Kaplan MCAT 45 Q's: Physics, Ochem

[CrimsonMirage]

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Hi,
I was working on the practice sections in Kaplan's MCAT 45 book and ran across some things in the answer key that did not make sense...especially for the following two questions:

OCHEM:
Q8) What are the products of base-promoted hydrolysis of N-propylbutanamide?
a) propanoic acid, 1-butanamine
b) propanoate ion, 1-butanamine
c) butanoic acid, 1-propanamine
d) butanoate ion, 1-propanamine

The answer key says it's C, but I don't see why you wouldn't have the formation of a carboxylate ion (answer choice D) if it's base-promoted hydrolysis. Any ideas?

PHYSICS:
Q45) Eight positively charged spheres (+1 C each) are distributed at equally spaced intervals around a circular hoop whose radius is 1 m. What is the electrostatic potential at the center of the hoop?
a) 0
b) 4k
c) 4(sqrt2)k
d) 8k

The answer key says A. However, the answer explanation says D (and they use the formula for potential kq/r to come up with it), arguing that potential cannot cancel if all the charges are positive. D sounds right to me as well, but let me know!

Any help appreciated..Thanks!

 

[AZFutureDoc]

I know that 2 is def A. Just look at the picture. (Or draw your own). It will pretty much look like a stop sign. Now. Whatever force one causes on another, that other charge exerts the same force. You could go to town and draw the 7 forces acting on each charge, but that is unnecessary. Since it is symmetrical, Newton's 3rd law says that the answer must be zero.

However, change the charge, or spacing of any of the charges, and this problem is pretty much impossible for the MCAT. It is a 5 minute physics problem in a basic college class, but not MCAT.

 

[spyderracing32]

For the O-chem, I'm assuming this takes place in an aqueous solution (base-promoted). The OH- ion is regenerated from the reaction of the propylamine ion that is formed from the elimination of the -NHCH2CH2CH3 group with water. Assuming the concentration of water is much greater than the concentration of the base the carboxylic acid with not be in the carboxylate form.

 

[BerkReviewTeach]

Hi,
two questions:

The answer key says it's C, but I don't see why you wouldn't have the formation of a carboxylate ion (answer choice D) if it's base-promoted hydrolysis. Any ideas?

The answer key says A. However, the answer explanation says D (and they use the formula for potential kq/r to come up with it), arguing that potential cannot cancel if all the charges are positive. D sounds right to me as well, but let me know!

Any help appreciated..Thanks!

For the first question, you are correct. It's just like saponification, the base-driven hydrolysis of an ester. Until you neutralize the system, the product must assume that charge. The base is stoichiometric, not catalytic, so for every hydroxide added, you generate a carboxylate.

Also, for the caroxylic acid to form, the carboxylate (RCO2-) would have to deprotonate water to form OH- and RCO2H, which is thermodynamically unfavorable.

As for the second question, consider the potential to be based on the impact of the forces. If you have eight equal charges symmetrically about you, then the vectors will cancel out. The charge above you is doing the same thing as the charge below you, but in the opposite direction. The charge to the left of you is doing the same thing as the charge to the right of you, but in the opposite direction. The net result is that there is no net force on the charge in the middle, so when released at that point it would not move. The best answer is choice A.

The moral to the story is that all MCAT prep books have typos and if you feel something is wrong, most of the time you are right. You need to trust yourself. In the future, if you post a question, don't list what the right answer is supposed to be. Unfortunately, it is in our nature to accept the answer as being correct and then try to explain its correctness. That leads to biased replies. Had you asked what answer people got, you would have likely seen D and A (respectively) being given by most posters.