Exam 2 (Reactions)

Question 1

Dow Process (3 steps)

Answer 1

Start with benzene ring with halogen on it. 1)-OH attacks H on C beneath halogen and creates a triple bond between the H and Halogen, kicking off the halogen. 2) -OH attacks again at the bottom of the newly formed triple bond and attaches there creating a lone pair and (-) charge where the halogen was. 3)The new (-) charge attacks a Hydrogen from H2O to cancel out the negative charge which is the final product. A benzene ring with an OH group attached.

Question 2

SNAr (4-5 steps)

Answer 2

Start with benzene ring with halogen and NO2. 1) -OH,NH2, or OR attack at the top of the bond to the halogen (Cl) and connects to that C with the Halogen and kicks the double bond between Cl and NO2 onto NO2 giving a (-) charge on the C attached to NO2. 2) The negative charge is pulled down to create a double bond with the N of NO2 and the double bond between the N and O is broken placing the negative charge on the O. 3) The double bond to the N is transferred back to the ring moving the other double bonds over 1 C. 4) Repeat until the negative charge is pushed to Carbon left of the OH and Cl carbon which will kick the Cl off and yield the final product of a benzene ring with OH and NO2 on Carbons next to each other

Question 3

KMnO4

Answer 3

Adds O=C-OH Carboxylic Acid

Question 4

Zn(Hg)/HCl added to a ring with O=C-R

Answer 4

Removes the O

Question 5

Ring with NO2 + Fe/HCl

Answer 5

Ring with NH2

Question 6

Ring with CH3 attached + 3 eq. Br2 or Cl2/hv

Answer 6

Ring with CBr3 or CCl3 attached

Question 7

Ring + HNO3/H2SO4

Answer 7

Ring with NO2 attached

Question 8

Ring with NO2 attached +tertbutyl attached to O=C-Cl

Answer 8

Ring with NO2 and ketone with tertbutyl as one of the R groups attached

Question 9

Clemenson Reduction

Answer 9

Zn(Hg)/HCl - - Used in Acylation

Question 10

Acylation is better than Alkylation because

Answer 10

Alkylation doesn’t always allow you add to a benzene ring with something on it

Question 11

Alkylation (No Steps)

Answer 11

Start with Benzene ring add something like CH3Cl/AlCl3. End with Benzene ring with CH3 attached.

Question 12

Acylation (No Steps)

Answer 12

Start with Benzene ring with NO2 (or other group) and add Halogen according to directing rules (i.e. ortho, para or meta)

Question 13

Nitration (2 Steps)

Answer 13

Start with Benzene ring + NO2. 1) Double bond attacks NO2 attaching NO2 to the ring and placing a (+) charge one carbon down from NO2 and removing the double bond that attacked. 2) O from H2O attacks the H on the NO2 carbon reforming the double bond that was removed. Final product is Benzene ring with NO2 attached + H(+)

Question 14

Sulfonation (3 Steps)

Answer 14

Start with benzene ring + Something contain Sulfur like SO3. 1) Double bond from ring attacks Sulfur and attached the SO3. 2) Add H2SO4. The (-) charge attacks the H on the same bond the sulfur is attached to recreating the double bond on the benzene ring. 3) Protinate SO3 with H3O. End with Benzene ring with SO3H attached.

Question 15

EAS (5 Steps)

Answer 15

Start with Benzene ring and add a + electrophile like Br2/FeBr3 (Br2 is the electrophile). 1) Double bond from ring attacks Electrophile placing a (+) on the Carbon below where the Electrophile is added on the ring. 2-4) Move + around the ring until just above the electrophile. 5) A (-) base attacks the H attached to the same C as the electrophile and reforms the double bond. End with Benzene ring with Electrophile attached + H+

Question 16

Diels-Alder (Cycloaddition)

Answer 16

Start with Diene and Dienophile. The two come together to make a 6 membered ring with 1 double bond, creating a transition state in the middle with A 6 member ring with a dotted circle in the middle.

Question 17

Adding HBR to 1-butene

Answer 17

Start with 1 butene. Br attacks one carbon over from double bond and attached. Then rearranges over one Carbon on the double bond and removes the double bond.

Question 18

Adding HBr to a diene

Answer 18

Br attacks at the top of one of the double bond and at the bottom of that same bond yielding two products, Br attached to the top + Br attached to the bottom

Question 19

Addition of two dienophiles

Answer 19

= a square

Question 20

Williamson Ether Synthesis

Answer 20

Nuc attack R that is attached to X and yields Nuc-R
X must be a good leaving group (I, Br, OTs OMs)
R must be undhindered (primary or secondary)
Will not work with aromatics because this SN2 reaction attacks from opposite side of leaving group

Question 21

Alkoxy Mercuration (2 Steps)

Answer 21

Start with Double bond + 1)Hg(OAc)2,H20/2)NaBH4. 1) OH attacks at bottom of double bond and attaches removing double bond and placing Hg at the top of the double bond thats been removed. 2) Hg is replaced with H
Remember the substrate (OH in this case) is added Markov.

Question 22

Jones Reagent

Answer 22

H2CrO4 + Acetone: oxidizes alcohol to most oxidized point

Question 23

PCC

Answer 23

Ring with 1 carbon replaced with NH + CrO3Cl

1 step oxidation that stops at aldehyde or ketone

Question 24

Most common reducing agents

Answer 24

H-, NaBH4 (only reduces aldehydes and ketones), LAH

Question 25

Grignard

Answer 25

(-) charge on carbon that attaches MgBr to ring attacks, the H3O protonates to an alcohol
Aldehydes being attacked make secondary alcohols
Ketones make tertiary alcohols
Esters make tertiary alcohols