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When sulphonilic acid (p-H(2)NC(6)H(4)SO...

When sulphonilic acid `(p-H_(2)NC_(6)H_(4)SO_(3)H)` is treated with excess of bromine water the product is:

A

tribromo product

B

dibromo product

C

monobromo product

D

tetrabromo product

Text Solution

AI Generated Solution

The correct Answer is:
To determine the product formed when sulphonilic acid `(p-H₂NC₆H₄SO₃H)` is treated with excess bromine water, we can follow these steps: ### Step 1: Identify the Structure of Sulphonilic Acid Sulphonilic acid has the following structure: - An amine group `-NH₂` - A sulfonic acid group `-SO₃H` - The structure can be represented as: ``` NH₂ | C₆H₄ | SO₃H ``` ### Step 2: Understand the Directing Effects of the Functional Groups The `-NH₂` group is an ortho-para directing group, which means it will direct electrophilic substitution reactions to the ortho and para positions relative to itself. ### Step 3: Reaction with Bromine Water When sulphonilic acid is treated with bromine water, bromine acts as an electrophile. The `-NH₂` group will increase the electron density of the aromatic ring, making it more reactive towards bromination. ### Step 4: First Bromination 1. The `-NH₂` group directs bromination to the ortho and para positions. 2. The first bromination will occur at the para position (due to steric hindrance at the ortho position) leading to the formation of para-bromosulphonilic acid. ### Step 5: Second Bromination 1. The newly formed para-bromosulphonilic acid can undergo further bromination. 2. The `-SO₃H` group is also ortho-para directing, allowing for bromination at the ortho position relative to the `-SO₃H` group. ### Step 6: Final Product Formation 1. The second bromination will occur at the ortho position relative to the `-SO₃H` group. 2. This results in a tribrominated product where: - One bromine is at the para position relative to `-NH₂` - Two bromines are at the ortho position relative to `-SO₃H` ### Final Structure The final structure of the product will be: ``` Br | NH₂ | C₆H₂Br₂ | SO₃H ``` This indicates that the product is a tribromo derivative of sulphonilic acid. ### Conclusion When sulphonilic acid is treated with excess bromine water, the product formed is a **tribromo derivative** of sulphonilic acid. ---
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MOTION-AROMATIC COMPOUND-Exercise - 2 (Level-I)
  1. Which of the following will undergo sulphonation at fastest rate ?

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  2. Which of the following is most reactive towards sulphonation ?

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  3. When sulphonilic acid (p-H(2)NC(6)H(4)SO(3)H) is treated with excess o...

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  4. Ring nitration of dimethyl benzene results in the formation of only on...

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  5. If p-methoxy toluene is nitrated, the major product is :

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  6. For the electrophilic substitution reaction involving nitration, which...

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  7. Identify the correct order of reactivity in electro- philic substituti...

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  8. The orbital picture of a singlet carbene (CH(2)) can be drawn as

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  9. The orbital picture of a triplet carbene can be drawn as

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  10. It is believed that chloroform and hydroxide ion react to produce an e...

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  11. CHF(2) Br overset(OH^(-))(rarr) (A) [Intermediate] overset("Trans -2-b...

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  12. Ph-underset(14)overset(O)overset("||")(C)-CHN(2) underset(H(2)O)overse...

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  13. Which of the following will not give carbylamine reaction

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  14. (A) Product (A) will be

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  15. overset(I)overset("|")(PhCHBr) overset(PhC equiv CPh)underset("EtOK")(...

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  16. Complete the following reaction

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  17. Intermediate produced during reimer tiemann’s formlylation will be

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  18. Which of the following compound doesn’t gives Hoffmann’s carbyl amine ...

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  19. CH(3) – NH(2) + CHCl(3) + KOH rarr major product will be :

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  20. What is the product of the following reaction ?

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