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Chemistry: Post your doubts here!
- Thread starter XPFMember
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There are three double bonds. Each can have Cis or trans. Here are the possibilities:
C...C...C...
C...C...T...
C...T...C...
C...T...T...
T...C...C...
T...C...T...
T...T...C...
T...T...T...
Where T represents a trans and C a cis bond arrangement.
In general, if there are x number of double bonds capable of making cis -trans isomerism (each), there will be a total of 2^x geometrical isomers of that compound.
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Why is it A?
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I'm getting B ... NaClO will be formed right?
But then how is it A?
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Why can it not be C?
Alcohols don't react with Na metal either right?
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6NaOH + 3Cl2 --> 5NaCl + NaClO3+3H2O
The mole ratio of Cl2:NaClO3 is 3:1, so 0.6/3=0.2 mol of NaClO3 will be formed.
Mr of NaClO3 = 106.5
Mass = 0.2*106.5 = 21.3g
Remember, NaClO is only formed with cold sodium hydroxide solution
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Alcohols react with sodium metal to form a salt and hydrogen.
For more information visit: http://tinyurl.com/bawcd94
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This is a tricky one, but it can be done by elimination. B, C and D suggest that the compound may be MgCl2, PCl5 or NaCl.
However, the reactions of the above with water are as such:
MgCl2 + 6H2O --> 2Cl- + [Mg(H2O)6]2+ ; pH 6.7
PCl5 + 4H2o -->H3PO4 + 5HCl ; pH 2
NaCl + H2O --> Na+ + Cl- +H2O ; pH 7
As we can see, none of these fit the information which states that the pH in solution is 12, so Option A is correct.
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Question 22:
After polymerising the alkenes they become saturated and so they will not undergo reactions such as hydration and oxidation. Therefore, the answer is C.
Question 27:
Options A and B are incorrect because they do not have CH2CO2CH3 parts in their compounds.
Option D is also incorrect because the left side of the compound shows a carbon atom combined to two other carbon atoms, thus unable to form a double bond with O and a single bond with H which represents an aldehyde group.
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Question 29:
With the addition of cold dilute KMnO4, two alcohol groups will be added to the carbon-carbon double bond which will in turn form a single bond, hence creating 2 new chiral carbons.
The addition of hot concentrated KMnO4 will oxidise the secondary alcohol group on the left to a ketone, which causes that carbon atom to no longer be chiral.
With the addition of cold dilute KMnO4, two alcohol groups will be added to the carbon-carbon double bond which will in turn form a single bond, hence creating 2 new chiral carbons.
The addition of hot concentrated KMnO4 will oxidise the secondary alcohol group on the left to a ketone, which causes that carbon atom to no longer be chiral.
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Thanks i do understand 27 and 29 now but can you please explain 22 a bit more? thanks again
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http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf
Q29 ( ans is B)
30 (C) nd 20 (C) pleasee
Q29 ( ans is B)
30 (C) nd 20 (C) pleasee
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A - polymerizing will form an alkane. Hydrolyzing alkanes will give CO2 and H2O. so A is wrongThanks i do understand 27 and 29 now but can you please explain 22 a bit more? thanks again
B - polymerizing will form an alkane. KMnO4 reacts with alkenes not alkanes to form alcohols. so B is wrong.
D - same as B; KMnO4 reacts with alkenes not alkanes to form alcohols. so D is wrong
C - polymerizing will form a chloro-alkane (there will be Cl in place of the OH groups)
Hydrolysis will replace the Cl with OH as HCl will be formed.
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O29 : X is wrong as if you chose any one carbon atom and draw a 3D structure around it, you can see that all the 4 different substituents are in different planes. So since the C has another C as one of the 4 substituents, they are in different planes.
Y is wrong because for cis-trans, you need 2 same substituents on the 2 carbons, in this case there are not.
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29 -All the carbon atoms bonded in the ring have tetrahedral structure, and are sp3 hybridised. this is because every carbon is attached to four groups. So like when you draw optical isomers, in fact, every carbon with 4 substituents have a 3 dimensional tetrahedral structure. Therefore, they cannot be on the same plane. Also, there is no Cis trans isomerism for a cycloalkene made up of less than 8 carbon atoms, since the ring constricts bond rotation. So neither students are correct.
30: the compound already given is:
CH3CH2CH2CO2CH3
Now just play around:
- CH3CH2CH2CO2CH3
- CH3CH(CH3)CO2CH3
- CH3CH2CO2CH2CH3
- CH3CO2CH2CH2CH3
- CH3CO2CH(CH3)CH3
- HCO2CH2CH2CH2CH3
- HCO2CH(CH3)CH2CH3
- HCO2CH2CH(CH3)CH3
- HCO2C(CH3)3
20. Again just play around
- CH3CH2CH2CH2CHO
- CH3CH2C*H(CH3)CHO
- OPTICAL ISOMER OF ABOVE
- CH3CH(CH3)CH2CHO
- C(CH3)3CHO
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Thanks a lot guys.. ^_^
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http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf
Q19: Why is B incorrect?
Q19: Why is B incorrect?