2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (I) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OCH_(3) (II) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2) (III) C_(2)H_(5)CH=underset(CH_(3))underset(|)(C)-CH_(3)

2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol

1.	2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (I) C_(2)H_(5)CH_(2)underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-OCH_(3) (II) C_(2)H_(5)CH_(2)underset(CH_(3))underset(\|)(C)=CH_(2) (III) C_(2)H_(5)CH=underset(CH_(3))underset(\|)(C)-CH_(3)
Answer» all of these I and III III only I and II Solution :2-Chloro-2-methylpantane being a `3^(@)` alkyl halide can undergo either `S_(N)1` or `E_(1) ` REACTION. Further, the intermediate carbocation (I) can undergo elimination in two DIFFERENT ways, to give 2-methyl-2-pentene (major) and 2-methyl-1-pentene (minor) `underset("2-Chloro-2-methylpentene")(CH_(3)CH_(2)CH_(2)-underset(Cl)underset(\|)overset(CH_(3))overset(\|)(C)-CH_(3))underset(-Cl^(-))overset("Ionization")to underset(3^(@)" carbocation (I)")(CH_(3)CH_(2)CH_(2)-underset(+)overset(CH_(3))overset(\|)(C)-CH_(3))` `(I) underset(S_(N)1)overset(CH_(3)ON a)to underset("2-Methoxy-2-methylpentane (Substitution PRODUCT)")(CH_(3)CH_(2)-underset(OCH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-CH_(3))` `(I)underset(-H^(+))overset(E_(1))to underset("2-Methyl-2-pentene (major product)")(CH_(3)CH_(2)CH=overset(CH_(3))overset(\|)(C)=CH_(2))+underset("2-Methyl-1-pentene (minor product)")(CH_(3)CH_(2)CH_(2)-overset(CH_(3))overset(\|)(C)=CH_(2))`