Answer
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Hint: Lucas test is used to differentiate among the primary, secondary and tertiary alcohols with the help of anhydrous zinc chloride on treating it with conc. Hydrochloric acid. In this reaction we see that the chloride present in the zinc chloride bond is being replaced with the hydroxyl group of the alcohol.
Complete step by step answer:
The answer within the ethanol tubing should go cloudy and so a yellow precipitate of tri-iodo-methane (iodoform) should be seen. This incorporates a distinct ‘antiseptic’ smell. The methanol tubing should remain clear.
The iodoform reaction is given by compounds with a methyl next to a group. Secondary alcohols with a $CH_3$ on the carbon carrying the OH (eg propan-2-ol) which will be oxidised to carbonyl compounds of this sort, also provides a positive iodoform test. (NB carboxylic acids do not)
So, the correct answer is Option B.
Additional information:
Procedure:
Practice (over a sink) producing single drops of ethanol from the pipette.
Add 10 drops of methanol to 1 tube.
Add 10 drops of ethanol to the opposite tubing.
Add 25 drops of iodine solution to every alcohol.
Add 10 drops of hydrated oxide solution to every alcohol.
Gently swirl the test tubes some times. The dark colour of the iodine should start to fade.
After two minutes carefully observe the 2 test tubes.
Note: The only primary alcohol which has the tendency to give the iodoform test is the ethanol. The methyl ketones have the tendency to give the haloform test as the methyl ketones and secondary alcohols have the ability to get oxidised to methyl ketones such as the isopropanol. The ethanol and acetaldehyde are the alcohol and aldehyde to undergo such reactions.
Complete step by step answer:
The answer within the ethanol tubing should go cloudy and so a yellow precipitate of tri-iodo-methane (iodoform) should be seen. This incorporates a distinct ‘antiseptic’ smell. The methanol tubing should remain clear.
The iodoform reaction is given by compounds with a methyl next to a group. Secondary alcohols with a $CH_3$ on the carbon carrying the OH (eg propan-2-ol) which will be oxidised to carbonyl compounds of this sort, also provides a positive iodoform test. (NB carboxylic acids do not)
So, the correct answer is Option B.
Additional information:
Procedure:
Practice (over a sink) producing single drops of ethanol from the pipette.
Add 10 drops of methanol to 1 tube.
Add 10 drops of ethanol to the opposite tubing.
Add 25 drops of iodine solution to every alcohol.
Add 10 drops of hydrated oxide solution to every alcohol.
Gently swirl the test tubes some times. The dark colour of the iodine should start to fade.
After two minutes carefully observe the 2 test tubes.
Note: The only primary alcohol which has the tendency to give the iodoform test is the ethanol. The methyl ketones have the tendency to give the haloform test as the methyl ketones and secondary alcohols have the ability to get oxidised to methyl ketones such as the isopropanol. The ethanol and acetaldehyde are the alcohol and aldehyde to undergo such reactions.
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