Chemistry
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New answer posted
a year agoContributor-Level 9
C? H? → possible structure of C? H? are; CH? -C (CH? )=CH? and CH? -CH=CH-CH?
(Unsaturation factor = 1)
Ozonolysis of CH? -C (CH? )=CH? gives acetone (B) and formaldehyde.
Oxidation of CH? -C (CH? )=CH? with KMnO? /H? gives acetone (B) and CO?
The question states it gives C? H? O (B). Hence the alkene is 2-Methylpropene (A).
[Reaction scheme showing ozonolysis and permanganate oxidation of 2-methylpropene]
New answer posted
a year ago
Contributor-Level 10
All carbonyl compound (aldehyde and ketone) give orange precipitate with 2,4-dinitrophenyl hydrazine (Brady's reagent)
New answer posted
a year agoContributor-Level 10
CrO? ²? (Cr? ) → Cr? ³, Total oxidation number change = 3
MnO? (Mn? ) → Mn? ², Total oxidation number change = 5
Cr? O? ²? (Cr? ) → 2Cr? ³, Total oxidation number change = 6
C? O? ²? (C? ³) → 2CO? (C? ), Total oxidation number change = 2
New answer posted
a year agoContributor-Level 10
(A) [PtCl? (NH? )? ], Number of GI = 2
(B) [Ni (CO)? ], Number of G.I = 0 as it is tetrahedral.
(C) [Ru (H? O)? Cl? ] Number of G.I = 2 (Facial and Meridional)
(D) [CoCl? (NH? )? ]? Number of G.I = 2 (Cis and trans)
New answer posted
a year agoContributor-Level 9
B.D.E H-H < B.D.E D-D
B.D.E H-H = 435.9 kJ / mole
B.D.E D-D = 443.4 kJ / mole
E H ≈ E D - 7.5
New answer posted
a year agoContributor-Level 9
Gabriel phthalimide is used for the preparation of 1° aliphatic amine not 1° aromatic amines since 1° aromatic amines do not undergo nucleophilic substitution reaction.
New answer posted
a year agoContributor-Level 10
Rotamers or conformers arises due to free rotation along σ - bond.
New answer posted
a year agoContributor-Level 10
Barfoed test:-
In this test, copper acetate in dilute acid is reduced in 30 seconds by mono saccharides while disaccharides takes several minutes.
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