The major product of the following reactions {CH}₃{CH}₂{CH}₂{CH}₂{COCH}₃ [{array}{c} {iii) Red P, } {Br}₂ \\ {iv) } {H}₂{O} {array}]{{array}{c} {(i)NaOBr} \\ {ii) } {H}₃{O}^+ {array}}

Correct Answer is : CH3CH2CH2CHBrCO2H{CH}₃{CH}₂{CH}₂{CHBrCO}₂{H}CH3CH2CH2CHBrCO2H

{CH}₃{CH}₂{CH}₂{CH}₂{CHBrCO}₂{H}

{CH}₃{CH}₂{CH}₂{CHBrCO}₂{H}

{CH}₃{CH}₂{CH}₂{CHBrCOBr}^{-1\,{CH}₃

Correct Answer is : CH3CH2CH2CHBrCO2H{CH}₃{CH}₂{CH}₂{CHBrCO}₂{H}CH3CH2CH2CHBrCO2H

Test Index

TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Chemistry

Question : 159 of 160

Marks: +1, -0

\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{COCH}_3 \xrightarrow[\begin{array}{c} \text{iii) Red P, } \mathrm{Br}_2 \\ \text{iv) } \mathrm{H}_2\mathrm{O} \end{array}]{\begin{array}{c} \text{(i)NaOBr} \\ \text{ii) } \mathrm{H}_3\mathrm{O}^+ \end{array}}

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHBrCO}_2\mathrm{H}$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHBrCO}_2\mathrm{H}$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHBrCO}_2\mathrm{H}$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHBrCOBr}^{-1\,\mathrm{CH}_3}$

Solution:

(i), (ii) 2-hexanone undergoes haloform reaction to form bromoform and a carboxylate ion.

This carboxylate ion undergoes hydrolysis to give pentanoic acid.

\mathrm{CH}_3\mathrm{CH}_2{\underset{\text{2-hexanone}}{\underset{\mathrm{CH}_2\mathrm{H}_2\mathrm{C}}{\underset{\mathrm{O}}{\underset{\parallel}{\mathrm{C}}}}}}-\mathrm{CH}_3 \xrightarrow{(i)\,\mathrm{NaOBr}}

(iii), (iv) Pentanoic acid, react with bromine in presence of red-phosphorus to give

\alpha

-bromocarboxylic acid. This reaction is Hell-Volhard-Zelinsky reaction.

\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \xrightarrow[\text{(iv) }\mathrm{H}_2\mathrm{O}]{\text{(iii) Red }\mathrm{P.Br}_2} \underset{\mathrm{CH}_3\mathrm{CH}_2\mathrm{H}_2\mathrm{C}}{\underset{\mathrm{Br}}{\underset{\vert}{\mathrm{H}-\mathrm{COOH}}}}

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