JEE MAIN - Chemistry (2023 - 8th April Morning Shift - No. 13)
Choose the halogen which is most reactive towards $$\mathrm{S}_{\mathrm{N}} 1$$ reaction in the given compounds (A, B, C & D)
$$\mathrm{A}-\mathrm{Br}_{(\mathrm{b})} ; \mathrm{B}-\mathrm{I}_{(\mathrm{a})} ; \mathrm{C}-\mathrm{Br}_{(\mathrm{a})} ; \mathrm{D}-\mathrm{Br}_{(\mathrm{a})}$$
$$\mathrm{A}-\mathrm{Br}_{(\mathrm{a})} ; \mathrm{B}-\mathrm{I}_{(\mathrm{a})} ; \mathrm{C}-\mathrm{Br}_{(\mathrm{a})} ; \mathrm{D}-\mathrm{Br}_{(\mathrm{a})}$$
$$\mathrm{A}-\mathrm{Br}_{(b)} ; \mathrm{B}-\mathrm{I}_{(b)} ; \mathrm{C}-\mathrm{Br}_{(b)} ; \mathrm{D}-\mathrm{Br}_{(\mathrm{b})}$$
$$\mathrm{A}-\mathrm{Br}_{(\mathrm{a})} ; \mathrm{B}-\mathrm{I}_{(\mathrm{a})} ; \mathrm{C}-\mathrm{Br}_{(b)} ; \mathrm{D}-\mathrm{Br}_{(\mathrm{a})}$$
Explanation
Organic compounds are more reactive towards SN 1 , if the carbocation is stable compared to others. Reactivity in reactions depends upon the stability of the carbocation intermediate.
_8th_April_Morning_Shift_en_13_2.png)
$\operatorname{Br}(\mathrm{a})$, Becuase formed intermediate carbocation get stable by conjugation with phenyl ring
_8th_April_Morning_Shift_en_13_3.png)
$ \mathrm{I}(\mathrm{a})$, Becuase formed intermediate carbocation become more stable by conjugation
_8th_April_Morning_Shift_en_13_4.png)
$ \operatorname{Br}(\mathrm{b})$, Becuase we can't remove $B r(a)$ from bridge head carbon (Bredt's rule)
_8th_April_Morning_Shift_en_13_5.png)
$ \operatorname{Br}(\mathrm{a})$, because formed intermediate $3^{\circ}$ carbocation is more stable (stability of carbocation $3^{\circ}>2^{\circ}$ $\left.>1^{\circ}\right)$
_8th_April_Morning_Shift_en_13_2.png)
$\operatorname{Br}(\mathrm{a})$, Becuase formed intermediate carbocation get stable by conjugation with phenyl ring
_8th_April_Morning_Shift_en_13_3.png)
$ \mathrm{I}(\mathrm{a})$, Becuase formed intermediate carbocation become more stable by conjugation
_8th_April_Morning_Shift_en_13_4.png)
$ \operatorname{Br}(\mathrm{b})$, Becuase we can't remove $B r(a)$ from bridge head carbon (Bredt's rule)
_8th_April_Morning_Shift_en_13_5.png)
$ \operatorname{Br}(\mathrm{a})$, because formed intermediate $3^{\circ}$ carbocation is more stable (stability of carbocation $3^{\circ}>2^{\circ}$ $\left.>1^{\circ}\right)$
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