Illustrate the following reactions giving a suitable example for each?

(i)Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime. CH3COONa + Soda-lime (mixture of NaOH + CaO in ratio 3:1) → CH4 + Na2CO3
(b)(i)Pentan-2-one and pentan-3-one can be distinguished by iodoform test. Pentan-2-one is a methyl ketone. Thus, it responds to this test. But pentan-3-one, not being a methyl ketone, does not respond to this test. C3H7-CO-CH3 + 3NaOI → C3H7-COONa + CHI3 + 2NaOH
C2H5-CO-C2H5 + NaIO → No yellow ppt of Iodoform
(ii) Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) can be distinguished by iodoform test. Acetophenone, being a methyl ketone, on treatment with I2/NaOH undergoes iodoform reaction to give a yellow ppt. of iodoform. On the other hand, benzaldehyde does not give this test.
C6H5COCH3 + 3NaOI → C6H5COONa + CHI3 + 2NaOH
C6H5CHO + NaOH → No yellow ppt of iodoform
(iii) Phenol and benzoic acid can be distinguished by ferric chloride test. Ferric chloride test: Phenol reacts with neutral FeCl3 to form ferric phenoxide complex giving violet coloration.