How do you convert the following? Sodium benzoate to Benzene
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Solution and Explanation
Decarboxylation of a carboxylic acid salt.
Step 2: Meaning
The carboxylate group ($-COONa$) is removed as a molecule of sodium carbonate.
Step 3: Analysis
Heating sodium benzoate strongly with soda lime (a 3:1 mixture of $NaOH$ and $CaO$) causes the complete removal of the side chain, leaving the bare benzene ring.
Final Answer: By heating sodium benzoate with soda lime (a mixture of $NaOH$ and $CaO$). $C_6H_5-COONa + NaOH \xrightarrow{CaO, \Delta} C_6H_6 \text{ (Benzene)} + Na_2CO_3$
Top CBSE CLASS XII Chemistry Questions
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylanilineAccount for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Top CBSE CLASS XII Carboxylic Acids Questions
- Assertion (A): The pK$_a$ of O$_2$N–CH$_2$–COOH is lower than that of CH$_3$–COOH. Reason (R): The –NO$_2$ group shows an electron-withdrawing effect, which increases the acidic character of O$_2$N–CH$_2$–COOH.
- Complete the following reaction: \[ \text{C}_6\text{H}_5\text{N}_2^+Cl^- \xrightarrow{\text{(i) HBF}_4} \xrightarrow{\text{(ii) NaNO}_2/\text{Cu}, \Delta} \quad ? \]
- Write the structures of A and B in the following reaction: \[ \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{NH}_3, \Delta} A \xrightarrow{\text{Br}_2/\text{NaOH}} B \]
- Assertion (A): Benzoic acid does not undergo Friedel – Crafts reaction.
Reason (R): Carboxyl group is deactivating and the catalyst aluminium chloride gets bonded to the carboxyl group. - Which one of the given compounds is a stronger acid and why?
CH$_2$FCH$_2$COOH or CH$_3$CHFCH$_2$COOH
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