Destroyer Q 31

[BiomajorPreDent]

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Destroyer Q 31:

This is confusing me..

I actually have 2 questions:

1. I hate skeleton structures but I am pretty sure I am reading this right.

Choice A: 2-IodoPENTANE is reacting with NaOCH3 in CH3OH and the answer is showing it yields 2-BUTENE...shouldnt it be 2-Pentene?

Anyway, I just noticed that problem as I was typing this question. My actual question was..

2. is it an elimination reaction because its a strong base? Because it is a 2ndary alkyl halide that I know can go either way but its in a protic polar solvent. So the strong base over-rules the solvent?

 

[nze82]

Destroyer Q 31:

This is confusing me..

I actually have 2 questions:

1. I hate skeleton structures but I am pretty sure I am reading this right.

Choice A: 2-IodoPENTANE is reacting with NaOCH3 in CH3OH and the answer is showing it yields 2-BUTENE...shouldnt it be 2-Pentene?

Anyway, I just noticed that problem as I was typing this question. My actual question was..

2. is it an elimination reaction because its a strong base? Because it is a 2ndary alkyl halide that I know can go either way but its in a protic polar solvent. So the strong base over-rules the solvent?

Whenever you see a strong base like NaOCH3 with the corresponding Alcohol (CH3OH), the reaction proceeds via E2.

 

[BiomajorPreDent]

okay cool

does anyone else have destroyer 09 handy? took look at the product formed butene vs pentene?

 

[nze82]

okay cool

does anyone else have destroyer 09 handy? took look at the product formed butene vs pentene?

It should be 2-Pentene. Elimination reactions don't add/remove C from the initial reactant, they simply convert alkanes to the corresponding alkenes.

 

[Tommy43087]

your reading it wrong its 2-iodobutane..I had to check cuz your question made no sense at all....better brush up on counting carbons right!

 

[BiomajorPreDent]

 

[Tommy43087]

First off, your doing something wrong for sure because you put 2-iodopentane but u have 2-bromopentane posted up. Second i am simply reading off what #31 is on my destroyer(2009) it is 2-iodobutane. Maybe we have different versions, idk.

 

[BiomajorPreDent]

Yes yes I know I accidently put Br on there it shud be an I. Here its fixed now.. BUT..my point is that there are clearly 5 carbons in the structure and this is the exact same skeleton structure in the destroyer book (2009)

 

[Tommy43087]

If we are talking about DAT destroyer 2009 edition Organic chemistry section than yes it is definitely 2-iodobutane i have no doubt about it. Maybe some crazy thing happened where you got a different copy/version than me or maybe your reading it wrong, but i doubt you have been reading it wrong this many time so i really dont know.

 

[BiomajorPreDent]

why dont you photocopy your page or something and put it up here, I drew what I saw, im curious to see yours

 

[Tommy43087]

why dont you photocopy your page or something and put it up here, I drew what I saw, im curious to see yours

Um i actally dont have a photocopier i never felt i needed it but its just simply 2-iodobutane, like i said maybe sometype of misprint happened on your version because it is supposed to be 2-iodobutane because the answer is 2-butene.

 

[Shinpe]

my 2009 31 ochem also shows 2-Iodobutane

 

[dennisDDS]

OP is not crazy

mine has 2-iodopentane too --> 2 butene

probably just a messed up batch.

 

[dennisDDS]

here you go OP lol..

 

[BiomajorPreDent]

dennis!!! you are the man!

Thank you. So the whole point of this craziness is to figure out if its a mis-print.

I am worried that I am studying stuff that is incorrect. I try to check everything against my textbook but every once in a while stuff like this pops up...🙁

 

[dennisDDS]

dennis!!! you are the man!

Thank you. So the whole point of this craziness is to figure out if its a mis-print.

I am worried that I am studying stuff that is incorrect. I try to check everything against my textbook but every once in a while stuff like this pops up...🙁

No problem! as long as you can say that you feel something is wrong then you must be studying good👍

Im going through Ochem Odyssey..

 

[joonkimdds]

here you go OP lol..

Can someone explain to me the product for E?
CH3OH is a strong base and we see tertiary halide so shouldn't it be E2 reaction?

But the answer sheet says it undergoes SN1 reaction.

Someone please help me~~~!!!

 

[nze82]

Can someone explain to me the product for E?
CH3OH is a strong base and we see tertiary halide so shouldn't it be E2 reaction?

But the answer sheet says it undergoes SN1 reaction.

Someone please help me~~~!!!

CH3OH is not a strong base (strong Nuc/Base almost always carry a negative charge). So, since we don't have a strong Nuc/Base and because the Carbon bearing the leaving group is tertiary, the reaction must be first order. Now, how do we determine whether it's SN1 or E1.
Elimination reactions are almost always accompanied by heat (promotes the departure of the leaving group), which is absent here. So, the reaction must be SN1.

 

[joonkimdds]

CH3OH is not a strong base (strong Nuc/Base almost always carry a negative charge). So, since we don't have a strong Nuc/Base and because the Carbon bearing the leaving group is tertiary, the reaction must be first order. Now, how do we determine whether it's SN1 or E1.
Elimination reactions are almost always accompanied by heat (promotes the departure of the leaving group), which is absent here. So, the reaction must be SN1.

I thought anything with OR or OH are considered strong base, am I wrong?

 

[nze82]

I thought anything with OR or OH are considered strong base, am I wrong?

I always considered species with a negative charge strong Nuc/Bases and that helped me distinguish between 1st order and 2nd order reactions. Water molecule is really H-O-H; it contains the -OH component you're talking about; yet, it's not a strong Nuc/Base!

 

[joonkimdds]

I always considered species with a negative charge strong Nuc/Bases and that helped me distinguish between 1st order and 2nd order reactions. Water molecule is really H-O-H; it contains the -OH component you're talking about; yet, it's not a strong Nuc/Base!

but isn't OH negatively charged and thus strong nuc/base?

For example, from the above example, CH3OH becomes CH3O(negative) + H(positive) and the CH3O(negative charge) is being substituted.

 

[nze82]

but isn't OH negatively charged and thus strong nuc/base?

For example, from the above example, CH3OH becomes CH3O(negative) + H(positive) and the CH3O(negative charge) is being substituted.

Like I said buddy, I don't want to tell you something I'm not sure of. Personally, if I see the negative charge I consider it strong Nuc/Base (if you look at E2 reactions they always show something like RONa/ROH, which essentially means that RONa breaks into RO- and Na+, so you can consider RO- a strong base) otherwise, I'd go with week Nuc/Base!
Sorry! I wish I could provide you with a more accurate explanation.

 

[UCB05]

CH3OH is a weak base
CH3O- is the strong base

If all they give is the unprotonated acid, assume weak base. It's only strong when the question gives something like CH3OH/CH3O-Na+. The salt increases the aqueous equilibrium concentration of the ionized species.

CH3OH by itself has such a low dissociation constant that it's nearly all protonated and therefore weak, as is the case in this question.