Hinsberg reagent is used to distinguish between :

- C H O, > C = O

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- C H_{2} O H, > C H O H, \to C - O H

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- O -, - O H

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- N H_{2}, - N H -, \to N

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Solution

The correct option is D $- N H_{2}, - N H -, \to N$
Hinsberg reagent (benzene sulphonyl chloride) is used to distinguish between primary, secondary and tertiary amine.
Primary amine, such as methylamine ( a primary amine) will dissolve in an alkali solution after it has undergone reaction with Hinsberg reagent.

$C_{6} H_{5} S O_{2} C l + C H_{3} N H_{2} \to C_{6} H_{5} S O_{2} N H C H_{3} K O H - -- \to C_{6} H_{5} S O_{2} N K C H_{3} (w a t e r s o l u b l e)$

Secondary amine forms alkyl sulphonamide which does not form salt with alkali and hence remains insoluble (solid) in alkali solution.

$C_{6} H_{5} S O_{2} C l + R_{2} N H \to C_{6} H_{5} S O_{2} N R_{2} water insoluble, ether soluble K O H - -- \to No action$

Tertiary amine do not react with Hinsberg reagent.

$C_{6} H_{5} S O_{2} C l + R_{3} N \to$ No reaction.

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Similar questions

P h - O H | C H - C H_{3} P C C \to (A) N H_{2} - N H - O | | C - N H_{2} ⟶ (B)

Product (B) is?

Q. In the given reaction
$C_{6} H_{5} - O | | C - O | | C - C_{6} H_{5} \frac{(i) C H_{3} M g B r (e x c e s s)}{(i i i) H O H / H^{@}} P r o d u c t$

Product will be:

$(a) C_{6} H_{5} C - O H | C | C H_{3} - O H | C | C H_{3} - C_{6} H_{5}$

$(b) C_{6} H_{5} C - O H | C | C H_{3} - O | C - C_{6} H_{5}$
$(c) C_{6} H_{5} - C O O H$
$(d) C_{6} H_{5} - C H O H - C H O H - C_{6} H_{5}$

Which of the following is not correct?

$O r d e r o f b a s i c i t y O r d e r o f n u c l e o p h i l i c i t y (a) F^{-} > C l^{-} > B r^{-} B r_{-} > C l^{-} > F^{-} (b) H - O - O^{-} > O H^{-} H - O - O^{-} > O H^{-} (c) C H_{3} > N H_{2}^{-} > O H - C H_{3} > N H_{2}^{-} > O H^{-} (d) H_{2} O > H_{2} S H_{2} O < H_{2} S$

$G l u c o s e A g_{2} O \to X$

$G l u c o s e C o n c . H N O_{3} \to Y$

X. Y and X respectively are:

Q. Which of the following compound can show Fehling's test?
(I)

P h - N H - O H

(II)

P h - C H O

(III)

H - O | | C - O H

(IV)

O H | C H_{2} - O | | C - C H_{3}

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