更新1:
A) NaH B) NaOH/H2O C) Br2 in CCl4 D) Cold conc. H2SO4 E) CrO3 in H2SO4 could any one explain the reason please
Which of the reagents listed below would serve as the basis for a simple chemical test to distinguish between (CH3)3COH and (CH3)2CHCH2OH ?
回答 (2)
2016-12-13 11:05 am
The answer: E) CrO₃ in H₂SO₄
For A), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH react with NaH to give (CH₃)₃CONa and (CH₃)₂CHCH₂ONa respectively as well as hydrogen gas.
For B), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH do not react with NaOH/H₂O.
For C), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH do not react with Br₂/CCl₄.
For D), both of of (CH₃)₃COH and (CH₃)₂CHCH₂OH react with cold and concentrated H₂SO₄ to give (CH₃)₃COSO₃H and (CH₃)₂CHCH₂OSO₃H respectively.
For E), (CH₃)₂CHCH₂OH is oxidized by CrO₃ in H₂SO₄ on heating to give (CH₃)₂CHCHO and then (CH₃)₂CHCOOH. CrO₃ in H₂SO₄ is reduced to Cr³⁺ ions and thus the reaction mixture changes from orange to green. On the other hand, (CH₃)₃COH does not react with CrO₃ in H₂SO₄ and thus there is no observable change.
For A), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH react with NaH to give (CH₃)₃CONa and (CH₃)₂CHCH₂ONa respectively as well as hydrogen gas.
For B), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH do not react with NaOH/H₂O.
For C), both of (CH₃)₃COH and (CH₃)₂CHCH₂OH do not react with Br₂/CCl₄.
For D), both of of (CH₃)₃COH and (CH₃)₂CHCH₂OH react with cold and concentrated H₂SO₄ to give (CH₃)₃COSO₃H and (CH₃)₂CHCH₂OSO₃H respectively.
For E), (CH₃)₂CHCH₂OH is oxidized by CrO₃ in H₂SO₄ on heating to give (CH₃)₂CHCHO and then (CH₃)₂CHCOOH. CrO₃ in H₂SO₄ is reduced to Cr³⁺ ions and thus the reaction mixture changes from orange to green. On the other hand, (CH₃)₃COH does not react with CrO₃ in H₂SO₄ and thus there is no observable change.
2016-12-13 10:55 am
E. Can tertiary alcohols be oxidized?
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