OChem Q from Achiever

[Dentist2be]

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Hi, I am having trouble figuring out which one is a weaker acid and which one is stronger.. I think I understand the concept, but when it comes to the problem solving, I get stuck. 🙁 Could any of you explain why the answer is supposed to be E?


Q. Which statement CORRECTLY explains the following acid-base reaction at equilibrium?

CH3H2C:- + HOC(CH3)3 D CH3CH3 + -:Ö:C(CH3)3

A. Weaker acid, HOC(CH3)3, and weaker base, CH3H2C:-, will be present in abundance.
B. Stronger acid, HOC(CH3)3, and stronger base, CH3H2C:-, will be present in abundance.
C. Weaker acid, CH3CH3, and stronger base, -:Ö:C(CH3)3, will be present in abundance.
D. Stronger acid, CH3CH3, and weaker base, -:Ö:C(CH3)3, will be present in abundance.
E. Weaker acid, CH3CH3, and weaker base, -:Ö:C(CH3)3, will be present in abundance.


And the second question:

Q.Which product will be formed in abundance through the following halogenation reaction?

(CH3)2CHCH2CH3 + Cl2 ---hv, 35degrees C---> ?

A. (CH3)2CClCH2CH3
B. (CH3)2CHCHClCH3
C. (CH3)2CHCH2CH2Cl
D. (CH3)2CHCH2CH3
E. All of the above

The answer should be C, right? (since free radical chlorinations are most likely to replace primary hydrogens.) But it says the answer is A, which is pretty opposite. Wanted to confirm that the answer given was wrong.

Thank you for your help!

 

[dat_student]

CH3H2C:- + HOC(CH3)3 D CH3CH3 + -:Ö:C(CH3)3

Re-write the above rxn...

And the second question:

Q.Which product will be formed in abundance through the following halogenation reaction?

A. (CH3)2CClCH2CH3
B. (CH3)2CHCHClCH3
C. (CH3)2CHCH2CH2Cl
D. (CH3)2CHCH2CH3
E. All of the above

The answer should be C, right? ...

Wrong!
A is the right answer because Cl is put on the tertiary radical which has the lowest activation energy.

 

[BenignDMD]

Wrong!
A is the right answer because Cl is put on the tertiary radical which has the lowest activation energy.

Yea, the answer is A because this is not a chlorination in the presence of UV or peroxides. If it were then this would follow Anti-Markonikov, but its not

 

[dat_student]

Yea, the answer is A because this is not a chlorination in the presence of UV or peroxides. If it were then this would follow Anti-Markonikov, but its not

Wrong!

The carbon structure is completely saturated. I can't think of any way to halogenate without forming radicals (i.e. in the absence of UV/peroxides).

 

[Dentist2be]

Re-write the above rxn...

Sorry, just change the 'D' in the middle of the equation to an arrow pointing both directions.

 

[Dentist2be]

Yea, the answer is A because this is not a chlorination in the presence of UV or peroxides. If it were then this would follow Anti-Markonikov, but its not

Actually it IS a chlorination in the presence of UV. I left out the equation:

(CH3)2CHCH2CH3 + Cl2 ---------> ?

(there are 'hv' above the arrow and 35 degrees celsius below the arrow.)


Sorry guys. I should have double-checked what I wrote. I think I was too tired/sleepy last night. 🙁

 

[taloddar2002]

The correct answer is E for the first question. Acid and Base reactions will want to produce the weakest acid and weakest base. CH3CH2- is an extremely strong base so it will pull the hydrogen off tert-butanol. This will lead to a weaker acid (CH3CH3) and a weaker base (-OC(CH3)3) than what you originally started with. The negative charge is much more stable on the oxygen atom than the carbon atom.

The correct answer for question 2 is A. A chlorine radical will pull of the hydrogen on the tertiary carbon to produce the most stable radical.

 

[Dentist2be]

The correct answer is E for the first question. Acid and Base reactions will want to produce the weakest acid and weakest base. CH3CH2- is an extremely strong base so it will pull the hydrogen off tert-butanol. This will lead to a weaker acid (CH3CH3) and a weaker base (-OC(CH3)3) than what you originally started with. The negative charge is much more stable on the oxygen atom than the carbon atom.

Thanks for your explanation! I think it's very clear. 🙂

The correct answer for question 2 is A. A chlorine radical will pull of the hydrogen on the tertiary carbon to produce the most stable radical.

I am still not convinced because of this: "Free-radical chlorination reactions are likely to replace primary hydrogens because of their abundance, despite the relative instability of primary radicals (pg.355 Kaplan blue book)"

Could anyone please clarify this?

 

[armorshell]

Thanks for your explanation! I think it's very clear. 🙂



I am still not convinced because of this: "Free-radical chlorination reactions are likely to replace primary hydrogens because of their abundance, despite the relative instability of primary radicals (pg.355 Kaplan blue book)"

Could anyone please clarify this?

Notice that a radical halogenation is a multiple step process, not a single step like an SN2...

