Question

What would be the product D in the above reaction sequence?

• A. 1° alcohol
• B. 2° alcohol
• C. 3° alcohol
• D. cyclohexene

Answer 1

Answer: (C)

3° alcohol

Answer 2

The reaction sequence is as follows:

1. Cyclohexene reacts with NBS (N-bromosuccinimide). NBS is a reagent for allylic bromination. In cyclohexene, the allylic positions are the carbons adjacent to the double bond. Both allylic positions are equivalent due to symmetry. So, allylic bromination occurs. Cyclohexene + NBS -> 3-bromocyclohexene (A)

2. 3-bromocyclohexene (A) reacts with Mg. This is the formation of a Grignard reagent. 3-bromocyclohexene + Mg -> Cyclohex-2-en-1-ylmagnesium bromide (B)

3. Product B (Cyclohex-2-en-1-ylmagnesium bromide) reacts with methyl acetate ($\text{CH}_3\text{COOCH}_3$) followed by treatment with $\text{H}^+$ to form C. The reaction of a Grignard reagent with an ester involves the addition of two equivalents of the Grignard reagent to the ester, followed by hydrolysis, to form a tertiary alcohol.

   First addition: Cyclohex-2-en-1-ylMgBr adds to the carbonyl carbon of methyl acetate. A tetrahedral intermediate is formed, which then eliminates the methoxide group to form a ketone.

   $\text{CH}_3\text{COOCH}_3$ + Cyclohex-2-en-1-ylMgBr -> $\text{CH}_3\text{-C}(-\text{OMgBr})(\text{OCH}_3)\text{-Cyclohex-2-en-1-yl}$ -> $\text{CH}_3\text{-C}(=\text{O})\text{-Cyclohex-2-en-1-yl}$ + $\text{MgBr}(\text{OCH}_3)$

   Second addition: The ketone reacts with another equivalent of the Grignard reagent.

   $\text{CH}_3\text{-C}(=\text{O})\text{-Cyclohex-2-en-1-yl}$ + Cyclohex-2-en-1-ylMgBr -> $\text{CH}_3\text{-C}(-\text{OMgBr})(\text{Cyclohex-2-en-1-yl})\text{-Cyclohex-2-en-1-yl}$

   Hydrolysis: Treatment with $\text{H}^+$ hydrolyzes the alkoxide to a tertiary alcohol.

   $\text{CH}_3\text{-C}(-\text{OMgBr})(\text{Cyclohex-2-en-1-yl})\text{-Cyclohex-2-en-1-yl}$ + $\text{H}^+$ -> $\text{CH}_3\text{-C}(-\text{OH})(\text{Cyclohex-2-en-1-yl})\text{-Cyclohex-2-en-1-yl}$ (C)

   So, C is a tertiary alcohol where the carbon bearing the hydroxyl group is bonded to a methyl group and two cyclohex-2-en-1-yl groups.

4. Product C is treated with (i) Trace of Grignard reagent (B) and (ii) $\text{H}^+$ to form D. If C is a tertiary alcohol, it has an acidic proton on the hydroxyl group. A Grignard reagent is a strong base and will react with this acidic proton, forming an alkoxide.

   $\text{CH}_3\text{-C}(-\text{OH})(\text{R})\text{-R}$ + R'MgBr -> $\text{CH}_3\text{-C}(-\text{O}^-)\text{R}\text{-R} \text{MgBr}^+$ + R'H (where R = Cyclohex-2-en-1-yl and R' = Cyclohex-2-en-1-yl)

   Since only a trace of Grignard reagent is added, it will deprotonate some of the alcohol molecules. Subsequent treatment with $\text{H}^+$ will protonate the alkoxide back to the alcohol.

   $\text{CH}_3\text{-C}(-\text{O}^-)\text{R}\text{-R} \text{MgBr}^+$ + $\text{H}^+$ -> $\text{CH}_3\text{-C}(-\text{OH})\text{R}\text{-R}$ + $\text{MgBr}^+$

   Thus, the reaction of the tertiary alcohol C with a trace of Grignard reagent followed by acid workup will essentially result in the recovery of the tertiary alcohol C. Therefore, D = C.

Product D is the tertiary alcohol (Cyclohex-2-en-1-yl)$_2$-$\text{C}(-\text{OH})\text{CH}_3$. This is a tertiary alcohol.

Let's consider the possibility that C is the ketone $\text{CH}_3\text{-C}(=\text{O})\text{-Cyclohex-2-en-1-yl}$. Then reaction with a trace of Grignard reagent B followed by $\text{H}^+$ would lead to the tertiary alcohol formed by the addition of one equivalent of B to the ketone. This is the same tertiary alcohol as C obtained when two equivalents of B react with the ester.

Based on the typical reaction of Grignard reagents with esters, C is the tertiary alcohol. The reaction from C to D with a trace of Grignard reagent and acid workup leads to the same tertiary alcohol. Therefore, D is a tertiary alcohol.