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Haloalkanes and Haloarenes Test - 66

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Haloalkanes and Haloarenes Test - 66
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  • Question 1
    1 / -0
    The total number of monohalogenated products formed by halogenation of $$2, 4, 4$$-trimethylhexane are:
    Solution

    Total number of monohalogenated products of $$2, 4, 4$$-trimethylhexane are six.

    $$(i)$$ $$2$$ monohalogenated products by replacing two $$-C{ H }_{ 3 }$$ groups at $${ C }_{ 2 }$$ and $${ C }_{ 4 }$$ position.

    $$(ii)$$ $$4$$ monohalogenated products by replacing $$H$$ by $$Cl$$ at $${ C }_{ 2 },{ C }_{ 3 },{ C }_{ 5 },{ C }_{ 6 }$$ positions. Thus, total $$6$$ isomers are possible.

  • Question 2
    1 / -0
    The reaction of 1-bromo-3-chlorocyclobutane with metallic sodium in dioxane under reflux condition gives:
    Solution

  • Question 3
    1 / -0
    In the following reaction, identify the product.

    Solution

    $$\begin{array}{l}\text { Ethyl Magnesium Bromide undergoes S}_N {2} \text { type of reaction with the epoxy bond in the prescence } \\\text { of ether medium. Hence option (C) is correct. }\end{array}$$

  • Question 4
    1 / -0
    Bromo ethane and isoproply chloride with metallic sodium in ether forms :
  • Question 5
    1 / -0
    Which of the following applies in the reaction?

    $${ CH }_{ 3 }CH(Br){ CH }_{ 2 }{ CH }_{ 3 }\xrightarrow [  ]{ Alc.KOH }$$ ?

    (I) $${ CH }_{ 3 }CH=CH-{ CH }_{ 3 }$$ (major product)
    (II) $${ CH }_{ 2 }=CH{ CH }_{ 2 }{ CH }_{ 3 }$$ (minor product)
    Solution

    This reaction is governed by Saytzeff's rule. According to this rule, the elimination of $$\beta$$-hydrogen atom takes place from the carbon having the less number of H-atoms or in other words, a stable alkene is formed (More substituted alkene is more stable)

  • Question 6
    1 / -0
    Which one of the following will show optical isomerism?
    Solution

    Hint:

    The full form of IUPAC is the International Union of Pure and Applied Chemistry.

    Explanation:

    The core carbon atom of $$2-hydroxypropanoic-acid$$ is connected to four distinct groups. The molecule is known as a chiral molecule, and the carbon atom is known as a chiral carbon atom. As a result, optical isomerism is observed in $$2-hydroxypropanoic acid.$$

    Final answer:

    The correct answer is option (B).

  • Question 7
    1 / -0

    Which of the given molecules are optically active?

    Solution

  • Question 8
    1 / -0
    Anisole can be prepared by the action of methyl iodide on sodium phenate. The reaction is called:
    Solution
    Anisole can be prepared by the action of methyl iodide on sodium phenate. The reaction is called Williamson's reaction.
    $$ \displaystyle \underset {  \displaystyle  \text { phenol } }{ C_6H_5-OH} + NaOH \rightarrow \underset {  \displaystyle  \text { sodium phenoxide } }{ C_6H_5-ONa} \xrightarrow {\underset {  \displaystyle  \text { methyl iodide  } }{ CH_3-I}}  \underset {  \displaystyle  \text { anisole } }{ C_6H_5-O-CH_3}$$

    Williamson's reaction can be used for the preparation of simple as well as mixed ethers.
  • Question 9
    1 / -0
    Which one of the following shows $${ S }_{ N }1$$ reaction most readily?
    Solution
    $${ S }_{ N }1$$ reaction is most favorable for tertiary halide. (Since the tertiary carbocation is more stable)

  • Question 10
    1 / -0
    In electrophilic aromatic substitution reactions of chlorobenzene, the ortho/para-directing ability of chlorine is due to its _______.
    Solution
    In electrophilic aromatic substitution reactions of chlorobenzene, the $$ortho /para$$ ability of chlorine is due to its positive resonance effect ($$+R$$).
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