborane adds to propene so that boron, being the electron-deficient atom, attaches to the less hindered terminal carbon while hydrogen goes to the internal carbon. Oxidation with alkaline $H_2{O}_2$ then replaces the carbon-boron bond with a carbon-hydroxyl bond while retaining position, so the hydroxyl ends up on the terminal carbon, giving propan-1-ol. Propan-2-ol is wrong because it is the Markovnikov product expected from acid-catalysed hydration, not from hydroboration. Propane-1,2-diol would require dihydroxylation across the double bond, for example with cold permanganate, so it is incorrect. Propanal is an aldehyde and would require oxidation beyond the alcohol stage, which does not occur under these mild conditions, so it does not apply. The regioselectivity originates partly from steric factors, since the bulky boron prefers the less hindered terminal carbon, and partly from the electronic polarity of the boron-hydrogen bond, in which the more electropositive boron behaves as the electrophilic end. Because the mechanism is a concerted four-centre addition, hydrogen and boron add to the same face, giving overall syn stereochemistry that is retained through oxidation. This reaction is the NCERT-highlighted route to anti-Markovnikov alcohols. A consistency check: because boron adds to the carbon bearing more hydrogens, the OH must end up on the terminal carbon, confirming propan-1-ol.