Orgo question

[charmstot]

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Which is the most basic? I don't understand why the answer is D
A. C6H5NH2
B. BrCH2NH2
C. FCH2NH2
D. CH3NH2
E. ClCH2NH2

 

[HBomb]

We're talking about a Lewis base here (lone pair on N). A, B, D, E all have electron withdrawing groups because of electronegativity or the benzene ring, so these molecules are less likely to give up the electrons on N.

 

[Eddie830]

Edit: oops, wasn't playing attention that A contains a benzene 🙄

 

[DrTacoElf]

The resonance possibilities for an aromatic group allow it to serve either as an electron donating substituent when the attached atom needs electrons (carbocation) or as an electron withdrawing substituent when the attached atom has an unshared pair to share (oxygen or nitrogen).

A little more info. This is why A is wrong.

The lone e- pair is less available to act as a lewis base, than it is with the alkyl counterpart.

 

[charmstot]

Correct me if I am wrong, I was under the impression that the benzene would be more stable if an electron pair was lost, therefore making it a better base than the alkyl groups

 

[DrTacoElf]

Correct me if I am wrong, I was under the impression that the benzene would be more stable if an electron pair was lost, therefore making it a better base than the alkyl groups

The whole basis behind why A is wrong is the fact that the lone e- pair on nitrogen is cycled around the ring in different resonance strucutures. Therefore this e-pair which is dontated to something (making it a base) is less available than with the alkyl group attached to the NH2.

 

[charmstot]

Ok, it makes sense now.

 

[HBomb]

The whole basis behind why A is wrong is the fact that the lone e- pair on nitrogen is cycled around the ring in different resonance strucutures. Therefore this e-pair which is dontated to something (making it a base) is less available than with the alkyl group attached to the NH2.

Exactly! Good job DTE. I think you're taking the DAT right around now. This is proof that YOU ARE READY!!!

 

[ems5184]

charmstot,

I think the advice being discussed on here is productive, but I recommend you try my way. Instead of trying to figure out which is most basic, I find it easier to figure out which is least acidic (this is equivalent, but sometimes easier to think in terms of this).

For this problem, if you do it this way, it is obviously the one with the methyl group attached to the amine group, since it has the least amount of substituents to stabilize the positive charge (if the N lost a proton).

Great discussion though everyone!!!

 

[rocknightmare]

i followed that similar reasoning.. i knew everythhing else could accomaidate a gain of electron from the electron withdrawing groups,