Hey, let me see if I can help clarify some concepts. (You are right to question Kaplan Orgo, btw! There are many mistakes in the orgo section.)
#8) Pi bonds are formed by which of the following orbitals?
2 s
2 p
1 s 1 p
all
answer is 2p but then in the explanation it says that sigma bonds can be 2 s, 2p, or 1s and 1 p overlaps. how can that be if 2p overlaps= Pi bond?
Pi bonds are formed when two p orbitals are parallel to each other (we define 'overlapping' in this way). When p orbitals are parallel, the electrons can be distributed across them to form a pi orbital. The bonds that make up the connectivity of the overall molecule are sigma bonds. These bonds are usually formed when orbitals
hybridize.
When you read that sigma bonds can be made up of s and p orbitals, this is referring to
hybrid orbitals. You musn't literally think of an overlapping s and 2 p orbitals in an sp2 hybridized orbital. An sp2 hybridized orbital is a rearrangement of an s and 2 p orbitals to create a *new* sp2 orbital. For example, take benzene. Each C is sp2 hybridized in order to form 3 bonds (to C, C, H). This means each C has a leftover p orbital, and since benzene is planar, all these p orbitals are parallel to each other, allowing for free flow of electrons between them (creating the pi bonds).
Ch4 KBB ochem, #7
R3C-X ------> R3C+------> R3C-Y
fill in for the arrows (-X, and + Y-)
so this is obviously sn1 and the answer is X: tosylate good LG Y: CN- good nuc
i always thought that good nuc's esp CN- is for SN2, and SN1 can handle weak nuc's?
Good nuc's definitely favor SN2, but remember that STERICS dictate whether an SN2 will actually happen. If the LG is tertiary, such as X in the example above, these are poor reaction conditions for SN2. However, a good, tertiary LG is excellent for SN1 because the carbocation intermediate will be better stabilized. If the reagents for the above rxn were HCN/NaCN, you better believe that once a carbocation is formed, the excellent CN- nucleophile will attack it! Nucleophiles don't discriminate between good electrophiles
🙂
If SN2 and SN1 reactions are equally favored, SN2s always prevail. The reason you probably learned SN1 reactions can only 'handle' weak nuc's was to emphasize that with proper sterics (primary LG), a strong nuc will favor an SN2. It's helpful to remember that SN2/E2 usually occur in basic conditions, and SN1/E1 in acidic or neutral.
Which of the following is the most stable carbanion?
X3C-
X2HC-
XH2C-
H3C-
(CH3)3C-
X- phenyl group
I said H3C- b/c its primary....
I thought for carbanions its 1>2>3 for stability....opposite of carbocations
but the answer is X3C- [x= phenyl group]
Carbanions follow the same stability as carbocations: 3>2>>1>>>Me. Think of radical mechanisms, and how the radical always wants to be on the most substituted atom. With three phenyl groups to shift around the negative charge, the carbanion is way more stable than a carbanion where the negative charge is localized to one atom.
Resonance might be the case, but the way i see it is when something is negative you dont wana push more electrons to it. the phenyl groups are e- rich and exactly do that. Kaplan is wrong man. i know my stuff.
With resonance, you are not stuffing more electrons onto the negative charge; you are shifting the electrons around to delocalize the negative charge. Draw out X3C and arrow the electrons on the central C moving to create a pi bond with one of the phenyls. When you do this, the pi bonds in that phenyl now shift until one of the C's in the phenyl house the extra electrons, and therefore the negative charge. You might be confusing activating/deactivating groups in electrophilic substitution rxns, where we typically treat phenyl substituents as an electron donating group (though it is deactivating).
Hope this helps! Understanding hybridization, SN2/SN1, and resonance are very important topics to understand for the DAT
🙂
