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CAn you explain of X againW: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3
Notice that the H on the centre Carbon is ultimately replaced by CN?CAn you explain of X again
okNotice that the H on the centre Carbon is ultimately replaced by CN?
This would require us to change the H with a halogen (either Br or Cl), one way to do it would be free radical substitution.
Then the halogen alkane intermediate will undergo a nucleophilic substitution by adding CN- (using NaCN or KCN)
Daaaang. Thanks again mateW: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3
No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.Daaaang. Thanks again mate
No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.
Would like to get an understanding of students exams schedule in other countries, so in Malaysia, both the A and O levels (?) are done in May and not November?
C40H82 ---> C16H34 + 2C12H24
How would you calculate the bond energy of this reaction? (bond energy C-C 350, C=C 610, C-H 410)
Is this the actual question?
We'll need to make some reasonable assumptions here that the C12H24 is an alkene rather than a cycloalkane. The C12H24 would be having one C=C bond as it is "missing" 2 hydrogen
Energy taken in to break bonds of C40H82
82 (C-H) and 39 (C-C) bonds
= 82 (410) + 39 (350)
=47270 kJ
Energy given out in forming bonds of C16H34 and 2 C12H24
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 34 (410) + 15 (350) + 2 (610) + 20 (350) + 48(410)
=46680 kJ
Enthalpy change = +47270 - 46680 = +590kJ/mol
Quite a bit of numbers at this late hour, hope I didn't make a mistake. There is a slightly shorter way, but I think this way should be more understandable.
Hello ! can someone plz explain the following : Q 4 , 5, 14,39 in
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
Thanks. The actual question is Q4b(iii) in the following link. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
In the mark scheme (http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_ms_4.pdf)
they used a method i don't understand to get a different answer. Can you check again please?
Try to draw the lewis structure of CN-
Ok, I spotted my mistake, it was ok till one part.
Energy given out in forming bonds of C16H34 and 2 C12H24
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 34 (410) + 15 (350) + 2 (610) + 20 (350) + 48(410)
=47090 kJ (instead of 46680 kJ in my first attempt)
Enthalpy change = +47270 - 47090 = +180kJ/mol
The marking scheme actually skips a lot of steps , unless we are very used to solving the question, it is not likely to do it in 1 single step like they did.
Number of bonds broken
82 (C-H) and 39 (C-C) bonds broken
Number of bonds formed
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 82 (C-H) and 35(C-C) and 2 (C=C) formed
Comparing the bonds broken vs formed, we have an overall of 4 (C-C) bonds broken and 2 (C=C) bonds formed.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Please can someone explain Q.6, 9, 14, 18 and 30?
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