27) Propose a synthesis for each reaction.

a)

Br

Br2

hv

Br

Br2

FeBr3

Br

Br

Br

Br

+

K+-OR

Br

Formation of alkenes from alkyl halides

Br

H2C

Br

Br

H OR

+ KBr + HOR

Br2

FeBr3

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.

b)

OH

Br2

hv

Br OH

NaOH

c)

O

Br2

hv

Br

K+-OR

mClPBA

O

O

O

OH

Cl

m-chloroperoxybenzoic acid

d) OH

Br2

hv

Br

K+-OR

H2O2

-OH

OH

BH3

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 (H2O2, ¯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.

H2C

Br

Br

H

BH2

H

BH2

OR

H O O H OH HOO + H2O

BH2HOOBH2

O

HO

OH

+ BH3

28) Propose a synthesis for each reaction.

a)

CH2CH2CH2CH2CH3

ClCOCH2CH2CH2CH3

AlCl3

O

Zn(Hg) HCl

Use 2 steps to avoid the 1,2 H shift during the Friedal-Crafts alkylation.

CH3CH2CH2CH2CH2Cl

AlCl3

AlCl3H2C Cl

HC

H

H3CH2CH2C CH2HC

H

H3CH2CH2C

+ AlCl4

CH3HCH3CH2CH2C

CH

CH3

CH2CH2CH3

H

AlCl4

CH

CH3

CH2CH2CH3

+ AlCl3 + HCl

CH2CH2CH2CH2CH3Not the same.

b)

CH2C(CH3)3

OClCOC(CH3)3

AlCl3

Zn(Hg) HCl

2 steps to avoid the methyl migration during alkylation.

ClCH2C(CH3)3

AlCl3

AlCl3H2C ClC

CH2C

+ AlCl4

CH2C

C

CH3

CH2CH3

H

AlCl4

C

CH3

CH2CH3

+ AlCl3 + HCl

CH3

CH3

Not

29) Draw out the synthesis to produce this intermediate in the synthesis of ibuprofen from benzene.

O

OClCOCH(CH3)2

AlCl3 Zn(Hg)

HCl

Use two steps due to 1,2 H shift that would occur if using the alkylation.

ClCH2CH(CH3)2

AlCl3

AlCl3H2C ClC

CH2C

+ AlCl4

HH

CH3C

C

H

AlCl4

C

+ AlCl3 + HCl

ClCOCH3

AlCl3

O

O

+

Then would have to separate using column chromatography.

30)Draw a synthesis for each reaction.

a)

CO2H

CH3Cl

AlCl3KMnO4 CO2H

Friedal-Crafts alkylation followed by oxidation.

b)NH2

HNO3

H2SO4NO2 H2 NH2

Pd-C

Nitration followed by a reduction.

c)CO2H

Br

ClCH3

AlCl3

Br2

FeBr3

Br

Br

+

KMnO4

CO2H

Br

Alkylation followed by bromination followed by oxidation.

This order b/c CO2H is a meta director and wouldn’t give you the right product.

32. Predit the prodcut.

d)OH

NO2

Sn

HCl

OH

NH2

e)OH

O R

NH2NH2

-OH

OH

R

C N

R

RNH2

R C

O

R

+ NH2 NH2

C N

R

RNH..

C N

R

RNH

H

C N

R

RN

H

..

....

..: C

R

R H

N N C

R

R H

H

: :..

....

.... ..

..:

H OHH O

H

:

O H

O H

:

:

..

..

..

..

MECHANISM OF THE WOLFF-KISHNER REACTIONMECHANISM OF THE WOLFF-KISHNER REACTION

hydrazone

-

-

-

-

-

NaOH

gas

ketone

alkane

C=Oremoved

high bp solvent

(you are not required to memorize this mechanism)

33d and e) Proditct the product.

CN

RCl

AlCl3

No reaction

ROCl

AlCl3No 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.

AlCl3No reaction

ROCl

O

R

N(R)2

AlCl3No reaction

ROCl

HN

O

AlCl3

ROClHN

O

O R

HN

OO

R

+

35. Predict the products.

a)

NO2HO

HNO3

H2SO4

NO2HO NO2HO

O2N NO2

NO2HO

Second, -OH is a very strong activator and dominates.

First, no substitution between meta substituents.

no

yes yes

b)

OH

SO3

H2SO4

HOHO

SO3H

SO3H

OH

-OH is a much stronger activator than the methyl group so it dominates, so…

yes

yes

c)

O

O

Cl

O

CH3CH2ClO

Cl

AlCl3O

O

Cl

One substituent has an oxygen with a lone pair connected to the ring so it is a strong activator and dominates

yes

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; RCO3H; BH3, H2O2 and –OH; Zn(Hg) and HCl;

KMnO4; H2 and Pd-C; NH2NH2 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.