


Let's analyze the given reaction to identify the products A and B.
The reaction involves two steps starting from a given alkene:
Hence, the products are:
Therefore, the correct answer matches the chemical structures shown in the option:
(1) Hydration Reaction:
\(CH_3 - CH = CH_2 + H^+ \rightarrow CH_3 - \overset{+}{C}H - CH_3 \; \text{(More stable)}\)

(2) Hydroboration Oxidation Reaction:
\(3CH_3 - CH = CH_2 + B_2H_6 \xrightarrow{\text{THF}} 2(CH_3CH_2CH_2)_3B\)
\((CH_3CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{\text{OH}^-} 3CH_3CH_2CH_2OH + H_3BO_3\) (B)
Thus the correct answer is Option 3.
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
