PhCHO- CH3CO2 O A HBr- BThe product B is-

PhCHO+ (CH_3CO)_2 O (A)Ph CH = CH COOH (A) (B) nbsp;PhBr|CHCH_2COOH Product (A) is getting formed through Perkin reaction. HBr added up to (A) following Ma ...

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Byju's Answer

Standard XII

Chemistry

Hell-Volhard-Zelinski Reaction

PhCHO+ CH3CO2...

Question

$P h C H O + (C H_{3} C O)_{2} O (1) C H_{3} C O O N a - --------- \to (2) Hydrolysis, Δ A H B r ⟶ B$

The product $B$ is:

P h C H = C H C H_{2} B r

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P h C | B r H C H_{2} C O O H

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P h C H_{2} C H (B r) C O O H

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P h C H = C H - C O B r

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Solution

The correct option is B $P h C | B r H C H_{2} C O O H$
$P h C H O + (C H_{3} C O)_{2} O (1) C H_{3} C O O N a - --------- \to (2) Hydrolysis, Δ P h - C H = C H - C O O H (A)$

$(A) H B r - -- \to P h C | B r H C H_{2} C O O H (B)$

Product (A) is getting formed through Perkin reaction. $H B r$ added up to (A) following Markovnikov pathway.

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P h C H O + (C H_{3} C O)_{2} O C H_{3} C O O N a - ------- \to A H B r - -- \to B

The product B is :

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PhCHO+(CH3CO)2O(1) CH3COONa−−−−−−−−−−→(2) Hydrolysis, ΔAHBr⟶B The product B is:

The correct option is B PhC|BrHCH2COOH PhCHO+(CH3CO)2O(1) CH3COONa−−−−−−−−−−→(2) Hydrolysis, ΔPh−CH=CH−COOH(A) (A)HBr−−−→PhC|BrHCH2COOH(B) Product (A) is getting formed through Perkin reaction. HBr added up to (A) following Markovnikov pathway.

$P h C H O + (C H_{3} C O)_{2} O (1) C H_{3} C O O N a - --------- \to (2) Hydrolysis, Δ A H B r ⟶ B$

The product $B$ is: