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Chemistry: Post your doubts here!

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can anyone explain this to me how is the answer B
In the first termination reaction the two free radicals are Ch3ChCh3 and Ch3Ch2Ch2 which are both possible
In the second one the two radicals are Ch3ChCh3 and Ch3ChCh3 which are both possible
But in the third reaction Ch3 Ch2, which is possible, but the second one is Ch3ChCh2Ch3, which IS NOT produced when propane is dissociates homolytically, but would be possible if it was butane!

Hope this helps you:)
 
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o
In the first termination reaction the two free radicals are Ch3ChCh3 and Ch3Ch2Ch2 which are both possible
In the second one the two radicals are Ch3ChCh3 and Ch3ChCh3 which are both possible
But in the third reaction Ch3 Ch2, which is possible, but the second one is Ch3ChCh2Ch3, which IS NOT produced when propane is dissociates homolytically, but would be possible if it was butane!

Hope this helps you:)


yea thanks got it forgot to consider the free radical
 
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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
Tysm!
 
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Can anyone help me out on the Physics Questions I posted in the physics thread?? < Had to post it here as no one was answering>
 
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Q34
Only 1 is correct because according to Le Chateliers principle Temp increase favours endothermic reaction.
an equilibrium mixture N2F4(g) 2NF2(g); ∆H positive (ENDOTHERMIC)
Q40
Again only 1 is correct as esters react with -COOH groups (Hydrolysis)
With alcohols (ethanol) X X
NaHCO3 is not a strong enough base to either hydrolyze or deprotonate an ester.
 
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I am confused wouldnt alcohol react with ester group as well forming carboxylic acid salt and alcohol? :/
From whatever I've learnt ..I've never come across a reaction in which an alcohol turns an ester in to a Carboxylic acid salt
( where did the salt come from? There was no metal or any electronegative element except for O ..only hydrocarbons...
for e.g ; CH3COOCH3 + CH3CH2OH ---> ?)



Where did you get this from? Plz post a link>
 
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Which pair of reactions could have the same common intermediate?
W CH3CH2CH3 → intermediate → (CH3)2CHCN
X CH3CH(OH)CH3 → intermediate → (CH3)2C(OH)CN
Y CH3CH=CH2 → intermediate → CH3CH(OH)CH3
Z CH3CO2CH2CH2CH3 → intermediate → CH3CH2CH2Br
A W and X B W and Y C X and Z D Y and Z

Ans is B
 
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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

Good work boy !
 
Messages
150
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Which pair of reactions could have the same common intermediate?
W CH3CH2CH3 → intermediate → (CH3)2CHCN
X CH3CH(OH)CH3 → intermediate → (CH3)2C(OH)CN
Y CH3CH=CH2 → intermediate → CH3CH(OH)CH3
Z CH3CO2CH2CH2CH3 → intermediate → CH3CH2CH2Br
A W and X B W and Y C X and Z D Y and Z

Ans is B
The intermediate for
W: CH3CHCH3+ (Hydrogen is lost from central carbon )
X: CH3C(OH)CH3+ (Hydrogen is lost from central carbon)
Y:CH3CHCH3+ (Hydrogen is ADDED to Ch2 due to markonikov's rule)
Z:CH3CH2CH2OH (Alcohol from ester hydrolysis)

So W and Y have the same intermediates!

hope this helped :)
 
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