Question Details

Identify the correct reagents that would bring about the following transformation.

Options

(i) $H_{2} O / H^{+}$

(ii) $C r O_{3}$

(i) $B H_{3}$

(ii) $H_{2} O_{2} / \overset{⊖}{O} H$

(iii) PCC

(i) $B H_{3}$

(ii) $H_{2} O_{2} / \overset{⊖}{O} H$

(iii) $a l k . K M n O_{4}$

(iv) $H_{3} O^{\oplus}$

(i) $H_{2} O / H^{+}$

(ii) PCC

Show Answer

Correct Answer :

Option B

(i) $B H_{3}$

(ii) $H_{2} O_{2} / \overset{⊖}{O} H$

(iii) PCC

(i) BH3, (ii) H2O2 / OH-, (iii) PCC

Solution :

The correct option is:
(i) $B H_{3}$
(ii) $H_{2} O_{2} / O H^{-}$
(iii) PCC

Step-by-Step Explanation:

1. Analysis of the Transformation:
Based on the provided reaction scheme, the starting material is allylcyclohexane (a cyclohexane ring attached to a $- C H_{2} - C H = C H_{2}$ group) and the product is 3-cyclohexylpropanal (a cyclohexane ring attached to a $- C H_{2} - C H_{2} - C H O$ aldehyde group).

To convert a terminal alkene to a terminal aldehyde, a two-step conversion sequence is required: hydration to a primary alcohol followed by selective oxidation.

2. Step 1: Anti-Markovnikov Hydration of the Alkene
If we use direct acid-catalyzed hydration (e.g., $H_{2} O / H^{+}$ ), the addition follows Markovnikov's rule, placing the hydroxyl group ( $- O H$ ) on the more substituted carbon (C2) to form a secondary alcohol. Subsequent oxidation of this secondary alcohol would yield a ketone rather than the desired aldehyde.
To obtain the primary alcohol at the terminal position, we must perform an anti-Markovnikov hydration using hydroboration-oxidation:
(i) $B H_{3}$ adds to the double bond.
(ii) Treatment with $H_{2} O_{2} / O H^{-}$ oxidizes the organoborane intermediate to yield 3-cyclohexylpropan-1-ol ( $C_{6} H_{11} - C H_{2} - C H_{2} - C H_{2} O H$ ).

3. Step 2: Selective Oxidation of the Primary Alcohol
To convert the primary alcohol to an aldehyde, we must use a mild oxidizing agent that does not over-oxidize the aldehyde into a carboxylic acid. Strong oxidizing agents like chromic acid ( $C r O_{3}$ under acidic conditions) or alkaline $K M n O_{4}$ would oxidize the alcohol directly to a carboxylic acid.
Using Pyridinium chlorochromate (PCC), a mild and anhydrous oxidizing agent, selectively oxidizes the primary alcohol to the desired aldehyde (3-cyclohexylpropanal).