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Isomerism · Question 10

NEET 2016 (Phase 1)MCQMedium📈 Topic in 0% papers+4−1
1

Which of the following biphenyls is optically active ?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]

Answer

(4) [figure]

Solution

A substituted biphenyl shows optical activity (atropisomerism) when:

(i) bulky groups at the ortho positions restrict free rotation about the central C–C bond, so the two rings stay non-planar, and

(ii) each ring carries two different ortho groups, so the molecule has no plane of symmetry.

(1) Only one ortho group (I) on each ring ⇒ rotation is not sufficiently hindered and a symmetric conformation is accessible ⇒ inactive.

(2) Second ring is unsubstituted, and the first ring has identical ortho groups (CH3, CH3) ⇒ plane of symmetry ⇒ inactive.

(3) Only meta substituents ⇒ free rotation ⇒ inactive.

(4) Each ring has two different bulky ortho groups (Br and I) ⇒ rotation is restricted and the molecule is non-superimposable on its mirror image ⇒ optically active.

Answer: (4)

NEET 2016 (Phase 1)

Keep practising — Isomerism

2

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is

  1. the eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain
  2. the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain
  3. the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain
  4. the eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain.
NEET 2016 (Phase 1)
3

The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon, is

  1. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation
  2. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism
  3. a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol
  4. a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration.
NEET 2016 (Phase 1)
4

Which among the given molecules can exhibit tautomerism? [figure]

  1. III only
  2. Both I and III
  3. Both I and II
  4. Both II and III
NEET 2016 (Phase 2)
5

The possible stereo-structures of CH3CHOHCOOH, which are optically active, are called

  1. atropisomers
  2. enantiomers
  3. mesomers
  4. diastereomers
AIPMT 2015
6

In an SN1 reaction on chiral centers, there is

  1. inversion more than retention leading to partial racemisation
  2. 100% retention
  3. 100% inversion
  4. 100% racemisation
AIPMT 2015
7

The number of structural isomers possible from the molecular formula C3H9N is

  1. 5
  2. 2
  3. 3
  4. 4
AIPMT 2015
8

Given : [figure] Which of the given compounds can exhibit tautomerism?

  1. II and III
  2. I, II and III
  3. I and II
  4. I and III
AIPMT 2015 (Cancelled)
9

Which of the following compounds will undergo racemisation when solution of KOH hydrolyses? [figure]

  1. (i) and (ii)
  2. (ii) and (iv)
  3. (iv)
  4. (i) and (iv)
AIPMT 2014
10

The order of stability of the following tautomeric compounds is [figure]

  1. II > I > III
  2. II > III > I
  3. I > II > III
  4. III > II > I
NEET 2013
11

Given: [figure] I and II are

  1. identical
  2. a pair of conformers
  3. a pair of geometrical isomers
  4. a pair of optical isomers
NEET 2013 (Karnataka)

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