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Identity Z in the sequence of reactions :

\text{CH}_{3}\text{CH}_{2}\text{CH}=\text{CH}_{2} \xrightarrow{{\text{HBr}/\text{H}_2 \text{O}_2 }}\text{Y}\xrightarrow{{\text{C}_{2}\text{H}_5 \text{ONa}}}\text{Z}

  • Option 1)

    CH3 - (CH2)3 - O - CH2CH3

  • Option 2)

    (CH3)2CH2 - O - CH2CH3

  • Option 3)

    CH3(CH2)4 - O - CH3

  • Option 4)

    CH3CH2 - CH (CH3) - O - CH2CH3

 

Answers (1)

best_answer

As we learnt in

Antimarkovnikov Rules / Peroxide effect / Kharash effect -

In presence of peroxide addition of HBr to unsymetrical alkene takes place opposite to Markovnikov's Rule.

- wherein

 

 

CH_{3}CH_{2}CH=CH_{2}\xrightarrow[H_2O_{2}]{HBr}(Y)\xrightarrow{}{C_{2}H_{5}\bar{O}Ca^{+}}(Z)

Y = CH3—CH2—CH2—CH2—Br

Z = CH3—CH2—CH2—CH2—O—CH2—CH3


Option 1)

CH3 - (CH2)3 - O - CH2CH3

this is the correct option

Option 2)

(CH3)2CH2 - O - CH2CH3

this is the incorrect option

Option 3)

CH3(CH2)4 - O - CH3

this is the incorrect option

Option 4)

CH3CH2 - CH (CH3) - O - CH2CH3

this is the incorrect option

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Plabita

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