When propanoic acid is treated with aqueous sodium bicarbonate, CO₂ is liberated. The ‘C’ of CO₂ comes from:

1. Carboxylic acid group

2. Methylene group

3. Bicarbonate

4. Methyl group

The major product is
\(\xrightarrow{LiAlH_4}\)

1.		2.
3.		4.

Cyanohydrin of 'X' gives lactic acid on hydrolysis. The compound 'X' will be:

1. Acetone                                 

2. Acetaldehyde

3. Propanal                                 

4. HCHO

In this reaction,

\(CH_{3}CHO \ + \ HCN \ \xrightarrow[]{} \ CH_{3}CH(OH)CN \ \xrightarrow[]{\textbf{H.OH}} \\ CH_{3}CH(OH)COOH\)

an asymmetric compound is generated. The acid obtained would be:        

1. 50% D + 50% L-isomer

2. 20% D + 80% L-isomer

3. D-isomer

4. L-isomer

Ethyl ester $\underset{\mathrm{Excess}}{\overset{{\mathrm{CH}}_3\mathrm{MgBr}}{\to }}$ P.

The product P will be

1.		2.
3.		4.

Compound (A) C₅H₁₀O forms a phenyl hydrazone and gives negative Tollen's and iodoform tests. Also, compound (A) on reduction gives n-pentane. Compound (A) is-

What is the chemical reaction of lachrymator or tear gas?

1. C₆H₅COCl                               

2. C₆H₅OC₆H₅

3. C₆H₅COCH₂Cl                         

4. C₆H₅COCH₃

A compound that gives two different alcohols on reduction with ${\mathrm{LiAlH}}_4$ is:

1. ${\mathrm{CH}}_3{\mathrm{COOCH}}_3$                         

2. ${\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5$

3. ${\mathrm{CH}}_3{\mathrm{COCH}}_3$                           

4. ${\mathrm{CH}}_3\mathrm{CHO}$

What is the major organic product formed from the following reaction?

1.		2.
3.		4.

What is present in acetone's enolic form?

1. 9 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 2 lone pairs

2. 8 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 2 lone pairs

3. 10 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 1 lone pair

4. 9 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 1 lone pair