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Alcohols Test - 7
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Alcohols Test - 7
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  • Question 1/10
    1 / -0.25

    Which of the following tests can be used to distinguish between two isomeric ketones: 3- pentanone and 2- pentanone?

    Solutions

    2- pentanone is a methyl ketone. It undergoes haloform reaction with iodine in presence of sodium hydroxide, to form yellow precipitate of iodoform. 3- pentanone is not a methyl ketone. It does not undergo haloform reaction. Hence, 3- pentanone and 2- pentanone can be distinguished by reaction with iodine in presence of NaOH.

  • Question 2/10
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    An organic compound X (C5 H12 O) gives effervescence with NaH. X on treatment with acidic CrO3 solution turns solution to blue-green forming C5 H10 O2 . Also X can be resolved into enantiomers. The correct statement regarding X is

    Solutions

    X is a primary, chiral alcohol.

  • Question 3/10
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    An organic compound C4 H10 O (X) on reaction with I2 /red-P give s C4 H9 I which on further reaction with AgNO2  gives C4 H9 NO2 (Y). Y on treatment with HNO2 forms a blue solution which turns to red on making solution slightly alkaline. The possible identity of X is

    Solutions

    X must be a primary alcohol as indicated by Victor Meyer 's test. Both (a) and (b) are primary satisfy the condition.

  • Question 4/10
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    An organic compound X (C7 H16 O) on treatm ent with concentrated HCI solution forms immediate turbidity even in the absence of ZnCl2 . Also, X is resolvable into enantiomers. The correct statement concerning X is

    Solutions

  • Question 5/10
    1 / -0.25

    Which reagent given below does not produce any visible change when added to ethylene glycol?

    Solutions

    H2 CO forms cyclic acetal with ethylene glycol but the change cannot be observed visually, in all other cases, change can be observed visually-effervescence with NaNH2 and NaH while colour change with CrO3 -H2 SO4 .

  • Question 6/10
    1 / -0.25

    Which of the following is least likely to form turbidity with HCI in the presence of ZnCI2 at room temperature?

    Solutions

    Although it is secondary alcohol, it forms a very unstable carbocation (Cl is electron withdrawing group), hence it does not give turbidity with Lucas reagent in cold condition.

  • Question 7/10
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    The correct statement regarding 3-ethyl-3-hexanol is

    Solutions

    All alcohols react (1 °, 2 °and 3 °) with cerric nitrate and a colour change from yellow to red is observed.

  • Question 8/10
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    An alcohol has molecular formula C6 H12 O  X and it gives immediate turbidity with cold, concentrated HCI even in the absence of ZnCI2 . X can also be obtained by treatment of an ether with excess of CH3 MgBr followed by acid hydrolysis. Hence, the correct statement regarding X is

    Solutions

  • Question 9/10
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    One or More than One Options Correct Type

    Direction : This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

    Q. 

    Which reagent(s) can be used to differentiate between 2-pentanol and  1-pentanol?

     

    Solutions

     

     

    2-pentanol forms yellow precipitate of CHI3 with I2 /NaOH but 1-pentanol does not. 2-pentanol give immediate turbidity with Lucas reagent but 1 -pentanoi does not gives turbidity with Lucas reagent at room temperature. Both alcohols (1 °and 2 °) change colour of chromic acid solution from orange to blue green and both give effervescence of NH3 (g) with NaNH2 , hence these reagents cannot be used for distinction between 1 °and 2 °alcohols.

     

     

  • Question 10/10
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    Alcohols given below that behaves like 1 °-aliphatic alcohol in Lucas test is/are

     

    Solutions

     

     

    Both have electron withdrawing groups, destabilises carbocation, do not form turbidity with Lucas reagent at room temperature like primary alcohols. Option (b) and option (d) have electron donating groups, stabilises benzylic carbocation, forms immediate turbidity with Lucas reagent like 2 °and 3 °alcohols.

     

     

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