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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Answer : The partial pressure of and
is, 216.5 mmHg and 649.5 mmHg
Explanation :
According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.
Formula used :
So,
where,
= partial pressure of gas
= mole fraction of gas
= total pressure of gas
= moles of gas
= total moles of gas
The balanced decomposition of ammonia reaction will be:
Now we have to determine the partial pressure of and
Given:
and,
Given:
Thus, the partial pressure of and
is, 216.5 mmHg and 649.5 mmHg