Chemistry: Post your doubts here!

[Mstudent]

Anyone????

 

[Holmes]

Anyone????

Nitrogen atom has two lone pairs and one single electron in it (N atomic no. 7 )
when CH3NCO is formed then
a double bond is formed between N and C .... here nitrogen's one lone pair is utilized in making bond,
and ONE single electron is used to make a Single bond with CH3 .
Now , N has only one lone pair that is not involved in any bonding.
so the figure is sth like:
bond pairs = 2 Lone pair = 1 we will count a Double bond as a SINGLE bond ( or one bond pair)
so this figure will give us an angle of 119' or 120'
Aswer = C
Hope it make sense!

 

[Mstudent]

Thank you very much!!, I confused it mistakenly with water, I didnt realise water had 2 lone pairs

 

[Mstudent]

ANYONE!!

 

[SohaibButt]

Please anyone?

 

[Mstudent]

Anyone could help me with q.9
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s14_qp_11.pdf

 

[amina1300]

Anyone could help me with q.9
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s14_qp_11.pdf

ONLY THE LAST ONE CANCELS OUT.

 

[Mstudent]

View attachment 62368
ONLY THE LAST ONE CANCELS OUT.

Thank you very much amina1300 .

 

[amina1300]

Please anyone?

Q10 Ans = C
Kc>1 tells you that the equilibrium will lie to the right,
H2(g) + I2(g) <==> 2HI(g) .................. Kc = 60.
2M .......0.3M ............?M

Kc = [HI]² / ([H2] [I2])
60 = x² / 2 / 0.3
x = 6
[HI] = 6M
moles of HI = 6

Q5
Q=VxCxdeltaT
total volume is 25+25 = 50 cm3
= 50 x 4.3 x (2.5)
=525
now we shud find enthalpy change fore one mole
number of moles = vol. x conc.
25/1000 x 0.35 = 8.75x10^-3


525 Joules was for 8.75x10^-3 moles
525 J = 8.75x10^-3 moles
? = 1 mole
525/(8.75x10^-3) = 60000 Joules/mole

TIP: If we need to guess an option without calculation, the heat of neutralization of a strong acid and strong alkali should be in the region of ~58 kJ mol–1.

Last Q :
Kc = x/(7.6x10^-6)^4*x
Kc = 2.997..x10^20 Ans = A

 

[amina1300]

ANYONE!!

View attachment 62361

"N" + ½O2 → NO nitric oxide (gas)
NO + ½O2 →fast→ NO2 nitrogen dioxide
NO is diatomic
And it is polar as it contains electro-negative N

I think A is correct .... ?

 

[Mstudent]

"N" + ½O2 → NO nitric oxide (gas)
NO + ½O2 →fast→ NO2 nitrogen dioxide
NO is diatomic
And it is polar as it contains electro-negative N

I think A is correct .... ?

Yeah you're right, thank you very much!

 

[amina1300]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w16_qp_13.pdf
Q37 and 20 Anyone??

 

[Mstudent]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w16_qp_13.pdf
Q37 and 20 Anyone??

For q. 37

1 will react to form propanone and 2-methyl propanoic acid
2 will also form a Carboxyllic aid and a ketone
3.will also for a carboxyllic acid and a ketone

so I suppose its A?

 

[amina1300]

For q. 37

1 will react to form propanone and 2-methyl propanoic acid
2 will also form a Carboxyllic aid and a ketone
3.will also for a carboxyllic acid and a ketone

so I suppose its A?

How do you know that what will form ..can you explain in detail? :/ I'm really bad at Organic Chemistry.

 

[Mstudent]

How do you know that what will form ..can you explain in detail? :/ I'm really bad at Organic Chemistry.

Oh. OK so here's the rule

IF a C=C is broken by:

1. COLD KMNO4-------> two OH groups will BE ADDED (Electrophillic addition) and form a Diol
2. HOT CONCENTRATED KMNO4---------> 3 Conditions
A. If NO Carbon groups joined to C in C=C------> CO2 and H2O
B. If ONE Carbon group joined to C in C=C ---------> Aldehyde---------> Carboxyllic acid (Write only the Carboxyllic acid in exams)
C. If TWO Carbon groups joined to C in C=C-------------> ketone

Thats it!

hope that helps

 

[amina1300]

Oh. OK so here's the rule

IF a C=C is broken by:

1. COLD KMNO4-------> two OH groups will BE ADDED (Electrophillic addition) and form a Diol
2. HOT CONCENTRATED KMNO4---------> 3 Conditions
A. If NO Carbon groups joined to C in C=C------> CO2 and H2O
B. If ONE Carbon group joined to C in C=C ---------> Aldehyde---------> Carboxyllic acid (Write only the Carboxyllic acid in exams)
C. If TWO Carbon groups joined to C in C=C-------------> ketone

Thats it!

hope that helps

OMg that was extremely helpful I can't thank you enough!
I couldn't comprehend this in question but thanks to you now I just realized that we look at both sides of the C=C

 

[Mstudent]

OMg that was extremely helpful I can't thank you enough!
I couldn't comprehend this in question but thanks to you now I just realized that we look at both sides of the C=C

You are welcome anytime. I know you lost your Organic chem notes, so if you have any question regarding Organic chem, feel free to ask!

