JEE MAIN - Chemistry (2019 - 9th January Morning Slot - No. 7)
The correct decreasing order for acid strength is :
NO2CH2COOH > FCH2COOH > CNCH2COOH > ClCH2COOH
FCH2COOH > NCCH2COOH > NO2CH2COOH > ClCH2COOH
CNCH2COOH > O2NCH2COOH > FCH2COOH > ClCH2COOH
NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH
Explanation
After releasing proton (H+) the more stable anion is more acidic nature. The 4 anions are $$ \to $$
_9th_January_Morning_Slot_en_7_1.png)
Here $$-$$NO2, $$-$$ CN, $$-$$Cl, $$-$$F all of them have $$-$$ I effective and all have same distance from the anion ($$-$$ O$$-$$).
So depending on which one have more power $$-$$ I effect will be more stable.
Acidic nature $$ \propto $$ $$-$$ I power.
$$-$$ I power order :
$$-$$ NO2 > $$-$$ C $$ \equiv $$ N > $$-$$ F > $$-$$ Cl
So, acidic strength order $$ \to $$
NO2 $$-$$ CH2 $$-$$ COOH > NC $$-$$ CH2 $$-$$ COOH
> F $$-$$ CH2 $$-$$ COOH > Cl $$-$$ CH2 $$-$$ COOH
Here $$-$$NO2, $$-$$ CN, $$-$$Cl, $$-$$F all of them have $$-$$ I effective and all have same distance from the anion ($$-$$ O$$-$$).
So depending on which one have more power $$-$$ I effect will be more stable.
Acidic nature $$ \propto $$ $$-$$ I power.
$$-$$ I power order :
$$-$$ NO2 > $$-$$ C $$ \equiv $$ N > $$-$$ F > $$-$$ Cl
So, acidic strength order $$ \to $$
NO2 $$-$$ CH2 $$-$$ COOH > NC $$-$$ CH2 $$-$$ COOH
> F $$-$$ CH2 $$-$$ COOH > Cl $$-$$ CH2 $$-$$ COOH
Comments (0)
