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A white crystalline solid X give followi...

A white crystalline solid `X` give following chemical test :
(i) it liberates `CO_(2)` with `NaHCO_(3)`
(ii) it forms a coloured dye on diazotisation and coupling with `beta` - naphthol
(iii) with `Br_(2)` water it forms white precipitate of `2,4,6` - tribromo aniline .
`'X'` can be identified as :

A

B

C

D

Text Solution

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To identify the white crystalline solid `X`, we will analyze the given chemical tests step by step. ### Step 1: Analyzing the first test The first test states that `X` liberates `CO₂` with `NaHCO₃`. This reaction indicates that `X` must contain a carboxylic acid group (`-COOH`) or a sulfonic acid group (`-SO₃H`). Both of these functional groups can react with sodium bicarbonate to release carbon dioxide. **Hint**: Look for compounds that contain either a carboxylic acid or a sulfonic acid group. ### Step 2: Analyzing the second test The second test indicates that `X` forms a colored dye upon diazotization and coupling with `β-naphthol`. This suggests that `X` must contain an amino group (`-NH₂`), as diazotization is a characteristic reaction of aniline derivatives. The reaction involves converting the amino group into a diazonium salt, which can then couple with `β-naphthol` to form a colored compound. **Hint**: Identify compounds that have an amino group (`-NH₂`) attached to a benzene ring. ### Step 3: Analyzing the third test The third test states that `X` reacts with `Br₂` water to form a white precipitate of `2,4,6-tribromoaniline`. This reaction is characteristic of aniline derivatives, where the amino group is highly reactive towards electrophilic substitution. The presence of the amino group allows for bromination at the ortho and para positions, leading to the formation of `2,4,6-tribromoaniline`. **Hint**: Confirm that the compound can undergo bromination at the ortho and para positions due to the presence of the amino group. ### Step 4: Evaluating the options Now, we can evaluate the options based on the information gathered: - **Option A**: Contains a `-COOH` group but lacks an `-NH₂` group. - **Option B**: Contains both `-COOH` and `-NH₂` groups, making it a suitable candidate. - **Option C**: Contains an `-NHCO` group instead of `-NH₂`, which is not suitable. - **Option D**: Similar to A, it lacks the necessary `-NH₂` group. Based on the evaluations, **Option B** is the correct identification of the compound `X`. ### Final Answer `X` can be identified as **Option B**.
To identify the white crystalline solid `X`, we will analyze the given chemical tests step by step. ### Step 1: Analyzing the first test The first test states that `X` liberates `CO₂` with `NaHCO₃`. This reaction indicates that `X` must contain a carboxylic acid group (`-COOH`) or a sulfonic acid group (`-SO₃H`). Both of these functional groups can react with sodium bicarbonate to release carbon dioxide. **Hint**: Look for compounds that contain either a carboxylic acid or a sulfonic acid group. ### Step 2: Analyzing the second test ...
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