(Solved): 2.4. (b) The products in each case, like the starting materials, are a saturated heteroatom co ...

2.4. (b)

The products in each case, like the starting materials, are a saturated heteroatom compound and a nucleophile, yet these reactions do not readily reverse. Explain.

2.4. (c)

a) 1-iodobutane with hydroxide \\[ \\begin{array}{l} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{I}^{-} \\\\ \\text {(alcohol) } \\\\ \\end{array} \\] b) 1-iodobutane with ethoxide \\[ \\begin{array}{r} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}+\\mathrm{I}^{-} \\\\ \\text {(ether) } \\end{array} \\] (ether) c) 1-iodobutane with amide \\( \\left(\\mathrm{NH}_{2}^{-}\\right) \\)anions \\[ \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I}+\\mathrm{NH}_{2}^{-} \\rightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}+\\mathrm{I}^{-} \\] (amine)\r\n\r\nExplain why the optical rotation of a solution of (+)-2-phenyl-2-pentanol goes to zero degrees when the compound is warmed with aqueous acid to give ( \\pm\\( )-2 \\)-phenyl-2chloropentane. (HINT: 2-phenyl-2-pentanol is a tertiary alcohol. Would it react by \\( S_{N} 1 \\) or \\( S_{N} 2 \\) ?)