## 15.79

$K = \frac{k_{forward}}{k_{reverse}}$

RyanTran2F
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### 15.79

Hi can someone explain how the answer key got the answer in Part A? I did not quite understand the explanation provided. The question asks, "In the reaction of HBr with the reactive intermediate CH3CH=CHCH2+, at low temperatures the predominant species is CH3CHBrCH=CH2, but at high temperatures, the predominant species is CH3CH=CHCH2Br. Which product is formed by the pathway with the larger activation energy?" Thanks :)
Last edited by RyanTran2F on Tue Mar 13, 2018 5:18 pm, edited 1 time in total.

Sungyoon_Baek_1A
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### Re: 15.79

At high temperatures, the predominant product is CH3CHCHCH2Br. A higher temperature means that there is more energy to overcome a higher activation energy. The high temperatures and there for greater energy allow the reaction to produce CH3CHCHCH2Br. For the product CH3CHBrCHCH2, we do not need a high temperature (we do not need the greater energy to overcome a high activation energy), so we know that it has a lower activation than the pathway to form CH3CHCHCH2Br.

Chem_Mod
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### Re: 15.79

Please state what the question is so we can better respond to your post.

Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.79

For 15.79 part A, the answer states that CH3CH=CHCH2Br is the product because "the positive charge on the reactive intermediate is on a primary carbon atom". How can you determine that it is a product of higher temperature from this information?