2-bromopentane is heated with potassium ethox | vedclass

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(C) $CH_3-CH(Br)-CH_2-CH_2-CH_3 + C_2H_5OK \xrightarrow{C_2H_5OH} CH_3-CH=CH-CH_2-CH_3$ (Pent$-2-$ene).When an alkyl halide reacts with an alcoholic b

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trans pent$-2-$ene

Vedclass · Answer

(C) $CH_3-CH(Br)-CH_2-CH_2-CH_3 + C_2H_5OK \xrightarrow{C_2H_5OH} CH_3-CH=CH-CH_2-CH_3$ (Pent$-2-$ene).When an alkyl halide reacts with an alcoholic base like potassium ethoxide,it undergoes a dehydrohalogenation reaction (elimination reaction) following Zaitsev's rule.The reaction produces a mixture of pent$-1-$ene and pent$-2-$ene. Pent$-2-$ene is the more substituted alkene and is thus the major product. Between the geometric isomers of pent$-2-$ene,the $trans$ isomer is more stable than the $cis$ isomer due to reduced steric hindrance between the alkyl groups. Therefore,$trans$ pent$-2-$ene is the major product.

$2-$bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

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