joonkimdds

Senior Member
10+ Year Member
7+ Year Member
Jun 30, 2005
2,780
2
Status
Pre-Dental
According to the road map, primary alcohol becomes carboxylic acid when K2Cr2O7 is used.

Road map 2 shows that R-CH2-CH3 becomes R-COOH.
Does this mean it change ethane into COOH?

and if so, what about the combination of both ethane and primary alcohol such as CH3-CH2-OH?

=============================================

My 2nd question is...
which is major product when we have a choice between ortho and para position?
 
Last edited:

Danny289

Member
10+ Year Member
Dec 2, 2006
1,523
4
Status
Pre-Dental
According to the road map, primary alcohol becomes carboxylic acid when K2Cr2O7 is used.

Road map 2 shows that R-CH2-CH3 becomes R-COOH.
Does this mean it change ethane into COOH?

and if so, what about the combination of both ethane and primary alcohol such as CH3-CH2-OH?

=============================================

My 2nd question is...
which is major product when we have a choice between ortho and para position?
yes K2Cr2O7 is oxidation agent, oxidize alcohol to acid,. in road map two you have Ar not R. Ar is Aromatic and yes it will give you benzoic Acid .
for second question you have two orto positions and one para, usually para is more stable, in DAT they are not going to confuse you don't worry.
 
Last edited:

doc3232

10+ Year Member
7+ Year Member
Feb 15, 2008
3,809
10
Status
Dental Student
For your second question, para is usually preferred even though there are two ortho positions. If the R group is small like a methyl then ortho is preferred, other than that, para is major product. Like Dan said though, probably something DAT won't ask with the intention to trip you up.
 
About the Ads

yankees27th

10+ Year Member
7+ Year Member
Nov 3, 2008
276
1
Status
Pre-Dental
yes K2Cr2O7 is oxidation agent, oxidize alcohol to acid,. in road map two you have Ar not R. Ar is Aromatic and yes it will give you benzoic Acid .
for second question you have two orto positions and one para, usually para is more stable, in DAT they are not going to confuse you don't worry.

Wait I'm confused. Why would on oxidizing agent turn an ethyl group into a COOH group? I think I vaguely remember seeing that from my ochem class but I'm not sure...
 

zuma35

10+ Year Member
5+ Year Member
Sep 6, 2008
128
0
Status
Dental Student
Oxidizing agent like KMnO4 and K2Cr2O7 will oxidize alkyl groups on a benzene ring, to carboxy acids.
 

yankees27th

10+ Year Member
7+ Year Member
Nov 3, 2008
276
1
Status
Pre-Dental
Oh yeah I think I remember now. As long as the alpha carbon has at least one hydrogen any length chain is changed to -COOH right?
 

joonkimdds

Senior Member
10+ Year Member
7+ Year Member
Jun 30, 2005
2,780
2
Status
Pre-Dental
I have an additional question.

You said an aromatic ring + alkyl group will make the alkyl group into COOH.
Does it matter how many CH are in the alkyl group?
For example if there is a ring with CH2CH2CH2CH2CH2CH3, how will this change?
is it still gonna be a ring with COOH? or will it be a ring with CH2CH2CH2CH2CH2COOH?
 

zuma35

10+ Year Member
5+ Year Member
Sep 6, 2008
128
0
Status
Dental Student
I have an additional question.

You said an aromatic ring + alkyl group will make the alkyl group into COOH.
Does it matter how many CH are in the alkyl group?
For example if there is a ring with CH2CH2CH2CH2CH2CH3, how will this change?
is it still gonna be a ring with COOH? or will it be a ring with CH2CH2CH2CH2CH2COOH?
Evene a longer alkyl chain would become just COOH. Also, as yankees pointed out, there must be at least one H on the carbon directly attached to the benzene ring.
 

yankees27th

10+ Year Member
7+ Year Member
Nov 3, 2008
276
1
Status
Pre-Dental
Is this only with K2CrO7 or do other oxidizing agents do the same thing?
 

gentile1225

10+ Year Member
5+ Year Member
May 12, 2008
383
0
New York
Status
Pre-Dental
K2Cr207/H+ will only burn the substituents off into a carboxy group only if there are benzyllic hydrogens available. Anything from the first carbon on, in the presence of a benzyllic hydrogen, will all burn off and you will only be left with the carboxy acid. From my understanding, K2Cr207 is the only oxidizing agent that will do this.

For example, if you have a benzene ring with an ethyl group and a t-butyl group on it and you apply the K2Cr2O7, then only will the ethly group burn off into a carboxy acid and the t-butyl group will be left alone since there are no benzyllic hydrogens.

As far as the o/p directing going, the para is usually more stable, depending on how sterically hindered the group that is doing the directing is.
 
About the Ads