Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that
each bin will hold only one item, and not all of the given reagents or structures will be used. (Me = methyl, CH3; Et = ethyl, CH3CH2.)
1 answer:
Answer:
The three-step synthesis of trans-2-pentene from acetylene is as follows.
<u>Step -1:</u> Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.
<u>Step -2:</u> Formation terminal alkyne to nonterminal alkynes.
<u>Step -3:</u> Formation of trans-pent - 2-pent-ene by reduction.
Explanation:
Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.
The chemical reaction of each step of chemical reactions is as follows.
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The pH of the solution is 2.54.
Explanation:
pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.
The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.
So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the
[HB] is the concentration of base.
Then
So the pH of the solution is 2.54.