Qvault ochem #8 quetions

[mario0923]

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the image 11 is
when you have 2-iodine-4-methylpentane reacting with NaOMe
I thought since it is a strong base (Chad said negative charge on an atom other than oxygen tends to be strong nucleophile, so I thought negative charge on oxygen will make the molecule a strong base) reacting with secondary group, the reaction is E2 not Sn2
But why Qvault says it is a strong nucleophile reacting with a good leaving group thus Sn2...???

the image 12 is
heat of combustion question. I thought heat of combustion is greater when alkane is less stable. Of ABCDE, when you draw the chair conformation, B has both groups on axial whereas C has 1 group in axial and 1 group in equatorial. B has 2 groups on axial so it must be the least stable so greatest heat of combustion, am I wrong?

In both of the questions the answer choice that I chose has the majority so is Qvault wrong?

thanks guys🙂

 

[n27s]

The first one is not a good question, i might not be right but i think it could be sn2 or e2 since its a secondary halide with a strong nu/strong base

Both groups in B would be in equatorial, you always start by placing the bulkiest substituent at equatorial to find the most stable chair conformation.

 

[Kratoz24x]

If it can be SN2 or E2 it's gonna prefer SN2 because SN2 is faster

 

[mario0923]

so the first question is wrong?
it should be Sn2 right?

for 2nd question, so I need to look at the most stable conformation? not the least stable conformation possible?
B can technically exist in both axial but it is dominantly in equatorial so that is why C (which has 1 axial and 1 equatorial) is the answer?

 

[GoBlue24]

If what you say is true, kratoz, doesn't that mean every secondary halide with a strong nug/strong base will always under go SN2 unless under protic solvent....?

For question two, your reasoning is right. The more unstable compound would produce the highest heat of combustion so in this case, the two largest groups should be in axial positions (which is B). IMO.

 

[theleatherwalle]

Hey if memory serves me correct you want the least stable isometric form will have the highest heat of combustion.

 

[n27s]

so the first question is wrong?
it should be Sn2 right?

for 2nd question, so I need to look at the most stable conformation? not the least stable conformation possible?
B can technically exist in both axial but it is dominantly in equatorial so that is why C (which has 1 axial and 1 equatorial) is the answer?

Yes exactly. Although your reasoning is correct, B will exist predominantly as equatorial since there is a bulky substituent and I remember chad saying to always start by assuming the bulky group is in equatorial and then look at everything else to assign them based on their position relative to the bulkiest substituent.