When $\mathrm{CH}_{3} \mathrm{CH}_{2} \ddot{\mathrm{S}} \mathrm{CH}_{2} \mathrm{CH}_{2} \ddot{\mathrm{S}} \mathrm{CH}_{2} \mathrm{CH}_{3}$ reacts with two equivalents of $\mathrm{CH}_{3} \mathrm{I},$ the following double sulfonium salt precipitates:
(a) Give a curved-arrow mechanism for the formation of this salt.
(b) Upon closer examination, this compound is found to be a mixture of two isomers with melting points of $123-124^{\circ} \mathrm{C}$ and $154^{\circ} \mathrm{C}$, respectively. Explain why two compounds of this structure are formed. What is the relationship between these isomers? (Hint: See Sec. $6.9 \mathrm{~B} .$ )