Determine the intermediates and the product for the following reaction. (11 (CH)Cul Br2 CHyCH2OH LiBr LizCO,, DMF 2) H2O CHC [11 (CH,) Cui 21 H2O KOC(CHjh
The Correct Answer and Explanation is:
To accurately determine the intermediates and the product of the given reaction, we first need to parse and understand the reaction conditions and reagents:
Reaction Summary (Interpreted Format):
- Starting Material: 11-bromoalkane (likely a compound with a Br at the end of a long carbon chain).
- Reagents in Step 1:
- (CH₃)₂CuLi (Gilman Reagent): Organocuprate
- Br₂
- CH₃CH₂OH (ethanol)
- LiBr
- Li₂CO₃ (lithium carbonate)
- DMF (dimethylformamide)
- Step 2:
- H₂O
- CHCl₃ (chloroform)
- Step 3:
- KOC(CH₃)₃ (potassium tert-butoxide)
Likely Reaction Pathway:
Step 1: Substitution via Gilman Reagent
Organocuprates (R₂CuLi) perform nucleophilic substitution (especially SN2) on primary alkyl halides (R–Br). The Gilman reagent (CH₃)₂CuLi substitutes the Br with a methyl group.
Intermediate 1: 11-methylalkane (R–CH₃ replacing R–Br)
Step 2: α-Halogenation or Hydrolysis
The use of Br₂ in ethanol, water, and carbonate in DMF indicates oxidative conditions—possibly forming an α-bromo ketone or alcohol from a methyl-substituted ketone.
Alternatively, the organocuprate could add to a compound to form a ketone intermediate that hydrolyzes in water (step 2) to give a carbonyl-containing intermediate.
Intermediate 2: Likely a ketone or alcohol at the terminal end.
Step 3: E2 Elimination
KOC(CH₃)₃ (a strong, bulky base) promotes E2 elimination, especially from secondary or tertiary alcohols or alkyl halides, favoring formation of alkenes.
Final Product: Internal alkene (possibly a terminal or internal double bond depending on the location of the leaving group).
Explanation (300 words):
This reaction proceeds through organocuprate chemistry, nucleophilic substitution, oxidation/hydrolysis, and elimination steps. In the first step, the Gilman reagent (CH₃)₂CuLi, a lithium diorganocuprate, attacks the alkyl halide (a bromoalkane) in a typical SN2 reaction, replacing the bromine atom with a methyl group. This yields an alkane with a new C–C bond.
In the second step, the reaction mixture contains Br₂, ethanol, lithium bromide, and lithium carbonate in DMF. This indicates oxidative conditions, likely converting the methylated intermediate into a carbonyl compound such as a ketone. Water is then used to hydrolyze any intermediates to stabilize the ketone or alcohol. The presence of CHCl₃ could be inert or used as a solvent.
In the final step, treatment with potassium tert-butoxide—a strong, non-nucleophilic base—leads to E2 elimination. This base preferentially abstracts β-hydrogens anti-periplanar to leaving groups (such as Br or OH), leading to the formation of a double bond (alkene).
Overall, the transformation starts with a halogenated alkane, installs a methyl group through organocuprate substitution, undergoes oxidative conversion to a carbonyl compound, and finally forms an alkene via elimination. Each step highlights classic reactivity patterns: SN2 substitution by Gilman reagents, oxidation/hydrolysis of alkyl chains, and strong base-induced elimination. The product is likely a substituted alkene (internal), derived from the original bromoalkane via C–C bond formation and elimination.
