Question

Consider the following sequence of reactions. 
\[ \mathrm{C}_6\mathrm{H}_5\mathrm{COONa} \xrightarrow{\text{NaOH/CaO}, \Delta} \text{X} \xrightarrow{\text{CO} + \text{HCl}, \text{Anhyd. AlCl}_3} \text{Y} \xrightarrow{\text{NaOH}} \text{A} + \text{B} \] 
If A is the reduction product of Y, what is B?

• A. Sodium formate
• B. Sodium phenoxide
• C. Sodium salt of benzoic acid
• D. Sodium salt of salicylic acid

Hint:

Remember these key reactions: Soda-lime decarboxylation:} R-COONa \(\xrightarrow{NaOH/CaO}, \Delta}\) R-H + Na\(_2\)CO\(_3\). Used to prepare alkanes from carboxylic acid salts. Gattermann-Koch reaction:} Benzene + CO + HCl \(\xrightarrow{Anhy. AlCl}_3}\) Benzaldehyde. Used to introduce a formyl group to an aromatic ring. Cannizzaro reaction:} Aldehydes without \(\alpha\)-hydrogens undergo disproportionation (self-oxidation and reduction) in the presence of concentrated base to yield an alcohol and a carboxylic acid salt.

Answer 1

The Correct Option is C

Step 1: Reaction of \( \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} \) (Sodium benzoate) with \( \operatorname{NaOH/CaO} \) and heat (\( \Delta \)) to form X.
This is a soda-lime decarboxylation reaction. Sodium benzoate is the sodium salt of benzoic acid. When a sodium salt of a carboxylic acid is heated with soda lime (a mixture of NaOH and CaO), it undergoes decarboxylation, removing the carboxylate group as \( \operatorname{Na}_2\operatorname{CO}_3 \) and forming a hydrocarbon.
$$ \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} + \operatorname{NaOH} \xrightarrow{CaO, \Delta} \operatorname{C}_6\operatorname{H}_6 + \operatorname{Na}_2\operatorname{CO}_3 $$ So, X is benzene (\( \operatorname{C}_6\operatorname{H}_6 \)).
Step 2: Reaction of X (\( \operatorname{C}_6\operatorname{H}_6 \)) with \( \operatorname{CO} + \operatorname{HCl} \) in the presence of anhydrous \( \operatorname{AlCl}_3 \) to form Y.
This is a Gattermann-Koch reaction. Benzene reacts with carbon monoxide and hydrogen chloride in the presence of a Lewis acid catalyst like anhydrous aluminum chloride (\( \operatorname{AlCl}_3 \)) to form benzaldehyde.
$$ \operatorname{C}_6\operatorname{H}_6 + \operatorname{CO} + \operatorname{HCl} \xrightarrow{\text{Anhy. AlCl}_3} \operatorname{C}_6\operatorname{H}_5\operatorname{CHO} + \operatorname{HCl} $$ So, Y is benzaldehyde (\( \operatorname{C}_6\operatorname{H}_5\operatorname{CHO} \)).
Step 3: Reaction of Y (\( \operatorname{C}_6\operatorname{H}_5\operatorname{CHO} \)) with \( \operatorname{NaOH} \) to form A + B.
Benzaldehyde (\( \operatorname{C}_6\operatorname{H}_5\operatorname{CHO} \)) is an aldehyde that does not possess an \( \alpha \)-hydrogen atom. When such aldehydes are treated with a concentrated solution of a strong base (like NaOH), they undergo a disproportionation reaction called the Cannizzaro reaction. In this reaction, one molecule of the aldehyde is oxidized to a carboxylic acid (which forms its sodium salt in the presence of NaOH), and another molecule is reduced to an alcohol.
$$ 2\operatorname{C}_6\operatorname{H}_5\operatorname{CHO} + \operatorname{NaOH} \rightarrow \operatorname{C}_6\operatorname{H}_5\operatorname{CH}_2\operatorname{OH} + \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} $$ Here:
- The reduction product is benzyl alcohol (\( \operatorname{C}_6\operatorname{H}_5\operatorname{CH}_2\operatorname{OH} \)). - The oxidation product (in the form of its sodium salt) is sodium benzoate (\( \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} \)).
Step 4: Identify B based on the given condition.
The problem states that A is the reduction product of Y.
So, A = \( \operatorname{C}_6\operatorname{H}_5\operatorname{CH}_2\operatorname{OH} \) (benzyl alcohol).
Consequently, B must be the other product, which is the oxidation product, sodium benzoate (\( \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} \)). Sodium benzoate is the sodium salt of benzoic acid.
Step 5: Compare B with the given options.
(1) Sodium formate (\( \operatorname{HCOONa} \)) - Incorrect.
(2) Sodium phenoxide (\( \operatorname{C}_6\operatorname{H}_5\operatorname{ONa} \)) - Incorrect.
(3) Sodium salt of benzoic acid (\( \operatorname{C}_6\operatorname{H}_5\operatorname{COONa} \)) - Correct.
(4) Sodium salt of salicylic acid - Incorrect.
The final answer is \( \boxed{\text{Sodium salt of benzoic acid}} \).