The compound (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

The reason for this is due to the nature of the SN1 mechanism. In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The leaving group leaves first, forming a carbocation, which is then attacked by the nucleophile.

The ease of formation of the carbocation depends on the stability of the leaving group. A good leaving group is one that can stabilize the negative charge that results when it leaves. In this case, iodine (I) is a better leaving group than bromine (Br) because it is larger and can better stabilize the negative charge due to its larger electron cloud. This makes the carbocation formation step easier and thus makes the overall SN1 reaction faster.

Therefore, (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

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The compound (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

The reason for this is due to the nature of the SN1 mechanism. In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The leaving group leaves first, forming a carbocation, which is then attacked by the nucleophile.

The ease of formation of the carbocation depends on the stability of the leaving group. A good leaving group is one that can stabilize the negative charge that results when it leaves. In this case, iodine (I) is a better leaving group than bromine (Br) because it is larger and can better stabilize the negative charge due to its larger electron cloud. This makes the carbocation formation step easier and thus makes the overall SN1 reaction faster.

Therefore, (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

Knowee AI · Accepted Answer

The compound (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

The reason for this is due to the nature of the SN1 mechanism. In an SN1 reaction, the rate-determining step is the formation of a carbocation intermediate. The leaving group leaves first, forming a carbocation, which is then attacked by the nucleophile.

The ease of formation of the carbocation depends on the stability of the leaving group. A good leaving group is one that can stabilize the negative charge that results when it leaves. In this case, iodine (I) is a better leaving group than bromine (Br) because it is larger and can better stabilize the negative charge due to its larger electron cloud. This makes the carbocation formation step easier and thus makes the overall SN1 reaction faster.

Therefore, (CH3)3C–I is more reactive towards SN1 reactions than (CH3)3C–Br.

Out of (CH3)3C–Br and (CH3)3C–I, which one is more reactive towards SN1and why

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