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Chemistry: Post your doubts here!
- Thread starter XPFMember
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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!
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can you post the paper no.?
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thank you for your help but could you explain more iam still confused
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Q3. e
first write a balanced equation of reaction between CH3CO2H and NaOH
CH3CO2H + NaOH ----------> CH3CO2-Na+ + H2O
we have 0.25 mol/dm3 of CH3CO2H and 0.1 mol/dm3 of NaOH
since the buffer is of acid and i's salt, all of the alkali will be converted to salt and the acid will be in excess
conc. of salt = 0.1 (same as the alkali)
conc. of acid reacted with alkali: since the mole ratio according to the equation is 1 : 1, so the conc. that reacted will also be same as the alkali, 0.1
conc. of acid remainig: 0.25 - 0.1 = 0.15 mol/dm3
now use the two conc. in the following formula
pH= pKa + log([salt]/[acid])
you have the pKa value and the conc. of salt and acid so just substitute it
pH= 4.76 + log([0.1]/[0.15]) = 4.58
that is your answer
as for Q5 b im unable to answer it
Q7 c
use the formula
no. of carbons = (100/1.1) x ([m+1]/[m])
you have the values of m:m+1 as 5.9:0.2
add them to the formula
no. of C = (100/1.1) x ([0.2]/[5.9]) = 3.08 = 3
so we have 3 carbons
the question says that our structure have an M+2 peak, which is formed due to either Cl37 /Cl35 or Br79/Br81
it aslo states that the peak is about the same height as the M peak
since there height is same we can conclude that it is Br because if it was Cl the difference in the peaks will be of ratio 3:1, meaning one of the two peaks will be more then double the length
as to how many Br are there it's only one because there is an other peak that is M+4 in which there are 2 atoms of higher and lower mass (isotopes)
and since they did not mention this peak here means that there is only one atom of Br
so we have 3 C and 1Br so the rest will be H, which in this case will be 7
the NMR spectrum shows us that there are 6H in one environment and 1 H in a different environment
so we can come up with a structure in which 1 H will be all alone
the structure will be
CH3-CH(Br)-CH3
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i hope this helps
if not then you can still ask again
(but i hope you get it this time) Attachments
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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
Q 5c(iii) & (iv)
which reaction exactly r they referring to in iii ??
please explain iv
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Thank you so much for your help
i hope this helps
if not then you can still ask again
- Messages
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i hope this helps!
Attachments
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thanksss alott man!! (Y) got it real good... hopefully wont forget it either...
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(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!
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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
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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
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thank u
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sooo... we're supposed to know that we remove the double bond in benzene to make it harder?
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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.
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* A restriction for the bonds to rotate, thus becoming less flexible.
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ummm .... sorry didn't get it... can u elaborate on that plz??