Chemistry : P1 and P2 doubts only (Only AS allowed)

[syed1995]

dont we heat the reaction mixture with reflux

All reactions of Haloalkanes are heat under reflux.. that's not the issue.. does the reaction with water takes place or not?

 

[A star]

PRODUCTS^moles/REACTANTS^moles

[CH3CO2C2H5][H2]/[CH3COOH][C2H5OH]

Kc = 4
Initial
Reactants=moles Ethanol=Ethanoic Acid = 0.5 moles
Products=moles of Water=Ester = 0.1 moles

Final would be
Reactants - x = ethanol-x , acid-x = (0.5-x) for both
Products +x = water+x,ester +x = 0.1+x for both

Now substitute that in the equation above...

(0.1+x)^2/(o.5-x)^2 = 4

take square root on both sides..

o.1+x/0.5-x = 2
0.1+x=1-2x
3x=0.9
x=0.3

Now calculate the no of moles.. by substituting x ..

Reactants = 0.5-x = 0.2 each
Products = 0.1+x = 0.4 each

Answer correct?

yupo me getting same

 

[A star]

Haloalkane + aq NaOH is a reaction to make alcohol.. but is there a reaction between Haloalkanes and Water to make alcohol when heated?

not under normal conditions i guess

 

[A star]

All reactions of Haloalkanes are heat under reflux.. that's not the issue.. does the reaction with water takes place or not?

you need a autocatalyst here

 

[MustafaMotani]

CH3CO2H + C2H5OH ----> CH3CO2C2H5 + H2O
0.5 ---------- 0.5 ---------------1.0-------------1.0

------- are just to separate values
Now we know that ratio is 1:1 if one mole of ethanoic Acid reacts it produce 1 mole of ester

so at Equilibrium :
CH3CO2H + C2H5OH ----> CH3CO2C2H5 + H2O
0.5-x --------- 0.5-x------------1+x-------------1+x

x moles of ethanoic acid reacted to give x moles of ester

Kc ={1+x}^2/{0.5-x}

Kc = 4 just solve it to get x
then yu can just substitue the value of x to get moles of respective substances ..
hope u get it

 

[syed1995]

you need a autocatalyst here

Can ya get me all the Organic reactions of Amine/NH3 .. I don't remember them at all... except for that I know pretty much all the basic reactions... revising the conditions now though...

The two I know are Haloalkane + NH3 --> Can ya give me the equation for this reaction and the final product?

And The nitrile to amine is addition of H2 under Ni catalyst yea?

 

[syed1995]

CH3CO2H + C2H5OH ----> CH3CO2C2H5 + H2O
0.5 ---------- 0.5 ---------------1.0-------------1.0

------- are just to separate values
Now we know that ratio is 1:1 if one mole of ethanoic Acid reacts it produce 1 mole of ester

so at Equilibrium :
CH3CO2H + C2H5OH ----> CH3CO2C2H5 + H2O
0.5-x --------- 0.5-x------------1+x-------------1+x

x moles of ethanoic acid reacted to give x moles of ester

Kc ={1+x}^2/{0.5-x}

Kc = 4 just solve it to get x
then yu can just substitue the value of x to get moles of respective substances ..
hope u get it

isn't it 0.1 :\ ?

 

[A star]

Can ya get me all the Organic reactions of Amine/NH3 .. I don't remember them at all... except for that I know pretty much all the basic reactions... revising the conditions now though...

The two I know are Haloalkane + NH3 --> Can ya give me the equation for this reaction and the final product?

And The nitrile to amine is addition of H2 under Ni catalyst yea?

2CH3X +2NH3 ---> CH3NH2 +H2

 

[syed1995]

2CH3X +2NH3 ---> CH3NH2 +H2

isn't CH3NH2 a radical which would react again with CH3X? And how did a Haloalkane reaction happen without the production of HX ?

 

[A star]

isn't CH3NH2 a radical which would react again with CH3X? And how did a Haloalkane reaction happen without the production of HX ?

sorry am a bit sleepy
CH3X +NH3 ---> CH3NH3+ Cl-
CH3NH3 X+ NH3 ---> CH3NH2 + NH4X

 

[strangerss]

n

i don't think that they will ask this but this seems like a competition between weak and strong oxidising agent so i think that hot concentrated will give the reaction.... anyone confirm?

No i didnt mean if they were put together i meant if the carbon was bonded to a halogen instead of a carbon and hydrogen what would the product be formed?

 

[Manobilly]

PRODUCTS^moles/REACTANTS^moles

[CH3CO2C2H5][H2]/[CH3COOH][C2H5OH]

Kc = 4
Initial
Reactants=moles Ethanol=Ethanoic Acid = 0.5 moles
Products=moles of Water=Ester = 0.1 moles

Final would be
Reactants - x = ethanol-x , acid-x = (0.5-x) for both
Products +x = water+x,ester +x = 0.1+x for both

Now substitute that in the equation above...

(0.1+x)^2/(o.5-x)^2 = 4

take square root on both sides..

o.1+x/0.5-x = 2
0.1+x=1-2x
3x=0.9
x=0.3

Now calculate the no of moles.. by substituting x ..

Reactants = 0.5-x = 0.2 each
Products = 0.1+x = 0.4 each

Answer correct?

JazakAllah ! Thanks a lot! U saved my life

 

[Manobilly]

U

isn't it 0.1 :\ ?

Ur answer was right

 

[Manobilly]

Have a question ,any1 awake here?

 

[PANDA-]

Have a question ,any1 awake here?

ask

 

[Manobilly]

Drawing the dipoles or lone pairs like this question,it is from may/June 2012 paper 23

 

[PANDA-]

Drawing the dipoles or lone pairs like this question,it is from may/June 2012 paper 23

Depends on the reaction... Whether it is neucleophilic substitution or electrophilic addition.
However, I cannot draw it right now, I'm struggling to prepare myself

 

[Manobilly]

When can U? I am struggling to atm

 

[PANDA-]

When can U? I am struggling to atm

I can't really... Sorry.
But check here http://www.chemguide.co.uk/mechanisms/nucsub/hydroxide.html#top

 

[daredevil]

Someone plz give me the reaction between CO and NO