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Aldehydes And Ketones Test - 16
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Aldehydes And Ketones Test - 16
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  • Question 1/10
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    The major product formed when cyclohexanecarbaldehyde reacts with Tollens’ reagent is-

    Solutions

    Hint: Test to differentiate between aldehyde and ketone via selective oxidation of aldehyde functional group into Carboxylic acid

    Explanation:

    Step 1: Do the oxidation on the carbonyl carbon of aldehyde to form carboxylate anion

    Step 2: Neutralisation of carboxylate anion occur with H2O to give oxidizing product cyclohexane carboxylic acid

     

  • Question 2/10
    1 / -0

    Iodoform test is not given by-

    Solutions

    Hint: Characteristic test for Methyl ketone group

    Explanation:

    The structures of the compounds are as follows:

    The first second and third compound contains  group hence they can show the iodoform test. But structure fourth does not contain it hence, it will not give an iodoform test.

     

  • Question 3/10
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    An esters that gets hydrolyzed most easily under alkaline conditions is -

    Solutions

    Hint: Basic hydrolysis of an ester using aqueous NaOH and KOH

    Explanation:

    Step 1: Electron withdrawing group attach to the benzene ring increases the reactivity of an ester towards nucleophilic substitution reaction. Since -NO2 group is a strong electron-withdrawing group. Hence, in basic medium ester-containing -NO2 group will be hydrolyzed more easily.

    Acidic hydrolysis of an ester gives a carboxylic acid and alcohol. Basic hydrolysis of an ester gives a carboxylate salt and alcohol. 

    Step 2: The mechanism is as follows:

     

  • Question 4/10
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    The product formed when acetone is treated with excess ethanol in the presence of hydrochloric acid is -

    Solutions

    Hint: The formation of hemiketal is a nucleophilic addition reaction.

    When carbonyl compounds like ketone are treated with alcohol, they form hemiketal (hemiketal and acetal/ketal.). Then hemiketal reacts with an excess of ethanol and forms ketal. The reaction is as follows:

     

  • Question 5/10
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    Chemical test used to distinguish between Benzoic acid and Ethyl benzoate is -

    Solutions

    Hint: Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test.

    Step 1: Sodium bicarbonate test: 

    Acids react with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas. Benzoic acid being an acid responds to this test, 

    Step 2:

    But ethylbenzoate does not.

     

  • Question 6/10
    1 / -0

    The electrophile involved in the below reaction is -

    Solutions

    Hint: Carbene

    The given reaction is Reimer Tiemann Reaction which involves formation of chlorocarbene which act as an electrophile. The electrophile formation reaction is as follows:

     

  • Question 7/10
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    Given the reaction:

    The correct statement among the following is -

    Solutions

    Hint: The reaction is Fries rearrangement

    Product "a" is more volatile than product "b" because in "a" intramolecular hydrogen bonding occurs and in "b" intermolecular hydrogen bonding occur. Intramolecular hydrogen bonding makes the compound more volatile than intermolecular hydrogen bonding.

    The boiling point is less.

    Vapour pressure is more and compound is more volatile

    The boiling point is more.

    Vapour pressure is less and compound is less volatile

     

  • Question 8/10
    1 / -0

     A, X, Y, and Z in the below mentioned reaction are -

    Solutions

    As given, a positive silver mirror test is attained with A therefore, A is an aldehyde. Further, on reaction with dilute alkali (OH-) i.e., aldol condensation, it is indicated that A is an aldehyde. This is because the aldol condensation of ketones is a reversible process and must be carried out under special conditions.

    Then A reacts with semicarbazide and a semicarbazone (Z) is formed. All the reactions are as follows:

     

  • Question 9/10
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    There is only one –NH2 group involved in semicarbazone formation out of two –NH2 group. It is due to-

    Solutions

    Hint: Resonance of nitrogen lone pair make it less reactive

    Explanation:

    Step 1:

    Semicarbazone formation on the carbonyl carbon of ketone and aldehyde as derivative. This semicarbazone formation is a nucleophilic addition type reaction mechanism based on the electrophilic character of carbonyl carbon of aldehyde and ketone.

    Semicarbazide undergoes resonance involving only one of the two −NH2 groups, which is attached directly to the carbonyl carbon atom.

    Step 2: 

    Therefore, the electron density on −NH2 group involved in the resonance also decreases. As a result, it cannot act

    as a nucleophile. Since the other −NH2 group is not involved in resonance; it can act as a nucleophile and can

    attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

     

     

  • Question 10/10
    1 / -0

    A strong base can abstract an α- hydrogen from-

    Solutions

    Hint: Ketone and aldehyde alpha hydroge is acidic.

    It is always easy to abstract the alpha H of a carbonyl compound like aldehyde or ketone. The conjugate base after removing alpha hydrogen is stabilized via resonance in case of aldehyde and ketone. Hence, aldehyde and ketone alpha hydrogen is acidic. 

    The reaction is as follows:

     

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