Organic Chem Questions - please help!!

[chb64]

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Hey all,

I was wondering if anyone could please help me out with some of these organic chemistry questions. These are actually from a practice OAT exam, but the content is very similar to the MCAT organic chem (I'm currently studying for both!). My exam is on Wednesday July 31 but I haven't been able to get help elsewhere so far 🙁 Can someone please help me out?

Exam questions are here: https://www.ada.org/oat/oat_sample_test.pdf

These questions are confusing me:

84. not sure which NMR signals go to which carbons

85. why not A??? This one would form a tertiary alkyl radical...

86. I only can count 6 pi orbitals (answer D). Where's the 7th??

89. What is the difference between aromaticity and resonance??

91. Can't answer choice B work too? I thought you just needed acid/water to form the alcohol from an alkene.

93. In my mind, the only way I can see a N atom with 2 pi bonds and 2 sigma bonds is if it's double bonded to two Oxygens (two double bonds). Making that sp hybridization

100. Just a general question, how do you tell which hydrogen is most acidic?

Any orgo gurus out that that can spare a few moments to help me out I'd hugely appreciate it! Thanks!!

 

[Synapsis]

84. Since this is H-NMR, the signals apply to hydrogens, not carbons. Narrow it down to B or E because only they have 14 hydrogens, like the question says. Then pick B because two hydrogens are equivalent ( the tertiary ones) and the other 12 are equivalent. So you get two signals. One from the 2H's, one from 12H's.

85. Not entirely sure. I would pick A. Since methyl radical is least stable, its proton is the hardest to abstract.

86. The answer to this should be 6. The N lone pair should be in an sp2 hybridized orbital.

89. A quick way to find if something is aromatic is to see if it has 6, 10, or 14 pi electrons in a ring. It can actually be more (the rule is 4n+2, where n =1, 2, 3..) but usually it's one of those. The electrons have to be in a ring, and the ring has to be planar. Resonance just refers to delocalization of pi electrons. So aromaticity implies resonance, but resonance doesn't imply aromaticity.

91. If you pick B, you will first protonate the alkene, giving a carbocation. Water will attack that carbocation and you get a cycloalkane with one alcohol instead of two. Of those, only osmium tetroxide can give you a cis diol (good thing to just memorize).

100. There's a lot of things to consider, but the most important is stability of the conjugate base. The conjugate base is more stable if it has resonance forms that can delocalize the negative charge (delocalized charge is always more stable - concentrated charge is highly unfavorable). Check out #3. If you take that H, you get three resonance forms. One with a carbanion (minor), one with the carbonyl on the left, one with the carbonyl on the right. That proton likely has a pKa of about 9, making it relatively acidic compared to most others.

Hope this helps some.

 

[chb64]

Thank you so much!! Much appreciated 🙂

For 86 - I'm still a bit confused with the hybridization/pi electrons of N. I count 5 pi orbitals. The lone pair on nitrogen exist in a pi orbital too? (so including N, there's a total of 6 pi orbitals?)

 

[Heday]

Thank you so much!! Much appreciated 🙂

For 86 - I'm still a bit confused with the hybridization/pi electrons of N. I count 5 pi orbitals. The lone pair on nitrogen exist in a pi orbital too? (so including N, there's a total of 6 pi orbitals?)

Yes, each carbon has a pi orbital (total of 5) + the nitrogen's pi orbital (totaling 6). The fact that each of these atoms have pi orbitals (in the same orientation) is what allows pyridine to be aromatic.

Nitrogen has three groups attached to it, correct? Two carbons and a lone pair so it's hybridization is sp2 (one s orbital and two p orbitals are hybridized, giving three new sp2 orbitals of "equal energy" and a fourth, leftover pi orbital). Each of the carbons are bonded to nitrogen house their electrons in sp2 hybridized orbitals. The pi orbital is involved with the double bond. This leaves a sp2 hybridized orbital for the remaining electrons.

Edit: This is a terrible question. Hybridization isn't real.

 

[FeinMS]

Yes, each carbon has a pi orbital (total of 5) + the nitrogen's pi orbital (totaling 6). The fact that each of these atoms have pi orbitals (in the same orientation) is what allows pyridine to be aromatic. However, the lone pair on nitrogen also exist in their own pi orbital, giving us a total of 7 pi orbitals in pyridine.

No, the lone pair resides in sp2 orbital. According to you, N is sp hybridized and should be linear in shape. That will cause too much ring strain.
6 is the right answer

 

[Heday]

No, the lone pair resides in sp2 orbital. According to you, N is sp hybridized and should be linear in shape. That will cause too much ring strain.
6 is the right answer

Yeah, you're right. I had edited my response right after submitting it (while researching), I guess you had already submitted your response. However, hybridization is only a tool to help us predict certain characteristics of molecules. Can't really say that orbitals really hybridize at the atomic level.