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Question: Identify the \( Z \) in the sequence of reaction \( C{H_3}C{H_2}CH = C{H_2}\underrightarrow {HBr/{H_...

Identify the ZZ in the sequence of reaction CH3CH2CH=CH2HBr/H2O2→YC{H_3}C{H_2}CH = C{H_2}\underrightarrow {HBr/{H_2}{O_2}}Y C2H5ONa→Z\underrightarrow {{C_2}{H_5}ONa}Z
(A) (CH3)2CH2−O−CH2CH3{\left( {C{H_3}} \right)_2}C{H_2} - O - C{H_2}C{H_3}
(B) CH3(CH2)4−O−CH3C{H_3}{\left( {C{H_2}} \right)_4} - O - C{H_3}
(C) CH3CH2−CH(CH3)−O−CH2CH3C{H_3}C{H_2} - CH\left( {C{H_3}} \right) - O - C{H_2}C{H_3}
(D) CH3−(CH2)3−O−CH2CH3C{H_3} - {\left( {C{H_2}} \right)_3} - O - C{H_2}C{H_3}

Explanation

Solution

Hint : Alkenes have the tendency to undergo an addition reaction via Markonikov’s rule or anti-Markovnikov's rule depending upon the solvent system used to carry out the reaction. Alkenes undergoes anti-Markovnikov's reaction with alkyl halide in presence of hydrogen peroxide.

Complete Step By Step Answer:
According to anti-Markovnikov's rule of addition when an alkene react with alkyl halide in presence of hydrogen peroxide solvent then the hydrogen of alkyl halide attach with the carbon with less number of hydrogen and halide attach to carbon having large number of hydrogen atoms.
When the given compound undergoes anti Markonikov’s reaction its terminal double bond breaks and bromine atoms attach to terminal carbon to form n−n - butyl bromide.
CH3CH2CH=CH2HBr/H2O2→C{H_3}C{H_2}CH = C{H_2}\underrightarrow {HBr/{H_2}{O_2}} CH3CH2CH2CH2BrC{H_3}C{H_2}C{H_2}C{H_2}Br
n−n - butyl bromide further treated with sodium ethoxide to form a new compound. When primary alkyl halide is treated with a base like sodium ethoxide it undergoes substitution reaction via the SN2S{N_2} mechanism to form ether and salt of sodium. This reaction is known as Williamson ether synthesis.
Hence, when n−n - butyl bromide is treated with sodium ethoxide it will undergo substitution reaction by removing the bromide ion to form sodium bromide and the rest of the molecule attach to n−n - butyl to form the ether.
CH3CH2CH2CH2BrC{H_3}C{H_2}C{H_2}C{H_2}Br C2H5ONa→\underrightarrow {{C_2}{H_5}ONa} CH3CH2−CH(CH3)−O−CH2CH3C{H_3}C{H_2} - CH\left( {C{H_3}} \right) - O - C{H_2}C{H_3} +NaBr+ NaBr
Hence, the final product of the reaction is an ether CH3CH2−CH(CH3)−O−CH2CH3C{H_3}C{H_2} - CH\left( {C{H_3}} \right) - O - C{H_2}C{H_3} whose IUPAC name is ethyl butyl ether.
Therefore, option (C)\left( C \right) is the correct option.

Note :
If hydrogen peroxide is not used in this reaction it will follow Markonikov’s rule to form 2−2 - bromobutane.
Laboratory conditions for Williamson synthesis include temperature in the range of 50−10050 - 100 ∘c\circ c . Alkoxide are very reactive species hence they are meant to be prepared immediately before the reaction starts.