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Alcohols Phenols and Ethers Test - 21

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Alcohols Phenols and Ethers Test - 21
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Weekly Quiz Competition
  • Question 1
    1 / -0
    (I) $$1,2$$-Dihydroxybenzene (II) $$1,3$$-Dihydroxybenzene (III) $$1,4$$-Dihydroxybenzene (IV) Hydroxybenzene
    The increasing order of boiling points of above-mentioned alcohols is:
    Solution
    With an increase in the molecular weight, the boiling point increases. With an increase in $$OH$$ groups, the extent of hydrogen bonding increases, and boiling point increases.

    Due to above two reasons, (IV) Hydroxybenzene has a lowest boiling point.

    In case of (I) 1,2-dihydroxybenzene, since two $$OH$$ groups are in the ortho position, there is intramolecular hydrogen bonding. 

    In (II) 1,3-dihydroxybenzene and (III) 1,4-dihydroxybenzene, intermolecular hydrogen bonding is present which leads to the molecular association. Hence (I) 1,2-dihydroxybenzene has a lower boiling point than (II) 1,3-dihydroxybenzene and (III) 1,4-dihydroxybenzene.

    The increasing order of boiling points of above-mentioned alcohols is  (IV) Hydroxybenzene < (I) 1,2-dihydroxybenzene < (II) 1,3-dihydroxybenzene  < (III) 1,4-dihydroxybenzene.
  • Question 2
    1 / -0
    2- phenyl propene on acidic hydration gives :
    Solution
    Hence, option $$A$$ is correct.

  • Question 3
    1 / -0
    How many structures of F are possible?

    Solution

    $$3$$ isomers of compound F are possible. They are shown above. Protonation in acidic conditions of the hydroxyl group of secondary alcohols converts the hydroxyl group into a leaving group, which departs as water. 

    The carbocation intermediate can rearrange and it will then be deprotonated in a $$E_1$$ elimination mechanism. This is the reverse of acid-catalyzed hydration of alkenes.

    Hence, the correct option is $$D$$

  • Question 4
    1 / -0
    Compound $$A, C_{8}H_{10}O$$, is found to react with $$NaOI$$ (produced by reacting $$Y$$ with $$NaOH$$) and yields a yellow precipitate with characteristic smell.
    $$A$$ and $$Y$$ are respectively are:
    Solution
    A and Y are 1-phenylethanol and iodine respectively.
    NaOI oxidizes 1-phenylethanol to methyl ketone (acetophenone) which gives positive iodoform test. The yellow precipitate is of iodoform $$(CHI_3)$$
  • Question 5
    1 / -0
    The compound $$A$$ on treatment with $$Na$$ gives $$B$$, and with $$PCl_{5}$$ gives $$C. B$$ and $$C$$ react together to give diethyl ether. $$A, B$$ and $$C$$ are in the order.
    Solution

    The compound A is ethanol $$\displaystyle C_{2}H_{5}OH$$.  The compound A on treatment with $$Na$$ gives sodium ethoxide $$\displaystyle  (C_{2}H_{5}ONa)$$ which is compound B.

    $$\displaystyle C_{2}H_{5}OH \xrightarrow [-H_2]{Na} C_{2}H_{5}ONa$$

    The compound A on treatment with $$PCl_5$$ gives ethyl chloride $$\displaystyle ( C_{2}H_{5}-Cl)$$ which is compound C.

    $$\displaystyle C_{2}H_{5}OH  \xrightarrow {PCl_5} C_{2}H_{5}-Cl+HCl+POCl_3$$

    B and C react together to give diethyl ether.

    $$\displaystyle C_{2}H_{5}ONa+C_{2}H_{5}Cl \rightarrow C_{2}H_{5}OC_{2}H_{5}+NaCl$$

    A,B and C are in the order $$\displaystyle C_{2}H_{5}OH, C_{2}H_{5}ONa, C_{2}H_{5}Cl$$.

    Option C is correct.
  • Question 6
    1 / -0
    Which of the following is not the product of dehydration of :

    Solution
    Product D cannot be obtained by the dehydration of the given alcohol as it will involve one tert carbocation into second tert carbocation.  There is no significant difference in the stability of these carbocations. These two carbocations are shown below.

  • Question 7
    1 / -0
    The reaction can be classified as:

    Solution
    The reaction can be classified as Williamson ether synthesis reaction.
    In this reaction, sodium alkoxide reacts with alkyl halide to form an ether.
    In the given reaction, cyclopentanol reacts with sodium hydride to form sodium salt of cyclopentanolate. It then reacts with methyl iodide to form methoxycyclopentane.
  • Question 8
    1 / -0
    Among the following sets of reaction which one produces anisole?
    Solution
    Among the following sets of reaction, the reaction B produces anisole:

    (A) Acetaldehyde reacts with grignards reagent followed by acidic hydrolysis to form secondary alcohol.

    (B) Phenol reacts with NaOH to form sodium phenoxide. It reacts with methyl iodide in $$SN^2$$ reaction to form
    anisole (methoxy benzene).
    $$C_6H_5- OH + NaOH \rightarrow C_6H_5 - ONa\underset {S_N}{\xrightarrow {CH_3I}}\underset
    {(Anisole)}{C_6H_5-OCH_3}$$ (Williamson's synthesis)

    (C) Phenol reacts with neutral ferric chloride solution to form violet colored complex.
     $$3C_6H_5OH + neutral FeCl_3\rightarrow \underset {\text {Violet colour complex} }
    {(C_6H_5O)_3Fe}+3HCl.$$

    (D) Toluene reacts with acetyl chloride in presence of anhydrous $$AlCl_3$$ to form p-methyl acetophenone. This is friedel craft's reaction.
    $$C_6H_5-CH_3+CH_3COCl+AlCl_3\rightarrow \underset { \text {Para- methyl aceto phenone} } {
    p-{CH_3-C_6H_4- COCH_3}}$$

  • Question 9
    1 / -0
    How many structures of X is possible ?

    Solution
     
    3 structures are possible.

    Option D is correct.

  • Question 10
    1 / -0
    The strongest acid among the following is:
    Solution
    m-Methoxy phenol is the strongest acid. 

    Methoxy group at meta position exerts an only-I effect.

     Due to this, it withdraws electrons from all the nuclear position of the benzene ring including the one holding the $$O-H$$ group consequently, electron density in the $$O-H$$ bond decreases. 

    Hence, m-methoxy phenol becomes more acidic than o and p-methoxy phenols.

    Option B is correct.

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