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Aldehydes Ketones and Carboxylic Acids Test - 72

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Aldehydes Ketones and Carboxylic Acids Test - 72
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  • Question 1
    1 / -0
    Best starting material to synthesize 2-methylpentan-2-ol is
    Solution
    The easiest method to prepare 2-methylpentan-2-ol can be prepared by hydrolysis of Ethyl magnesium bromide with Butan-4-one.

    Hence, Option "D" is the correct answer.
  • Question 2
    1 / -0
    Nucleophilic addition reaction will be most favoured in:
    Solution

  • Question 3
    1 / -0
    The synthesis of crotonaldehyde from acetaldehyde is an example of........... reaction.
    Solution
    Crotanaldehyde is produced by the aldol condensation of acetaldehyde
    crotanaldehyde Aldehydes are carbonyl molecules that have a carbon attached to oxygen with a double bond, another group, and a hydrogen atom. ... Aldehydes and ketones undergo nucleophilic addition reactions due to the polarity in the carbonyl bond that makes them vulnerable to a nucleophile, an atom that donates electrons.
  • Question 4
    1 / -0
    In the reaction,

        $${ C }_{ 2 }{ H }_{ 5 }OH\xrightarrow [  ]{ P{ I }_{ 3 } } (A)\xrightarrow [  ]{ KCN } (B)\xrightarrow [  ]{ Hydrolysis } (C)$$

    The product (C) is:
    Solution

  • Question 5
    1 / -0
    Which of the following statements is not correct? 
    Solution

  • Question 6
    1 / -0
    Which of the following compounds will give carboxylic acid with nitrous acid $$ (HNO_2)? $$
    Solution
    $$RCONH_2 + HNO_2 \rightarrow RCOOH$$
    Hence option (B) is correct.
  • Question 7
    1 / -0
    Oil of vanilla bean is
    Solution
    Vanillin is an organic compound with the molecular formula $$C_8H_8O_3$$. It is a phenolic aldehyde. Its functional groups include aldehyde, hydroxyl, and ether. It is the primary component of the extract of the vanilla bean.

    Option A is correct

  • Question 8
    1 / -0

    Directions For Questions

    A carbonyl compound $$P$$, which gives positive iodoform test, undergoes reaction with $$MeMgBr$$ followed by dehydration to give an olefin $$Q$$. Ozonolysis of $$Q$$ leads to a dicarbonyl compound $$R$$, which undergoes intramolecular aldol reaction to give predominantly $$S$$.
    $$P \underset {\underset {3. H_{2}SO_{4}, \triangle}{2.H^{+}, H_{2}O}}{\xrightarrow {1. MeMgBr}} Q \underset {2. Zn, H_{2}O}{\xrightarrow {1. O_{3}}} R \underset {2.\triangle}{\xrightarrow {1. OH^{-}}} S$$.

    ...view full instructions

    The structures of the products $$Q$$ and $$R$$, respectively, are
    Solution

  • Question 9
    1 / -0

    Directions For Questions

    A carbonyl compound $$P$$, which gives positive iodoform test, undergoes reaction with $$MeMgBr$$ followed by dehydration to give an olefin $$Q$$. Ozonolysis of $$Q$$ leads to a dicarbonyl compound $$R$$, which undergoes intramolecular aldol reaction to give predominantly $$S$$.
    $$P \underset {\underset {3. H_{2}SO_{4}, \triangle}{2.H^{+}, H_{2}O}}{\xrightarrow {1. MeMgBr}} Q \underset {2. Zn, H_{2}O}{\xrightarrow {1. O_{3}}} R \underset {2.\triangle}{\xrightarrow {1. OH^{-}}} S$$.

    ...view full instructions

    The structure of the carbonyl compound $$P$$ is
    Solution

  • Question 10
    1 / -0

    Compound ' P ' of the following reaction sequence can be:

    Solution

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