A sample of 2,3-dibromo-3-methylpentane is he | vedclass

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(A) Step $1$: Dehalogenation of $2,3-$dibromo$-3-$methylpentane with zinc dust $(Zn)$ leads to the formation of an alkene via elimination of $ZnBr_2$.

Vedclass · Accepted Answer

$CH_3-CH_2-CH(CH_3)-CH_2-CH_3$

Vedclass · Answer

(A) Step $1$: Dehalogenation of $2,3-$dibromo$-3-$methylpentane with zinc dust $(Zn)$ leads to the formation of an alkene via elimination of $ZnBr_2$.$CH_3-CH(Br)-C(Br)(CH_3)-CH_2-CH_3 + Zn 	o CH_3-CH=C(CH_3)-CH_2-CH_3 + ZnBr_2$.Step $2$: The resulting alkene ($3-$methylpent$-2-$ene) is heated with $HI$ in the presence of red phosphorus $(Red \ P)$. Red phosphorus and $HI$ act as a strong reducing agent,which reduces the alkene to the corresponding alkane.$CH_3-CH=C(CH_3)-CH_2-CH_3 + 2HI \xrightarrow{Red \ P} CH_3-CH_2-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentane).Thus,the final product is $3-$methylpentane.

$A$ sample of $2,3-$dibromo$-3-$methylpentane is heated with zinc dust. The resulting product is isolated and then heated with $HI$ in the presence of red phosphorus. Indicate the structure that represents the final organic product formed in the reaction.

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