C — all others endanger you.
It would be d.
Reason being said...
the electron configuration normally goes like this...
1s2 2s2 2p6 3s2 3p6 4s2....
until you hit the transition metals..remember those have a special rule..
even though you are in the 4 sublevels for the orbitals ... it goes down 1
Making it 3d..(1,2,3,4,5,6,7,8,9,10)
Going on...
at 5s2 then, 4d1, 4d2, 4d3, 4d4, etc..
at 6s2 then, 5d1, 6d2, 6d3, 6d4, etc..
Thus, D orbital is your answer.
Answer:Methane gas is evolved from the reaction mixture.
Explanation:
When ethyl acetoacetate is treated with grignard reagent a carbanion is generated.
There are two acidic hydrogens which are present on the carbon which is in between the ester and the ketone group in ethylacetoacetate.
These two protons are also called active methylene protons and they are very acidic in nature due to the presence of two electron withdrawing substituents that is an ester and ketone.
CH₃MgBr is grignard reagent and it is an organo-metallic copmpound . Carbon here in CH₃MgBr exists as carbanion CH3⁻ which is basic enough to abstract the acidic protons present on ethylacetoacetate.
As CH3⁻ abstracts an acidic proton from ethylacetoacetate it become CH₄ which is methane. As methane is a gas so it is methane gas which is evolved from the reaction mixture.
As the acidic proton is abstracted from ethylacetoacetate which leads to generation of carbanion and this carbanion is very stable as it can be delocalized on to the two carbonyl groups . As we add aqueous acid to the reaction mixture the carbanion can again be protonated and its protonation would lead to the generation of ethylacetoacetate again.
Hey There!:
number of moles = 2 moles KI
Volume = 1 L
Therefore:
M = n /V
M = 2 / 1
M = 2.0 mol/L
hope this helps!
