Question

Acetone reacts with iodine $({I_2})$ to form iodoform in the presence of
A.$CaC{O_3}$
B.NaOH
C.KOH
D.$MgC{O_3}$

Answer 1

Hint: Iodoform is basically an organ iodine compound with the chemical formula $CH{I_3}$. It belongs to the family of organic halogen compounds. Moreover, it is a crystalline pale-yellow substance which is volatile in nature.

Complete step by step answer:
Iodoform is manufactured by the electrolysis of aqueous solution containing acetone, inorganic iodides and sodium carbonate. Moreover, it is also known as tri-iodomethane, carbon tri-iodide and methyl tri-iodide. The names are generally due to the presence of three iodine molecules in the compound.
Now, acetone has a chemical formula ${(C{H_3})_2}CO$ . It is also known as propanone or dimethyl ketone and it is the smallest and the simplest ketone. When it reacts with iodine in presence of NaOH, it forms iodoform. The reaction is as shown:
 $C{H_3}COOH + 3{I_2} + 4NaOH \to CH{I_3} + 3NaI + C{H_3}COONa + 3{H_2}O$
So, we can see that iodoform is formed in the presence of NaOH.
Moreover, iodoform test is used to check the presence of carbonyl compounds with the structure $R - CO - C{H_3}$ or alcohols with the structure $R - CH(OH)C{H_3}$ in a given unknown substance. The reaction of iodine, a base and a methyl ketone give a yellow precipitate. The compounds that give a positive iodoform test are acetaldehyde, ethanol, methyl ketones etc.

Hence, option B is correct.

Note: Iodoform is used as a disinfectant on a small scale. It was also used as a component in medicines for healing and antiseptic dressing of wounds and sores in the 20th century. Moreover, there are many side effects associated with iodoform and therefore its use has now been suppressed with the evolution of new antiseptics.