MHT CET · Chemistry · Carboxylic Acid Derivatives
The compound from which formic acid cannot be prepared is
- A methyl alcohol
- B carbon monoxide \(+\mathrm{NaOH}\)
- C glycerol
- D methyl magnesium bromide
Answer & Solution
Correct Answer
(D) methyl magnesium bromide
Step-by-step Solution
Detailed explanation
Formic acid (HCOOH) cannot be prepared from methyl magnesium bromide \(\left(\mathrm{CH}_{3} \mathrm{MgBr}\right)\) while from all other reagents, it is obtained as
(a) \(\quad \mathrm{CH}_{3} \mathrm{OH} \quad \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{HCHO} \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{HCOOH}\)
methyl alcohol
(b) \(\mathrm{CO}+\mathrm{NaOH} \longrightarrow \mathrm{HCOONa} \frac{\mathrm{NaHSO}_{4}}{\Delta}\)
\(\mathrm{HCOOH}+\mathrm{Na}_{2} \mathrm{SO}_{4}\)
(c) Glycerol + Oxalic acid \(\frac{383 \mathrm{~K}}{\mathrm{H}_{2} \mathrm{O}}\) HCOOH
\(+\) glycerol
(a) \(\quad \mathrm{CH}_{3} \mathrm{OH} \quad \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{HCHO} \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{HCOOH}\)
methyl alcohol
(b) \(\mathrm{CO}+\mathrm{NaOH} \longrightarrow \mathrm{HCOONa} \frac{\mathrm{NaHSO}_{4}}{\Delta}\)
\(\mathrm{HCOOH}+\mathrm{Na}_{2} \mathrm{SO}_{4}\)
(c) Glycerol + Oxalic acid \(\frac{383 \mathrm{~K}}{\mathrm{H}_{2} \mathrm{O}}\) HCOOH
\(+\) glycerol
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