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chem homework question
An organice compouds A containing one halogen atom was formed to have the following compostion by mss; carbon 39.9%, hydrogen 7.3%. bromine 52.9%.
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
Scroll to see replies
39.9/12 : 7.3/1 : 52.9/80
3.325 : 7.3 : 0.66
Divide thro by 0.66 = 5.037 : 11.06 : 1
C5H11Br
hot aq KOH dehydrohalogenates to give an alcohol and /or alkenes
there are two distinct ones (structural) and one geometric isomer
the fact that two distinct ones arise tells you that the Br is at least secondary
reaction of C with HBr is hydrohalogenation of an alkene - logically there are two possible orientations of the H-Br that adds. C is therefore an alkene.
D is an alcohol and as the oxidation product cannot be oxidised further (no Tollens test) it must be a secondary alcohol - the reaction with NaBH4 merely confirms that it is reducible back to the alcohol.
C-C-C-C-C is a possible chain
if the Br were on the central carbon there would only be two products of the first reaction with KOH (one alcohol and one alkene as it is symmetrical)
Br is then on carbon number 2
they could be...
A is CH3CH(Br)CH2CH2CH3
B is CH3CH=CHCH2CH3
C is CH2=CHCH2CH2CH3
D is CH3CH(OH)CH2CH2CH3
E is CH3COCH2CH2CH3
reaction of C with HBr follows Markovnikovs rule and produces mainly the 2- bromo product A.
reaction of B with HBr forms mainly the 3 bromo product F (the isomer of A) by Markovnikov
but you are told that B and C do not have stereoisomers so then
B must be CH3CH=C(CH3)2 which has no cis trans (geometric) isomers
C must be CH2=CHCH2(CH3)2 neither does this.
therefore
A is CH3CHBr-CH(CH3)2
D is CH3CH(OH)CH(CH3)2
E is CH3COCH(CH3)2
3.325 : 7.3 : 0.66
Divide thro by 0.66 = 5.037 : 11.06 : 1
C5H11Br
hot aq KOH dehydrohalogenates to give an alcohol and /or alkenes
there are two distinct ones (structural) and one geometric isomer
the fact that two distinct ones arise tells you that the Br is at least secondary
reaction of C with HBr is hydrohalogenation of an alkene - logically there are two possible orientations of the H-Br that adds. C is therefore an alkene.
D is an alcohol and as the oxidation product cannot be oxidised further (no Tollens test) it must be a secondary alcohol - the reaction with NaBH4 merely confirms that it is reducible back to the alcohol.
C-C-C-C-C is a possible chain
if the Br were on the central carbon there would only be two products of the first reaction with KOH (one alcohol and one alkene as it is symmetrical)
Br is then on carbon number 2
they could be...
A is CH3CH(Br)CH2CH2CH3
B is CH3CH=CHCH2CH3
C is CH2=CHCH2CH2CH3
D is CH3CH(OH)CH2CH2CH3
E is CH3COCH2CH2CH3
reaction of C with HBr follows Markovnikovs rule and produces mainly the 2- bromo product A.
reaction of B with HBr forms mainly the 3 bromo product F (the isomer of A) by Markovnikov
but you are told that B and C do not have stereoisomers so then
B must be CH3CH=C(CH3)2 which has no cis trans (geometric) isomers
C must be CH2=CHCH2(CH3)2 neither does this.
therefore
A is CH3CHBr-CH(CH3)2
D is CH3CH(OH)CH(CH3)2
E is CH3COCH(CH3)2
papz_007
An organice compouds A containing one halogen atom was formed to have the following compostion by mss; carbon 39.9%, hydrogen 7.3%. bromine 52.9%.
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
A is C5H11Br
Are you sure its aquous KOH not ethanolic?
If its aqueous then i dont think any isomers would be formed.
papz_007
An organice compouds A containing one halogen atom was formed to have the following compostion by mss; carbon 39.9%, hydrogen 7.3%. bromine 52.9%.
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
Reaction of C with HBr forms mainly A but reacton of B with HBr forms mainly F, a structual isomer A.
Compound D undergoes reaction with acidified potassuim dichromate (VI) to form E.
Compound E did not react with Tollens reagent, but does react with NaBH4 to form D.
Draw structual formula for A to F, cleary explaining any essential reasoning.
A is C5H12BR
B i think is C5H12OH
C i think is C5(OH)H12??
dont no the rest any help?
E is i supose a ketone
In light of a post after this, i will correct this post.
D must be pentan-2-ol, as it is the only moleclue with a chiral centre (therefore it has stereoisomers). This means that E is pentan-2-one.
C with HBr goes to A, so if C is Pentan-1-ol, then A is 1-bromopentane
B with HBr goes to F, so if B is Pentan-3-ol, then F is 3-bromopentane.
B and C are interchangeable.
David Baker
A can be 3 things, all with structural formula of C5H11BR
1- Bromo pentane, 2- bromo pentane or 3- bromopentane.
