Oswal Practice Papers CBSE Class 12 Chemistry Solutions (Practice Paper - 9)

Section-A

1. (b) Pd/BaSO₄

Explanation :

A catalyst used in Rosenmund’s reaction can be prepared by reduction of palladium(II) chloride solution in the presence of BaSO₄.
The palladium catalyst used in Rosenmund’s reduction must be poisoned with BaSO₄ because the untreated catalyst Palladium (Pd) is too reactive and will undergo over reduction forming other products.

2. (a) (a)-(2), (b)-(1), (c)-(4), (d)-(5), (E)-(3)

Explanation :

Lanthanoid oxide is used in TV screens, Lanthanoid are used in production of iron alloy, Misch metal consists of lanthanoid metal + iron, Mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint and mixed oxides of lanthanoids are employed in petroleum cracking.

3. (c) ‘a’ carbon of glucose and ‘b’ carbon of fructose.

Explanation :

In a cyclic structure of glucose or fructose, C atom which is adjacent to oxygen atom is known as anomeric carbon. In the given structures a and b on C which are present adjacent to O atom are anomeric. Both anomeric carbons differ in the configuration of -OH group.

4. (b) CH₃CH₂NHCH₃

Explanation :

The reduction of isonitrile with hydrogen in presence of Ni or Pt is known as catalytic hydrogenation. The —NC is known as isonitrile. In isonitrile, the negative charged carbon work as a nucleophile. It attacks hydrogen and gets protonated. The triple bond shifts to the nitrogen and the lone pair of the nitrogen attacks on another hydrogen and the breaking of the N = C bond results in the secondary amine. Thus, the reduction of CH₃CH₂NC with hydrogen in presence of Ni or Pt as catalyst gives CH₃CH₂NHCH₃.

$$CH_3CH_2NC\xrightarrow{H_2/Ni or Pt}CH_3CH_2NHCH_3$$

5. (d) 1-chloro-2, 2-dimethyl propane

Explanation :

According to IUPAC Nomenclature, the name of the given compound is 1-chloro-2,2- dimethyl propane.

6.(c) CH₃COOH

Explanation :

Ester can be identified by fruity smell at the end of reaction. When alcohol is treated with acid, alcohol gets protonated via H₂SO₄ and attacks acid thus, formation of Ester and removal of water molecule can take place. Ester can be identified by fruity smell at the end of reaction.
$$\underset{Acetic acid}{CH_3COOH}+\underset{Ethanol}{C_2H_5OH}\xrightleftharpoons{abc}\underset{Ethylacetate(fruity smell)}{CH_3COOC_2H_5}+H_2O$$

7. (a) A

Explanation :

Lower the reduction potential better will be reactivity or chemical activity. Among A, B, C and D, Standard electrode potential i.e., reduction potential of A is minimum (–3.05V) i.e., its oxidation potential is maximum which implies ‘A’ has maximum chemical reactivity.

8. (d) Potassium hexacyanidoferrate(III)

Explanation :

According to the IUPAC nomenclature, the IUPAC name of K₃[Fe(CN)₆] is potassium hexacyanidoferrate (III).

9. (d) 3

Explanation :

aG + bH → products
Suppose order of reaction = n
When concentration of both C and H doubled then rate increase by eight times.
Rate = K(reactants)ⁿ
(8) = K(2)ⁿ
(2)³ = K(2)ⁿ
n = 3
When concentration of G is doubled keeping the concentration of fixed, the rate is double.
Then, rate ∝ [G]¹
rate ∝ [G]¹ [H]².

10. (c) 3-methyl-2-butanone

Explanation :

The longest chain is of four carbon. The functional group here is keto group. The numbering will start from left carbon as the functional group should also have the lowest numbering.
The substituent methyl group will be at third carbon.
Hence, the IUPAC name of this structure is 3-methyl-2-butanone.

11. (d) Fe²⁺

Explanation :

Let us compare the electronic configuration of these elements:
Mg²⁺ : 1s²2s²2p⁶ : no unpaired electron.
Ti³⁺ : 1s²2s²2p⁶3s²3p⁶3d¹ : one unpaired electron.
V³⁺ : 1s²2s²2p⁶3s²3p⁶3d² : two unpaired electrons.
Fe²⁺ : 1s²2s²2p⁶3s²sp⁶3d⁶ : four unpaired electrons.
Thus, we see that Fe²⁺ has the most unpaired electrons.

