



Reactivity towards SN1 depends upon the stability of the carbocation.

Let's figure out which of these compounds will react the fastest via an $ \text{S}_\text{N}1 $ mechanism. Remember, the rate-determining step in an $ \text{S}_\text{N}1 $ reaction is the formation of a carbocation. The more stable the carbocation intermediate, the faster the reaction will proceed.
Analyzing the Carbocations Formed:
(1) Cyclohexylmethyl carbocation:
This is a primary carbocation ($ \text{1}^\circ $). Primary carbocations are generally unstable due to the lack of hyperconjugation and inductive stabilization from more alkyl groups.
(2) Cyclohexyl carbocation:
This is a secondary carbocation ($ \text{2}^\circ $). Secondary carbocations are more stable than primary carbocations due to more hyperconjugative and inductive effects.
(3) Bromobenzene:
If the bromine were to leave, it would form a phenyl carbocation. Phenyl carbocations are highly unstable because the positive charge resides in an $ sp^2 $ hybridized orbital, which is held more tightly to the nucleus, and the carbocation cannot be stabilized by resonance in the usual sense.
(4) 1-Bromo-1-phenylethane:
Loss of bromine would result in the formation of a benzylic secondary carbocation ($ \text{2}^\circ $ benzylic). Benzylic carbocations are particularly stable due to resonance with the benzene ring, which delocalizes the positive charge over several carbon atoms. Furthermore, it's also a secondary carbocation, benefiting from hyperconjugation and inductive effects from the methyl group.
Comparing the Stabilities of the Potential Carbocations:
The order of stability is as follows:
Benzylic $ \text{2}^\circ $ > $ \text{2}^\circ $ alkyl > $ \text{1}^\circ $ alkyl >> phenyl
Conclusion:
The carbocation formed from compound (4) is the most stable. This means that compound (4) will undergo $ \text{S}_\text{N}1 $ reaction at the fastest rate.
Final Answer:
The compound that reacts the fastest via an $ \text{S}_\text{N}1 $ mechanism is: $ \boxed{(4)} $
Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and C are not the same.
C. $Ar, K^{+}, Cl^{–}, Ca^{2+} and S^{2–}$ are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si $>$ Al $>$ Mg $>$ Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below:
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
Among the following, choose the ones with an equal number of atoms.
Choose the correct answer from the options given below:
Identify the suitable reagent for the following conversion: $Ph-C(=O)-OCH_3$ $\longrightarrow$ $Ph-CHO$
List-I | List-II | ||
| (A) | 1 mol of H2O to O2 | (I) | 3F |
| (B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
| (C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
| (D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
List-I | List-II | ||
| (A) | [Co(NH3)5(NO2)]Cl2 | (I) | Solvate isomerism |
| (B) | [Co(NH3)5(SO4)]Br | (II) | Linkage isomerism |
| (C) | [Co(NH3)6] [Cr(CN)6] | (III) | Ionization isomerism |
| (D) | [Co(H2O)6]Cl3 | (IV) | Coordination isomerism |


A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
Which of the following statements are true?
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and C are not the same.
C. $Ar, K^{+}, Cl^{–}, Ca^{2+} and S^{2–}$ are all isoelectronic species.
D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si $>$ Al $>$ Mg $>$ Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
Choose the correct answer from the options given below: