Resonance structures of a molecule should have :
Resonating structures differ in bonding pattern.
Phenol is less acidic than :
Higher the stability of the corresponding anion, more will be the acidic character of the parent compound.
Higher stability of acetate ions than phenoxide ion is due to equivalent resonating structures in the former
Dipole moment is shown by :
1, 4-Dichlorobenzene (p-dichlorobenzene) and trans1, 2-dichloroethene have zero dipole moment because of their symmetrical structures.
Only two isomeric monochloro derivatives are possible for:
In n-butane, Cl can add at either the first or second carbon giving two isomers.
will give three isomers with Cl group at either of the CH3 groups, second C-atom and 3rd C-atom.
Benzene forms only one single derivative.
Option (d) : will again give two isomers with Cl at either one of the CH3 groups or on the central C-atom
Which of the following have asymmetric carbon atom?
An asymmetric carbon atom is one which is attached with 4 different groups. Hence (c) & (d) are correct.
What is the decreasing order of strength of the bases
TIPS/Formulae : Conjugate base of strong acid is weak while conjugate base of a weak acid is stronger.
Acidic strength of acids, HOH > CH ≡ CH > NH3 > CH3.CH3 Hence the order of strength of bases,
Which of the following compounds will show geometrical isomerism?
Only 2-butene and 1-phenylpropene can show geometrical isomerism (cis-and trans-isomers).
Among the following compounds, the strongest acid is
Order of acidic strength CH3OH > CH ≡ CH > C6H6 > C2H6 ; CH3OH is most acidic because O is more electronegative than C and capable of accommodating negative charge in CH3O– Although alcohols are neutral towards the litmus paper.
Tautomerism is exhibited by
For a carbonyl compound to show tautomerism, it must have at least one H at the α – carbon atom. (a), (c) and (d) show tautomerism.
Tautomerism is not possible
An aromatic molecule will
NOTE : An aromatic species will have :
(4n + 2) p electrons (by Huckel’s Rule)
planar structure (due to resonance)
cyclic structure (due to presence of sp2 - hybrid carbon atoms).
The correct statements(s) concerning the structures E,F and G is (are) –
E and F ; and also E and G differ in position of atom (H), so these are tautomers (not resonating structures.
Geometrical isomers are also diastereomers).
The correct statement(s) about the compound given below is (are)
The given molecule although posseses neither centre of symmetry nor a plane of symmetry (hence optically active) but it has an axis of symmetry (Cn).
NOTE : A Cn axis of symmetry is an axis about which the molecule can be rotated by 360°/n to produce a molecule indistinguishable from the original molecule.
The correct statement(s) about the compound
H3C(HO)HC–CH = CH – CH(OH)CH3 (X) is(are)
In the Newman projection for 2,2-dimethylbutane
X and Y can respectively be
Structural formula of 2, 2-dimethylbutane is
(I) Newman projection using C1–C2 bond
(II) Newman projection using C3–C2 bond
Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are)
Which of the following molecules, in pure form, is (are) unstable at room temperature ?
b and c, being antiaromatic, are unstable at room temperature.
Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct ?
Converting all the structures in the Fischer projection
M and N are diastereoisomers M and O are identical M and P are enantiomers M and Q are diastereoisomers Hence, the correct options are a, b, c.
The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to
In tert butyl cation, carbon bearing positive charge has one vacant p-orbital hence it is σ–p (empty) conjugation or hyperconjugation.
In 2-butene, hyperconjugation is between σ→π* bond.