Regardless of the specific technique used (GC, HPLC, TLC, etc.), what is the fundamental mechanism required for all chromatographic separations to occur?

1. Use of molecules that are soluble in water

2. Use of inert carrier gas

3. Calculation of Rf value for the molecule separated

4. Use of a mobile and a stationary phase

Subtopic:  Quantitative Analysis of Organic Compounds |
 68%
Level 2: 60%+
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The number of σ and π bonds in the molecule C6H6 are -

1. 6 C – C sigma ( σ C - C )  bonds, 5 C–H sigma ( ( σ C - H )  bonds, and 3 C=C pi ( π C - C ) 
2. 6 C – C sigma ( σ C - C )  bonds, 5 C–H sigma ( ( σ C - H )  bonds, and 2 C=C pi ( π C - C ) 
3. 6 C – C sigma ( σ C - C )  bonds, 6 C–H sigma ( ( σ C - H )  bonds, and 3 C=C pi ( π C - C ) 
4. 6 C – C sigma ( σ C - C )  bonds, 6 C–H sigma ( ( σ C - H )  bonds, and 2 C=C pi ( π C - C ) 

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 88%
Level 1: 80%+
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The number of σ and π bonds in the molecule C6H12 are:

1. 7 = (σC-C) , 11 =  (σC-H), and 0= π 

2. 6 = (σC-C) , 12 =  (σC-H), and 0= π 

3. 12 = (σC-C) , 6 =  (σC-H), and 1= π 

4. 5 = (σC-C) , 13 =  (σC-H), and 1= π 

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 64%
Level 2: 60%+
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The number of σ and π bonds in the molecule CH2Cl2 are -

1.  2 = (σC-Cl) , 1= (σC-H) , and 1= π

2. 2 = (σC-Cl) , 2= (σC-H) , and 0= π

3. 2 = (σC-Cl) , 1= (σC-H) , and 1= π

4. 2 = (σC-Cl) , 1= (σC-H) , and 1= π

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 93%
Level 1: 80%+
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Chlorine atom can be classified as:

1. Carbocation 2. Nucleophile
3. Electrophile 4. Carbanion
Subtopic:  Nucleophile & Electrophile |
Level 3: 35%-60%
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C1H2=C2=O

The correct hybridization states of carbon atoms in the above compound are - 

1. C1= sp , C2= sp3 2. C1= sp, C2= sp
3. C1= sp, C2= sp 4. C1= sp , C2= sp2
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 92%
Level 1: 80%+
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The correct hybridization states of carbon atoms marked as 1,2,3 in the following compound are:
\({ }^1 \mathrm{CH}_3-\mathrm{ }^2{C} \mathrm{H}={}^3{\mathrm{C}} \mathrm{H}_2 \)

1. C1= sp , C2= sp, C3= sp2 2. C1= sp2 , C2= sp, C3= sp3
3. C1= sp3 , C2= sp, C3= sp2 4. C1= sp3 , C2= sp, C3= sp3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 94%
Level 1: 80%+
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The correct hybridization states of carbon atoms in the following compound are -     C1H2=C2H-C3N
 

1. C1= sp , C2= sp, C3= sp2 2. C1= sp2 , C2= sp,  C 3 = sp3
3. C1= sp2 , C2= sp,  C 3 = sp 4. C1= sp3 , C2= sp,  C 3 = sp3
Subtopic:  Hybridisation & Structure of Carbon Compounds |
 90%
Level 1: 80%+
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The correct hybridization states of carbon atoms in C6H6 (cyclic compound) is/are:

1.  sp

2. sp

3. sp3

4.  All of the above

Subtopic:  Hybridisation & Structure of Carbon Compounds |
 84%
Level 1: 80%+
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During estimation of nitrogen present in an organic compound, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H2SO4. The percentage of nitrogen in the compound is-

1. 46.0 %

2. 51.0%

3. 56.0%

4. 49.0%

Subtopic:  Quantitative Analysis of Organic Compounds |
 76%
Level 2: 60%+
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