IIT-JEE 2011 Paper 1 Offline | Basics of Organic Chemistry Question 27 | Chemistry

IIT-JEE 2011 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Among the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any) is(are)

H₂C = C = 0

H₂C = C = CH₂

IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

In the Newman projection for 2,2-dimethylbutane, X and Y can, respectively, be

IIT-JEE 2010 Paper 1 Offline Chemistry - Basics of Organic Chemistry Question 11 English

H and H

H and C$$_2$$H$$_5$$

C$$_2$$H$$_5$$ and H

CH$$_3$$ and CH$$_3$$

IIT-JEE 2009 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

The correct statement(s) about the compound $$\mathrm{H_3C(HO)HC-CH=CH-CH(OH)CH_3~~(X)}$$ is (are)

The total number of stereoisomers possible for X is 6.

The total number of diastereomers possible for X is 3.

If the stereochemistry about the double bond in X in $$trans$$, the number of enantiomers possible for X is 4.

If the stereochemistry about the double bond in X is $$cis$$, the number of enantiomers possible for X is 2.

IIT-JEE 2008 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

The correct statement(s) concerning the structures E, F and G is(are) :

IIT-JEE 2008 Paper 1 Offline Chemistry - Basics of Organic Chemistry Question 4 English

E, F and G are resonance structures.

E, F and E, G are tautomers.

F and G are geometrical isomers.

F and G are diasteromers.

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