
Color of tertiary alcohol in victor Meyers test isA. BlueB. GreenC. RedD. No color
Answer
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Hint: Alcohol or $–OH$ functional groups are acidic in nature and can be prepared by various methods like hydration of alkenes, in the presence of acid and by the action of Grignard reagent. Dehydration of alcohols gives alkenes.
Complete step by step answer:
Note: Tertiary alcohols are resistant to oxidation therefore it does not oxidize like primary and secondary alcohol to give aldehydes and carboxylic acid. Though we get no colour for tertiary alcohol in victor Meyers test but it gives white turbidity instantly under Lucas test.
Complete step by step answer:
Alcohol can be primary, secondary and tertiary depending on the degree of the carbon it is attached to, therefore we can detect the presence of –OH functional group by lucas test and victor Meyers test. Both of these tests help in detection of primary, secondary and tertiary alcohol.
Let us understand the details of Victor Meyers test and find out the results that different degrees of –OH group gives with this test. The compound with the presence of –OH group is treated with red phosphorous in presence of iodide first to give iodoalkane and then it is treated by silver nitrate followed by treating the obtained nitroalkane under nitrous acid made by the mixture of HCl and $ NaN{O_2} $ to give nitrolic acid which is made alkaline by adding sodium hydroxide to it and the ultimate product obtained is a sodium salt of nitrolic acid if the compound is a primary alcohol and is red in colour.
If the compound is a secondary alcohol it forms a pseudo nitrol which gives blue colouration on treatment with alkali.
Whereas a tertiary alcohol does not react after forming nitroalkane by the action of silver nitrite and gives no coloration.
$ {1^\circ}{R}CH_2OH\xrightarrow{{P/{I_2}}}{R}CH_2I\xrightarrow{{AgN{O_2}}}{R}CH_2N{O_2}\xrightarrow{{HN{O_2}}}{R}C(NOH)N{O_2} $
${2^\circ}{R_2}CHOH\xrightarrow{{P/{I_2}}}{R_2}CHI\xrightarrow{{AgN{O_2}}}{R_2}CHN{O_2}\xrightarrow{{HN{O_2}}}{R_2}C(NO)N{O_2} $
${3^\circ}{R_3}COH\xrightarrow{{P/{I_2}}}{R_3}CI\xrightarrow{{AgN{O_2}}}{R_3}CN{O_2}\xrightarrow{{HN{O_2}}} $
Hence from the above explanation we can conclude that tertiary alcohol gives no colour under victor Meyers test. Hence option (D) is correct.
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