# How does C−Br bond obtain energy through solvation?

I came across a question related to SN1 mechanism. In an example of SN1 reaction, i.e. hydrolysis of tert-butyl bromide to form tert-butyl alcohol, the following was given in NCERT Chemistry: Textbook for class XII, Part II [1, p. 304]:

Step I is the slowest and reversible. It involves the C—Br bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent. Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.

Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo SN1 reaction very fast because of the high stability of 3° carbocations. We can sum up the order of reactivity of alkyl halides towards SN1 and SN2 reactions as follows:

$$\def\CL{{\large\text{Tertiary halide; Secondary halide; Primary halide;}~\ce{CH3X}}} \ce{->[\text{For}~\mathrm{S_N2}~\text{reaction}][\CL]} \\ \ce{<-[\hphantom{[\CL}][\text{For}~\mathrm{S_N1}~\text{reaction}]}$$

How does C-Br bond obtain energy through Solvation of bromide ion ?

### References

1. NCERT. Chemistry: Textbook for Class XII. Part II, Reprint: August 2019 Bhadrapada 1941; M. Siraj Anwar, Shveta Uppal, Arun Chitkara, Bibash Kumar Das, R.N. Bhardwaj, Mukesh Gaur, Series Eds.; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-716-7. (PDF)