Hint: In this reaction, I^– will attack the less substituted carbon.

Explanation:

Since mixed ether is given in this reaction, note that iodide will bond with the smaller alkyl group to minimise the steric repulsion as iodide ion is bulky itself.

The reaction is as follows:

${\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{O}-{\mathrm{CH}}_2-{\mathrm{CH}}_3+\mathrm{HI}\stackrel{∆}{\to }{\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}$

Hint: The compound containing methyl ketone and secondary alcohol gives the iodoform test.

The compound containing  and gives the iodoform test.

Step 1:

Compounds having ${\mathrm{H}}_3\mathrm{C}-\mathrm{CO}-$ group and ${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right)-$ gives iodoform when heated with Iodine and NaOH.

Step 2:

Among the following compounds CH₃CH₂OH, CH₃COCH₃  and propan-2-ol  will give iodoform test.
Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.

Hence, the second option is the correct answer.

Hint: The first reaction is an oxidation reaction and the second reaction is an electrophilic aromatic substitution reaction.

Alkaline KMnO₄is a strong oxidizing agent and therefore it oxidizes the whole carbon chain directly attached to the benzene nucleus into the –COOH group.

Furthermore, the Br₂ group in the presence of a halogen carrier causes bromination whereas ethyl alcohol in an acidic medium results in esterification. 

HINT : Glycol is an antifreeze agent.

Explanation:

Antifreeze is a liquid that prevents the radiator in cars from freezing or overheating. It’s also known as engine coolant. It’s key ingredient is ethylene glycol, which lowers water’s freezing point and raises its boiling point. 

So, option (1) is the correct answer.