Question

PROBLEM 6-26 Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate. сH, ОСН, -CH-CH, + Cн, —С—СН—СH, CH I (a) CH,ОН CH, C- CH,—С—СН— СH, heat HOCH, CH CH3 biod Iyinsgoon OCH,CH CI CH,CH,OH (b) + llt noi obimt OCH,CH, bouCH heat CH CH3

please explain your answers

Answer 1

(a)

CH33 CH3 H HOCHZ ÇHz ö. H3c-༦-g-cH CH? —CH A Hyc—c—e cH3 H3G—༩— CH3 པ . Hyc—༦– -¢-CH₃ cH ） པ་ ཀ ཁ པ ཡ ་ག ་ | ག ་ག ་ :6—H H2C" H CH3 H3G H3G།83 cH3 H3༦།༠ cH CH3 Học–ẽ g–CH, CH3 H3c——g—CH H3c—————cH CH? CH₃

In this solvolysis reaction, first the iodide leaves from the molecule to give a secondary carbocation. The secondary carbocation thus formed can directly react with the methanol nucleophile to give the first product (follow the red arrow). The secondary carbocation can undergo alkyl shift to give the more stable tertiary carbocation (follow the blue arrow) which then reacts with methanol nucleophile to give the second product.

H, CH3 CH3 H 0 Ун Сн, CHZ a-o-o-oh azahod H CH3 Co? CH3 4.ÖZH Hd

In this solvolysis reaction, first the chloride leaves from the molecule to give a secondary carbocation. The secondary carbocation thus formed can directly react with the methanol nucleophile to give the first product (follow the red arrow). The secondary carbocation can undergo hydride shift to give the more stable tertiary carbocation (follow the blue arrow) which then reacts with methanol nucleophile to give the second product.