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Chemistry: Post your doubts here!

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The t
can u indicate on the picture the chiral carbon and also how u identified A and B to be cis trans isomers. I find these structures very confusing nd can't really find the isomers in them.
thankyou ۔
the terminal carbon in the chain has 4 different group attached to it showing it as chiral carbon.
Trans isomers always have same group in different quadrant. like in this "H" is in different quadrant
 
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In both option C and D we have a chiral carbon and a acid group too :/
 

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In the third compound, there is a lenghty carbon chain and a primary alcohol (they are attached)
Now when CH3CH2(CH3)CH2- is oxidised, a ketone is formed and when a primary alcohol is oxidised, a carboxilic acid is formed.
Therefore, oxidation(gain of oxygen) breaks this compound into two compounds.
 
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Well, I just checked the marking scheme and the answer is A :s
In the third compound, there is a lenghty carbon chain and a primary alcohol (they are attached)
Now when CH3CH2(CH3)CH2- is oxidised, a ketone is formed and when a primary alcohol is oxidised, a carboxilic acid is formed.
Therefore, oxidation(gain of oxygen) breaks this compound into two compounds.
 
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First question
upload_2014-11-25_13-11-4-png.49279


The answer, as the mark scheme says, is B. I chose A and got it wrong..Can anybody kindly explain a bit about that please!!!


The Second one
upload_2014-11-25_13-12-57-png.49280


Could anybody explain how to decide the bond angle in this case? D can be ruled out because N has lone pairs, but for the remaining three , I feel hard.

The third, most important one
upload_2014-11-25_13-14-49-png.49281


Answer is A. What@!? HSO3- act as a base in the second reactoin???? I feel just totally baffled...

OK, the last one, thanks guys for being so patient
upload_2014-11-25_13-15-39-png.49282


The answer is A... But I feel D seems to be more correct. Potassium forms covalent bonding in most cases, thus it should show the greatest tendency... Not sure if I have misunderstood the question however.
 
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1st one -_- too lazy to read all tht sorry :S

2nd one is : the nitrogen, having a double bond to the adjacent carbon and a single bond to the methyl group, is sp^2 hybridized and, therefore, the C-N-C bond angle would be 120°.

3rd one : the HNO3 - acts as a base as it neutralises the H+ (acidic)

4th one : its aluminium since its cation is Al3+ it has 3 electrons free to make covalent bonds unlike Mg2+ Ne (noble gas) K+
rexsun
 
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Messages
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First question
upload_2014-11-25_13-11-4-png.49279


The answer, as the mark scheme says, is B. I chose A and got it wrong..Can anybody kindly explain a bit about that please!!!


The Second one
upload_2014-11-25_13-12-57-png.49280


Could anybody explain how to decide the bond angle in this case? D can be ruled out because N has lone pairs, but for the remaining three , I feel hard.

The third, most important one
upload_2014-11-25_13-14-49-png.49281


Answer is A. What@!? HSO3- act as a base in the second reactoin???? I feel just totally baffled...

OK, the last one, thanks guys for being so patient
upload_2014-11-25_13-15-39-png.49282


The answer is A... But I feel D seems to be more correct. Potassium forms covalent bonding in most cases, thus it should show the greatest tendency... Not sure if I have misunderstood the question however.

PS i think for quest 1 its A
 
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papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20%289701%29/9701_w05_qp_4.pdf

can sum1 pls help wid q2 d..........
 
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1st one -_- too lazy to read all tht sorry :S

2nd one is : the nitrogen, having a double bond to the adjacent carbon and a single bond to the methyl group, is sp^2 hybridized and, therefore, the C-N-C bond angle would be 120°.

3rd one : the HNO3 - acts as a base as it neutralises the H+ (acidic)

4th one : its aluminium since its cation is Al3+ it has 3 electrons free to make covalent bonds unlike Mg2+ Ne (noble gas) K+
rexsun
Thanks so much! Though I still don't quiet understand what's meant by SP^2 and why it's 120 degree. Anyway I will find out.Thanks
 
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Yes please..
I just reviewed my notes and you're right.
I drew the first one correct.
The second compound is an aldehyde and aldehydes, under harsh conditions can be oxidised into Carboxilic Acids, so if Y is (CH3)3CCH2CH2OH, (C6H14O) Z would be (CH3)3CCH2COOH (C6H12O2).Same reason goes for the third compound.
 
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