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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Give one chemical test to distinguish between the following pair of compounds. Aniline and benzylamine

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प्रश्न

Give one chemical test to distinguish between the following pair of compounds.

Aniline and benzylamine

रासायनिक समीकरणे/रचना
फरक स्पष्ट करा
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उत्तर

Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

\[\ce{\underset{Benzylamine}{C6H5CH2 - NH2} + HNO2 ->[NaNO2 + HCl] \underset{(Unstable)}{[C6H5CH2 - \overset{+}{N2}C\overset{-}{l}]} ->[H2O] N2 ^ + \underset{Benzy alcohol}{C6H5CH2 - OH} + HCl}\]

On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved.

\[\ce{C6H5NH2 ->[NaNO2 + HCl][273 - 278 K] C6H5 - \overset{+}{N2}C\overset{-}{l} + NaCl + 2H2O}\]

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पाठ 9: Amines - Exercises [पृष्ठ २७८]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 9 Amines
Exercises | Q 9.2 (iv) | पृष्ठ २७८

संबंधित प्रश्‍न

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(a) C6H12N4

(b) C6H24H4                    

(c) C6H12N4O2

(d) C6H24N4O2


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\[\ce{C6H5NH2 + H2SO4 (conc.)}\]


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\[\ce{C6H5N2Cl + C2H5OH ->}\]


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\[\ce{C6H5NH2 + (CH3CO)2O ->}\]


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\[\ce{C6H5N2Cl ->[(i) HBF4][(ii) NaNO2/Cu, \Delta]}\]


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\[\ce{\underset{(Acetyl chloride)}{CH3COCl}->[H2][Pd-BaSO4]A + HCl}\]


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