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Post by Mr Kwok on Feb 27, 2009 4:17:48 GMT -5
Hi Everyone, I think you must be wondering this too.... Can we used acidified potassium permanagate to disguish chloromethylbenzene and 4-chloro methylbenzene?  Please share your views here! Regards Mr Kwok |
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Post by belinda on Feb 28, 2009 0:59:37 GMT -5
I guess the answer is yes. PURPLE KMnO4 will decolourise when added to B but will remain purple when added to A. I think only methyl substituent can be oxdised to -COOH substituent on the benzene ring.
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Post by alicia on Feb 28, 2009 23:06:05 GMT -5
i think we can.
For A, oxidation can take place as long as the alpha carbon has a H. hence it will be oxidised to C6H5COOH and HCl(g) would be produced.
For B, oxidation takes place too. C6H5COOH and H20 would be formed.
therefore, can distinguish A and B by checking for the presence of HCl(g).
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kay
New Member
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Post by kay on Feb 28, 2009 23:06:09 GMT -5
i think we can differentiate using the chemical test.
For A, upon using acidified potassium permanganate, products formed will be C6H5COOH and HCL. The compound B will have the CH3 chain oxidized to -COOH but it will not form HCL. Hence, both will decolourise but acidic fumes will be seen upon heating A under reflux but no acidic fumes will be seen upon heating B, thus we can differentiate between the 2 compounds.
kay
2SB2
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Post by daniel on Mar 1, 2009 2:47:28 GMT -5
i feel that the answer is yes, this is because the 2 C-H bonds and the C-Cl bond will break when potassium permanganate is reacted wuth chloromethylbenzene. Forming the carboxylic acid group and HCl gas will be evolved. when 4-chloro methylbenzene is reacted instead, the carboxylic acid will still be formed, however, the gas evolved would be H2 instead. Therefore, the simple chemical test can be to test for the presence of HCl gas to differentiate between chloromethyl benzene and 4-chloro methylbenzene.
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Post by seahyimin on Mar 1, 2009 5:37:44 GMT -5
I think this method can be used. Both A and B will get oxidised, so solution decolourises for both. For A we get benzoic acid, water and ClO2 gas(Cl is oxidised), which is reddish-yellow in colour so it can be seen. For B we get 4-Chlorobenzoic acid and water, and no gas is produced.
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Post by angela on Mar 1, 2009 7:49:11 GMT -5
I think we can do that. A will be oxidised to form benzoic acid and hydrogen chloride as Cl is oxidised too. B will also be oxidised to form 4-chlorobenzoic acid but no hydrogen chloride. Thus, the two can be differentiated by looking out for the production of hydrogen chloride which appears as white fumes.
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Post by joanna on Mar 1, 2009 9:01:11 GMT -5
should be can. A benzene ring attached to a COOH group would be formed for A after reacting with KMnO4, leaving the Cl to combine with H to form HCl. However, for B, the benzene ring will be attached to COOH group and Cl atom. i think for A white fumes would be produced when hydrogen chloride gas is formed.
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Post by davidlim on Mar 1, 2009 13:15:07 GMT -5
Yes.
Observation: White fumes of acidic vapour is releasd by the reaction with A but not with B.
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Post by liling on Mar 2, 2009 6:30:08 GMT -5
I think it is possible to differentiate them by using acidified KMnO4.
Chloromethylbenzene will react and form benzoic acid with gaseous HCl which can be observed as white fumes whereas 4-chloromethyl benzene will react and from 4 chlorobenzoic acid and water.
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Post by Mr Kwok on Mar 4, 2009 21:04:52 GMT -5
Hi everyone,
I think theoretically heating with acidified KMnO4 should be fine. I have not done that experiment before and neither could I source a book which describes that such a test was done. My guess that it is probably more efficient to use the "nucleophilc substitution then precipitation" test. More efficient and more clear cut.
Having said that if we heat both compounds to acidified KMnO4, what is the possible outcome.
In the case of 4-chloromethylbenzene, we will observe the decolorisation of purple KMnO4, since the -CH3 substituent has been oxidised to -COOH.
In the case of chloromethylbenzene, my sense is that the following occurs:
1) chloromethylbenzene can undergo SN1 with H2O as the nucleophile to form a phenylmethanol (primary alcohol).
2) the alcohol would be oxidised - Hence, purple KMnO4 decolourises.
3) Cl- will also be oxidised by the acidified KMnO4 - giving Cl2 (g), a pale yellowish-green gas, which will bleach the litmus paper. Hence, this is the observation which helps to distinguish the two compounds.
I noticed some of you mentioned that white fumes of HCl appears. Well, if there is the absence of water and heat is present, yes HCl will evolved. However, if the reaction is carried out in aqueous medium, we will not observe it as the HCl will dissolve in water readily.
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grace
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Post by grace on Aug 20, 2013 8:05:59 GMT -5
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grace
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Post by grace on Oct 26, 2013 8:50:36 GMT -5
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grace
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Post by grace on Nov 10, 2013 11:27:11 GMT -5
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grace
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Post by grace on Nov 25, 2013 10:02:06 GMT -5
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