common questions Flashcards

1
Q

Give reasons:
(i) Oxidation of aldehydes is easier than ketones
(ii) Benzoic acid is a stronger acid than acetic acid

A

(i) As aldehydes contain H atoms on the carbonyl group but ketones do not.
Cleavage of C—H bond in aldehydes is easier than cleavage of C—C bond in
ketones.
(ii) This is because of greater electronegativity of sp2
hybridised carbon to which
carboxyl carbon is attached in benzoic acid. or any other suitable answer

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2
Q

Why cannot vitamin C be stored in our body?

A

As vitamin C is water soluble, therefore, it is readily excreted in urine and hence
cannot be stored in the body.

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3
Q
A
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4
Q

What do you mean by rate constant?

A

Rate Constant: It is equal to the rate of reaction when molarity is 1M

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5
Q

(a)Why are low spin tetrahedral complexes not formed
(b)[Fe(H2O)6]
3+is strongly paramagnetic where as [Fe(CN)6]
3-
is weakly
paramagnetic.

A

(a) Low spin tetrahedral complexes are rarely observed because orbital splitting
energies for tetrahedral complexes are not sufficiently large for forcing pairing.
(b)In both the cases, Fe is in oxidation state +3. Outer electronic configuration of
Fe3+is:

In the presence of CN–
, the 3d electrons pair up leaving only one unpaired electron.
The hybridisation involved is d2sp3
forming inner orbital complex which is weakly
paramagnetic. In the presence of H2O (a weak ligand), 3d electrons do not pair up.
The hybridisation involved is sp3d
2
forming an outer orbital complex. As it contains
five unpaired electrons so itis strongly paramagnetic

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6
Q

Give reasons:
(a) n-Butyl bromide has higher boiling point than t-butyl bromide.
(b) Racemic mixture is optically inactive.
(c) The presence of nitro group (-NO2) at o/p positions increases the reactivity of
haloarenes towards nucleophilic substitution reactions.

A

(a) n-Butyl bromide has higher boiling point than t-butyl bromide because it has larger
surface area hence have more Van der Waals’ forces.
(b)Rotation due to one enantiomer is cancelled by another enantiomer.
(c)The presence of nitro group (-NO2) at ortho and para positions withdraws the
electron density’ from benzene ring and thus facilitating the attack of nucleophile.

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7
Q

Give reasons for the following: -
(a) Alcohols are more soluble in water than the hydrocarbon of comparable
molecular masses.
(b)Phenoxide ion is more stable than phenol.
(c) Ortho nitro phenol is more acidic than Ortho-methoxyphenol.

A

(a)Alcohols make H bond with water molecules whereas hydrocarbons do not..
(b) in phenoxide ion delocalisation of negative charge while in phenol separation of
charges.
(c)Ortho nitrophenol is more acidic due to electron withdrawing effect of nitro group
which facilitates release of proton.

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8
Q

Give reasons:

(i) Alpha hydrogen of aldehydes and ketones are acidic in nature.
(ii)Propanone is less reactive than ethanal towards addition of HCN.
(iii) Benzoic acid does not give Friedel-Crafts reaction.

A

(i) due to the strong withdrawing effect of the carbonyl group and resonance
stabilisation of the conjugate base.
(ii) the presence of two methyl groups in propanone hinders more the approach of
nucleophile to carbonyl carbon than in ethanal having one methyl group and two
methyl groups reduce the positivity of the carbonyl carbon more effectively in
propanone than in ethanal.
(iii) Carboxyl group is strongly deactivated and catalyst AlCl3 is a lewis acid that gets
bonded to the carboxyl group strongly.

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9
Q

Name the linkages which join the amino acids in proteins?

A

peptide linkage

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10
Q

(i) While performing practicals it is directed that the bottle of liquid ammonia is to be
5
Page 6 of 6
cooled before opening seal but Suman forgot to cool it before opening. What
consequence she faced ?
(ii)Sia’s father is suffering from high blood pressure but he is advised to consume
less quantity of common salt. Why?
(iii) Out of 1M sucrose and 1M urea solution which has more osmotic pressure?
(iv) A compound CuSO4.5H2O undergoes complete dissociation in water. What will
be its value of Van’t Hoff factor?
(v) If molality of dilute solution is doubled,what will be the value of molal elevation
constant(Kb)?

A

(i)It is cooled to lower down the pressure of ammonia else the gas will bump out of the
bottle.
(ii)More salt use will increase ions in the body fluid which increases blood pressure
(iii)Both have same osmotic pressure.
(iv) i=2
(v) Kb doesn’t depend upon ‘m’ so remain unchanged.

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11
Q

(a) Mn2O7 is acidic whereas MnO is basic.
(b) Among divalent ions of 3d transition series,Mn2+ exhibits maximum
paramagnetism
(c)Actinoids show wide range of oxidation states.
(d) is stable in aqueous solution inspite of having configuration. Why?
(e) The E° values of Mn and Zn is more negative. Give reason.
(f) The transition metals are generally paramagnetic in nature why?
(g) Scandium is a transition element but Zinc is not. Why?

A

(a) Mn has + 7 oxidation state in Mn
metal, valence electrons involved in bonding and are not available. Instead effective
nuclear charge is high and hence it accept electrons and behave as aLewisacid.
(b)Because Mn2+ contain 5 unpaired electrons
(c) This is due to comparable energies of 5f. 6d and 7s
(d)The high energy to transform Cu(s) to
enthalpy.
(e) The stability of the half-filled d sub
dଵ଴configuration in Znଶାare related to their more negative E° V values.
(f) due to incomplete d orbitals and presence of unpaired electron.
(g)Because of fully filled d orbitals in zinc
33
(A)(i) Stronger the base lesser the pKb values. C
(C2H5)2NH
(ii) (C2H5)2NH > CH3NH2 > C6H5
(B)
A= benzoic acid, B= benzamide, C= aniline
+ 7 oxidation state in Mn2O7 and +2 in MnO. In higher oxidation state of the
metal, valence electrons involved in bonding and are not available. Instead effective
nuclear charge is high and hence it accept electrons and behave as aLewisacid.
contain 5 unpaired electrons in their 3d sub shells.
(c) This is due to comparable energies of 5f. 6d and 7s-orbitals.
(d)The high energy to transform Cu(s) to Cuଶା (aq) is balanced by its hydration
filled d sub-shell in Mnଶା and the completely filled
are related to their more negative E° V values.
(f) due to incomplete d orbitals and presence of unpaired electron.
tals in zinc and in Zn2

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12
Q
A
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