Destroyer org # 44

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Mstoothlady2012

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People who have destroyer & went over the org section....how is the answer e for # 44? I searched the whole Mcmurry book & my notes from org II....& the only answer i get is that e is not aromatic? How is e aromatic? SN1 prefers tertiary carbocation & if you have aromatic carbocation then that is great! hence i would choose B has the answer....because we would get tertiary carbocation & B is definitely aromatic....I am so frustrated!!

Help please! thanks
 
A benzyl carbocation is stable, a phenyl is not so B is a poor choice. E is aromatic because it follows all the rules of aromaticity (all carbons sp2, conjugation, huckle's rule, etc.)

Let me know if this doesnt make sense.
 
Heptyne is not aromatic (becuase in order to be aromatic, you need to have a ring). If you look at choice E for #44, the carbon bonded to Cl is SP3. One of the rules for aromaticity is an empty p orbital available for conjugation, which the sp3 carbon can not provide. For this reason, all carbons in the ring have to be sp2. Does that help?
 
sorry i changed my question....i didnt mean heptyne... i just got confused on naming hehe...basically i am asking will something that looks like choice e but without Cl & no positive charge be aromatic? or is 1,3,5 cycloheptene aromatic?
 
No, because there will be one carbon that is sp3 hybridized and will break the conjugation.
 
If you look at choice E for #44, the carbon bonded to Cl is SP3. One of the rules for aromaticity is an empty p orbital available for conjugation, which the sp3 carbon can not provide. For this reason, all carbons in the ring have to be sp2.
Don't mean to drag this one back from the dead, but I could use some clarification - could you explain to me the process you used to determine that the C attached to the Cl is sp3? That would help me a ton!
 
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