Q.Hydroboration-oxidation of propene gives.....

Acetaldehyde + $CH_3MgBr$, then hydrolysis, gives isopropanol (a secondary alcohol).

A Grignard reagent adds across the polar $C=O$ of the aldehyde to give a magnesium alkoxide, which on hydrolysis (with dilute acid/water) yields the alcohol:

$CH_3CHO + CH_3MgBr \xrightarrow{dry\ ether} CH_3-CH(OMgBr)-CH_3 \xrightarrow{H_2O/H^+} CH_3-CH(OH)-CH_3 + Mg(OH)Br$

Since an aldehyde reacting with a Grignard reagent gives a secondary alcohol, the product is propan-2-ol (isopropanol).

Hydroboration–oxidation adds $-OH$ to the LESS substituted (terminal) carbon, giving the anti-Markovnikov, primary alcohol product.

Hydroboration of propene ($CH_3-CH=CH_2$) with diborane ($B_2H_6$) proceeds with boron attaching to the less hindered, terminal carbon (anti-Markovnikov regiochemistry), forming a trialkylborane. Subsequent oxidation with alkaline hydrogen peroxide replaces the C–B bond with a C–OH bond with retention of position, giving the primary alcohol:

$CH_3-CH=CH_2 \xrightarrow{B_2H_6} (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} CH_3-CH_2-CH_2-OH$

So the product is propan-1-ol, not the Markovnikov-expected propan-2-ol.

Baeyer's reagent is cold, dilute, alkaline $KMnO_4$, used to test for unsaturation and to convert alkenes into vicinal diols.

Baeyer's reagent is a cold, dilute, alkaline (basic) solution of potassium permanganate. It is used as a test for carbon-carbon unsaturation (alkenes/alkynes): the purple colour of $KMnO_4$ is decolourised as it oxidatively adds across the double bond, converting an alkene into a vicinal diol (a syn-dihydroxylation), e.g. propene $\rightarrow$ propane-1,2-diol.