Question

Which base would not effectively deprotonate acetylene? (ch3)2nli ch3och2mgbr lioch3 ch3li kh?

Answer 1

Acetylene has a chemical formula which can be written as:

C2H2

 

We can see that there are two positive ions, H+. Now what deprotonation means is that the H+ is removed from acetylene to form acetylene ion and water. In this case, I believe that the answer would be:

LiOCH3

Answer 2

LiOCH₃ would not effectively deprotonate acetylene

Further explanation

The equilibrium reaction can be determined if the pKa or Ka values ​​of the acid and conjugate acids (acids in the product) are known.

So it can be concluded that the reacting acid can protonate the base or vice versa base compounds can deprotonating the acid in the reaction, so that the reaction can proceed to the right to form a product or not

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid

Acids and bases according to Bronsted-Lowry

• Acid = donor (donor) proton (H + ion)

• Base = proton (receiver) acceptor (H + ion)

If the acid gives (H⁺), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H⁺), then the base is formed can release protons and is called the conjugate acid from the original base.

The value of the equilibrium constant (K)

Can be formulated:

K acid-base reaction = Ka acid on the left / K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K

$\large{\boxed{\bold{K\:=\:10^{-pK}}}}$

K value> 1) indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, the pKa value of the conjugated acid from the base of the compounds above

• 1. (CH₃)₂NLi, pKa = 38

• 2. CH₃OCH₂MgBr, pKa = 50

• 3. LiOCH₃, pKa = 16

• 4. CH₃Li, pKa = 50

• 5. KH, pKa = 50

Whereas the pKa of Acetylene (C₂H₂) itself is = 25

From the conjugated acid pKa value of some of the bases above shows only LiOCH₃ bases that cannot deprotonate acetylene because the pKa value is smaller than the pKa acetylene

The reactions that occur are:

LiOCH₃ + HC ≡ CH ---> HOCH₃ + LiC = CH

The value of the equilibrium constant K is

pK = pKa acetylene - pKa HOCH₃

pK = 25-16

pK = 9

$K\:=\:10^{-pK}$

$K\:=\:10^{-9}$

K values ​​<1 indicate a reaction cannot occur

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Keywords : pKa,  acetylene, deprotonate, the conjugate acid