Question

Compound ‘A’ on heating with soda lime gives propane. Identify the compound 'A'.

• A. \( CH_3-CH_2-CH_2OH \)
• B. \( CH_3CH_2CO_2Na \)
• C. \( CH_3CH_2CH_2CO_2Na \)
• D. \( CH_3COCH_3 \)
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Hint:

Decarboxylation using sodalime removes one carbon from the carboxylate, forming the corresponding alkane.

Answer 1

The Correct Option is C

Step 1: Understanding Decarboxylation with Sodalime
- Sodalime (\( NaOH + CaO \)) is used to decarboxylate carboxylates (RCOO\( ^- \)), removing CO\(_2\) and forming an alkane.
- The general reaction is: \[ RCO_2Na + NaOH \xrightarrow{\text{sodalime}} RH + Na_2CO_3 \] Step 2: Identifying the Correct Compound
- (1) \( CH_3-CH_2-CH_2OH \): This is propanol, which does not undergo decarboxylation.
- (2) \( CH_3CH_2CO_2Na \): This is sodium propanoate. Decarboxylation would give ethane, not propane.
- (3) \( CH_3CH_2CH_2CO_2Na \): This is sodium butanoate, which upon decarboxylation gives propane. Correct.
- (4) \( CH_3COCH_3 \): This is acetone, which does not undergo decarboxylation. c
Step 3: Applying the Decarboxylation Reaction
\[ CH_3CH_2CH_2CO_2Na + NaOH \xrightarrow{\text{sodalime}} CH_3CH_2CH_3 + Na_2CO_3 \] Thus, the correct answer is sodium butanoate (\( CH_3CH_2CH_2CO_2Na \)), which gives propane upon heating with sodalime.
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