Solutions_Manual_for_Organic_Chemistry_6th_Ed 188

Solutions_Manual_for_Organic_Chemistry_6th_Ed 188 - 8-56...

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Unformatted text preview: 8 -56 Once the bromonium ion i s formed, i t can be attacked by either nuc\eophile, bromide or chloride, leadi ng to the mi xture of products. Q H H Br -- Br � ~ H Br ��-� � 'i---Y H �.. HBr .- :Cl : H Cl 8-57 Two possible orientati ons of attack of bromi ne radical are possible: (A) anti -Markovnikov (B) Markovnikov >= + • Br -- / Br 3 ° radical \. C, The first step in the mechani s m i s endothermic and rate determi ning. The 3° radical produced i n anti­ Markovnikov attac k (A) of bromine radical is several kl/mole more stable than the 1° radical generated by Markovnikov attac k ( B ) . The Hammond Postulate tells us that it is reasonable to assume that the activation energy for anti -Markovnikov addition is lower than for Markovnikov addition. This defines the first half of the energy di agram. The relative stabi l i ties of the final products are somewhat difficult to predict. (Remember that stabi lity of final products does not necessari l y reflect rel ative stabi lities of intermedi ates; this is why a thermodynamic product can be different from a ki netic product.) From bond dissoc i ation energies (kJ/mole) in Table 4-2 : anti -Markovnikov H t0 3°C Br to 1° C 3 81 285 666 kJ/mo\e Markovnikov H to 1 ° C Br t0 3°C 410 272 682 kl/mole >= + • Br -- 1° radical 'l- r � H2 HBr � h + • Br Br y; + • Br H If it takes more energy to break bonds in the Markovnikov product, it must be lower in energy, therefore, more stable-OPPOSITE OF STABILITY OF THE INTERMEDIATES! Now we are ready to construct the energy di agram; see the next page . 18 1 ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.

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