2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:
(a) `C_(2)H_(5)CH_(2)overset(CH_(3))overset(|)underset(CH_(3))underset(|)(C)-OCH_(3)` (b) `C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2)` (c) `C_(2)H_(5)CH_(2)=underset(CH_(3))underset(|)(C)-CH_(3)`

Options

A. all of these
B. I and III
C. III only
D. I and II

Answer

A

Explanation

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))`