12. (c) Cu₂O

Explanation :

Fehlings solution is a tartaric acid complex of cupric ions.
When acetaldehyde is heated with Fehling’s solution, it gives a red precipitate of Cu₂O.
$$CH_3CHO+2Cu^{2+}+5OH–\xrightarrow{\Delta}CH_3COO–+Cu_2O↓+3H_2O\\\text{Acetaldehyde is oxidised to acetate ion and cupric ions are reduced to cuprous oxide.}$$

13. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

Due to the presence of lone pair of electrons on oxygen atom. Ethers behave as base and form stable oxonium salts with mineral acid. Thus, both assertion and reason are correct statements and reason is the correct explanation of assertion.

14. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

In the presence of enzyme, substrate molecule can be attacked by a reagent effectively because active sites of enzymes hold the substrate molecule in a suitable position. So, enzyme catalysed reactions are stereospecific reactions. Thus, both assertion and reason are correct statements and reason is the correct explanation of assertion.

15. (d) A is false, but R is true.

Explanation :

According to Kohlrausch’s law the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.
$$\mu^∞=\mu\lambda_+^∞+n\lambda_{–}∞\\\text{r = molar conductance at infinite dilution.}\\\lambda_+,\lambda_–=\text{conductance of cations and anions.}\\\text{m, n = number of ions formed.}$$

So, Kohlrausch law helps to find molar cooductivity of weak electrolyte and infinite dilutions. The conductance of weak electrolytes is low and even at infinite dilution the dissociation of electolyte is not complete, so the molar conductivity of a weak electrolyte at infinite dilution cannot be determined
experimentally. Thus, assertion is wrong statement but reason is correct statement.

16. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

Tetrahedral complexes do not show geometrical isomerism because relative positions of the unidentate ligands attached to the central metal atom are same with respect to each other. This type of isomerism arises in heteroleptic complexes. Thus, both assertion and reason are correct statements and reason is the correct explanation of assertion.

Section-B

17. Electronic configuration of Ce : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹ 5d¹ 6s².
Ce³⁺ : [Xe] 4f¹
Magnetic moment can be calculated as:
μ = √n(n+2)
Where,
n = Number of unpaired electrons
For
Ce³⁺ = ₅₄[Xe]4f ¹(only one unpaired electron)
i.e., n = 1
According to ‘spin-only’ formula,
Magnetic moment of μ = √n(n+2)BM
=√1(1+ 2) BM = √3 BM = 1.73 BM

18. (a)

(b) There are haloalkanes that can undergo elimination in two different ways resulting in two different products. Alkenes with less number of hydrogens on the double-bonded carbon atoms are the preferred product. This process is known as Saytzeff’s rule. According to Saytzeff’s rule in
dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

(a) Benzalchloride, undergoes hydrolysis and forms benzaldehyde.

(b) A compound with no point of symmetry and which is non-superimposable on its mirror image is called a chiral compound and the property is called chirality.

19. Aniline is obtained from nitrobenzene by reduction with Sn/Conc. HCl. The chemical reaction can be represented as follows:

20. According to Nernst equation,

$$E_{cell}=E°_{cell}\frac{−2.303RT}{nF}log\frac{[Products]^x}{[Reactants]^y}\\E°_{cell}=\frac{−2.303RT}{nF}log\space K_c\\\text{Where,}\\K_c=\text{equilibrium const.}=\frac{[Products]^x} {[Reactants]^y}\\E_{cell}=0,\text{at equilibrium and at 298 K temperature,}\\E°_{cell}\frac{0.0591}{nF}log\space K_c\\\text{Hence, equilibrium constant}(K_c)\text{ is related to}E°_{cell}\text{and not to}E_{cell.}$$

21. (a)

(b)

Section-C

22. (a)
$$\underset{Methylchloride}{CH_3Cl}\xrightarrow{KCN}CH_3CN\xrightarrow[Catalyst]{H_2NH}\underset{Ethylamine}{CH_3CH_2NH_2}\xrightarrow{NHO_3}CH_3CH_2OH\xrightarrow{PCl_5}CH_3CH_2Cl$$
$$(b)\space\space\space\underset{Ethyl chloride}{CH_3CH_2Cl}\xrightarrow{KCN}CH_3CH_2CN\xrightarrow[\Delta]{H^+/H_2O}\underset{Propanoic acid}{CH_3CH_2COOH}$$

23. (a) Hofmann’s bromamide reaction: When amide is treated with bromide in alkaline solution, and amide yields an amine containing one carbon less than the starting amide.

