10, 13) Nauman Questions with Solved
Solutions.
2 consecutive eliminations
Products: 2 C=C or 1 Carbon Carbon triple bond
Reagents: NaNH2, THF/DMF, heat - Answer Synthesis of Alkynes via consecutive elimination
Form alkane from alkene
syn addition
Reagents: H2, Pd/Pt/Ni - Answer Addition of H2 to alkenes--hydrogenation
1. Alkyne to alkane
Reagents: Excess/ H2, Pd/C
2. Lindlar's Catalyst (cis-alkene)
Reagents: H2, Lindlar's Catalyst
3. Na or Li, NH3 (liq), -78 degrees C (trans-alkene)
Reagents: Na or Li, NH3 (liq), -78 degrees C - Answer Addition of H2 to alkynes
**Do NOT have to know mechanism**
Multistep Synthesis
Reagents: NANH2, 2 degree, 1 degree or methyl halides
Sn2 Reaction - Answer Alkyne Synthesis
Carbocation intermediate
Carbocation rearrangement
Markovnikov
Reagents: HCl, HBr, HI
H--> attaches to C with greatest number of H's
X--> attaches to C with least number of H's - Answer Addition of hydrogen halides (HX)
**Know Mechanism**
Carbocation intermediate
Carbocation rearrangement (hydride or methyl shift)
, Markovnikov
Reagents: H3O+, H2O/ H+, [H2SO4]
most substituted alcohol produced - Answer Addition of water (H2O+/ H+) and alcohols
(ROH/ H+)
**Know Mechanism**
Bromonium intermediate
anti-addition
cis/ trans symmetrical
Reagents: Cl2/ CH2Cl2 or Br2/ CH2Cl2)
Halohydrin-->molecules with OH group and halide
Reagents: Cl2 or Br2/ H2O
OH goes on more substituted Carbon
Markovnikov
No Carbocation rearrangement
Anti-addition
bromonium intermediate - Answer Addition of Halogens (Cl2/ CH2Cl2 or Br2/ CH2Cl2)
Markovnikov
Mercuronium intermediate
2 Steps
1. Oxymercuration--> Reagent: Hg(OAc)2
2. Demercuration--> Reagent: NaBH4 - Answer Oxymercuration-Demercuration
Anti-Markovnikov--> H goes to C with least number of H's
syn-addition
Forms 2 products when cyclic
Reagents: BH3, THF/ H2O2, OH-, H2O - Answer Hydroboration-Oxidation
Anti-Markovnikov
Free radical addition
3 steps
1. Initiation
2. Propagation
3. Termination