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Alcohols Phenols and Ethers Test - 41

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Alcohols Phenols and Ethers Test - 41
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  • Question 1
    1 / -0
    The correct decreasing order of dehydration of following alcohols with conc. $$H_2SO_4$$ is:

    Solution
    Dehydration order will include D first because after dehydration aromatic compound(stable) is formed . In this way, the order is arranged according to the stability of the compound formed.
  • Question 2
    1 / -0
    Methylphenyl ether can be obtained by reacting :
    Solution
    This reaction helps in the formation of methyl phenyl ether.

  • Question 3
    1 / -0
    Which of the following is best method to prepare phenyl-t-butyl ether?
    Solution
    This reaction is  williamson ether synthesis.

  • Question 4
    1 / -0
    Rate of dehydration when given compound is treated with conc, $${ H }_{ 2 }{ SO }_{ 4 }$$.

  • Question 5
    1 / -0
    Correct representation of 3- methylpent-3-en-2-ol is :
    Solution

  • Question 6
    1 / -0
    The best method to prepare 3-methylbutan-2-ol from 3-methylbut-1-ene is:
    Solution
    To prepare 3-methylbutan-2-ol from 3-methylbut-1-ene shouls involve Markovnikov's addition of $$H-OH$$ across double bond
    A. Addition of water in presence of dil. $$H_2SO_4$$ follows Markovnikov's addition of $$H-OH$$ across the alkene
    B. Addition of $$HCI$$ followed by reaction with dil. $$NaOH$$ to an alkene can give substitution and elimination product after alkyl halide formation
    C. Hydroboration-oxidation reaction follows Anti-Markovnikov's addition of $$H-OH$$
    D.Reimer - Tiemann reaction.is a chemical reaction used for the ortho-formylation of phenols
    Thus the best method to prepare 3-methylbutan-2-ol from 3-methylbut-1-ene is the addition of water in presence of dil. $$H_2SO_4$$.

  • Question 7
    1 / -0
    Product of the above reaction is:

    Solution

  • Question 8
    1 / -0
    A hydrocarbon $$'A' (C_{3}H_{8})$$ on treatment with chlorine in presence of sunlight yielded compound $$'B'$$ as the major product. Reaction of $$'B'$$ with aqueous $$KOH$$ gave $$'C'$$ which on treatment with concentrated $$H_{2}SO_{4}$$ yielded $$'D'$$. Hydrogenation of $$'D'$$ gave back $$'A'$$. The sequence of reactions involved in the above conversion is:
    Solution
    (i) $$C_{3}H_{8} + Cl_{2}\xrightarrow {hv}  C_{3}H_{7}Cl + HCl [Substitution]$$
    (ii) $$C_{3}H_{7}Cl + Aq\ KOH \xrightarrow {hv} C_{3}H_{7}OH + KCl [Substitution]$$
    (iii) $$C_{3}H_{7}OH \xrightarrow {dil\ H_{2}SO_{4}} C_{3}H_{6} + H_{2}O [Dehydration]$$
    (iv) $$C_{3}H_{6} + H_{2} \rightarrow C_{3}H_{8} [Addition]$$.
  • Question 9
    1 / -0
    Ethyl alcohol obtained by fermentation of starch or molasses is called wash, and what is it’s purity?
    Solution

  • Question 10
    1 / -0
    The correct IUPAC name of the compound is ?

    Solution
    -OH group is given preference over Cl group so, 1 number to carbon attach to -OH group
    so,4-chloro-3 ethyl-cyclohexan -1-ol

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