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:
The reaction you are describing appears to be a two-step transformation involving an organocuprate reagent, a brominated substrate, and a nucleophilic substitution or reductive work-up. Based on your description, here is the cleaned-up version of the reactants and conditions (interpreted as best as possible from the formatting):
Step 1:
- Reagents: (CH₃)₂CuLi (Gilman reagent), Br₂CHCH₂OH, LiBr, Li₂CO₃, DMF
- Conditions: Organocuprate substitution in DMF
Step 2:
- Reagents: H₂O, CH₃CN (possibly), KOC(CH₃)₃ (a strong non-nucleophilic base, potassium tert-butoxide)
Step-by-step Explanation:
Step 1 – Nucleophilic Substitution with Gilman Reagent:
(CH₃)₂CuLi is a Gilman reagent, a type of organocuprate that can undergo nucleophilic substitution with alkyl halides, particularly primary halides like BrCH₂CH₂OH (2-bromoethanol).
The organocuprate replaces the bromine atom with a methyl group: BrCH2CH2OH+(CH3)2CuLi→CH₃CH2CH2OH\text{BrCH}_2\text{CH}_2\text{OH} + (CH₃)₂CuLi \rightarrow \text{CH₃CH}_2\text{CH}_2\text{OH}
So the intermediate is propanol (CH₃CH₂CH₂OH).
Step 2 – Dehydration or Elimination Reaction:
In the second step, treatment with a strong base like potassium tert-butoxide (KOtBu) in a protic solvent leads to elimination of water (E2 elimination), forming an alkene: CH₃CH2CH2OH→KOtBu / HeatCH2=CHCH3+H2O\text{CH₃CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{KOtBu / Heat}} \text{CH}_2=\text{CHCH}_3 + H₂O
So the final product is propene (CH₂=CHCH₃).
Summary:
- Intermediate: Propanol (CH₃CH₂CH₂OH)
- Final Product: Propene (CH₂=CHCH₃)
This reaction sequence involves a two-step synthetic transformation commonly encountered in organic chemistry involving organocuprate reagents and elimination reactions.
In the first step, the organocuprate reagent, dimethylcuprate [(CH₃)₂CuLi], reacts with 2-bromoethanol (BrCH₂CH₂OH). Organocuprates are known for their ability to perform nucleophilic substitutions on primary alkyl halides, displacing the halide with an alkyl group. In this case, one of the methyl groups from the Gilman reagent replaces the bromide atom on the substrate, yielding propanol (CH₃CH₂CH₂OH) as the intermediate. The reaction is conducted in polar aprotic solvents like DMF, with LiBr and Li₂CO₃ aiding in halide displacement and neutralization.
In the second step, this intermediate undergoes elimination. Potassium tert-butoxide (KOtBu) is a strong, non-nucleophilic base that favors E2 elimination. Upon heating, it abstracts a β-hydrogen from propanol, promoting elimination of water to form a double bond. The product of this elimination is propene (CH₂=CHCH₃). This step illustrates a common method for converting alcohols into alkenes through base-induced dehydration.
Thus, the reaction sequence efficiently converts a brominated alcohol precursor into an alkene via nucleophilic substitution followed by elimination. The key transformations include SN2 displacement by an organocuprate and E2 elimination under basic conditions.
