JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 10)
For the following reactions
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Ks and Ke, are respectively, the rate constants for substitution and elimination and $$\mu = {{{k_s}} \over {{k_e}}}$$ the correct options is
Ks and Ke, are respectively, the rate constants for substitution and elimination and $$\mu = {{{k_s}} \over {{k_e}}}$$ the correct options is
$${\mu _B} > {\mu _A}$$ and Ke(A) > Ke(B)
$${\mu _A} > {\mu _B}$$ and Ke(B) > Ke(A)
$${\mu _B} > {\mu _A}$$ and Ke(B) > Ke(A)
$${\mu _A} > {\mu _B}$$ and Ke(A) > Ke(B)
Explanation
For 1o alkayl halide rate of substitute reaction is more than rate of elimination reaction.
When Z = A, then ks $$>$$ ke and
when Z = B, then ks $$<$$ ke
$$ \therefore $$ $$\mu $$A $$>$$ $$\mu $$B
As B is strong base than A so for B more elimination reaction happens.
$$ \therefore $$ Ke(B) $$>$$ Ke(A)
When Z = A, then ks $$>$$ ke and
when Z = B, then ks $$<$$ ke
$$ \therefore $$ $$\mu $$A $$>$$ $$\mu $$B
As B is strong base than A so for B more elimination reaction happens.
$$ \therefore $$ Ke(B) $$>$$ Ke(A)
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