Prepare for your Organic Chemistry 1 final exam with these practice questions and answers. This covers nomenclature, functional groups, reactions, and reaction mechanisms.
Q: H2SO4
Answer: Weak base, weak nucleophile
Q: MeOH
Answer: Weak base, weak Nu
Q: OH-
Answer: Strong base, strong Nu
Q: tBu
Answer: Strong base, weak Nu
Q: I-
Answer: weak base, strong Nugood leaving group
Q: SN1 reaction
Answer: weak base/ weak Nu or weak base/ strong Nuresults in mixture of isomerspolar protic solventstepwise rxn2 ‘ or 3’ alkyl halide
Q: SN2 reaction
Answer: strong base/ strong Nu1′ alkyl halidesbackside attack- inversion of stereochemistryconcerted rxnpolar aprotic solvent
Q: O3
Answer: Ozonolysis of an alkene2 products (oxidative cleavage)
Q: O3, H2O
Answer: Ozonolysis of alkynes2 products (oxidative cleavage)
Q: OsO4, NaHSO4, H2O
Answer: syn diol from alkeneConcerted reaction
Q: H2, Lindlar’s
Answer: product is a cis alkene
Q: LiAlH4 (LAH), H2O
Answer: epoxides -> alcoholshalides -> alkanesSN2- like reaction: inversion of stereochemistry/ backside attack
Q: mCPBA
Answer: direct epoxidation of alkenesproduct is a trans diol
Q: KMnO4, KOH, H2O
Answer: product is a syn diol
Q: Pd/ C2H2
Answer: reduces an alkyne to an alkane
Q: Na deg./ NH3
Answer: alkyne to a trans alkene
Q: TsCl, pyrimidine
Answer: replaces OH with TsCl
Q: HBr, H2O2, hv
Answer: alkene -> alkaneadds Brno stereochemistry
Q: 1) mCPBA, 2) H2O, H+
Answer: epoxide intermediate, OH group opens ringproduct is a trans diol
Q: SOCl2, pyrimidine
Answer: replaces an OH with ClSN2-like reaction
Q: POCl3, pyrimidine
Answer: E2 reaction on an alcohol
Q: PBr3
Answer: SN2 reaction with Br- on an alcohol
Q: 1′ R-OH + H-X
Answer: R- H2O + X-, forms carbocation intermediate + X- + H2OIdentity of X- matters: strong Nu (I-, Br-, Cl-)= SN1, weak Nu (HSO4-, TsO-)= E1
Q: Nu strength: periodic trends
Answer: Size: smaller/ more linear= more nucleophilic (can react at more sites)Electronegativity: more electronegative= less nucleophilic (C > N > O > F)Polarizability: more polarizable = more nucleophilic (I > Br > Cl > F)
Q: 2H2, PdC
Answer: reduces alkyne to an alkane4 new C-H bonds formed
Q: 1′ alkyl halide + KI, DMSO
Answer: SN2 rxnreplace alkyl halide with Iinversion of sterechemistry
Q: Factors that stabilize charge
Answer: 1. resonance- charge over greater volume2. induction- slight positive/ negative charges stretched out across molecule3. periodic trends- electronegativity stabilizes charge (O- > N- > C-); size- bigger atoms = more stable
Q: Nucleophile strength
Answer: stable anion= weak bases (strong acid), potentially weak Nuunstable anion= strong base (weak acid), potentially strong Nu
Q: Retrosynthesis strategies
Answer: 1. functional group transformations?2. moving functional groups?3. changing carbon framework?4. stereochemistry changes? stays the same?
Q: Radical chain reaction mechanism
Answer: 1. initiation- make radicals2. propagation- make product and more radicals3. termination step- destroy radicals
Q: Makes an alkene
Answer: Elimination of R-X or R-OH (E1 or E2)Reduction of an alkyne (Pd, C2H2-> alkane; Lindlar’s, H2 -> cix alkene; Na deg., NH3 -> trans alkene)
Q: Making an alkyne
Answer: Di-halide + NaNH2
Q: Cis alkenes
Answer: always form alkynesalkene (trans) -> vicinal dihalide -> alkyne -> alkene (cis)
Q: Leaving group ability
Answer: F < Cl < Br < IGood leaving groups are stabilized anions/ neutral
Q: Good Leaving groups
Answer: TsCl, H2O, Cl-, Br-, I-