x arrow ii h pow h2o i naoh100 pow circ c the major product

Releted MCQ Question on
Organic Chemistry >> Carboxylic Acid

Releted Question 1

When acetaldehyde is heated with Fehling’s solution it gives a precipitate of

A. $$Cu$$

B. $$CuO$$

C. $$C{u_2}O$$

D. $$Cu + C{u_2}O + CuO$$

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Releted Question 2

In the Cannizzaro reaction given below,
\[2PhCHO\xrightarrow{^{-}OH}PhC{{H}_{2}}OH+PhCO_{2}^{-},\]
the slowest step is

A. the attack of $$^ - OH$$ at the carbonyl group,

B. the transfer of hydride to the carbonyl group,

C. the abstraction of proton from the carboxylic acid,

D. the deprotonation of $$PhC{H_2}OH.$$

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Releted Question 3

When propionic acid is treated with aqueous sodium bicarbonate, $$C{O_2}$$ is liberated. The $$'C'$$ of $$C{O_2}$$ comes from

A. methyl group

B. carboxylic acid group

C. methylene group

D. bicarbonate

View Answer

Releted Question 4

Benzoyl chloride is prepared from benzoic acid by

A. $$C{l_2},hv$$

B. $$S{O_2}C{l_2}$$

C. $$SOC{l_2}$$

D. $$C{l_2},{H_2}O$$

View Answer

\[\xrightarrow[\left( ii \right)\,{{H}^{+}}/{{H}_{2}}O]{\left( i \right)\,NaOH/{{100}^{\circ }}C}?\]
The major product in the above reaction is

Releted MCQ Question on
Organic Chemistry >> Carboxylic Acid

When acetaldehyde is heated with Fehling’s solution it gives a precipitate of

In the Cannizzaro reaction given below,
\[2PhCHO\xrightarrow{^{-}OH}PhC{{H}_{2}}OH+PhCO_{2}^{-},\]
the slowest step is

When propionic acid is treated with aqueous sodium bicarbonate, $$C{O_2}$$ is liberated. The $$'C'$$ of $$C{O_2}$$ comes from

Benzoyl chloride is prepared from benzoic acid by

Practice More Releted MCQ Question on
Carboxylic Acid

Practice More MCQ Question on Chemistry Section

\[\xrightarrow[\left( ii \right)\,{{H}^{+}}/{{H}_{2}}O]{\left( i \right)\,NaOH/{{100}^{\circ }}C}?\] The major product in the above reaction is

Releted MCQ Question on Organic Chemistry >> Carboxylic Acid

When acetaldehyde is heated with Fehling’s solution it gives a precipitate of

In the Cannizzaro reaction given below, \[2PhCHO\xrightarrow{^{-}OH}PhC{{H}_{2}}OH+PhCO_{2}^{-},\] the slowest step is

When propionic acid is treated with aqueous sodium bicarbonate, $$C{O_2}$$ is liberated. The $$'C'$$ of $$C{O_2}$$ comes from

Benzoyl chloride is prepared from benzoic acid by

Practice More Releted MCQ Question on Carboxylic Acid

Practice More MCQ Question on Chemistry Section

\[\xrightarrow[\left( ii \right)\,{{H}^{+}}/{{H}_{2}}O]{\left( i \right)\,NaOH/{{100}^{\circ }}C}?\]
The major product in the above reaction is

Releted MCQ Question on
Organic Chemistry >> Carboxylic Acid

In the Cannizzaro reaction given below,
\[2PhCHO\xrightarrow{^{-}OH}PhC{{H}_{2}}OH+PhCO_{2}^{-},\]
the slowest step is

Practice More Releted MCQ Question on
Carboxylic Acid