polarity

[datdat]

this is from topscore2. q71

which one is less polar?(so that more soluble in pentane)

1. H2N-benzene-NH2
2. H3CO-benzene-OCH3
(both are attached in 1,4 potision)

answer is 2. but don't understand why...looks like they don't have net dipole for me

anyone knows why?

---

Members don't see this ad.

 

[americanpierg]

this is from topscore2. q71

which one is less polar?(so that more soluble in pentane)

1. H2N-benzene-NH2
2. H3CO-benzene-OCH3
(both are attached in 1,4 potision)

answer is 2. but don't understand why...looks like they don't have net dipole for me

anyone knows why?

not quite sure. anyone? only reason i got it right was cuz all the other choices were obviously wrong, and nh2 was more water soluable than ethers. i would like to know the right reasoning jus in case.

 

[UCB05]

not quite sure. anyone? only reason i got it right was cuz all the other choices were obviously wrong, and nh2 was more water soluable than ethers. i would like to know the right reasoning jus in case.

Notice that N and O are both normally polar elements of functional groups, but N is attached to -H whereas O is attached to -CH3, a stronger electron-donating group, which decreases the polarity of the methoxy group compared to the amine group. With both choices having the same attachments and orientation on benzene, the less polar molecule would be #2

 

[americanpierg]

Notice that N and O are both normally polar elements of functional groups, but N is attached to -H whereas O is attached to -CH3, a stronger electron-donating group, which decreases the polarity of the methoxy group compared to the amine group. With both choices having the same attachments and orientation on benzene, the less polar molecule would be #2

If the same effect is produced on both sides, they cancel each other out and the net would be 0 anyways. If you're right and even if the CH3 does donate electrons to oxygen, that would make the methoxy group more polar than it already is (giving an already electronegative atom more electrons = making the overall effect even more polar)

Plus NH3 is is a stronger electron-donating group than OCH3 so that reasoning doesnt make sense if youre looking at their effect on the benzene. The only reasoning i can think of is that the NH3 groups make the benzene more "negative" compared to the ether.

 

[UCB05]

If the same effect is produced on both sides, they cancel each other out and the net would be 0 anyways. If you're right and even if the CH3 does donate electrons to oxygen, that would make the methoxy group more polar than it already is (giving an already electronegative atom more electrons = making the overall effect even more polar)

Plus NH3 is is a stronger electron-donating group than OCH3 so that reasoning doesnt make sense if youre looking at their effect on the benzene. The only reasoning i can think of is that the NH3 groups make the benzene more "negative" compared to the ether.

I was thinking more along the lines of the polarity of the electronegative atoms in each and how that would affect interaction with the nonpolar pentane solvent. In this case, the N and O being electronegative will adversely affect nonpolarity. The CH3 in the methoxy group has an electron donating effect into O, which will in turn make it less electrophilic than N which is only differently attached to H's. This reduced electrophilicity on the electronegative "polar" element is what I am guessing makes the molecule as a whole less polar in pentane, despite teh fact that both O and N are electronegative. Given that the two choices have the same attachment locations to benzene, I would choose to ignore the central benzene as a consideration of polarity.

Not sure if i'm getting my idea across. No net molecular polarity due to symmetry, little difference in substituent ED into the benzene core, but reduced substituent electrophilicity makes the dimethoxy molecule less polar in nonpolar solvent.

 

[americanpierg]

I was thinking more along the lines of the polarity of the electronegative atoms in each and how that would affect interaction with the nonpolar pentane solvent. In this case, the N and O being electronegative will adversely affect nonpolarity. The CH3 in the methoxy group has an electron donating effect into O, which will in turn make it less electrophilic than N which is only differently attached to H's. This reduced electrophilicity on the electronegative "polar" element is what I am guessing makes the molecule as a whole less polar in pentane, despite teh fact that both O and N are electronegative. Given that the two choices have the same attachment locations to benzene, I would choose to ignore the central benzene as a consideration of polarity.

Not sure if i'm getting my idea across. No net molecular polarity due to symmetry, little difference in substituent ED into the benzene core, but reduced substituent electrophilicity makes the dimethoxy molecule less polar in nonpolar solvent.

in terms of the OCH3 itself, the O would be electrophilic, but if you're talking about the interaction between the OCH3 and any outside molecules, in this case pentane, then the O would be nucleophilic. Donating electrons to the O would increase its nucleophilicity, the same effect as donating electrons to the benzene ring itself.

In essence, what you are suggesting would be adding electron density to an electronegative atom, which would make it more "negative" and therefore even more polar.


anyways i think im right anyways. the NH2 is a stronger activating group than OCH3, because N is less electronegative than O, and therefore donates more electron density to the benzene, so the benzene ring would be more "negative" while the two NH2 groups are more "positive".


BTW even though a CH3 group does donate electron density a benzene ring, doesnt mean that its more polar than when an H is attached instead. H2O is a whole lot more polar than (CH3)2O. the difference in electronegativity of O and H much, much larger than the difference between the O and CH3, even if the CH3 slightly donates electron density to the O.