Convert Propanoic acid to Ethane
To convert propanoic acid to ethane, we perform a reduction reaction using a strong reducing agent like LiAlH.
The steps are: 1. Reduction of Propanoic acid: Propanoic acid is reduced by lithium aluminum hydride (LiAlH) to ethanol. \[ \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \] 2. Reduction of Ethanol to Ethane: Ethanol is then reduced further by a reducing agent like zinc in the presence of hydrochloric acid (Zn/HCl) to ethane. \[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Zn/HCl}} \text{CH}_3\text{CH}_3 \] Thus, Propanoic acid is first converted to ethanol and then ethanol to ethane.
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.
Acidified \(KMnO_4\) oxidizes sulphite to:
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]
Which element is a strong reducing agent in +2 oxidation state and why?