Q gives positive iodoform test and its 1H NMR spectrum shows singlets at 1.0 ppm (3H) and at 1.3 ppm (3H)
R gives positive iodoform test and its1H NMR spectrum shows singlets at 1.0 ppm (3H) and at 2.2 ppm (3H)
A bright yellow precipitate is formed when Q and R treated separately with 2,4 dinitrophenyl hydrazine
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The Correct Option isC, D
Solution and Explanation
To determine which statements about Q and R are correct, let's analyze each reaction and the tests involved.
Fehling’s Test: This test is used to identify reducing sugars and aldehydes, but Q and R are not clearly aldehydes, so neither gives a positive Fehling’s test.
Iodoform Test for R:
The iodoform test identifies compounds with the structure CH3CO- or CH3CHOH- groups. R can undergo this test showing it contains such groups.
R gives a positive iodoform test, confirming the presence of a methyl ketone or secondary alcohol with a methyl group.
The 1H NMR spectrum of R shows singlets at 1.0 ppm (3H) and 2.2 ppm (3H), aligning with typical chemical shifts for a methyl group and a methyl ketone.
2,4-Dinitrophenylhydrazine (DNPH) Test for Q and R:
DNPH reacts with carbonyl groups to form a yellow precipitate, which indicates the presence of an aldehyde or ketone.
Both Q and R likely contain carbonyl groups, given that they form bright yellow precipitates when treated with DNPH.
Based on this analysis, the correct statements are:
R gives a positive iodoform test and its1H NMR spectrum shows singlets at 1.0 ppm (3H) and at 2.2 ppm (3H).
A bright yellow precipitate is formed when Q and R are treated separately with 2,4-dinitrophenylhydrazine.