So in between steps you have suitable time for a radical re-arrangement to the more energetically stable form.

Also note that a fair amount of the n-chloro alkane will also be produced, just you'll get mainly the highly subsituted one...

 

[Dentist2be]

I still have a problem with this cholrination thing.. Here is another very similar question to the second one from my original post. Could anyone tell me why Cl is on the primary carbon this time? (can't draw benzene rings here.. I'll just write it out)

methylbenzene + Cl2 ---hv---> chloromethyl benzene

Feel like I really need to know this stuff. Can't afford to ignore..
Thanks!!

 

[armorshell]

I still have a problem with this cholrination thing.. Here is another very similar question to the second one from my original post. Could anyone tell me why Cl is on the primary carbon this time? (can't draw benzene rings here.. I'll just write it out)

methylbenzene + Cl2 ---hv---> chloromethyl benzene

Feel like I really need to know this stuff. Can't afford to ignore..
Thanks!!

You get the primary radical on the methyl benzene because it's a resonance stabilized radical, and is very very low energy

 

[RockstarDMD]

I still have a problem with this cholrination thing.. Here is another very similar question to the second one from my original post. Could anyone tell me why Cl is on the primary carbon this time? (can't draw benzene rings here.. I'll just write it out)

methylbenzene + Cl2 ---hv---> chloromethyl benzene

Feel like I really need to know this stuff. Can't afford to ignore..
Thanks!!

When you see hv think radical conditions, (anti-Markovnikov addition).In other words the Chlorine will attack the molecule to form the most stable radical. 😉

 

[Dentist2be]

When you see hv think radical conditions, (anti-Markovnikov addition).In other words the Chlorine will attack the molecule to form the most stable radical. 😉

yeah.. but there is also hv in the reaction for the first question. (I edited the question so it's correct now. Sorry for the confusion) Is it something to do with the temperature? Or I guess it's because a benzene is more stable and low in energy as armorshell said?

 

[RockstarDMD]

yeah.. but there is also hv in the reaction for the first question. (I edited the question so it's correct now. Sorry for the confusion) Is it something to do with the temperature? Or I guess it's because a benzene is more stable and low in energy as armorshell said?

You are trying to compare apples to oranges. The first question with heat chose the more substituted carbon to chlorinate, this is because in contrast with bromination which you are probably use to seeing, chlorine is much less selective.

The most recent question is basically acting like a NBS rxn (N-bromosuccinimide). However, this is tipically not a common rxn. I would need to see the structure to give you an exact answer. 👍 the purpose of this type of question is for you to dig deep for the answer which it looks like you are doing.

 

[Law0830]

Hey guys i cant agree with any of you here for that answer. I did some reasearch on this question and saw that Cl2 is not as selective as Br2 thus in this specific question the major product would be the chlorination of one of the 2 methyl groups on the left. Look anywhere and you will see that the tertiary position for this problem is not the correct answer. Their are 6 equal hydrogens on the methyls that is the reason why it goes on one of them.

 

[Mpower95]

Can someone please further explain this? These questions are from Achiever Orgo Test 3: # 81 (the first question) and #99 (second question).

I have always had the understanding that Cl2 + UV conditions = anti-Markovnikov addition. This logic works for question #81 with the chloromethyl benzene answer, but not for #99 with the saturated carbons and Cl2 + UV + 35 degrees C. Temperature/kinetics promote Markonikov for radical conditions?....really lost. Thanks for the help.

 

[Uracil]

ok, guys, here it is!

Facts:

* Bromine radicals are less reactive than chlorine radicals.
* Chlorination is LESS selective.
* Bromination IS selective for the R-H that gives the most stable radical.
* ORDER OF reactivity : F2 > CL2 > Br2 > I2 (only Br and Cl are used in the laboratory)
* Radical is indeed always ANTI-Markov

ok, so since Chlorine is much more reactive than Bromine, we can find 1-chloro____, 2-chloro____, 3-chloro____(whichever R chain is..)

The question is asking about which one will be formed in ABUNDANCE, meaning that you must choose the one that had the most stable radical. Remember that you are not taking away the fact that the other choices do not exist. They are formed but just not in abundance!

In exam krackers (which I listen to on my iPod ) say that the selectivity of the radical reactions can be predicted mathematically based on a combination of factors. If we use radical chlorination we could find 1-chloro, 2-chloro, 3-chloro, 4-chloro, etc. because Chlorine is LESS selective and more reactive and it's all over the place!
If we experimentally do a radical chlorination reaction, we would find that :
1- chloropropane [ for example ] (44% in abundance experimentally)
2- chloropropane (56% in abundance experimentally)

See? The most stable chlorine radical, that's the one in ABUNDANCE, but does NOT imply that the other forms are not part of the products.

*Armorshell is totally right.

 

[STACM]

I dont understand how CH3CH2- is considered a strong base. I thought OR-, OH-, or NH2 are the only ones considered strong?