haloalkane

Consideration of SN1 and SN2 includes：
1. structure of substrate around the carbon which is being attacked
2. concentration and reactivity of the nucleophile
3. stability of leaving group
4. nature of the solvent

The mechanism of (a) is SN2 while that of (b) is elimination.

Explanation:
Both (CH3)2CHONa and CH3ONa is an alkoxide anion, they are strong nucleophile and strong Lewis base(donate electron pair).

Since CH3Br and (CH3)2CHBr are haloalkane and secondary haloalkane respectively, SN2 reaction or elimination is feasible mechanism rather than SN1.

Now consider the structure of substrate around the carbon which is being attacked, in (a), elimination is impossible to take place due to the electropositive carbon atom does not have alkyl groups. In (b), (CH3)2CHBr is secondary haloalkane in which a carbon-carbon double bond can be formed by breaking any one C-H bond of the two methyl groups. Hence, (a) proceed in SN1 only while (b) undergoes elimination in majority.

2008-09-29 11:14:30 補充：
Hence, (a) proceed in SN1 only while (b) undergoes elimination in majority.
寫錯

Hence, (a) proceed in SN2 only while (b) undergoes elimination in majority.
才正確

2008-09-29 11:21:28 補充：
(b) 反應式給出的反應只是其中一個並且係minor反應
朋友仔你話(CH3)2CHBr冇atom可以被eliminate
錯喎你....它是secondary haloalkane, 係可以被eliminate只要nucleophile的basicity夠高
和high temperature

2008-09-29 11:48:14 補充：
http://hk.myblog.yahoo.com/wilson_007_35/article?new=1&mid=883
請上呢個網址睇我的補充解釋
放心無毒，因為呢個係ｙａｈｏｏ　ｂｌｏｇ

下面果位朋友都要黎睇阿

2008-10-01 01:00:19 補充：
網友2003,同埋下面果位朋友
我補充埋呢次
可能我用英文太癈寫出黎有人睇唔明，所以今次畫埋圖俾你地睇，剩係用口講冇用

http://hk.myblog.yahoo.com/wilson_007_35/article?new=1&mid=892

請上去參看

2008-10-01 01:05:56 補充：
講多無謂．．
一條唔太難的問題答到咁複雜．．你冇你咁得閒

http://hk.myblog.yahoo.com/wilson_007_35/article?new=1&mid=892

我的最後解釋係呢個網址

放心冇毒，因為呢個係ｙａｈｏｏ　ｂｌｏｇ

The 1st poster pointed out in the bginning that (a) is Sn2 but then said (a) is Sn1 at the end - not sure if that's a typing error.
Here's my answer:
(a) is Sn2.
(b) is also Sn2 according to the formula given. (b) CANNOT be elimination.
(a) is faster than (b). (b) would be slightly slower.
(a) is better because:
1) (CH3)2CHONa is a better nucleophile than CH3ONa because (CH3)2CHONa side chains are more electron donating (towards the Oxygen atom). So it's more likely to look for something '+ve' to attack, and it will attack at a faster rate than CH3ONa. It's a classic Sn2 reaction with alkyl halides.
Both alkoxides are good nucleophiles - but (CH3)2CHONa is better.
2) It's an Sn2 reaction between that requires the 'lowest energy level' for the reaction to occur. It's not Sn1 because 'Br' cannot just leave even though it's a good leaving group. It needs to be alot more electronegative for it to leave just like that. And that's impossible for Br to do.
(b) is not Sn1 because:
Same reason - you can't have (CH3)2CH+ because Br- cannot leave just like that. This cation requires far too high energy to form, and Br- is not electronegative enough.
(b) is not elimination according to the formula because it's one central carbon with 2 methyl side chains, and a Br leaving group. Nothing can eliminate!
(b) would be slower than (a) due to steric hinderance - because the central carbon of (CH3)2CHBr is quite congested.
Hope that's the answer you're after/

2008-09-29 19:56:04 補充：
I think Poster 1 have misunderstood my answer with regards to 'energy level'. I'm not referring to 'energy level' as what you might think as 'energy barrier'.

2008-09-29 19:56:15 補充：
My reference to 'energy level' is only refering to the fact that the cation FORMED (not TO form) would contain such a high energy that it is in fact 'unstable' and hence, not formed at all.

