Home
Class 12
CHEMISTRY
A solution of Benzylamine, Cyclohexanol ...

A solution of Benzylamine, Cyclohexanol and Picric acid in ethyl acetate was extracted initially with a saturated solution of `NaHCO_(3)` to give fraction A. The left over organic phase was extracted with chloroform with KOH give fraction B. The final organic layer was labelled as fraction C.
Fractions A, B and C contain respectively :

A

Picric acid, Benzylamine, Cyclohexanol

B

Benzylamine, Picric acid, Cyclohexanol

C

Picric acid, Cyclohexanol, Benzylamine

D

Benzylamine, Cyclohexanol, Picric acid

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the extraction process step by step, identifying which compounds are present in each fraction (A, B, and C). ### Step 1: Identify the Compounds The compounds involved in the extraction are: 1. Benzylamine (C6H5CH2NH2) - a weak base. 2. Cyclohexanol (C6H11OH) - a neutral alcohol. 3. Picric acid (2,4,6-trinitrophenol) - a strong acid due to the presence of three nitro groups. ### Step 2: Extraction with NaHCO3 The first extraction is done using a saturated solution of sodium bicarbonate (NaHCO3). Since NaHCO3 is a weak base, it will react with the acidic component in the solution. - **Picric acid** is the only acidic compound among the three. It will react with NaHCO3 to form the soluble sodium picrate and release carbon dioxide (CO2) and water. - **Benzylamine** and **Cyclohexanol** do not react with NaHCO3 because they are not acidic enough. **Result of Step 2:** - **Fraction A** contains sodium picrate (soluble in water). - **Leftover organic phase** contains Benzylamine and Cyclohexanol. ### Step 3: Extraction with Chloroform and KOH The leftover organic phase (Benzylamine and Cyclohexanol) is then extracted with chloroform (CHCl3) in the presence of potassium hydroxide (KOH). - **Benzylamine**, being a weak base, can react with KOH. In this case, KOH will deprotonate benzylamine, allowing it to form a soluble salt. - **Cyclohexanol** is neutral and does not react with KOH. **Result of Step 3:** - **Fraction B** will contain the salt of Benzylamine (phenyl isocyanide). - **Fraction C** will contain Cyclohexanol, which remains unreacted and is not soluble in KOH. ### Final Summary of Fractions - **Fraction A**: Contains **Picric acid** (sodium picrate). - **Fraction B**: Contains **Benzylamine** (as a salt). - **Fraction C**: Contains **Cyclohexanol**. ### Conclusion The fractions contain: - **Fraction A**: Picric acid - **Fraction B**: Benzylamine - **Fraction C**: Cyclohexanol

To solve the problem, we need to analyze the extraction process step by step, identifying which compounds are present in each fraction (A, B, and C). ### Step 1: Identify the Compounds The compounds involved in the extraction are: 1. Benzylamine (C6H5CH2NH2) - a weak base. 2. Cyclohexanol (C6H11OH) - a neutral alcohol. 3. Picric acid (2,4,6-trinitrophenol) - a strong acid due to the presence of three nitro groups. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

A solution of m - chloroaniline, m- chlorophenol and m - chlorobenzoinc acid in ethyl acetate was extracted initially with a saturated solution of Na HCO _ 3 to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C, contain respectively :

A solution of Benzyl chloride, Benzoic acid and Aniline was extracted initially with a HCl solution of to give fraction A. The left over organic phase was extracted with conc. NaOH solution to give fraction B. The final organic layer was labelled as fraction A, B and C contain respectively

The sum of mole fractions of A, B and C in an aqueous solution containing 0.2 moles of each A, B and C is

The sodium extract on acidification with acetric acid and then adding lead acetate solution gives a black precipitate. The organic compound contains.