 

[Dukula Jayasinghe]

ORGANIC SHORT NOTES
Alkanes
Undergoes free radical substitution
Ex - Ethane + Cl2
Conditions : UV light
Ex -
Initiation Cl2 --> 2Cl*
Propagation steps : (a) C2H6 + Cl* --> C2H5* + HCl (b) C2H5* + Cl2 --> C2H5Cl + Cl*
Termination steps : (a) C2H5* + Cl* --> C2H5Cl (b) Cl* + Cl* --> Cl2 (c) C2H5* + C2H5* --> C4H10

Alkenes
Undergoes electrophyllic addition
(a) With Hydrogen
Conditions : Nickel catalyst , 333K temperature
(b) With water
Conditions : H3PO4 catalyst,water, (273+330)K temperature
(c) With Hot concerntrared KMnO4
The >C=C< bond breaks. If one of the C in the bond have two Carbons attatched to it, it becomes a keytone. Otherwise, it will become a carboxyllic acid.
(d) With cold dilute KMnO4
The >C=C< bond breaks to form a diol. For an example CH2=CH2 will become CH2(OH)CH2(OH).
(e) With Br2 in CCl4
The double bond breaks and the two Bromines go and bond with the two carbons ex - C2H4 --> CH2BrCH2Br
(f) With Br2(aq)
The double bond breaks and one bromine atom bonds with one carbon atom while an OH group go and bond with the other. ex - C2H4 --> CH2BrCH2(OH)
(g) Electrophyllic addition of alkenes
Things to remember : This is a two step reaction.
The Markonikov addition principle : It states that the halogen should go and bond itself with the carbon containing more alkyl (CH3) groups whilst the hydrogen should go and bond with the carbon containg more H atoms. This principle is for HX (where X is a halogen) reaction with alkenes.
(h) Polymerization
Things to remember : Two types, addition polymerization and condensation polymerization
Addition polymerization is with alkenes.
Conditions : High temperature , Highh pressure or High temperature and Zigger-Nata catalyst

Alcohols
Things to remember : Tertiary alcohols cannot be oxidised.
(a) Dehydration
Add conc H2SO4 and heat gently. An alkene will form.
(b) Reaction with PCl5,PCl3 and SOCl2
Replaces OH with a Cl atom.
For PCl3 and SOCl2, it is major condition to heat but not with PCl5.
PCl5 + CH3OH --> CH3Cl + POCl3 + HCl
SOCl2 + 2CH3OH --> SO2 + CH3Cl + H2O
(c) Oxidation
Secondary alcohols become a keytone.
Primary alcohols become carboxyllic acid if heated under reflux. Becomes and aldehyde if it is heated and distilled off.
(d) With Na
2CH3OH + 2Na --> 2CH3ONa +H2
Effervesence etc is observed

Carboxyllic acids and esters
Carboxyllic acids react with NaOH, Na, Na2CO3 etc. Also it reacts with alcohols to make esters. Another thing is that it is partially soluble in water.
(a) Reaction with NaOH, Na and Na2CO3
CH3CO2H + Na --> CH3CO2Na + 1/2H2
CH3CO2H + NaOH --> CH3CO2Na + H2O
2CH3CO2H + Na2CO3 --> 2CH3CO2Na + CO2 + H2O
(b) esterification
Conditions : Add conc H2SO4 and the alcohol. Heat gently.
The C-OH bond in the carboxyllic acid breaks to bond with the alcohol.
CH3CO2H + CH3OH --> CH3CO2CH3 + H2O

ESTERS
(a) Reaction with acids and alkalines
i) acid --> CH3CO2CH3
The ester bond breaks to form the initial carboxyllic acid and alcohol.
ii) alkaline --> CH3CO2CH3
The ester bond breaks to form sodium--thoxide + alcohol
ex -
FOR ACIDS
CH3CO2H --> CH3CO2H + CH3OH
FOR ALKALINE
CH3CO2H ---> CH3CO2Na + CH3OH

 

[JESS1016K]

Cyanohydrins can be made from carbonyl compounds by generating CN–
ions from HCN in the
presence of a weak base.
In a similar reaction, –
CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases.

Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong
base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3

this question is from w09qp1. pls help. the ans is C but I dont know why

 

[Anas Abbal]

can anyone explain this to me how is the answer B