Question

Select the product from the following

I: II: CH3-CH-CH2-CH

OH III: CH3-CH-CH2-CH

OH OH Consider the given

• A. III
• B. I, III
• C. I, II
• D. II, III

Answer 1

Answer: (B)

(2) I, III

Answer 2

The reaction involves the Grignard reagent formed from bromobenzene and a hydroxyl-aldehyde compound.

Step 1: Formation of Grignard Reagent

Bromobenzene reacts with magnesium in ether to form phenylmagnesium bromide.

$\text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{ether}} \text{C}_6\text{H}_5\text{MgBr}$

Step 2: Reaction of Grignard Reagent with the Organic Substrate

The organic substrate given in the reaction scheme is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)CHO}$ (4-hydroxypentanal). However, the structures provided in options II and III, particularly III, suggest that the intended reactant might be $\text{CH}_3\text{CH(OH)CH}_2\text{CHO}$ (3-hydroxybutanal), which has a 4-carbon chain, matching the skeleton of options II and III. Assuming there is a typo in the reactant structure and it should be 3-hydroxybutanal, we proceed with this assumption as it leads to one of the given product options.

The reactant, $\text{CH}_3\text{CH(OH)CH}_2\text{CHO}$, contains two functional groups:

1. A hydroxyl group (-OH), which has an acidic proton.
2. An aldehyde group (-CHO), which is an electrophilic carbon for nucleophilic attack.

Grignard reagents are strong bases and strong nucleophiles.

• Acid-Base Reaction: Grignard reagents react very rapidly with acidic protons (like those in -OH groups).

  $\text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \rightarrow \text{C}_6\text{H}_6 + \text{CH}_3\text{CH(OMgBr)CH}_2\text{CHO}$

  This reaction produces benzene ($\text{C}_6\text{H}_6$), which is Product I. This reaction consumes one equivalent of the Grignard reagent.

• Nucleophilic Addition: After the acidic proton is removed (or if excess Grignard reagent is used), the Grignard reagent can attack the carbonyl carbon of the aldehyde.

  If excess $\text{C}_6\text{H}_5\text{MgBr}$ is used, after the first equivalent reacts with the -OH group, the remaining Grignard reagent will react with the aldehyde group of the resulting alkoxide:

  $\text{CH}_3\text{CH(OMgBr)CH}_2\text{CHO} + \text{C}_6\text{H}_5\text{MgBr} \rightarrow \text{CH}_3\text{CH(OMgBr)CH}_2\text{CH(OMgBr)C}_6\text{H}_5$

Step 3: Hydrolysis

The reaction mixture is then hydrolyzed with $\text{H}_3\text{O}^+$. This protonates the alkoxide intermediates back to alcohols.

$\text{CH}_3\text{CH(OMgBr)CH}_2\text{CH(OMgBr)C}_6\text{H}_5 \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH(OH)CH}_2\text{CH(OH)C}_6\text{H}_5$

This product is 1-phenylbutane-1,3-diol, which matches Product III.

Therefore, if excess Grignard reagent is used, both the acid-base reaction and the nucleophilic addition to the carbonyl will occur, leading to the formation of both benzene (I) and 1-phenylbutane-1,3-diol (III).

Let's re-verify the options based on the structures given:

I: Benzene (C6H6)

II: CH3-CH(OH)-CH2-CHO (3-hydroxybutanal) - This is the reactant (if we assume the typo). It's not a product of the reaction.

III: CH3-CH(OH)-CH2-CH(OH)-C6H5 (1-phenylbutane-1,3-diol) - This is the product of nucleophilic addition.

Since both I and III are formed under typical conditions where Grignard reagent reacts with a molecule containing both an acidic proton and a carbonyl group (assuming excess Grignard reagent), option (2) which states "I, III" is the correct choice.

Explanation of the solution:

1. Bromobenzene reacts with Mg/ether to form phenylmagnesium bromide (C6H5MgBr), a Grignard reagent.
2. The substrate, assumed to be 3-hydroxybutanal (CH3CH(OH)CH2CHO) due to options, has an acidic -OH group and an aldehyde -CHO group.
3. Grignard reagents are strong bases. They first react with the acidic -OH group, producing benzene (C6H6, Product I) and an alkoxide.
4. If excess Grignard reagent is used, the remaining Grignard reagent acts as a nucleophile and attacks the electrophilic carbon of the aldehyde group in the alkoxide.
5. Subsequent hydrolysis (H3O+) converts the intermediate alkoxides into alcohols, yielding 1-phenylbutane-1,3-diol (CH3CH(OH)CH2CH(OH)C6H5, Product III).
6. Therefore, both I and III are the products.