Because I (iodide) is better leaving group than Cl, so it will leave when this molecule is reacted with strong base (sodium tert-butyl oxide) giving the elimination product provided in picture<span />
Of the following, which can serve as the solvent in a solution?
1 answer:
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Answer:
Dissolve 226 g of KCl in enough water to make 1.5 L of solution
Explanation:
1. Calculate the moles of KCl needed
2. Calculate the mass of KCl
3. Prepare the solution
- Measure out 224 g of KCl.
- Dissolve the KCl in a few hundred millilitres of distilled water.
- Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.
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.
