O Levels Chemistry Question – Organic Chemistry, Ether

24 Jul

Posted by: sean in: Chemistry Notes & Tips

This is a Organic Chemistry question that came out in 2006 Nov GCE ‘O’ Level Examination Paper 2.

Check it out!

Question:

(Click image to have a better view)

PS: Leave your suggested answer in the “Comments Section” right below this blogpost

Click here for early bird registration!

What's Next?

Digg it

Stumble it

Twitter it

15 Responses

Sean Chua
31|Jul|2008 1

Hi.

Seems like many students are not answering this particular question on Organic Chemistry. Wonder why so…

Let me keep the momentum going, especially for those that are waiting for any suggested answers.

a) Name: methoxybutane
Formula: CH3OC4H9

bi) 30 degree Celsius
ii) As molecule gets bigger, the hydrocarbon content increases, and more van der waals Intermolecular Forces of attraction between the molecules. More energy is required to break these forces, thus BPs increases.

c) True, False, False, True

d) C4H10O + 6O2 –> 4CO2 + 5H2O

Hope this suggested answers help.

Sean
Master Trainer
http://www.SimpleChemConcepts.com

Reply
JO.
02|Aug|2008 2

Hello.
thank you for the suggested answers.
i was wondering if 38.8 deg C is acceptable for 6bi.
thanks.

Reply
sean
05|Aug|2008 3

Hi Jo.

That should be fine. It still falls within the range of acceptable answers.

Keep up the spirit!

Sean

Reply
Sivvy
10|Aug|2008 4

Hey Sir,
Just wondering;
How do you derive at 30 degree celsius for your answer to part (bi)?
Because I happened to calculate my answer to be 38.8 degrees.

For part (bii), is it acceptable to say that the boiling points increase due
to the presence of more carbon atoms?

For part (d), could you explain how you get the answer? Pleaseeeeeeeee?
I don’t even know how to work it out! :/

Thanks for your help,
Really really appreciate it (;

Cheers,
Sivvy.

Reply
sean
18|Aug|2008 5

Hi Sivvy,

Your answer 38.8 degree Celsius is a good acceptable answer.

For part bii) just saying more carbon atoms is not good enough thou. You need to mention that the intermolecular forces of attraction increases as Carbon length increases.

For Part d), you need to sum up all the C, H and O atoms together. And for combustions of such organic compounds, containing only C, H and O, the 2 products must be H2O and CO2. Balance your equation from there and you will get the answer.

Hope this helps.

PS: Where are you from?

PPS: Do write to me if you need more help.

Cheer
Sean
Founder & Master Trainer
SimpleChemConcepts.com

Reply
raihan
31|Aug|2008 6

why your answer to (d) is different with the answer i got from the answer sheet of past year paper? In the book it says that the answer is 2C2H5OC2H5 + 17O2 forms 8CO2 + 20H2O

Reply
sean
31|Aug|2008 7

Hi Raihan,

Which country are you based in and which ten years series are you using?

There is always a 1-5% chance that answers in ten years series might be incorrect.
These books are published by many publisher and the answers given are not standard answers from Cambridge examiner.

Let’s us check out your answer given:

2C2H5OC2H5 + 17O2 —> 8CO2 + 20H2O

To check if they are correct -> it means to check if elements on both side are balanced?

LHS:
8 C
36 O
20 H

RHS:
8 C
36 O
40 H

Can you see where is the error for your answer now????

You can verify my suggested answer as shown below:

C4H10O + 6O2 –> 4CO2 + 5H2O

PS: Let me know your thoughts on it.

Sean

Reply
raihan
02|Sep|2008 8

I’m from Brunei Darussalam. I got the answer from the REDSPOT PURE CHEMISTRY PAST YEAR PAPER 1998-2006. Thanks for the help, i really appreciate it. Ya, i think your answer is correct. Thank you very much.

Reply
fardowsa
09|Oct|2008 9

i would like to know the structure or graphic fromuale for all the isomer possible C5H12 o

Reply
sean Reply:
April 24th, 2009 at 1:57 pm
Hi fardowsa,

Experential Learning is the most effective.

How about showing me how many isomers you managed to draw for C5H12?

Hear from you.

To Your Academic Success in Chemistry,
Sean Chua
Master Trainer & Author
WINNERS Education Group

“Experience Learning with A Difference”

Reply
Jim Reeves
13|Nov|2008 10

Great Information.I found your site very interesting With the right information Thanks.

Reply
sean Reply:
April 24th, 2009 at 1:58 pm
Hi Jim,

Thank you for the positive comments.

You are doing well with interesting information also.

All the best!

Sean Chua
Master Trainer & Author
WINNERS Education Group

“Experience Learning with A Difference”

Reply
blah Lickoo
11|Mar|2009 11

HI!!!i realised that most of the questions on the website are really useful and i believe this is gonna to pull up mu grades by a lot.
Just a question for Mr Sean.

Why dont you just combine your questions in docuements to be downloaded like you have for the Mole Concept?I believe that it is easier to review rather than going to several links to check out the questions for a certain topic?

Thank You.

Reply
sean Reply:
April 24th, 2009 at 2:03 pm
Hi “blah Lickoo”,

Thank you for the comment. I am glad you are benefitting from this humble Chemistry website.

Readers will still continue to receive some consolidated questions on topical basis from me. I am not gog to send useless documents to readers each and everyday, because when “things come cheap and free…they deem it not useful”.

The best way of learning is Experiential Learning and it involves self-participation. By coming to this website and search for information that are useful to them, readers will learn more effectively.

Those are my thoughts based on my 12 years experience as a coach. If you have better suggestion on how my readers can benefit more from this blogsite, please do not hesitate to leave down your suggestions.

Thank you.

To Your Academic Success in Chemistry,

Sean Chua
Master Trainer & Author
WINNERS Education Group

“Experience Learning with A Difference”

Reply
WALLEED BIN QASIM
23|Nov|2009 12

so according to the answers are as follow………………
CH3C5H9

METHOXYBUTANE.

38.1 DEG.CELC(TAKING AVERAGE OF UPPER AND LOWER BOUND)

THE LINEAR FORCES BECOME MORE COMPACT AND THUS THE BOILING POINT INCR…

1……T F
2……T F
3. F T
4……T F

C2H5OC2H5+6O2—>4CO2+5H20

Reply