quick test q's

[queenskillers]

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hey guys i got a couple q's:

if an nonbonding electron pair in an aromatic ring participates in the aromatic ring (ex pyrole), is it sp3 and parallel to the plane (equatorial) wheras as if the pair does not participate in the ring (pyridine) it is sp2 and perpendicular (axial) (its all in the aromatic chapter p. 323 in kaplan white book). is this stuff accurate?

would butanol with heat/H+ yield 1 butene? (should it yield 2 butene because E1 has favors zaitsav more subst product)

in neopentylbromide with HBr/H+, the answer is 2 bromo-2methyl butane. i understand that there is a shift bit i dont understand how (is this an SN1 and how so since neopentylbromide is a primary halide?)

and if anyone has time, can they give the resonance rules in priority since i keep having trouble with establishing the correct lewis structure and then rearranging it to its correct resonant structures

thanksss

 

[queenskillers]

anybody?

 

[jpark9]

The nitrogen atom in pyrrole is sp3 hybridized because 3 bonds to other atoms (2C, 1H) and 1 pair of nonbonding electrons. However, in pyridine there is a double bond between 1 of the ring Carbon and the nitrogen which makes it a sp2 hybridization. The nonbonding electrons in pyrrole contributes to the aromaticity so it is conjugated with the double bonds in the ring. Imagine the electron cloud standing up on the ring. The nonbonding electrons in pyridine however doesn't contribute to the aromaticity so it sticks out of the ring and thus it is a Lewis base (e- donator).

for butanol, you are right. It favors more substituted double bond formation so the major product will be 2-butene.

To be honest, I do not have a good answer to your third question. but if the correct answer is what you posted, then H+ must add first to the bromine on the neopentane and now since -BrH+ carries a + charge, it must be even better leaving group than bromine.

Can you specify your question a little more? like give an example you are having trouble with such as resonance structure of nitrobenzene or something?

I hope it helps. Good luck!

 

[queenskillers]

thank you for your post i really appreciate it...........as far as the resonance q's go, i guess i wanted to know how to determine which resonance q is major (for example from I think the priority is to make sure each atom has a stable octet, and then look for the one with the least amount of formal charges and then see if the formal charges amount to a net charge of 0, and then finally the seperation of formal charges/neg charge on more electronegative atom)

 

[jpark9]

You are absolutely correct about the priority of choosing the major resonance contributor. at least I agree 🙂. Good luck!!

 

[Flipper405]

Regarding the neopentyl bromide question, I think it is SN1, and there is a methyl shift. Take a look at this (imagine that the reaction is HBr/H+ rather than what's in the picture.

That site also mentions : "Furthermore, β-alkyl substitution also decreases the rate of substitution, as witnessed by the failure of neopentyl bromide, (CH3)3CCH2-Br (a 1º-bromide), to react." (with regard to its SN2 reactivity)