Ph-C-N-Partial HydrolysisH3O-A-Br2-KOHBProduct B is-

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Reduction of Amides,Oximes and Azides with LAH

Ph-C≡N→Partia...

Question

$Ph - C \equiv N H_{3} O^{+} - --------- \to Partial Hydrolysis (A) {Br}_{2} + KOH - ---- \to (B)$
Product (B) is:

P h - C H_{2} - N H_{2}

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P h - O H

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P h - N H_{2}

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P h - C H_{3}

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Solution

The correct option is C $P h - N H_{2}$
$A i s P h O | | C N H_{2}$
$B i s P h N H_{2}$

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Similar questions

Ph - C \equiv N H_{3} O^{+} - --------- \to Partial Hydrolysis (A) {Br}_{2} + KOH - ---- \to (B)

Product (B) is:

H_{3} C - C \equiv C - C H_{3} H_{2} - ------ \to P d - B a S O_{4} B r_{2} - -- \to C C l_{4} M

H_{3} C - C \equiv C - C H_{3} N a - ---- \to l i q . N H_{3} B r_{2} - -- \to C C l_{4} N

Relationship between

M

and

N

is:

$I$ f ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} + {(\frac{Z}{c})}^{n} = 1$ , then $\frac{\partial z}{\partial x} =$
(here $z$ is a function of $x$ and $y$ )

Q. (A)

P h - O H

(B)

- C H_{2} - O H

(C)

P h - C O_{2} H

(D)

- C H_{2} - \oplus N H_{3}

Decreasing order of acid strength is:

Q. The major product in the following reaction is

P h - C l + B r_{2} Anhydrous F e B r_{3} - ---------- \to

Product

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Ph−C≡NH3O+−−−−−−−−−−→Partial Hydrolysis(A)Br2+KOH−−−−−→(B) Product (B) is:

The correct option is C Ph−NH2A is PhO||CNH2 B is PhNH2

$Ph - C \equiv N H_{3} O^{+} - --------- \to Partial Hydrolysis (A) {Br}_{2} + KOH - ---- \to (B)$
Product (B) is:

The correct option is C $P h - N H_{2}$
$A i s P h O | | C N H_{2}$
$B i s P h N H_{2}$