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The major product formed in the reaction given below will be:

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Fate of aliphatic diazonium ion - The diazonium ions of aliphatic amines are very unstable and produces carbocation immediately, which can produce different products.  - wherein      Fate of aromatic diazonium ion - Diazonium salts of aromatic amines are comparatively more stable and evolve nitrogen only on heating.These diazonium salts can be isolated at low temperatures. -             As...

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives:

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

    Coupling reaction of aromatic diazonium - Diazonium ions of aromatic amines also undergo coupling reaction with aromatic rings having a strong activating group to form diazo compounds. - wherein      Option 1)Option 2)Option 3)Option 4)

The major product in the following reaction is:

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

      Basicity of Aromatic Amines - The unshared pair of electrons at the nitrogen atom is in resonance with the benzene ring and hence not fully available for donation as in the case of aniline. -       Nature of group and basicity - Electron withdrawing groups ( -I and -M  effect ) decreases basicity, whereas Electron releasing groups ( +I and +M  effect ) increases basicity. -...

The major product 'Y' in the following reaction is :

  • Option 1)

  • Option 2)

     

  • Option 3)

  • Option 4)

Option 1)Option 2) Option 3)Option 4)

The major product of the following reaction is : 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 
( Nucleophilicity of  is greater than  ,  thus  will attack on ).      The final product is : .  option (2) is correct.   Option 1) Option 2) Option 3) Option 4)

Ethylamine (C_{2}H_{5}NH_{2}) can be obtained from N-ethylphthalimide 

on treatment with:

  • Option 1)

    NH_{2}NH_{2}

  • Option 2)

    CaH_{2}

  • Option 3)

    NaBH_{4}

  • Option 4)

    H_{2}O

 Reagent used is . So, option (1) is correct.Option 1)Option 2)Option 3)Option 4)

Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene ?

  • Option 1)

    H_{2}/Ni                   

  • Option 2)

     (i)\; \; LiAlH_{4}\; \; (ii)\; H_{3}O^{+}

  • Option 3)

     (i)\; SnCl_{2}+HCl\left ( gas \right )\; \; \; (ii)\; NaBH_{4}

  • Option 4)

    (i)\; HCl/H_{2}O\; \; (ii)\; NaBH_{4}

Option 1)                   Option 2) Option 3) Option 4)

Hinsberg's reagent is :

  • Option 1)

    C_{6}H_{5}COCl

  • Option 2)

    SOCl_{2}

  • Option 3)

    C_{6}H_{5}SO_{2}Cl

  • Option 4)

    \left ( COCl \right )_{2}

  The Hinsberg Test - This test can be used to demonstrate whether an amine is primary, secondary or tertiary.Primary amines react with benzene sulfonyl chloride to form N - substituted benzene sulfonamides.These, in turn, undergo acid-base reactions with the excess potassium hydroxide to form water soluble potasssium salt. Acidification of this solution will cause the water soluble in the next...

The peptide that gives positive ceric ammonium nitrate and carbylamine tests is :

  • Option 1)

    Ser-Lys

  • Option 2)

    Gln-Asp

  • Option 3)

    Lys-Asp

  • Option 4)

    Asp-Gln

  Test for Alcohol - Alcohol + Ceric Ammonium Nitrate   Red Solution  - wherein Alcohol is Present     carbylamnie Reaction - Product is isocyanide & this reaction is used for the detection of primary amines. - wherein                                                                                                positive ceric ammonium nitrate        gives  positive carbylamine tests         ...

Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0^{\circ}C. This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is :

  • Option 1)

      

  • Option 2)

      

  • Option 3)

      

  • Option 4)

       

                                                                                                                                                    Option 1)  Option 2)  Option 3)  Option 4)   

Coupling of benzene diazonium chloride with 1- naphthol in alkaline medium will give :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Option 1)Option 2)Option 3)Option 4)

Which of the following amines can be prepared by Gabriel phthalimide reaction?

  • Option 1)

    neo-pentylamine
     

  • Option 2)

    triethylamine

  • Option 3)

    n-butylamine

     

  • Option 4)

    t-butylamine

Gabriel phthalimide reaction is used for preparation of amine. So ans is :- n-butylamine amine  Option 1)neo-pentylamine  Option 2)triethylamineOption 3)n-butylamine  Option 4)t-butylamine

In the following compounds, the decreasing order of basic strength will be :

 

  • Option 1)

    C_{2}H_{5}NH_{2}>NH_{3}>(C_{2}H_{5})_{2}NH
     

  • Option 2)

    NH_{3}>C_{2}H_{5}NH_{2}>(C_{2}H_{5})_{2}NH

  • Option 3)

    (C_{2}H_{5})_{2}NH>C_{2}H_{5}NH_{2}>NH_{3}

     

  • Option 4)

    (C_{2}H_{5})_{2}NH>NH_{3}>C_{2}H_{5}NH_{2}

Option 1)  Option 2)Option 3)  Option 4)

 

  1. 1^{0} amine with same number of carbons

  2. 1^{0} amine with one carbon less

  3. 2^{0} amine with one carbon less

  4. 1^{0} amine with one carbon more

option 2

 

 it can be seen that the nitro-group withdraws electrons from the ringand hence it deactivates the benzene ring for further electrophilic substitution.
the nitration of benezene using mixed conc. H2SO4 and HNO3 If a large amount of KHSO4 is added to the mixture, the rate of nitration will be
Nitrobenzene is formed in the nitration process of benzene using concentrated and . The reaction is given by; If a large amount of  is added to the mixture, more  ion furnishes and rate of electrophilic aromatic reaction slow down i.e concentration of electrophile is decreased. Hence addition of a large amount of   to the mixture tends to slow down the rate of nitration.
organic.PNG
@sushma  Answer 4 The reason for the first option- All alkali metal( Na, K, etc.)  dissolves in liquid ammonia giving deep blue solutions which are conducting in nature. These solutions contain ammoniated cations and ammoniated electrons as shown below: M + ( x + y ) NH3 → M+(NH3 )x + e-(NH3)y The blue colour of the solution is considered to be due to ammoniated electrons which absorb energy...
Schotten-Baumann reaction. The Schotten-Baumann reaction is an organic reaction used to convert an acyl halide or anhydride to an amide if reacted with an amine and base, or an ester if reacted with an alcohol and base. Mechanism: The reaction with the alcohol to produce ester goes in a similar way.

 Among the following, the essential amino acid is :

  • Option 1)

    Alanine

  • Option 2)

     Valine

  • Option 3)

     Aspartic acid

  • Option 4)

     Serine

     

 
Option 1) Alanine Option 2)  Valine Option 3)  Aspartic acid Option 4)  Serine  
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