Chem doubts

[Phosu]

nov 09 pp1 var 1
que 4, 7,  21(how do we find the double bonds)?, 29 (please explain mechanism),

[ruby92]

attach d paper
i cant download it for sme reason :S:s

[Phosu]

ok here you go

[nid404]

4) lone pair-lone pair repulsion is maximum, followed by lone pair-bond pair, least is bond pair-bond pair

S had 2 lone pairs, 2 bond pairs,which is 104.5 degrees(x)
C has 4 bond pairs and no lone pairs, which gives rise to angle of 109.5 degrees(y)
N has 3 bond pairs and 1 lone pair, which is 107.5(Z)

so order is y>z>x
which is C

7)
in A ox state of Cr before is +6 and after rxn is also +6
x+ (4X-2)=-2
and later
x+ (3X-2)=0
x=+6
in B it goes from +6 to +6
I'll explain how
before reaction, the compound has a charge of -2
x+ (4X-2)=-2
x=+6
After rxn
2x+(7X-2)=-2
x=+6
In C
x+(2X-2)+(2X-1)=0
x=+6
after rxn
x+(2X-4)=-2
x=+6

In D
x+(2X-2)+(2X-1)=0
x=+6
After rxn
2x+(3X-2)=0
x=+3

Hence reduction in ox no in D

21)Total no of double bonds
3 in the cyclohexane ring
1 in the aldehyde (C=O)
1 is the alkene C=C between C11 and C12

so total of 5

cis isomer has the higher groups(R) on the same side.

Hence A

[nid404]

29) CN^- will replace Br
2-ethyl-3-methylbutanoic acid has a total of 7 carbons.
Carbon from CN contributes to this no.
Hence the reactant has 6 carbons point 1
This eliminates option C and D with 7 carbons

It can't be A...cause CN replaces Br which is on the 3rd carbon in the chain...It won't give an ethyl side chain on the 3rd carbon

Hence it's B  check the diagram

[Phosu]

great explanation, Thanks alot

[nid404]

 

[Phosu]

how do we know which structures will be formed?

[nid404]

1 n 2

not 3 because when u split 3, u get one ethyl radical and another butyl radical...
It should be a propyl-propyl radical joining together

[sweetie]

i have doubts tooo 

nov.06
Q40> Why is no.3 wrong? isnt c-o double bond a ketone?
Q37> PLZ XPLAIN
Q21>  PLZ XPLAIN
Q14, Q11, Q6, Q9

THANX IN ADVANCE 
                                     

[nid404]

after im done with my school work

[MaNi_DaDuDe]

after im done with my school work

Great work you doing Nid, as always.

+rep.

[sweetie]

Q9> da ans. is A, but if so why is da initial rate of da reaction lower than the original exp.?
Q.16> plzz xplain

Q.17> mentioned in da que. dat excess of oxygen is available so i thought dat da
CO----> CO₂ and da SO₂-------> SO₃
rrrrrrrrite???




plzzzzzzzzzzz reply

[nid404]

Q9)
I don't knw how the reaction rate came in here
Anyway I'll explain

Kc= [X₂Y]²
        [X₂]²[Y₂]

Kc=2
ratio of X2Y and X2 is the same....hence conc same....
they get cancelled in the eqn above...
therefore 1/[Y₂] =2
conc of [Y₂]= 1/2=0.5
for reaction two

 Ratio of Y₂: X₂: X₂Y
               1             : 2                   :2
Kc= [0.5]^1/2 [1]
        [1]
=1/root 2
    

[nid404]

Q16)  Cl is more electronegative than Iodine. This is because its radius is much smaller...HCl bond is therefore stronger, shorter bond length

Q17) Ok...the ans to that is..the conversion of SO2 to SO3 is a reversible reaction and is exothermic, at high temperature when burning in the air, it is bound not to happen...It require a catalyst too(V2O5 as in contact process)

hope that helps