The heats of hydrogenation for 3--methylbuten | vedclass

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Question

(B) Let the reaction be: $3-$-methylbutene $\rightarrow$ $2-$-pentene.We can use Hess's Law by constructing a cycle:$1.$ $3-$-methylbutene $+ H_2 \rig

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(B) Let the reaction be: $3-$-methylbutene $ightarrow$ $2-$-pentene.We can use Hess's Law by constructing a cycle:$1.$ $3-$-methylbutene $+ H_2 ightarrow 2-$-methylbutane,$\Delta H_1 = -30\, kcal/mol$$2.$ $2-$-methylbutane $ightarrow$ pentane,$\Delta H_2 = \Delta H_{comb}(2-$-methylbutane$) - \Delta H_{comb}(	ext{pentane}) = -784 - (-782) = -2\, kcal/mol$$3.$ Pentane $ightarrow 2-$-pentene $+ H_2$,$\Delta H_3 = -(\Delta H_{hydrog}(2-$-pentene$)) = -(-28) = +28\, kcal/mol$Adding these steps:$3-$-methylbutene $ightarrow 2-$-pentene$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -30 - 2 + 28 = -4\, kcal/mol$.

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