Transcript Document
27) Propose a synthesis for each reaction.
a)
Br 2
hv
Br Br Br Br 2 FeBr 3 Br K + OR + Br Br Br
Formation of alkenes from alkyl halides
H 2 C H OR Br + KBr + HOR Br Br So if we formed the double bond before doing the electrophillic aromatic substitution we would brominate the wrong double bond. This is due to the vinyl double bond being much more reactive than those on the aromatic ring. Br Br 2 FeBr 3
b)
Br 2
hv
OH Br NaOH OH
c)
O Br 2
hv
O O OH Cl
m-chloroperoxybenzoic acid
Br K + OR mClPBA O
d)
Br 2
hv
OH Br K + OR BH 3 H 2 O 2 -OH OH
Alkenes
Hydroboration —Oxidation
• •
Since alkylboranes react rapidly with water and spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H 2 O 2 , ¯ OH).
Oxidation replaces the C —B bond with a C—O bond, forming a new OH group with retention of configuration.
•
The overall result of this two-step sequence is syn addition of the elements of H and OH to a double bond in an “ anti-Markovnikov ” fashion.
Br H 2 C H OR Br BH 2 H BH 2
H O O H BH 2 HOO OH HOO HO O + H 2 O BH 2 OH + BH 3
28) Propose a synthesis for each reaction.
a)
CH 2 CH 2 CH 2 CH 2 CH 3 O ClCOCH 2 CH 2 CH 2 CH 3 AlCl 3 Zn(Hg) HCl
Use 2 steps to avoid the 1,2 H shift during the Friedal-Crafts alkylation.
CH 3 CH 2 CH 2 CH 2 CH 2 Cl AlCl 3 H 3 CH 2 CH 2 C H C H H 2 C Cl AlCl 3 H 3 CH 2 CH 2 C H C H + AlCl 4 CH 2
AlCl 4 H CH 3 CH CH 2 CH 2 CH 3 CH 3 CH + AlCl CH 2 CH 2 CH 3 3 + HCl H 3 CH 2 CH 2 C H C CH 3 Not the same.
CH 2 CH 2 CH 2 CH 2 CH 3
b)
ClCOC(CH 3 ) 3 AlCl 3 CH 2 C(CH 3 ) 3 O Zn(Hg) HCl
2 steps to avoid the methyl migration during alkylation.
ClCH 2 C(CH 3 ) 3 AlCl 3 C H 2 C Cl AlCl 3 C CH 2 + AlCl 4
AlCl 4 H C CH 3 CH 3 CH 2 CH 3 CH 3 C CH 3 CH 2 CH 3 + AlCl 3 + HCl Not C CH 2
29) Draw out the synthesis to produce this intermediate in the synthesis of ibuprofen from benzene.
O ClCOCH(CH3)2 O AlCl 3 Zn(Hg) HCl
Use two steps due to 1,2 H shift that would occur if using the alkylation.
ClCH 2 CH(CH 3 ) 2 AlCl 3 H C H 2 C Cl AlCl 3 H C CH 2 + AlCl 4
AlCl 4 H C C + AlCl 3 + HCl C CH 3
ClCOCH3 AlCl 3 + O O
Then would have to separate using column chromatography.
30)Draw a synthesis for each reaction.
a)
CO 2 H CH 3 Cl AlCl 3 KMnO 4 CO 2 H
Friedal-Crafts alkylation followed by oxidation.
b)
NH 2 HNO 3 H 2 SO 4 NO 2 H 2 Pd-C
Nitration followed by a reduction.
NH 2
c)
CO 2 H Br ClCH3 AlCl 3 Br 2 FeBr 3 Br CO 2 H KMnO 4 Br + Br
Alkylation followed by bromination followed by oxidation.
This order b/c CO 2 H is a meta director and wouldn’t give you the right product.
32. Predit the prodcut.
OH d) OH Sn HCl NO 2 NH 2 OH OH e) NH 2 NH 2 -OH R O R
MECHANISM OF THE WOLFF-KISHNER REACTION
(you are not required to memorize this mechanism) R : O C R ..
+ NaOH high bp solvent R R C ..
N ..
NH 2 : ..
O H ..
hydrazone ketone R R C H ..
N : ..
O H ..
R R C ..
N ..
..
H O H O H H R C R H ..
N R C : : N R H N : gas R C R H H C=O removed alkane
33d and e) Proditct the product.
CN RCl AlCl 3 No reaction ROCl AlCl 3 No reaction -CN is a meta director and a ring deactivator so no Friedal Crafts reactions will occur.
35 b,c and e) Predict the products.
O R ROCl AlCl 3 No reaction N(R) 2 ROCl AlCl 3 No reaction
H N O ROCl AlCl 3 + O R H N O H N O O R
35. Predict the products.
a) HO NO 2 HNO 3 H2SO 4 O 2 N HO NO 2 HO First, no substitution between meta substituents.
Second, -OH is a very strong activator and dominates.
yes yes HO NO 2 no NO 2 NO 2
b) SO 3 H SO 3 H2SO 4 HO HO OH SO 3 H -OH is a much stronger activator than the methyl group so it dominates, so… yes yes OH
c) Cl Cl O O O CH 3 CH 2 Cl AlCl 3 O One substituent has an oxygen with a lone pair connected to the ring so it is a strong activator and dominates Cl yes O O yes So para is already occupied so that leaves the ortho positions which are the same.
So what do you need to know… Mechanisms for the five types of electrophillic aromatic substitution.
And benzylic bromination.
Also know what the following reagents will do in both ways: -OR; -OH; RCO 3 H; BH3, H2O2 and –OH; Zn(Hg) and HCl; KMnO 4 ; H 2 and Pd-C; NH 2 NH 2 and –OH.
In other words if you have a starting material and a product be able to tell me what reagents will cause the reaction.
Also, if you have starting materials and reagents be able to give me the product.