Question

Grignard reagent RMgBr (P) reacts with water and forms a gas (Q). One gram of Q occupies 1.4 dm$^3$ at STP. (P) on reaction with dry ice in dry ether followed by H$_3$O$^+$ forms compound (Z). 0.1 mole of (Z) will weigh _________ g (Nearest integer).

Hint:

Grignard reagents with CO$_2$ always form carboxylic acids.

Answer 1

Correct Answer: 6

The question involves a Grignard reagent RMgBr, which reacts with water to produce a gas. We'll determine the nature of gas (Q), the compound (Z), and calculate the weight of 0.1 mole of (Z).
First, identify the gas (Q). Grignard reagents react with water to produce corresponding hydrocarbons. Hence, RMgBr + H₂O → RH + Mg(OH)Br. Thus, Q is the hydrocarbon RH.
We know from the problem that 1 gram of Q occupies 1.4 dm³ at STP. Molar volume at STP is 22.4 dm³/mol. Thus, the molar mass (M) of Q is: (1 g)/(1.4 dm³) × 22.4 dm³/mol = 16 g/mol. The only hydrocarbon with this molar mass is CH₄. Therefore, R is CH₃.
Now consider the reaction of Grignard reagent CH₃MgBr with CO₂ (dry ice) and H₃O⁺:
CH₃MgBr + CO₂ → CH₃COOMgBr followed by hydrolysis gives CH₃COOH.
Thus, compound (Z) is acetic acid (CH₃COOH) with a molar mass of 60 g/mol.
Finally, calculate the weight of 0.1 mole of (Z):
(0.1 mole) × (60 g/mol) = 6 g.
This value is within the expected range (6,6).