`
Self Studies

Amines Test - 8

Result Self Studies

Amines Test - 8
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    The molecular formula of ethyl acetate is

    Solution

    Its molecular formula is C4H8Oand its chemical formula is CH3COOC2H5.

  • Question 2
    1 / -0

    When methylamine reacts with HCl, the product is

    Solution

    Due to the presence of lone pair on nitrogen, methyl amine acts as a Lewis base and reacts with HCl, H+ ion from HCl forms an adduct (salt) methyl ammonium chloride, CH3NH3+Cl-.

  • Question 3
    1 / -0

    Direct nitration of aniline yields significant amount of meta derivative. To obtain more p – nitro derivative, one or more of the below can be done

    Solution

    Direct nitration of aniline yield significant amount of meta derivative, this is because the use of HNO3 during nitration of aniline causes the formation of anilinium ion(C6H5NH3+). Anilinium ion is responsible for the formation of metra nitro aniline. To prevent this, initial reaction of aniline with acetic anhydride acetylates -NH2 group.

    C6H5NH2 + CH3COOCOCH3 →  C6H5NHCOCH3.

    Now, -NHCOCH3 is an activating group, which on nitration followed by hydrolysis form para nitro aniline as a major product.

  • Question 4
    1 / -0

    Aniline does not undergo Friedel – Crafts reaction

    Solution

    AlCl3 being a lewis acid reacts with the lone pair of -NH2 group of aniline forming an adduct (C6H5NH2+AlCl3) which deactivates the benzene system hence no friedal craft reaction occurs.

  • Question 5
    1 / -0

    The base hydrolysis of an ester is called

    Solution

    Base hydrolysis of ester produces the salt of carboxylic acid and alcohols as the product, and soaps are salts of carboxylic acids. Therefore, this reaction is called as saponification reaction.

  • Question 6
    1 / -0

    Gabriel synthesis is used for the preparation of :

    Solution

    In Gabriel Pthalamide reaction, the sodium or potassium salt of pthalimide is N-alkylated with a primary alkyl halide to give the corresponding N-alkylphthalimideis for producing primary amines. This is because of the reaction of sodium or potassium salt of phthalimide with alkyl halide impure SN2 reaction.

  • Question 7
    1 / -0

    Which one of the following cannot be obtained by Gabriel phthalimide synthesis?

    Solution

    In Gabriel phthalimide reaction, a potassium salt of phthalimide is formed. It reacts readily with the primary alkyl halide to form the corresponding alkyl derivative. But aryl halide (C6H5X) does not react with potassium salt of phthalimide. Because C-X bond in haloarene (alkyl halide) is difficult to be cleaved due to a partial double bond character and hence, do not undergo SN2 reaction with potassium salt of phthalimide. So, aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now