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Chemistry: Post your doubts here!

B

Browny

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Can anyone please explain to me how to chose the element which forms a oxiding agent or a reducing agent when reacted with water, when the question gives part of the periodic table and asks this question?
 
Raiyan3

Raiyan3

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goodluckayesha said:
This stuff isn't in the syllabus or taught. This is precisely why it has shown you the 4 figures above the actual question. The examiner wants you to analyze those figures, see whats going on, and apply THAT to the question.

The question shows that all carbons in Benzene can be made to lie on the same plane, but not in cyclohexane. The question also shows that butane Carbons can be made to lie on the same plane, but methylpropane, also an alkane, cannot.
This tells us that if a compound has a ring, OR branching, it cannot be made to lie on the same plane.

All of them are co-planer apart from B.

C is coplaner because you don't have to consider the Oxygen atoms. The others are unbranched and two individual chains lying on the same plane.
Click to expand...

Righttt i kinda get it noww! Thank you very much. May the best of the luck be with you 'Ayesha' :)
 
daredevil

daredevil

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Browny said:
Can anyone please explain to me how to chose the element which forms a oxiding agent or a reducing agent when reacted with water, when the question gives part of the periodic table and asks this question?
Click to expand...
I think it goes like this:
if Na reacts with H2O it gives NaOH so it becomes oxidised. and we know that if an element itself becomes oxidised then it is a reducing agent.

similarly if talking about Cl2 then it reacts with H2O to form HCl so it becomes reduced. which means that it is an oxidising agent.

i hope u get it :)
 
N

NinjaInPyjamas

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AbbbbY said:
Everything, my dear Watson, has a logical explanation.


c)
S into T is hydration so I'd use steam + phosphoric acid catalyst @ 300C
S into U is di-ol so Cold Dilute Acidified KMnO4 in a heartbeat.
T into S is dehydration so I'd use conc H2SO4 @ 180C [Can also use H3PO4 @ 300C but that's not recommended]

T with Na (excess).
Replace all the OH's with O-+Na
NaOOCCH(ONA)CH2COONa

U with Na2CO3 (excess)
Replace the carboxylic OH's with O-+Na but not the alcohol OH's.

NaOOCCH(OH)CH(OH)COONa

Add the +'s and -'s on top. Was making it look all weird in typed format since I can't superscript it here.

e)
It's merely asking for cis trans isomers.

cis will be both COOH's on the same side of the two carbons and trans will be diagnol/opposite sides of the two carbons. If this doesn't make sense let me know and I'll draw it out for you.
Click to expand...

Thanks for the excellent explanation! :D
Sorry, I meant (f) instead, could you please explain that?
 
Suchal Riaz

Suchal Riaz

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NinjaInPyjamas said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_22.pdf

q 5 (d) why is it ethanoic acid?
Click to expand...
I will follow through the question from beginning to let you know how examiner give you hints.
Code: 
pattern:
-(information)
the deduction we can make from information

-Q is a sweet smelling liquid.
therefore it is an ester

-Q has Mr of 87.5
mass of C2H4O = 2*12+4+16=44
n=87.5/44~2
so the formula is C4H8O2

now we can see there can be three ester isomers. methyl propanoate, ethyl ethanoate and propyl methanoate. we will see from the following information which isomer it was and then we will go to the original question of finding R.
​

-the solution was hydrolysed. so the carboxyllic acid and alcohol are separated. there can be three kinds of acids and alcohols. methanol, ethanol, propanol, methanoic acid, ethanoic acid and propanoic acid.
- heating with an oxidising agent gives single reagent R
the alcohol will oxidise to produce aldehyde and then will produce carboxyllic acid. as a single compound is formed, both carboxylic acid from ester and oxidised form alcohol are same. the only option we are left with is ethyl ethanoate. which will hydrolyse to ethanol and ethanoic acid. ethanol will then oxidise to ethanal. ethanal will oxidise to ethanoic acid. as the test for C=O and CHO are negative there is no aldehyde. the only thing left is ethanoic acid.

i hope it clears the confusion and helps in reaching conclusions in similar questions.
good luck for the papers :)​
 
ZaqZainab

ZaqZainab

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