Practical Organic Chemistry

Phosphorus is detected in the form of canary yellow ppt on reaction with ammonium molybdate.

(1) The organic compound does not react with natural FeCl₃ solution. It means phenolic group is absent in the organic compound.

(2) The organic compound does not give Fehling solution test. It means aldehyde group (- CHO) is absent in the organic compound.

Lysine is an amino acid which gives nindydrin test.

Furfural has aldehyde group that is why it reacts with 1-napthol to give purple colouration.

Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give pink colouration.

Styrene reacts with KMnO₄ and gives benzoic acid and brown MnO₂.

$ \to $ 2, 4 $-$ DNP test is given by aldehyde on ketone

$ \to $ Iodoform test is given by compound having


group.

(1)  To react with FeCl₃, with benzene ring there should be a $-$OH group attached.

(2)  For carbylamine test there should be $-$NH₂ group in a compound.

(3)  NaHCO₃ reacts when there is $-$COOH group in a compound.

(4)  For Bayer's test there should be double bond or tripple bond between two carbon.

Ligand concentration is maximum when the reaction has not been initiated. At that point, absorption of light by ligands will be maximum. As the reaction starts, ligand concentration decreases so is the absorption of light till the point where all the metal ions are consumed and further complex formation does not take place.

The solution that can be prepared directly by dissolving an accurately weighed amount of the substance in water and making up the volume to a known amount by water is known as primary standard solution. Oxalic acid is an example of primary standard while NaOH is a secondary standard solution. Hence, strength of NaOH solution can be determined by titrating against oxalic acid. This process is known as standardisation.

In Kjeldahl’s method, ammonium ion or ammonia should generate from a compound.

So, those compounds which contain nitrogen in nitro and azo groups and nitrogen in the ring cannot generate ammonium ion or ammonia.

Among the given option only aniline can generate NH₃. So it is suitable far nitrogen estimation by Kjeldahl’s method.

Sodium fusion extract is heated with dilute HNO₃ to decompose NaCN or Na₂S (if present) to HCN and H₂S gases, respectively.

The presence of N or S interferes with the test of halogens by forming precipitate with AgNO₃.

Phenol gives a violet colour with neutral ferric chloride solution whereas benzoic acid does not give this test.

Moles of $HCl$ taken $ = 20 \times 0.1 \times {10^{ - 3}} = 2 \times {10^{ - 3}}$

Moles of $HCl$ neutralised by $NaOH$ solution

$ = 15 \times 0.1 \times {10^{ - 3}} = 1.5 \times {10^{ - 3}}$

Moles of $HCl$ neutralised by ammonia

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \times {10^{ - 3}} - 1.5 \times {10^{ - 3}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 3}}$

$\% $ of nitrogen $ = {{1.4 \times N \times V} \over {w.t.\,\,of\,\,Subs\tan ce}} \times 100$

$ = {{1.4 \times 0.5 \times {{10}^{ - 3}}} \over {29.5 \times {{10}^{ - 3}}}} \times 100 = 23.7\% $

If nitrogen is present in organic compound, then sodium extract contains NaCN.

Na + C + N $\buildrel {Fuse} \over \longrightarrow $ NaCN

FeSO₄ + 6NaCN $ \to $ Na₄[Fe(CN)₆] + Na₂SO₄

Na₄[Fe(CN)₆] changes to prussian blue Fe₄[Fe(CN)₆]₃ on reaction with FeCl₃.

FeCl₃ + Na₄[Fe(CN)₆] $ \to $ Fe₄[Fe(CN)₆]₃ + 12NaCl