Chemistry: Post your doubts here!

[daredevil]

can you post the paper no.?

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

 

[daredevil]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

Q 5c(iii) & (iv)

which reaction exactly r they referring to in iii ??

please explain iv

 

[Abdel Moniem]

Thank you so much for your help

i will first explain the wrong answers!
D = because it is a tertiary aldehyde so it wont react!
A = the basic CH3CO2CH3 structure is being disturbed, when reacting only the functional groups will react
B = it could have been the answer but in our question, the molecule of CH3CO2CH3 was ionised to -CH2CO2CH3, with H being removed from the 1st carbon but as you can see in this it has reacted with the 3rd carbon of the molecule so this one wont do either!
so our answer is C because it does not disturb the CH3CO2CH3 molecule and the reaction is taking place at 1st carbon!
hope you understood it!
i think it is a little confusing, if you dont get it i will try explaining again!

i hope this helps
if not then you can still ask again (but i hope you get it this time)

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

i hope this helps!

 

[daredevil]

i hope this helps!

thanksss alott man!! (Y) got it real good... hopefully wont forget it either...

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

Q 5c(iii) & (iv)

which reaction exactly r they referring to in iii ??

please explain iv

(iii) since our polymerisation is condensation polymerisation and our E and F are phenol and carboxylic acid, we have ester linkage!
phenol will react much more readily with acyl chloride then with carboxylic acid and will give a higher yield
so we will first convert carboxylic acid to acyl chloride (if we are starting with carboxylic acid) so that our reaction proceed more easily
we will also add phenol with NaOH so that we have (C6H6)O-Na+
now both of them will react more readily and it will be easier to carry out in a laboratory

(iv) sorry i dont understand this one myself!

 

[falak123]

plz help...i cant figure ot the organic molecules using he NMR spectra ...how to do it?

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_42.pdf

Q 5c(iii) & (iv)

which reaction exactly r they referring to in iii ??

please explain iv

i fouund the answer to part (iv)
benzene rings form rigid structures and make the polymer less flexible
so in order to make it flexible, make it into a cyclic acid or remove the double bond in the benzene

 

[sidbloom1995]

plz help...i cant figure ot the organic molecules using he NMR spectra ...how to do it?

you can try on youtube
i searched it and found a very good video recently
just type NMR spectroscopy and you will get a bunch of options, choose the one from the user

• Leah Fisch

 

[falak123]

you can try on youtube
i searched it and found a very good video recently
just type NMR spectroscopy and you will get a bunch of options, choose the one from the user

• Leah Fisch

thank u

 

[daredevil]

i fouund the answer to part (iv)
benzene rings form rigid structures and make the polymer less flexible
so in order to make it flexible, make it into a cyclic acid or remove the double bond in the benzene

sooo... we're supposed to know that we remove the double bond in benzene to make it harder?

 

[Namehere]

sooo... we're supposed to know that we remove the double bond in benzene to make it harder?

You are supposed to know that around double bonds there is a restriction of the carbon bonds and that for single bonds there isn´t.

 

[Namehere]

You are supposed to know that around double bonds there is a restriction of the carbon bonds and that for single bonds there isn´t.

* A restriction for the bonds to rotate, thus becoming less flexible.

 

[daredevil]

You are supposed to know that around double bonds there is a restriction of the carbon bonds and that for single bonds there isn´t.

ummm .... sorry didn't get it... can u elaborate on that plz??

 

[daredevil]

* A restriction for the bonds to rotate, thus becoming less flexible.

why is there a restriction for rotation??

 

[Snow Angel]

regarding chemistry as level practicals, how are we supposed to estimate the rough titre? and how to manage time with doing the experiment as well as doing the calculations? please answer as possible. thanks in advance.

 

[daredevil]

regarding chemistry as level practicals, how are we supposed to estimate the rough titre? and how to manage time with doing the experiment as well as doing the calculations? please answer as possible. thanks in advance.

for the time part be as deft as possible and should practice titration as much as u can in the laboratory along with the calculations becz there just is no other way to perfect it....

furthermore... where choosing the rough titration value u let the titre pour into the the titrant at a steady stream and keep looking at the titrant. as soon as it changes colour close the tap. the reading u get here is the rough titration value and is more often than not larger than the actual titration result. this reading gives u an idea about where ur actual value will lie. e.g. if the rough value = 27cm3 u can reduce the stream to drops when u reach let's say 20cm3 so that u can later close the tap exactly when the titrant changes colour. i hope it's not too confusing

 

[Namehere]

why is there a restriction for rotation??

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

 

[Namehere]

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

...and i guess that if you want that rotation to occur you will have to break the pi system, and that requires energy. I´m not sure about this, so if anyone knows the answer please do say it.

 

[sidbloom1995]

why is there a restriction for rotation??

Well, dont know if im right so if im not please do correct me!

Its because when you have double bonds like in ethene, the pi system (which has a pi lobe above and below the plane) doesn´t allow rotation because there is no space for it to rotate (since its occupied by the pi lobe)

yes the restriction is there because of pi-bond
think of it as a ball and it is inserted with a stick like in lollipop....but not stuck so if you spin it, it will spin in it's place but now if you attach two sticks with it, the ball wont spin any more because if it tries to rotate around one stick the other will stop it! so it's movement have been restricted!
that is the image i have of it in my mind