In the given sequence of reactions,identify P | vedclass

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(D) The complete reaction sequence is as follows:$1$. $CH_2=CH_2 + Br_2 \xrightarrow{P=Br_2} CH_2(Br)-CH_2(Br)$ (Electrophilic addition)$2$. $CH_2(Br)

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$Br_2$,Alc. $KOH, NaNH_2$,Red hot iron tube

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(D) The complete reaction sequence is as follows:$1$. $CH_2=CH_2 + Br_2 \xrightarrow{P=Br_2} CH_2(Br)-CH_2(Br)$ (Electrophilic addition)$2$. $CH_2(Br)-CH_2(Br) \xrightarrow{Q=Alc. KOH} CH_2=CH-Br$ (Dehydrohalogenation)$3$. $CH_2=CH-Br \xrightarrow{R=NaNH_2} CH \equiv CH$ (Dehydrohalogenation)$4$. $3 CH \equiv CH \xrightarrow{S=Red \ hot \ iron \ tube} C_6H_6$ (Cyclic polymerization)Thus,the reagents $P, Q, R$ and $S$ are $Br_2$,alc. $KOH, NaNH_2$,and Red hot iron tube respectively.

In the given sequence of reactions,identify $P, Q, R$ and $S$ respectively.
$CH_2=CH_2 + Br_2$ $\xrightarrow{P} CH_2(Br)-CH_2(Br)$ $\xrightarrow{Q} CH_2=CH-Br$ $\xrightarrow{R} CH \equiv CH$ $\xrightarrow{S} C_6H_6$

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The compound with molecular formula $C_6H_6$,which gives only one monobromo derivative and takes up four moles of hydrogen per mole for complete hydrogenation,has $................\pi$ electrons.

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In the given sequence of reactions,identify $P, Q, R$ and $S$ respectively.$CH_2=CH_2 + Br_2$ $\xrightarrow{P} CH_2(Br)-CH_2(Br)$ $\xrightarrow{Q} CH_2=CH-Br$ $\xrightarrow{R} CH \equiv CH$ $\xrightarrow{S} C_6H_6$