GS-1 Biological Sciences Question 18

[jnguyen07080]

Why is the electrophilic aromatic substitution between benzene and bromine considered to be first order rather than second order? I thought that the first step of the reaction is benzene attacking the bromine which requires both substrate concentrations in the rate law.

[admin]

Good news and bad news. The bad news is that it is a bit complicated. Essentially, however, because of the catalyst which supercharges the Br to a highly reactive cation, the concentration of the substrate essentially does not matter thus it acts like a first order reaction.

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/chembox_brom_arom-e.htm

The good news is that you can forget about it! Knowledge that depends on your understanding of aromatic chemistry is no longer required on the new MCAT. If you see the question on a test, quickly do your best and move on since it is likely to be a test question.

[makingmd]

Hello - I have seen 2 references (aromatic compounds in organic chenistry and E1/E2/Alkenes) mentioned as not being required information for the new MCAT. Is there a comprehensive list of what has been removed?

Thank you!!

[admin]

You mentioned everything exept one. Here is the list of topics that do not require previous knowledge since 2003 (ie the topics can be discussed but information to solve the problem must be included in the passage or question):

Alkenes, alkynes, ethers
Aromatics compounds including benzenes

[chronicida2879]

First, the link you cited uses an excess of an activated aromatic. Since the concentration of the aromatic remains relatively constant (because it's so large) during the reaction, it's pseudo first order. Clearly this is a different situation.

Second, they even say that if you didn't have an excess, "The rate of the reaction depends on the concentration of the bromine cBr and the aromatics cAr (reaction second order)."

Third, if this won't be on the MCAT, why not replace the question, especially in the free test that prospective customers will use to judge your products?

[chiez78849]

Quoted from the linked site.

"The rate of the reaction depends on the concentration of the bromine cBr and the aromatics cAr (reaction second order). Under the given conditions (see procedure - large excess of one of the reactants) the reaction is pseudo first order. "

Indeed, 2nd order; agree with Chronic.

[mlamon7062]

As chornic mentions, the referenced site does not apply since they are studying pseudo-first order reactions.

There are 3 steps in this reaction:

1) Formation of the electrophilic zwitterionic complex between Br2 and FeBr3. This is a bimolecular reaction

2) Formation of the sigma complex which is a bimolecular reaction between the benzene and FeBr5

3) Reformation of the aromatic which involves the base deprotonation of a hydrogen, again, a bimolecular reaction.

Whichever one of the steps above is the rate-limiting step, will yield a bimolecular, 2nd order reaction.

[mohsinalam5969]

The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is intuitively defined as how fast or slow a reaction takes place.
The reaction of benzene with bromine to form bromobenzene is an electrophilic substitution pseudo first order reaction.
Pseudo first order reaction is a second-order reaction in which one of the reactants is present in such great amounts that its effect is not seen. For example, the hydrolysis of an ester in water. If the water is present in great amounts, any change will be too small, relatively, to be noticed. The reaction then appears dependent on concentration of ester alone. To see the dependence on water, you'd have to dissolve the ester in acetone with a few % water.

For example we have a reaction

A + B ==> Product

the rate of reaction is .....

Rate= dP/dt .... which is also the same as ....

Rate = K.[A].[B]

The total order of reaction will equal to the sum of the orders of the reactants A + B = 1+1 = 2 ... so the above reaction is a second order reaction.


Now If we converted one of those reactants to be in excess concentration in the system, then we can get a pseudo first order reaction

example
ConcA = 0.1M
excess ConcB = 1M

after the reaction [A] = 0 and [B] = 0.99M

so reactant B wont contribute to the total order as it is in excess and viewed as a constant, this is known as pseudo first order.

so the new rate = K'.[A]

and the total order is = to the sum of the orders which is = to 1, as [A] is the only reactant contributing to the rate.

K' = K.[B]

but member even though [B] is assumed as a constant any large changes in [B] will effect K' and therefore effect the rate.

Measuring a second order reaction rate can be problematic: the concentrations of the two reactants must be followed simultaneously, which is more difficult; or measure one of them and calculate the other as a difference, which is less precise. A common solution for that problem is the pseudo first order approximation.