




Reaction mechanism: Lithium borohydride (LiBH$_4$) is a selective reducing agent that reduces esters (CO$_2$Et) to alcohols while leaving carboxylic acids (CO$_2$H) unchanged.
Step-by-step process:
Option (A) The ester group (CO$_2$Et) is reduced by LiBH$_4$ in ethanol (EtOH) to form a primary alcohol (-CH$_2$CH$_2$OH).
Option (B) The carboxylic acid group (CO$_2$H) remains unchanged during the reaction.
Option (C) Protonation with H$_3$O$^+$ ensures the stability of the final product.
The final product contains an unchanged carboxylic acid group and a newly formed primary alcohol group, corresponding to option (4).
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

Given below are two statements:
Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with \( \text{CH}_3\text{COCl} / \text{AlCl}_3 \) will give the product shown. 
Statement II: \( -\text{NO}_2 \) group is a meta-directing and deactivating group.
In the light of the above statements, choose the most appropriate answer from the options given below.