(b) Carbylamine reaction: Also called as isocyanide reaction which is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.
$$\underset{Primary amine}{R — NH_2}+\underset{Chloroform}{CHCl_3}+\underset{Potassium hydroxide}{3KOH (alc.)}\xrightarrow{\Delta}\underset{Carbylamine}{R — NC}+3KCl + 3H_2O\\\text{For example,}\\\underset{Methanamine}{CH_3 — NH_2}+CHCl_3 + 3KOH (alc.)\xrightarrow{\Delta}\underset{Methyl carbylamine or methyl isocyanide}{CH_3 — NC}+ 3KCl + 3H_2O$$

(c) A coupling reaction: Arene diazonium salts react with highly reactive aromatic compounds such as phenols and amines to form brightly coloured azo compounds.
Ar – N = N – Ar. This reaction is known as coupling reaction.

24. (a) Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.
For example, in a sucrose molecule two monosaccharide units, a-glucose and b-fructose, are joined together by a glycosidic linkage.

(b) Monosaccharides : Ribose, 2-deoxyribose, Sucrose Galactose, Fructose. Disaccharides : Maltose, Lactose.
(c) In α-helix structure of proteins, the peptide chains are coiled to form right handed helix involving hydrogen bonding. In β-pleated structure of proteins, the peptide chains lie side by side joined together by inter-molecular hydrogen bonding.

25. (a) Most of the transition metal ions are coloured both in the solid-state and in aqueous solutions. The colour of these ions is attributed to the presence of an incomplete (n – 1) d-subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colours of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain colour
is absorbed causing the promotion of d-electrons. This is known as d-d transitions. The remaining colours of white light are transmitted and the compound appears coloured.

(b) The electronic configuration of Mn²⁺ is [Ar] 3d⁵ which is half-filled and hence it is stable. Therefore, the third ionization enthalpy is very high, i.e., the third electron cannot be easily removed. In the case of Fe²⁺, the electronic configuration is 3d⁶. Therefore, Fe²⁺ can easily lose one electron to acquire
a 3d⁵ stable electronic configuration.

(c) According to the definition, transition elements are those which have partially filled d-subshell in their elementary state or in one of the oxidation states. Silver (Z = 47) can exhibit a + 2 oxidation state in which it has an incompletely filled d-subshell (4d⁹ configuration). Hence, silver is regarded
as a transition element.
On the other hand, zinc (Z = 30) has the configuration 3d¹⁰ 4s². It does not have partially filled d-subshells in its elementary form or in a commonly occurring oxidation state (Zn²⁺: 3d¹⁰). Therefore, it is not regarded as a transition element

26. The thermal decomposition of SO₂C_l2 at a constant volume is represented by the following equation :
$$SO_2Cl_{2(g)}→SO_{2(g)}+Cl_{2(g)}\\At\space t=0\space P_0\space\space 0\space\space 0\\At\space t=t\space P_0–p\space\space p\space\space p\\\text{After time t, total pressure,}P_t=(P_0–p)+p+p\\⇒P_t=P_0+p\\⇒p=P_t–P_0\\\text{Therefore,}\space P_0–P=P_0–(P_t–P_0)\\=2P_0–P_t\\\text{For a first order reaction,}\\k=\frac{2.303}{t}log\frac{p_0}{p_0-P}\\=\frac{2.303}{t}log\frac{p_0}{2p_0-P_t}\\\text{When t=100 sec, k =}\frac{2.303}{100}log\frac{0.5}{2×0.5-0.6}=2.231×10^{–3}s^{–1}.\\When P_t= 0.65atm,\\⇒P_0+p=0.65\\⇒p=0.65–P0=0.65–0.5\\=0.15 atm\\\text{Therefore, when the total pressure is 0.65 atm, pressure of}SO_2Cl_2 is\\SO_2Cl_2=P_0–p\\=0.5–0.15\\=0.35 atm\\\text{Therefore, the rate of equation, when total pressure is 0.65 atm, is given by}\\Rate = k (P_{SO_{2}}Cl_2)\\= (2.23×10^{–3} s^{–1}) (0.35 atm)\\= 7.8×10^{–4} atm s^{–1}.$$

27. It is given that T₁ = 298 K
T₂ = (298 + 10) K = 308 K
Now, substituting these values in the equation :
$$log\frac{K_2}{K_1}=\frac{E_0}{2.303R}\frac{[T_2-T_1]}{T_1T_2}\\\text{We get}\\log\frac{K_2}{K_1}=\frac{E_0}{2.303×8.314}[\frac{10}{298×308}]\\⇒log2=\frac{E_a}{2.303×8.314}[\frac{10}{298×308}]\\⇒E_a =\frac{2.303×8.314×298×308×log2}{10}\\=\frac{2.303×8.314×298×308×0.3010}{10}\\=52897.78 Jmol^{–1}\\=52.9kJmol^{–1.}$$

28. n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.
The order of increasing polarity is:
Cyclohexane < CH₃CN < CH₃OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH₃OH < CH₃CN < Cyclohexane

Section-D

29. (a) (CH₃)₃CCl is a tert-alkyl halide. It does not undergo nucleophilic substitution with C₂H₅O^⊕ Na^⊕ due to steric reason.
(b) Isobutene.
(c)

tert-butylethyl ether is prepared by the reaction between t-butoxide and ethyl chloride involving SN² mechanism.

