Bio GChem and Org Questions

[DonExodus]

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1) Do animal cells have vacuoles? Ive found a good 30 contradicting answers.
2) Do plant cells have centrioles? I thought no...
3) During competitive inhibition, the inhibitor can bind to the substrate site, OR the allosteric site, correct? With non-comp- its just the allosteric.

4) When can the bonding radius of an atom be larger than its atomic? I know Cl- is larger than Cl, but thats not the bonding radius.

Finally, this is from Kaplan-
5) Which of the following is the most stable carbanion? (C-)
Ph = Phenyl
a) (Ph)3C-
b) (Ph)2HC-
c) PhH2C-
d) H3C-
e) (CH3)3C-

I chose d, since a carbanion will be stabilized by something that withdraws electrons (also an activating substituent btw). Since all of those things donate electrons more then H, I chose D. The correct answer is A, because of resonance. The problem is, Im failing to see the resonance there. I can pop the - charge on the phenyl rings, but I cant figure out where to throw the existing bonds.
Thoughts?

Thanks
Whit

 

[wigglytooth]

1) Do animal cells have vacuoles? Ive found a good 30 contradicting answers.
Vacuoles are basically storage membranes. Transport in all cells use vacuoles. The big difference between plant and animal cells on the vacuole front is that a defining feature of plant cells is a large central vacuole.

2) Do plant cells have centrioles? I thought no...
No, they do not require them during cell division because the plant wall is rigid enough to maintain the shape of the cell. Spindle still forms though, just not anchored by centrioles. I don't remember the exact mechanism.

3) During competitive inhibition, the inhibitor can bind to the substrate site, OR the allosteric site, correct? With non-comp- its just the allosteric.
Don't remember..


4) When can the bonding radius of an atom be larger than its atomic? I know Cl- is larger than Cl, but thats not the bonding radius.
Don't remember..

Finally, this is from Kaplan-
5) Which of the following is the most stable carbanion? (C-)
Ph = Phenyl
a) (Ph)3C-
b) (Ph)2HC-
c) PhH2C-
d) H3C-
e) (CH3)3C-

I chose d, since a carbanion will be stabilized by something that withdraws electrons (also an activating substituent btw). Since all of those things donate electrons more then H, I chose D. The correct answer is A, because of resonance. The problem is, Im failing to see the resonance there. I can pop the - charge on the phenyl rings, but I cant figure out where to throw the existing bonds.
Thoughts?

-Throw the existing bonds on the carbons in the rings. You get a lot of resonance structures this way making A the most stable.

 

[slayerdeus]

Finally, this is from Kaplan-
5) Which of the following is the most stable carbanion? (C-)
Ph = Phenyl
a) (Ph)3C-
b) (Ph)2HC-
c) PhH2C-
d) H3C-
e) (CH3)3C-

I chose d, since a carbanion will be stabilized by something that withdraws electrons (also an activating substituent btw). Since all of those things donate electrons more then H, I chose D. The correct answer is A, because of resonance. The problem is, Im failing to see the resonance there. I can pop the - charge on the phenyl rings, but I cant figure out where to throw the existing bonds.
Thoughts?

Thanks
Whit

I think you had some confusion with withdrawing groups and activating groups. Activating/deactivating groups only deal with aromatic substituents. In this general case, activating groups donate electrons to benzene, this makes benzene even more nucleophilic and susceptable to attack by an electrophile. On the other hand deactivating groups in general withdraw electron density from benzene and thus cause electrophiles to attack less readily on the ring.

 

[DonExodus]

Yes, I understand that, I was just pointing it out randomly. The basis for stabilization is still there (resonance + partial charges, C+ stability). Something e- donating would make a C- less stable, and ph groups, being e- rich, are good e- donors.

Like I said, I still fail to see the resonance :/ I just cant keep all the bonds / e-s accounted for.

 

[lintydent]

3) During competitive inhibition, the inhibitor can bind to the substrate site, OR the allosteric site, correct? With non-comp- its just the allosteric.

Competitive inhibition mimics the substrate and so they both compete for the ACTIVE site. It doesn't bind to the substrate or the allosteric site.

Noncompetitive inhibition, binds to the enzyme at a location away from the active site...not necessarily the allosteric site...and alters the conformation of the enzyme so that the active site is no longer functional.

So what does bind to the allosteric site? An allosteric inhibitor or activator that acts like a nonreversible noncompetitive inhibitor in that they both change the shape of the enzyme...but the inhibitor stabilizes the inhibited conformation, inhibiting activity, and the activator stabilizes the active conformation.

 

[Envision]

1) Do animal cells have vacuoles? Ive found a good 30 contradicting answers.
2) Do plant cells have centrioles? I thought no...
3) During competitive inhibition, the inhibitor can bind to the substrate site, OR the allosteric site, correct? With non-comp- its just the allosteric.

4) When can the bonding radius of an atom be larger than its atomic? I know Cl- is larger than Cl, but thats not the bonding radius.

Finally, this is from Kaplan-
5) Which of the following is the most stable carbanion? (C-)
Ph = Phenyl
a) (Ph)3C-
b) (Ph)2HC-
c) PhH2C-
d) H3C-
e) (CH3)3C-

I chose d, since a carbanion will be stabilized by something that withdraws electrons (also an activating substituent btw). Since all of those things donate electrons more then H, I chose D. The correct answer is A, because of resonance. The problem is, Im failing to see the resonance there. I can pop the - charge on the phenyl rings, but I cant figure out where to throw the existing bonds.
Thoughts?

Thanks
Whit

Choice D is an intuitive trap answer. as you know, the negative charge on the carbanion is destabilized by electron donating groups such as alkyl groups. thus, H's are pretty good because it doesnt have much donating abilities (although there is still some sigma bond donation from the H's). the best answer however, is A, because benzene rings (electron withdrawing) can delocalize the carbanion's negative charge, stabilizing it. in addition, EDG and EWG are applied in aromatic ring addition and substitution but is not exclusive to this area of ochem. in this case, it applies.