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Organic Chemistry Some Basic Principles and Techniques Test - 14

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Organic Chemistry Some Basic Principles and Techniques Test - 14
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following is correct IUPAC name of compound?
    $$CH_{3}-CH_{2}-CH=CH_{}-CH_{3}$$
    Solution
    $$\overset {5}{C}H_{3}-\overset {4}{C}H_{2}-\overset {3}{C}H=\overset {2}{C}H-\overset {1}{C}H_{3}$$

    The IUPAC name of the given compound is Pent-2-ene.
  • Question 2
    1 / -0
    The name of the following compound is :
    $${ CH }_{ 3 }{ CH }_{ 2 }({ CH }_{ 3 }){ CHNH }_{ 2 }$$
    Solution
    The given compound is :
    $$CH_{3}CH_{2}(CH_{3})CHNH_{2}$$
    Its IUPAC name is :
    1-amino-1-methylpropane
    Hence, option D is correct.
  • Question 3
    1 / -0
    The correct IUPAC name of the alkyl group,
    $$ { CH }_{ 3 }-\overset { \overset { { C }_{ 2 }H_{ 5 } }{ | }  }{ CH } -{ CH }_{ 2 }-{ \overset { \overset { { C }_{ 2 }{ H }_{ 5 } }{ | }  }{ CH-C } H }_{ 3}$$
    Solution
    The correct IUPAC name of the given compound is 2,4-diethylpentane.
    Hence, option B is correct.
  • Question 4
    1 / -0
    IUPAC name of the compound 
    $$BrCH_2-CHCl-CHCl_2$$ is 
    Solution

  • Question 5
    1 / -0
    IUPAC nomenclature of 
    $$H_3C-\overset{\,\,CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH=\overset{\,\,\,\,CH_3}{\overset{|}{C}}-CH_3$$
    Solution
    According to IUPAC,
    a. The longest chain contains total 5 carbon atoms so the root word must be "pent".
    b. Double bond gets higher priority than the alkyl group. So the numbering should be done in such a way that the double bonded carbons get lower number. There are two methyl groups at the C4 and one methyl group at the C2. So the name of thr compound will be 2,4,4-trimethylpent-2-ene. So the correct answer is A
  • Question 6
    1 / -0
    The IUPAC name of $${\text{C}}{{\text{H}}_{\text{3}}} - {\text{CH}}\,{\text{ = }}\,{\text{CH}}\, - {\text{C}} \equiv \,{\text{CH}}$$ is :
    Solution
    According to IUPAC nomenclature,
    a. There are total 5 carbon atoms in the backbone chain. So the root word should be Pent.
    b. Triple will be given more priority than the double bond. But in the name of the compound "ene" will be written before "yne" irrespective of the priority.
    So the name of the compound is Pent-3-en-1-yne. So the correct answer is C 
  • Question 7
    1 / -0
    IUPAC name of $$ CH_3-C \equiv C-HC-(CH_3)_2 $$
    Solution

  • Question 8
    1 / -0
    What is the IUPAC name of the given compound
    $$ { CH }_{ 3 }-\underset { \underset { OH }{ | }  }{ CH } -{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }-\overset { \underset { | }{ Br }  }{ \underset { \overset { | }{ Br }  }{ C }  } -{ CH }_{ 3 } $$
    Solution

  • Question 9
    1 / -0
    The IUPAc name of 

    Solution

  • Question 10
    1 / -0
    The correct name of $$H_3C-CH_2-C\equiv C-CH=CH_2$$ is 
    Solution

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