DAT destroyer ochem error ?

[ddsshin]

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hello guys.
I am doing ochem problems in destroyer and I think there are couple errors...
28. When you do E2 elimination, you get more substituted alkene. But in here, you get less substituted alkene...

Also # 80. 2-chloro butane undergoes E2 elimination and ends up with less substituted alkene.... this is wrong right?>??


# 90
When you add HCl /ROOR to alkene, Cl should end up with less substituted.
But here... Cl goes to more substituted... Am I wrong????

Also, I have a question to ask..
What is reducing and non-reducing sugar... It says the sucrose is non reducing sugar.

Thanks!

 

[pbure9]

I don't know what version you have so I can't really help you out.

But for number 28 and 80, it could be E2 and be a very bulky base which would go Hoffman, where the double bond goes onto the less substituted side.

As for HCl and ROOR, peroxides only work with HBr.
http://forums.studentdoctor.net/showthread.php?t=508107 (only "proof" i could find, sorry)

 

[joonkimdds]

hello guys.
I am doing ochem problems in destroyer and I think there are couple errors...
28. When you do E2 elimination, you get more substituted alkene. But in here, you get less substituted alkene...

Also # 80. 2-chloro butane undergoes E2 elimination and ends up with less substituted alkene.... this is wrong right?>??


# 90
When you add HCl /ROOR to alkene, Cl should end up with less substituted.
But here... Cl goes to more substituted... Am I wrong????

Thanks!

HCl is diff from HBr.
E2 doesn't always result in more/less subs. It's about bulky and not.

 

[TooQuickSlick]

Is this the 2008 edition, I'm looking at the 2009 and I don't see these questions. BTW a reducing sugar is one that reduces an oxidizing agents (glucose, maltose, lactose, maltose), so sugars that cant reduce are non-reducing.

About the alkene when you do E2 you don't necessarily get the most substituted alkene do to steric hindrance and bulkiness of the base, and theres no rearrangement unlike in E1 which allows carbocation rearrangement.

For number 90 yes Br should add anti-markovnikov since it is in the presence of peroxide.

 

[nze82]

hello guys.
I am doing ochem problems in destroyer and I think there are couple errors...
28. When you do E2 elimination, you get more substituted alkene. But in here, you get less substituted alkene...
NO! This is not an error! Although, I don't have the actual question in front of me, I can tell you that E2 reactions can give two types of products:
1)Saytzeff product = More substituted Alkene
2)Hoffman product = Less substituted Alkene
The Hoffman product is favored over the Saytzeff product, when we use a Bulky base.

Also # 80. 2-chloro butane undergoes E2 elimination and ends up with less substituted alkene.... this is wrong right?>??
Same as above!
# 90
When you add HCl /ROOR to alkene, Cl should end up with less substituted.
But here... Cl goes to more substituted... Am I wrong????
Only HBr adds Antimarkovnikov in presence of ROOR. HCl adds in a markovnikov fashion, even when ROOR is present (This is clearly mentioned in the solution section of destroyer! I suggest you look at it)
Also, I have a question to ask..
What is reducing and non-reducing sugar... It says the sucrose is non reducing sugar.
Reducing sugars are sugars with one -H and one -OH group attached to the anomeric carbon (Carbon#1). For example, Glucose is reducing sugar. Non-reducing sugars on the other hand, don't have the -OH group; instead, an -OR group attached to the anomeric carbon. For example, Sucrose is a non-reducing sugar.

Thanks!

Hope this helped!

 

[ddsshin]

It says that grignard is powerful base and remove proton from any compound with an acidic proton. Isn't methylamine very basic???
Why is ether? isnt ether more acidic than amine???

 

[nze82]

It says that grignard is powerful base and remove proton from any compound with an acidic proton. Isn't methylamine very basic???
Why is ether? isnt ether more acidic than amine???

Ethers have a general structure as follows:

R-O-R

Do you see an acidic proton here? No!!

 

[ddsshin]

ok, in destroyer roadmap it says that CrO3, pyr CH2Cl2 acts same as PCC which oxidize 1 OH into aldehyde. Is this true because in my text CrO3 function same as KMno4, K2Cr2O7, Na2Cr2O7....
Also, why ether has more acidic proton than amine????

 

[nze82]

ok, in destroyer roadmap it says that CrO3, pyr CH2Cl2 acts same as PCC which oxidize 1 OH into aldehyde. Is this true because in my text CrO3 function same as KMno4, K2Cr2O7, Na2Cr2O7....
Also, why ether has more acidic proton than amine????

You keep saying ether has a more acidic proton than amine and I just showed you ethers don't have acidic protons!!
In fact, ethers are only slightly more acidic than alkynes!
Not sure why you keep repeating that question.😕

 

[bbwreckerz04]

In terms of acidity I think ethers are more acidic than amines due to the fact that amines are the most basic organic compounds therefore the bigger pKa so weaker acid.

 

[loveoforganic]

Just a note for E2 reactions - no one has yet mentioned that they proceed either syn- or anti-periplanar (preferably anti). This is a critical thing to consider.

 

[tdkyun]

For example, Lactose is a non-reducing sugar.

Lactose is a REDUCING sugar.

 

[nze82]

Lactose is a REDUCING sugar.

You're right! Tnx for correcting me!

 

[192LT192]

It says that grignard is powerful base and remove proton from any compound with an acidic proton. Isn't methylamine very basic???
Why is ether? isnt ether more acidic than amine???

grignard is such a powerful base that it can even rip off hydrogens from an amine. amine can act as an acid here. also, ethers do not have acidic protons. those are the only things (or one of the only things) that do not mess up a grignard.

 

[192LT192]

hello guys.
I am doing ochem problems in destroyer and I think there are couple errors...
28. When you do E2 elimination, you get more substituted alkene. But in here, you get less substituted alkene...

Also # 80. 2-chloro butane undergoes E2 elimination and ends up with less substituted alkene.... this is wrong right?>??

Thanks!

okay so there are strong bases to look for to see if the reaction is going to go E2:

1) NaNH2
2)CH3ONa
3)C2H5ONa
4)KOH or NaOH
5)(CH3)3CONa
6)LDA

for the last 2 bases, the resulting alkene would be the Hofman product (the less substituted one) when used because they are big and bulky. The zaitsev product (more substituted) would result from using the first 4 bases and doing an E2.