Properties of Carboxylic Acids and Their Derivatives Questions in English

(A) Step $1$: Carboxylic acids are resistant to reduction by catalytic hydrogenation ($H_2/Ni$ or $H_2/Pd$) because the $C=O$ bond in the carboxyl group is less reactive than in aldehydes or ketones.
Step $2$: $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent capable of reducing carboxylic acids to primary alcohols.
Step $3$: The reaction proceeds as: $RCOOH \xrightarrow{LiAlH_4 / H_3O^+} RCH_2OH$.
Step $4$: Therefore, $LiAlH_4$ is the correct reagent for this transformation.

(D) Step $1$: The acidity of a compound depends on the stability of its conjugate base.
Step $2$: The conjugate base of a carboxylic acid is the carboxylate ion $(RCOO^-)$, where the negative charge is delocalized over two highly electronegative oxygen atoms.
Step $3$: The conjugate base of a phenol is the phenoxide ion $(C_6H_5O^-)$, where the negative charge is delocalized over the carbon atoms of the benzene ring and one oxygen atom.
Step $4$: Since oxygen is more electronegative than carbon, the negative charge is more effectively stabilized in the carboxylate ion than in the phenoxide ion. Therefore, carboxylic acids are more acidic.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group attached to a carbon or hydrogen atom.
$(1)$ Ethanal $(CH_3CHO)$ contains the $CH_3CO-$ group.
$(2)$ Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group.
$(3)$ Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group.
$(4)$ Ethanoic acid $(CH_3COOH)$ does not contain the $CH_3CO-$ group attached to a carbon or hydrogen atom, as the carbonyl carbon is bonded to an $-OH$ group. Therefore, it does not give the iodoform test.