2008-09-29 19:57:48 補充：
We need to be very careful with the term 'energy level' here - in depth discussions with regards to the so called 'energy level' would include things like 'energy required to stablise an ion', not only the 'barrier'.

2008-09-29 19:58:36 補充：
You are right with regards to bond breaking in your other answer. Nothing wrong with that.

2008-09-29 19:59:45 補充：
However, you did mention here that there might be elimination to form a 'double bond' - but from what I understood is that the substrate has only one carbon, not two - so a double bond formation is highly impossible.

2008-09-29 20:01:24 補充：
Again, when I say 'lowest energy level' - it donesn't refer to energy barrier as a simple idea. It combines the bond breaking and bond forming actions. I should have termed it more like 'net energy' in the system.

2008-09-29 20:03:28 補充：
Poster 1 said:
In (b), (CH3)2CHBr is secondary haloalkane in which a carbon-carbon double bond can be formed by breaking any one C-H bond of the two methyl groups. <-- not possible to form a double bond, not that I know of with one central carbon!

2008-09-29 20:40:14 補充：
Poster 1 - i left a few comments/messages on your blog with regards to this and some other stuff. Please check.

2008-09-30 20:33:50 補充：
No - one must be very, very careful about the terms 'elimination' and 'substitution'.

2008-09-30 20:33:59 補充：
'Elimination' refers to reactions that lead to a double bond formation. 'Substitution' is simply a reaction that leads to Nu- attack and product formation, whether or not double bonds are involved. Check any university entry level textbooks for more.

2008-09-30 20:35:34 補充：
Therefore, it this particular question, there is only ONE central carbon in the substrate, so there's no way for a double bond formation.

You have to remember that an elimination reaction always involve a lost of H+, to form a double bond!

2008-09-30 20:37:10 補充：
It a common mistake to think of simple substitution reactions such as this, with an elimination process - because we think 'oh, Br is being "eliminated" from the starting material!'

One cannot take different terms a their face values - must investigate deeper.

2008-09-30 20:41:00 補充：
You said: (b) 反應式給出的反應只是其中一個並且係minor反應

No, that is also incorrect. The formula gives the MAJOR reaction. The MINOR reaction would be deprotonation due to the alkoxide ion, which is almost non-happening because Br is such a good leaving group, and the alkoxide is very basic (pKa ~16 to 20).

2008-09-30 20:42:22 補充：
This reaction can almost happen at room temperature, under anhydrous conditions - otherwise the alkoxide will be destoryed. Some heat, maybe reflux for 1hr should get it done.

Th reaction would be even quicker if iodide is the leaving group.

2008-09-30 20:47:37 補充：
Please see 'Comments/Suggestions' page for more.

2008-09-30 20:49:35 補充：
So that's why I keep saying both reactions are in fact Sn2.

2008-09-30 20:56:17 補充：
Re: Deprotonation due to alkoxide as a minor reaction (or 'side reaction' as more properly termed) - IF THIS DOES HAPPEN (BECAUSE IT'S IMPOSSIBLE!) you'd end up with a carboanion, which almost doesn't exist, and CH3OH (methanol) as a by-product.

2008-09-30 20:57:12 補充：
The Anion would be protonated to give the starting material again due to the presence of CH3OH or moisture in the solvent. And the alkoxide will also be destoryed.

That is why this reaction MUST be done in dry (anhydrous) conditions.

2008-10-01 19:25:45 補充：
Then again, you still haven't answered the question with regards to reaction rate anyways - which was the main part of the question.


And get your definition of 'energy' clear up before you attack. I don't need to be 'creative' - they are in textbooks.

2008-10-01 19:26:54 補充：
Oh and guess what - you have just created a carbon with 5 bonds in your other answer on your blog. Nicely done indeed.

2008-10-01 19:31:17 補充：
Oh yeah did you see the question we were give - it was:
(b) (CH3)2CHBr + CH3ONa --> (CH3)2CH一O一CH3 + NaBr

So tell me again that's an elimination reaction, with regards to this particular product in this particular question.

2008-10-01 19:37:43 補充：
Yes that's right - you have an intermediate with a 5-bond carbon, and a -ve charge on a 5-bond carbon. Both are impossible! Even if it is elimination it would happen in one step. Check before you attack.