30. (a) When the solute does not undergo any dissociation or association in the solution i.e., for nonelectrolyte solution the value of ‘i’ is equal to unity.

(b) The value of i for K2SO4 will be:
K₂SO₄ → 2K+ + SO₄^2–
i = 3.

(c) For an electrolytes if one molecule dissociate into ‘n’ particles and ‘a’ be the degree of dissociation then after dissociation particles formed from one particle are (1– a + na)
This is ‘i’.
So,
i = 1 – a + n α

When the solute undergo association in the solution ‘i’ becomes < 1 but when the solute undergoes dissociation in the solution ‘i’ becomes > 1.

Section-E

31. (a) Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of crosssection 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol κ. If ρ is resistivity, then we can write :
$$k=\frac{1}{\rho},\text{units are}\space Ω^{–1}cm^{–1}or\space S\space cm^{–1}$$

Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.
$$\varLambda_mk=\frac{A}{l},\text{Also,}\space\varLambda_mk=\frac{k×1000}{C}\text{where C = concentration}\\\text{Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).}\\\varLambda_mk=kV$$

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of Λ_m with √c for strong and weak electrolytes is shown in the following plot:

(b) (i) Cell reaction.
$$Cu_{(s)}+2Ag^+_{(aq)}→2Ag_{(s)}+Cu^{2+}_{(aq)}\\n=2.\\\text{Using Nernst equation;}\\E_{cell}=E^o_{cell}-\frac{0.0591}{n}log\frac{[Cu^{2+}]^{1}}{[Ag^+]^2}\space((T=298 K))\\i.e.,\space E_{cell}=(+ 0.80V– 0.34V)–\frac{0.0591}{2}(log)\frac{4}{(0.1)^2}\\=+0.46V-\frac{0.0591}{2}(log400)\\=+0.46 V–0.0295×2.6021\\=0.46V–0.0768V=0.3832 V\\Hence,\space E_{cell}=0.3832V.$$

Hence 3 moles of electrons are required.
i.e., Q = 3 × 1 F = 3 × 96500 C.
as, 1 F = charge of 1 mol of electron.
Also, Q = It
$$t=\frac{Q}{t}=\frac{3×96500}{2}second\\=144,750 sec.\space [1 hour = 3600 second]\\\text{Hence, time in hrs}\frac{144750}{3600}=40.20hrs.$$

Ans. (a) (i) E^o _Cr³⁺/Cr = – 0.74 (Given)
$$E^o_{Cd^{2+}/Cd}=–0.40 V\\\text{The galvanic cell of the given reaction is depicted as :}\\Cr_{(s)}|Cr^{3+}_{(aq)}||Cd^{2+}_{(aq)}|Cd_{(s)}\\\text{Now, the standard cell potential is}\\E^o_{cell}=E^o_R-E^o{_L}\\=–0.40–(–0.74)\\=+0.34V\\\Delta_rG=−nFE^°_{cell}\\\text{In the given equation,n=6}\\F=96487Cmol^{-1}\\E^o_{cell}=+0.34V\\Then,\Delta_rG^°=–6×96487\space C mol^{–1}×0.34 V\\=–196833.48 J mol^{–1}\\=–196.83 kJ mol^{–1}\\Again,\Delta_rG^°=–RTln K\\⇒Delta_rG^°=–2.303 RT log K\\⇒log K=-\frac{\Delta_rG}{2.303 RT}\\=\frac{196.83×10^3}{2.303×8.314×298}\\=34.496\\K=\text{antilog (34.496)}\\= 3.13×10^{34}$$

