FL1 C/P Passage 9 Suspected Typo and #49

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erichsuh94
Posts: 13
Joined: Wed May 23, 2018 4:16 am

FL1 C/P Passage 9 Suspected Typo and #49

Post by erichsuh94 » Wed May 23, 2018 4:59 am

Hello,

First, in regard to the suspected typo:

1) On table one and the column for delta H for H-> H+, isn't the value supposed to be 188.1 instead of 181.1?
Because heat of formation for H+ is 169.3 and that of H is -18.8, the delta H of H->H+ comes out to be (169.3) - (-18.8) = 188.1 .

Also, passage states RSE value is a difference between delta H of H deprotonation and delta H of X deprotonation.
And only way I get the value as listed (i.e. -6.3 -2.8 0.1 2.9 -260.4 ) is if I assume delta H of H is 188.1


Now my question in regard to #49:

I am having bit difficult time making sense out of the answer explanation- but I have came up with a reasoning on my own, I would greatly appreciate if you could confirm if it is logical.

So when enthalpy change for a reaction is
1) negative, the product is more stable than the reactant.
2) positive, the product is less stable than the reactant.

Enthalpy change for deprotonation of H is 188.1 , so this means the product that is formed is much less stable than the reactant.

When the para-substituent is OCH3, for example, enthalpy change for deprotonation of X is 181.8.

While it still means that X+ is much less stable than X, the degree to which the product of reaction becomes less stable than the pertinent reactant is less than when there is no para-substituent.

And this difference is what is being represented by RSE (in this case 181.8 - 188.1 = -6.3) and this is the reason why "more negative RSE corresponds to the more stabilizing effect the para-substituent brings".


Thank you!!
NS_Tutor_Katelyn
Posts: 157
Joined: Wed May 30, 2018 11:51 am
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Re: FL1 C/P Passage 9 Suspected Typo and #49

Post by NS_Tutor_Katelyn » Thu Jun 07, 2018 4:28 pm

erichsuh94 wrote:Hello,

First, in regard to the suspected typo:

1) On table one and the column for delta H for H-> H+, isn't the value supposed to be 188.1 instead of 181.1?
Because heat of formation for H+ is 169.3 and that of H is -18.8, the delta H of H->H+ comes out to be (169.3) - (-18.8) = 188.1 .

Also, passage states RSE value is a difference between delta H of H deprotonation and delta H of X deprotonation.
And only way I get the value as listed (i.e. -6.3 -2.8 0.1 2.9 -260.4 ) is if I assume delta H of H is 188.1


Now my question in regard to #49:

I am having bit difficult time making sense out of the answer explanation- but I have came up with a reasoning on my own, I would greatly appreciate if you could confirm if it is logical.

So when enthalpy change for a reaction is
1) negative, the product is more stable than the reactant.
2) positive, the product is less stable than the reactant.

Enthalpy change for deprotonation of H is 188.1 , so this means the product that is formed is much less stable than the reactant.

When the para-substituent is OCH3, for example, enthalpy change for deprotonation of X is 181.8.

While it still means that X+ is much less stable than X, the degree to which the product of reaction becomes less stable than the pertinent reactant is less than when there is no para-substituent.

And this difference is what is being represented by RSE (in this case 181.8 - 188.1 = -6.3) and this is the reason why "more negative RSE corresponds to the more stabilizing effect the para-substituent brings".


Thank you!!
Correct, the RSE represents the change in energy (i.e. change in stability) caused by each of the various substituents, so we can use the RSE values to predict the stability trend.
Katelyn Sawyer
Senior Tutor
katelyn@nextsteptestprep.com
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