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Organic Chemistry Some Basic Principles and Techniques Test - 53

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Organic Chemistry Some Basic Principles and Techniques Test - 53
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Weekly Quiz Competition
  • Question 1
    1 / -0
    $$CO + NaOH \rightarrow ?$$
    Solution
    $$CO+NaOH \xrightarrow[]{\Delta }HCOONa$$
    Option A is correct.
  • Question 2
    1 / -0
    How many $$\pi$$ bonds are present in naphthalene molecule?
    Solution
    $$5\pi$$ bonds are present in naphthalene.

  • Question 3
    1 / -0
    Which of the following is azo-group
    Solution
    ago group is $$RN= RN$$
    so correct match the group is option B is 
  • Question 4
    1 / -0
    Ethylene possess
    Solution
    Ethylene possess $$5\sigma$$ and $$1\pi$$ bond.   

  • Question 5
    1 / -0
    Cresol has
    Solution
    Cresols are organic compounds in which methyl and hydroxyl groups are directly attached to the benzene ring. 
    Hence, option $$B$$ is correct.

  • Question 6
    1 / -0
    Which of the following do not contain an acyl group?
    Solution
    Acid chloride is $$RCOCl$$, Amide is $$RCONH_2$$, Ester is $$RCOOR$$, and ether is $$ROR$$.
    Acyl group has the formula $$RCO-$$. Hence, ether do not contain acyl group.
    Option $$D$$ is correct.
  • Question 7
    1 / -0
    Number of $$\sigma$$ and $$\pi$$ bonds present in 1-buten-3-yne respectively are 
    Solution

  • Question 8
    1 / -0
    A carbon-carbon triple bond in ethyne $$(-C\equiv C-)$$ consists of 
    Solution
    A carbon-carbon triple bonds consist of one $$\sigma$$ bond and $$2\pi-$$ bonds.
  • Question 9
    1 / -0
    The number of $$\pi$$ bonds in 3-hexyne-1-ene is 
    Solution
    $$\overset{1}{CH_2}\overset{1\ \pi}=\overset{2}{CH}-\overset{3}{C}\overset{2\ \pi}{\equiv} \overset{4}{C}-\overset{5}{CH_2}-\overset{6}{CH_3}$$
                     3-hexyne-1-ene

    The number of $$\pi$$ bonds in 3-hexyne-1-ene is 3.
  • Question 10
    1 / -0
    Alchols may be represented by the general formula:
    Solution
    One of the H atoms of alkanes (having the molecular formula $$ C_{n}H_{2n+2} $$) is replaced by an $$-OH$$ group to form alcohols.
    Thus $$ C_{n}H_{2n+2}-(1H)+(1-OH)\rightarrow C_{n} H_{2n+1}OH $$

    Option C is the correct answer.
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