Question

the following

(cyclohexane ring with double bond to ch₂)
$xrightarrow{1.ce{b2h6/thf} \\ 2.ce{h2o2/oh-}}$ a

1. methyl cyclo hexene
2. methyl cyclo hexane carbaldehyde

Explanation:

To solve this problem, we analyze the reaction:

Step 1: Identify the Reaction Type

The given reaction uses $\ce{B2H6/THF}$ followed by $\ce{H2O2/OH^-}$, which is hydroboration - oxidation of an alkene. Hydroboration - oxidation follows anti - Markovnikov addition and adds $\ce{-OH}$ to the less substituted carbon of the double bond (syn addition, but regiochemistry is anti - Markovnikov).

Step 2: Analyze the Reactant

The reactant is a cyclohexene derivative with a $\ce{=CH2}$ (methylene) group: $\ce{C6H11 - CH=CH2}$ (cyclohexyl ethene).

Step 3: Predict the Product (A)

In hydroboration - oxidation:

• The $\ce{B2H6}$ (borane) adds to the double bond, with $\ce{H}$ attaching to the more substituted carbon and $\ce{B}$ to the less substituted carbon (anti - Markovnikov).
• Subsequent oxidation with $\ce{H2O2/OH^-}$ replaces $\ce{B}$ with $\ce{-OH}$, resulting in an alcohol.

For $\ce{C6H11 - CH=CH2}$:

• The double bond is between $\ce{CH2}$ (less substituted, 1° carbon) and the cyclohexyl - attached carbon (more substituted, 2° carbon).
• After hydroboration - oxidation, $\ce{-OH}$ adds to the $\ce{CH2}$ (less substituted) carbon.

Thus, the product (A) is cyclohexylmethanol (structure: $\ce{C6H11 - CH2 - OH}$).

Answer:

The product \( \boldsymbol{A} \) is cyclohexylmethanol (structure: \( \ce{C6H11 - CH2 - OH} \)).