OCHEM NAGORSKI EXAM 4 REAGENTS

How do i determine E/Z?

E is trans, Z is cis!

How to find degrees of unsaturation:

Determine # of C’s (nC), Halogens(nX), and Nitrogens(nN).

The number of Hydrogens (Saturated) would be

= 2nC + 2 -nX + nN

Then divide (Hsat-Hactual)/2 = Degrees of unsaturation

If you see a double bond → H2, and PdC/PtO2/Ni/Ra-Ni

Catalytic hydrogenation: adds a Hydrogen across a O=C Bond or C=C double bond,

is stereospecific, so both hydrogens will go to one face of the bonds. either cis wedge or cis dash.

If you see a double bond → HX

Addition of the halogen, will go to the most substituted side. (Markovinikov addition)

Double bond → HX over H2O2 and light/heat

Anti-markovinikov Addition, where the halogen is added to the LEAST substituted end.

Double bond → just H20 over acid

acid removes its proton, then H+ attacks the hydrogen on the double bond, then H2O is added to the carbocation, then you have a hydronium ion, which then loses a proton.

Joins in a markovinikov fashion, where the OH joins the most substituted end.

Double bond → H2SO4 then → H2O

Addition of a HSO4 LG, which is a great LG, can be worked up with H2O and make an alcohol.

Double bond → BH3 or THF → H2O2/H2O

Anti-markovinikov reaction where the H2O2 leaves an OH on the LEAST subtituted side

if you see a CH-CH bond with a Halogen and an R group on both C’s → Base (2 parts NaNH2/NH3 liquid) then → H2O

Creates a C-C triple bond, as one halogen leaves, makes a double bond, then the very strong base NaNH2 will take the proton, then the other halogen leaves, and the second NaNH2 takes that proton, leaving you with a c-c triple bond.

What if you se a c-c bond with a halogen on each end, but → a weaker base like NaOCH3 over HOCH3?

This will stop at the formation of a c=c rather than a c-c triple bond.

IF i start with a c-c triple bond → NaNH2 over NH3 liquid?

This will deprotonate one side of the c-c triple bond, leaving a carbocation intermediary which can be substituted by a second step.

If I see a C-C triple bond → H2 over PtO2 (Adams catalyst), Pd-C, Ra-Ni with a solvent like methanol, acetic acid, or ethyl acetate.

this will cause the C-C triple bond to be hydrogenated, making it a c-c single bonded alkane.

What if a see a c-c triple bond → H2/Lindlar’s Catalyst?

This stops at the alkene, or c=c bond. The hydrogen are added in a cis fashion, meaning any substituents are also in the Cis

c-c triple bond → Na*, NH3 Liquid (Sodium radical) then → H2O

This stops at a c=c double bond, but in a trans fashion in the final product

What if I see a c-c triple bond → 1HX?

Ex, Ch3CH2C-CH where C-CH is a triple bond

The halogen will attatch to the most substituted end of the c-c triple bond, ie Ch3-CH2-X-C=CH2.

both cis and trans products will be observed.

What if I see a c-c triple bond reacting with excess HX?

the halogen will substitute to the same side of the c-c triple bond twice!

What if I see a c-c triple bond → H+ with H2O solvent?

one of the carbons on the c-c triple bond will become a ketone alpha carbon (C=O with two R groups on the C), and the other becomes a CH2.

What if I see a terminal c-c triple bond, → H2O/H+ over HgSO4?

Then a R-C=O-CH3 will form

what if I see an internal c-c triple bond →H2O/H+ over HgSO4?

you get equal parts of a =O on each of the carbons across the triple bond

what if I see R-C-=-C-R’ triple bond → 1) Ozone 2)H2O?

The ozone will cleave the triple bond which then creates R-COOH (Carboxcylic Acid) or COOH-C-R’, where it clips one side of the triple bond off and replaces it with a COOH group.

What if I see a Primary Haloalkane → H2O ?

The Halide leaves and then forms an alcohol.

what if I have a strong leaving group on a secondary carbon with chirality → H2O

The leaving group is replaced by OH, and you see a racemixation of each.

which is the most oxidized; CO2, COOH, H2C=O, CH3OH, CH4?

CO2 is the most reduced.

What if I see a O=C double bond → H2 over a catalyst (PtO2, Pd-C, or Ni-C) ?

you have an addition of H on the Carbon, leading to a C-OH bond

What if I see a Ketone and Aldehyde → 1) NaBH4/EtOH 2)H+/H2O ?

Fully reduces ketones and aldehydes to an alcohol, but cannot reduce COOH groups.

What if I see a ketone, Aldehyde, or Carboxcylic Acid → LiAlH4/EtOEt then H+/H2O

It will full reduce them to an alcohol

What if i want to oxidize a molecule?

Use a chromium reagent, such as K2Cr2O7, Na2Cr2O7, or CrO3

What if i see a secondary alcohol (2R-CH-OH) → Na2Cr2O7/H2SO4/H2O

This will oxidize the secondary alcohol into a O=C

What if I see a primary alcohol (R-CH2-OH)→ K2Cr2O7 over H2SO4/H2O?

The OH will oxidize to a COOH.

What if I see a primary alcohol → PCC?

This will cause the oxidation to stop at a R-CH=O (aldehyde) rather than all the way to a COOH.

What will KMnO4 do?

Very strong oxidizer, similar to chromium reagents.

2* alcohols → Ketones

1* alcohols → COOH