$$(ii) E^oFe^{3+}/Fe^{2+}=0.77V\\E^o_{Ag^{+}/Ag}= 0.80 V\\\text{The galvanic cell of the given reaction is depicted as :}\\Fe^{2+}_{(aq)}|Fe^{3+}_{(aq)}||{Ag}^+_{(aq)}|Ag_{(s)}\\\text{Now, the standard cell potential is}\\E^o_{cell}=E^o_RE^o_L\\=0.80–0.77=0.03V\\Here,n=1.\\\Delta_rG°=−nFE°_{cell}\\=–1×96487C mol^{–1}×0.03V\\=–2894.61Jmol^{–1}\\=–2.89 kJ mol^{–1}\\Again,\space\Delta_rG^°=–2.303RT\space log\space K\\⇒ log K =\frac{-\Delta_rG}{2.303 RT}\\=\frac{2894.61}{2.303×8.314×298}\\\text{K=antilog} (0.5073)\\=\text{3.2(approximately)}.$$
$$(b)\space\space\space C=0.025 molL^{–1}\\\Lambda_m=46.1Scm^2mol^{–1}\\\lambda^o_{(H^+)}=349.6S.cm^2mol^{–1}\\\lambda^o_{(HCOO^−)}=54.6Scm^2mol^{–1}\\\lambda^o_m{(HCOOH)}=\lambda^o_{(H^+)}+\lambda^o_{(HCOO^-)}\\=349.6+54.6=404.2Scm^2mol^{–1}\\\text{Now, degree of dissociation:}\\\alpha=\frac{\Lambda_m{(HCOOH)}}{\Lambda^o_m{(HCOOH)}}=\frac{46 1}{404.2}\\\text{= 0.114 (approximately)}\\\text{Thus, dissociation constant:}\\k=\frac{c\alpha^2}{(1-\alpha)}\\=\frac{0.025mol\space L^{-1}(0.114)^2}{1-0.114)}\\=3.67×10–4molL–1.$$

32. (a) Geometrical isomers of complex [Co(en)₂Cl₂]⁺:

(b) In the presence of strong field ligand CO, the unpaired d-electrons of Ni pair up so [Ni(CO₄] is diamagnetic but Cl^– being a weak ligand is unable to pair up the unpaired electrons, so [Ni(Cl₄)]^2– is paramagnetic.
In [NiCl₄]^2–complex ion, nickel is in +2 oxidation state and the configuration is 3d⁸. Since the molecule is tetrahedral, it involves sp³-hybridisation as shown below:

The molecule is paramagnetic because it contains two unpaired electrons.
In [NiCl₄] nickel is in O oxidation state and has the configuration 4s² 3d⁴ or 3d¹⁰. The molecule is tetrahedral and involves sp³-hybridisation as given below:

Each CO donates a pair of electrons forming four Ni-CO bonds. The compound is diamagnetic since it contains no unpaired electron.

(c) Spectrochemical series is a series in which the ligands have been arranged in order of increasing field strength.
I^– < Br– < SCN < Cl^– < S^–2 < N3 ^– < F^– < ONO^– < OH^– < SO₄–2 < NO₄ ^– < C₂O₄–2< O^–2 < H₂O ∼ NCS^– <
EDTA^–4 < NH3 ∼ Py < en < bpy ∼ phen < NO₂^– < PR₃ < CH₃ < CN^– ∼ CO
From I– To H₂O are weak field ligands.
From NCS^– To CO are strong field ligands.
It is based on the interaction of metal with ligand. Greater the interaction of the ligand with metal, greater will be the splitting of d-orbitals and thus, strong pairing will take place. It is an experimentally determined series based on the absorption of light by complexes with different ligands.

(a) (i) [Co(NH₃) (Cl) (en)₂]²⁺ Ammine chlorido bis-(ethane-1, 2-diamine) cobalt(III) on Co(27).

Since there are no unpaired electrons, complex is diamagnetic.

(ii) [Ni(H₂O)₂(C₂O₄)₂]^2– Diaquadioxatatonickelate(II) ion Ni(28):

The complex has two unpaired electrons, therefore, it will be paramagnetic.

33. (a) Cyanohydrin: Cyanohydrins are organic compounds having hydroxyl (– OH) and cyano (CN) groups on the same carbon atom.

Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to give cyanohydrin. These reactions are known as cyanohydrin reactions.
$$\underset{Ketone}{RR′C = O + HCN}\xrightarrow{NaCN}\underset{Cyanohydrin}{RR′C(OH)CN}$$
Cyanohydrins are useful synthetic intermediates.

(b) Acetal : Acetals are gem-dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom.

When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal.

(c) Semicarbazone : Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Semicarbazones are useful for identification and characterisation of aldehydes and ketones.

(d) Aldol: A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base, atleast one of which either aldehyde should have an α-hydrogen atom.

(e) Hemiacetal: Hemiacetals are α-alkoxyalcohols or gem-alkoxyalcohols.

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of HCl gas.

(f) Oxime: Oximes are a class of organic compounds derived from aldehydes and ketones. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, then it is known as ketoxime.

On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes.
$$\underset{Hydroxylamine}{>C=O+H_2NOH}\xrightarrow{}>C=N–OH+H_2O$$

(g) Ketal: Ketals are gem-dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups.