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Author Question: Which compound can most easily undergo elimination to give an alkyne? ... (Read 90 times)

809779

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on: Aug 23, 2018
Which compound can most easily undergo elimination to give an alkyne?
 

Question 2

2-Hexyne is allowed to react with Li in NH3(liq). The obtained product is treated with Cl2 in CCl4. Which of the following isomer is main product?
 

Question 3

Which of the following molecules cannot be synthesized from propyne using only one reaction?
 

Question 4

What is the product of the following reaction?
 
 

Question 5

Which reaction sequence on an alkene will accomplish the same end result as treating the alkene with O3 followed by Zn/AcOH?
  1.RCO3H followed by KMnO4/H3O+
  2.BH3/THF followed by OsO4 followed by NaHSO3/H2O
  3.OsO4/NMO followed by HIO4
  4.Hg(OAc)2/H2O/THF followed by NaBH4 followed by KOH
  5.Br2/H2O followed by KOH, followed by H3O+

Question 6

What is the product of reaction of 1-methylcyclohexene with Hg(OAc)2/H2O/THF followed by NaBD4? (D = 2H)
 

Question 7

Which of the following compounds would yield trialkylborane shown below when treated with BH3/THF?
 
  1.2-methylbut-1-ene
  2.2-methylbut-2-ene
  3.3-methylbut-1-ene
  4.2-methylpropene
  5.(E)-2-pentene

Question 8

Which of the following reactions would not yield at least one alcohol functional group?
   
  1.hydroboration of an internal alkene, followed by H2O2/NaOH
  2.reaction of a mono-substituted alkene with NBS/H2O/DMSO
  3.osmium tetroxide oxidation of a tetrasubstituted cyclic alkene, followed by NaHSO3
  4.oxymercuration/demercuration of terminal alkene
  5.oxidative cleavage of a tetrasubstituted alkene using KMnO4/H+

Question 9

What is the best reagent to carry out the following reaction?
   


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xiaomengxian

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Reply #1 on: Aug 23, 2018
Answer to Question 1

4

Answer to Question 2

1

Answer to Question 3

4

Answer to Question 4

4

Answer to Question 5

3

Answer to Question 6

2

Answer to Question 7

1

Answer to Question 8

5

Answer to Question 9

4

Answer to Question 10

3


809779

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Reply #2 on: Aug 23, 2018
Thanks


xiaomengxian

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Reply #3 on: Aug 23, 2018
Welcome :)


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