When 20g of a compound (A) `(M.F. = C_(4)H_(10)O_(4))` reacts with excess of `CH_(3)MgBr`, 14.6 L of `CH_(4)` is obtained at STP. What is structure formula of (A) ?
Number of active hydrogen present in organic compound will be four based on mole concept calculation so the answer will be (a) and (b)
Topper's Solved these Questions
GRIGNARD REAGENTS
MS CHOUHAN|Exercise Level 1 (Q.61 To Q.66)|1 Videos
GENERAL ORGANIC CHEMISTRY
MS CHOUHAN|Exercise LEVEL- 1|1 Videos
ISOMERISM (STRUCTURAL & STEREOISOMERISM)
MS CHOUHAN|Exercise Subjective Problems|17 Videos
Similar Questions
Explore conceptually related problems
0.40 g of an organic compound (A), (M.F.-C_(5)H_(8)O) reacts with x mole of CH_(3)MgBr to liberate 224 mL of a gas at STP. With excess of H_(2) , (A) gives pentan-1-ol. The correct structure of (A) is :
0.092 g of a compound with the molecular formula C_(3)H_(8)O_(3) on reaction with an excess of CH_(3)MgI gives 67.00 mL of methane at STP. The number of active hydrogen atoms present in a molecule of the compound is :
How many isomer of C_(4)H_(10)O reacts with CH_(3)MgBr to evolve CH_(4) gas ? (Excluding stereoisomer)
When 2.86 g of a mixture of 1- butene, C_(4)H_(8) and butane C_(4)H_(10) was burned in excess of oxygen 8.80 g of CO_(2) and 4:14 g of H_(2)O were obtained. What is percentage by mass of butane in the mixture
A sample of 450 mg of unknown alcohol is added to CH_(3)MgBr when 168 of CH_(4) at STP is obtained the unknown alcohol is
MS CHOUHAN-GRIGNARD REAGENTS-Level 1 (Q.61 To Q.66)