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Chem doubts

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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

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