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Thuisopdrachten uitwerkingen OC 1 en 2 $5.96   Add to cart

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Thuisopdrachten uitwerkingen OC 1 en 2

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uitwerkingen van de thuisopdrachten van oc1 en oc2

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  • August 21, 2023
  • 15
  • 2023/2024
  • Case
  • -
  • 9-10
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, thuisopdracht 2 !

SNZ nucleofiele substitutie
in Htlzuigtcldeelectronen naar zich toe door verschil inelectronegativiteit


HI cl evenwicht naar rechts
-



8- → H
>

H H Cl
+ -

HÌÚH wanneer de Hvantkl vervangen wordt door Hsc treedt er NUCSUG OP
↳ ↳
-




/
H H Elektronegativiteit Cl - 3,160=2.55 & Nis een goed NUC .




Hi H H 2ᵉ orde kinetiek
, -10 '
"
H reactiesnelheid is afhankel k [NUC] als [ el ] # K' [HSN][Nice]
Hyn
c- van
H De
- -


> .




,
µ
' >
"
H
go.tl
> ekctroneg '
HO
'


F-
Hsc HZN >
- -


>
vertrekkende
>
H > neemt toe
"
nucleofiliteit
Li / C. N OF neemt
af ±
-




> por > a-
siecnste
nuc
groep reeks nucleofiliciteit
de ionstraal :
> reeks
neemt toe beste vertrekkende
Br
groep
I


SNI . nucleofiele substitutie

"

wanneer een elektrofiel centrum is omgeven door Grote alkylgroepen kan het nuc Niet via SNZ "




mecnreageren vanwege sterische hindering Sterische hindering veroorzaakt door repalsie tussen . de

elektronen van de gevulde orbitalen wanneer 2 moleculen elkaar naderen om te gaan reageren / Botsen

Mech : _




Rvt
RRtc-o-H-natbr-I-CY-BRR-ct.gr
Ri
R - C -

Br + Na '⊖ËH > -



>
nat -
H

R / :< ( Ekc ) (nac ) R / R
sterisch carbokation
snelheid isafh van [ electro / IE
gehinderd intermediair


F- Phe
H
'

Br
-


H
-
Br
H Br Ht µ
de ⑦ lading kan verplaatsen
+ H
H Brvertrekt + H H
H
) m n Hy "
"
) H
< > c


H
delokalisatie structuren

H Br
H
-
=
energetisch gunstig

AIIYI kation all lkation

AHYI Bromide


BCNZYI Kation ,
Met ✗ (=D , BNI ) als vertrekkende sroepkanereenbenzylkation ontstaan dat gestabiliseerd wordt door
delokalisatie en dat kan reageren op benz lische Positie nuc :
-




"


Ê
H "
H "
'
⊕ < >
⊕ "




H - "
H
H
nuc
-0 carbokation reeks

nuc
H
R H H
\
minst stabiel
2 R
p
H +

Rsc -

✗ + NUC : > R> C- MUCH -



R
tertiair secundair Primair

① Komt
nb : Hsc' niet voorin oplossing


ij
IJ
ijij ij ij

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