It doesnt really matter which you choose to be A as long as you are consistent.
For all intents and purposes, 2- bromo pentane is A.
Adding HBr to C gives A, therefore C must have been pentan-2-ol
B with HBr gives a structural isomer of A known as F. If we say that B is pentan-3-ol, then F is 3 - bromopentane. This results in D being pentan-1-ol.
When D is oxidised with H+/Cr2O7-, it is a primary alcohol, and so forms an aldehyde and then a carboxylic acid ( which can be reduced back to compound D with NaBH4 and does not react with tollen's reagent. ) This means that E is pentanoic acid.
The other option is to say that B is pentan-1-ol and F is 1- bromopentane. This results in D being pentan-3-ol and when oxidised, forms a ketone, which does not react with tollen's reagent and can be reduced with NaBH4. E is the ketone known as pentan-3-one.
Likewise you could switch compounds C and D and get the ketone pentan-2-one. It all depends on you being consistent. If you change C to pentan-3-ol, then you must also change A to 3-bromopentane.
I still dont have a clue what stereo-isomerism has to do with this question as there are no double bonds ( cis- trans ) and optical isomerism cannot happen since the carbon atoms do not have 4 different groups attached to it. D is also a structural isomer of B and C.
I do not think any alkenes are formed, the KOH is in aqeous, not anhydrous conditions.
1- Bromo pentane, 2- bromo pentane or 3- bromopentane.
It doesnt really matter which you choose to be A as long as you are consistent.
For all intents and purposes, 2- bromo pentane is A.
Adding HBr to C gives A, therefore C must have been pentan-2-ol
B with HBr gives a structural isomer of A known as F. If we say that B is pentan-3-ol, then F is 3 - bromopentane. This results in D being pentan-1-ol.
When D is oxidised with H+/Cr2O7-, it is a primary alcohol, and so forms an aldehyde and then a carboxylic acid ( which can be reduced back to compound D with NaBH4 and does not react with tollen's reagent. ) This means that E is pentanoic acid.
The other option is to say that B is pentan-1-ol and F is 1- bromopentane. This results in D being pentan-3-ol and when oxidised, forms a ketone, which does not react with tollen's reagent and can be reduced with NaBH4. E is the ketone known as pentan-3-one.
Likewise you could switch compounds C and D and get the ketone pentan-2-one. It all depends on you being consistent. If you change C to pentan-3-ol, then you must also change A to 3-bromopentane.
I still dont have a clue what stereo-isomerism has to do with this question as there are no double bonds ( cis- trans ) and optical isomerism cannot happen since the carbon atoms do not have 4 different groups attached to it. D is also a structural isomer of B and C.
I do not think any alkenes are formed, the KOH is in aqeous, not anhydrous conditions.
I think this guy has the right idea. From what i can rmember of A level chem, NaBH4 does not work on carboxylic acids. If A was 2 bromopentane, then there would be optical isomerism since there are 4 different groups attached, CH3, H, Br, and C3H7.
Oh yeh good spot, i was wondering why they mentioned optical isomerism, i just saw the carbon atom and disregarded it.
Carboxylic acids react with NaBH4 (or another strong reducing agent such as LiAlH4) to go back to primary alcohols. But this point was wrong anyway.
Carboxylic acids react with NaBH4 (or another strong reducing agent such as LiAlH4) to go back to primary alcohols. But this point was wrong anyway.
hello... knock knock ... is there anyone at home?
Take a look at the second post!
Take a look at the second post!
charco
hello... knock knock ... is there anyone at home?
Take a look at the second post!
Take a look at the second post!
Yeh this post confused me. The conditions were not anhydrous, they were aqueous. Hence, no elimination.
--------------
In light of a post after this, i will correct this post.
D must be pentan-2-ol, as it is the only moleclue with a chiral centre (therefore it has stereoisomers). This means that E is pentan-2-one.
C with HBr goes to A, so if C is Pentan-1-ol, then A is 1-bromopentane
B with HBr goes to F, so if B is Pentan-3-ol, then F is 3-bromopentane.
B and C are interchangeable.
I agree with charco; and just to note the fact that when hydroxide ions react with a haloalkane, both elimination and substitution occur simultaneously. The proportion of each occuring depends on the ollowing factors:
1) In ethanol there is greater amount of elimination than in aquaous.
2) For a higher temperature there is a greater amount of elimination.
3) When the molecule has more branching then there is greater elimination.
These factors explain why both alkenes and alcohols were formed.
1) In ethanol there is greater amount of elimination than in aquaous.
2) For a higher temperature there is a greater amount of elimination.
3) When the molecule has more branching then there is greater elimination.
These factors explain why both alkenes and alcohols were formed.
kyro8008
I agree with charco; and just to note the fact that when hydroxide ions react with a haloalkane, both elimination and substitution occur simultaneously. The proportion of each occuring depends on the ollowing factors:
1) In ethanol there is greater amount of elimination than in aquaous.
2) For a higher temperature there is a greater amount of elimination.
3) When the molecule has more branching then there is greater elimination.
These factors explain why both alkenes and alcohols were formed.
1) In ethanol there is greater amount of elimination than in aquaous.
2) For a higher temperature there is a greater amount of elimination.
3) When the molecule has more branching then there is greater elimination.
These factors explain why both alkenes and alcohols were formed.
This is A level not Phd level. The stuff you have mentioned is not in the A level specification (at least not for edexcel).
As far as A-level goes:
aq. KOH, heat under reflux gives the alcohol
ethanolic KOH, heat under refluc gives the alkene.
ba_ba1
This is A level not Phd level. The stuff you have mentioned is not in the A level specification (at least not for edexcel).
As far as A-level goes:
aq. KOH, heat under reflux gives the alcohol
ethanolic KOH, heat under refluc gives the alkene.
As far as A-level goes:
aq. KOH, heat under reflux gives the alcohol
ethanolic KOH, heat under refluc gives the alkene.
I am doing my A2's at the moment and its in the AQA specification for AS. On that note, ive seen alot of things around on other specifications ive never heard of before, I didnt realise how much some of it varied.
ba_ba1
The second post is a load of rubbish since adding KOH (aq) to a haloalkane does not give an alkene!
My, what a thoughtful analysis!
Organic reactions frequently give a mixture of products with similar conditions. Elimination occurs because the base abstracts a proton from the alpha carbon. As the hydroxide ion is a base (duh!) it is also capable of doing this albeit to a lesser extent that the ethoxide ion.
The ideal conditions in an organic reaction are those that give the greatest percentage of the required product NOT the product in exclusivity.
IDEAL conditions for nucleophilic substitution - hot aqueous KOH
IDEAL conditions for elimination hot ethanolic KOH
both sets of conditions will produce some of the undesired products.
please excuse the rubbish....
On treatment with hot, aqueouse pottasuim sodium hydroxide, a mixture of three compounds ( B,C,D compounds ). Compounds B and C are strucutal isomers and do not have steroisomers.
for those intent on alcohol only answers....
just how do you explain that nucleophilic substitution of ONE compound manages to produce three others, two of which are NOT stereoisomers but ARE structural isomers.
let's examine the possibilities
---------------------------------------------------------
B and C are structural and DO NOT have stereoisomers
CH3CH(OH)CH2CH2CH3 IS optically active (therefore stereoisomers, cannot be B or C)
CH3CH2CH(OH)CH2CH3 is NOT optically active (no stereoisomers, could be B OR C)
if it is B then what could C be (and vice versa)?
----------------------------------------------------------
B and C are structural and DO NOT have stereoisomers
CH3CH(OH)CH(CH3)2 IS optically active (therefore stereoisomers, cannot be B or C)
CH3CH2C(OH)(CH3)2 is NOT optically active (no stereoisomers, could be B OR C)
if it is B then what could C be (and vice versa)?
----------------------------------------------------------
nucleophilic substitution involves...
R-X plus Nu- ----> R-Nu + X-
How does R manage to change in this sequence?
For all answers please send a stamp addressed envelope to:
Frank Spotnitz, The X files, BBC television, London, UK
charco
My, what a thoughtful analysis!
Organic reactions frequently give a mixture of products with similar conditions. Elimination occurs because the base abstracts a proton from the alpha carbon. As the hydroxide ion is a base (duh!) it is also capable of doing this albeit to a lesser extent that the ethoxide ion.
The ideal conditions in an organic reaction are those that give the greatest percentage of the required product NOT the product in exclusivity.
IDEAL conditions for nucleophilic substitution - hot aqueous KOH
IDEAL conditions for elimination hot ethanolic KOH
both sets of conditions will produce some of the undesired products.
please excuse the rubbish....
Organic reactions frequently give a mixture of products with similar conditions. Elimination occurs because the base abstracts a proton from the alpha carbon. As the hydroxide ion is a base (duh!) it is also capable of doing this albeit to a lesser extent that the ethoxide ion.
The ideal conditions in an organic reaction are those that give the greatest percentage of the required product NOT the product in exclusivity.
IDEAL conditions for nucleophilic substitution - hot aqueous KOH
IDEAL conditions for elimination hot ethanolic KOH
both sets of conditions will produce some of the undesired products.
please excuse the rubbish....
Practically, yes there will be a mixture of products, Theoretically, no there will not be a mixture. The theory is that aqueous conditions give alcohols and anhydrous conditions will give alkenes, no matter what board you are on. Ive checked AS and A2 revision guides and text books, and nowhere does it say that you will get alkenes from aqueous conditions.
OK then, If you have read through all the possible A level texts I suggest you read the following link thoroughly
nucleophilic substitution versus elimination
the author, Jim Clark, is an A level examiner, author of several A level texts and an authority on the subject!
I quote from the website...
"In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors"
nucleophilic substitution versus elimination
the author, Jim Clark, is an A level examiner, author of several A level texts and an authority on the subject!
I quote from the website...
"In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors"
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