To help, I drew a diagram. This represents an ionic bond between Na and Cl. Na is giving his single electron to Cl, which is indicated by the arrow, to make Cl full with 8 electrons.
For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e
., the carbon atoms depicted with stereobonds). Show the product stereochemistry clearly. If there is more than one organic product, both products may be drawn in the same box. Ignore elimination involving C3 and any carbon atom other than C4.
1 answer:
Answer:
2-methyl-butene
Explanation:
For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)
Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.
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Answer:
1) Net ionic equation :
2) 0.765 M is the molarity of the carbonic acid solution.
Explanation:
1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:
..[1]
In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:
..[2]
In aqueous potassium carbonate , potassium ions and carbonate ion is present.:
..[3]
From one:[1] ,[2] and [3]:
Cancelling common ions on both sides to get net ionic equation :
2)
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
0.765 M is the molarity of the carbonic acid solution.