When ethyl acetoacetate (CH3COCH2CO2CH2CH3) is treated with one equivalent of CH3MgBr, a gas is evolved from the reaction mixtur
e, and after adding aqueous acid, ethyl acetoacetate is recovered in high yield. Identify the gas formed and explain why the starting material was recovered in this reaction. Be sure to answer all parts.
1 answer:
Answer:
Methane gas is formed in this reaction. After adding aqueous acid, ethyl acetoacetate is reformed through protonation of it.
Explanation:
acts as base towards ethyl acetoacetate. Because ethyl acetoacetate contains active methylene group.
Hence abstracts a proton from methylene group and being converted to methane gas.
After addition of aqueous acid, conjugate base of ethyl acetoacetate gets protonated and ethyl acetoacetate is reproduced.
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<h3>Explanation:
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⇒The composition , decomposition or displacement of molecules of matter during chemical change is called chemical reaction.
▪️
Various conditions bring about these changes. The chemical reactions are represented by chemicalequation. The compounds or elements that take part in chemical reaction are called reactant. They are written at the left side of an arrow that represent a change while the compound or elements that formed after the chemical change are called product. They are written at the right side of the arrow.
▪️
When nitrogen reacts with hydrogen to form ammonia :
Nitrogen + Hydrogen ⇒ Ammonia
N₂ + 3H₂ ⇒ 2NH₃
<u>Presentation </u><u>of </u><u>a </u><u>chemical </u><u>reaction </u><u>in </u><u>the </u><u>form </u><u>of </u><u>equation </u><u>is </u><u>called </u><u>chemical </u><u>equation </u>. <u>Chemical equation may be word equations or formula equations.</u>
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Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;
The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g
The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT
Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.