Step 1: Explanation of \( E^0 \) Value - The electrode potential (\( E^0 \)) depends on: - Atomization enthalpy (\( \Delta H^0_{\text{atom}} \)): The energy required to convert solid Cu to Cu\(^{2+}\) is high. - Hydration enthalpy (\( \Delta H^0_{\text{hydr}} \)): Cu\(^{2+}\) has low hydration energy, making it less stable in aqueous solution.
Step 2: Effect on \( E^0 \) Value - Due to low hydration enthalpy, the reduction of Cu\(^{2+}\) to Cu is not highly favored. - Hence, Cu\(^{2+}/\)Cu has a positive \( E^0 \) value of \( +0.34 \) V, indicating that Cu is less reactive than expected.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.