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: pH=12.69
Explanation:
Initial 0.12 0 0
Eqm 0.12-x x x
(neglecting small value of x in comparison to 0.12)
Moles of
0.06 moles of NaOH will give 0.06 moles of
Now moles of
will be neutralized by
moles of
and
moles of
will be left.
Molarity of
pH = 14 - pOH= 14 - 1.31 = 12.69
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
Equilibrium constant of the reaction :
K = 4
Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:
Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M