Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “P-Block Elements – Ammonia”.

1. What is the chemical formula of ammonia?
a) NH₂
b) NH₃
c) NH₄⁺
d) NH₅
View Answer

Answer: a
Clarification: Ammonia is a tri-hydride of nitrogen. It is a colorless gas with a highly pungent smell that causes nasal irritation. It is the most stable hydride of any element in group 15.

2. What kind of smell is ammonia recognized by?
a) Acidic
b) Sweet
c) Rotten
d) Pungent

Answer: d
Clarification: Ammonia has a highly pungent smell. With a sharp, pinching smell, it causes severe irritation in the nose and throat. This pungent smell of ammonia is used as a means to identify the presence of ammonium cation, NH₄⁺ in salt analysis.

3. How many unshared pair of electrons does an ammonia molecule have?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: The nitrogen atom in NH₃ (ammonia) bears one lone pair of electron. Each single electron of hydrogen is bonded to the nitrogen atom. Three of out five electrons of nitrogen are involved in bonding whereas the unbounded two electrons make up the single unshared pair of electron.

4. What happens when sodium is put in a solution of ammonia?
a) It generates a lot of heat
b) It does not dissolve
c) It produces deep blue color
d) Ammonia liquid evaporates due to heat
Answer: c
Clarification: When sodium is introduced to liquid ammonia, it separates into sodium ion and a single electron (Na → Na⁺ + e^–). The single electron produces the deep blue color due to solvation effect producing e^–[NH₃]_n.

5. What is one method of qualitatively analyzing a given salt for presence of ammonia?
a) Solution turns blue litmus red
b) Heating the salt causing decrepitation
c) Using a reagent to obtain dirty brown precipitate
d) Addition of NaOH causing white gelatinous precipitate
Answer: c
Clarification: Typically, Nessler’s reagent is used to test for ammonia. Nessler’s reagent is an alkaline solution of potassium tetraiodomercurate (II). When it combines with ammonia it gives dirty brown precipitate due to the formation of [OHg₂.NH₂]I. Another method is to add a strong base and heat the mixture. The gas given off is pungent and turns red litmus blue.

6. What is the most acidic of all?
a) NH₃
b) NaOH
c) KOH
d) Alkaline KMnO₄
Answer: a
Clarification: In the listed options, except ammonia, every other compound is a strong base which automatically makes ammonia the most acidic relative to the mentioned compounds. Although it is a weak base, compared to NaOH, KOH and alkaline KMnO₄ it is the most acidic amongst the four options.

7. In the Haber-Bosch process, what is formed by the reaction of natural gas and steam?
a) Ammonia
b) Nitrogen
c) Oxygen
d) Hydrogen
View Answer

Answer: d
Clarification: The reaction of natural gas and steam is known as the steam reformation of methane. It is an utmost important step in the Bosch-Haber (industrial process to produce ammonia), which produces hydrogen gas, H₂, one of the two reactants alongside nitrogen to produce ammonia in presence of iron catalyst.

8. Which of the following conditions would improve the yield of ammonia production from Bosch-Haber process?
a) High temperature, high pressure
b) High temperature, low pressure
c) Low temperature, low pressure
d) Low temperature, high pressure
Answer: d
Clarification: Following the Le Chatelier’s principle, since the forward reaction is exothermic (i.e. N₂ + 3H₂ → 2NH₃), low temperatures would increase the rate of reaction in turn producing more ammonia. In addition to this, since the total number of molecules on the product side lesser, high pressures will increase the rate of reaction.

9. Nitrogen in plants is taken in what form?
a) Ammonia
b) Amide
c) Nitrate
d) Nitrite
Answer: c
Clarification: Plants take up nitrogen predominantly in the form of nitrate ions, along side ammonium ions. This is made available to them through fertilizers, whose typical examples usually include Ammonium Phosphate, calcium ammonium nitrate.

Chemistry Multiple Choice Questions on “P-Block Elements – Chlorine”.

1. What is the molecular mass of chlorine?
a) 71.0 kg/kmole
b) 35.5 g/kmole
c) 71.0 g/kmole
d) 35.5 kg/mole
Answer: a
Clarification: In the periodic table, the atomic mass of chlorine atom is 35.5 AMU. Since chlorine exists in diatomic form, the molecular mass = 35.5g/mole x 2 = 71.0 g/mole. On multiplying and dividing by 1000, 71.0 kg/kmole.

2. Which of the following would instantly react with chlorine?
a) Copper
b) Magnesium
c) Iron
d) Aluminum
Answer: b
Clarification: Out of the four metals given, magnesium will readily lose electrons due to its relatively higher reactivity. Copper and iron are not affected by gaseous or liquid chlorine. Lastly, aluminum forms a tough, passive oxide layer which prevents it from reacting with external reagents.

3. What is the color of chlorine?
a) Yellow grey
b) Greenish brown
c) Green yellow
d) Yellow
Answer: c
Clarification: Every halogen bears a characteristic color which is used in identifying it. Fluorine is a bright yellow gas. Chlorine is greenish yellow, followed by bromine which is brown and iodine which is purple/violet. Astatine is a black solid which is radioactive in nature.

4. When excess chlorine reacts with methane in presence of sunlight, which is the major product?
a) Refrigerant – 40
b) Methylene chloride
c) Chloroform
d) Carbon tetrachloride
Answer: d
Clarification: In presence of sunlight, a free radical substitution reaction takes place between the reacting methane and chlorine molecules where in each elementary reaction a hydrogen atom is replaced by a chlorine atom in the new product. First, methane converts to 1-chloromethane which in turn produces dichloromethane followed by tri-chloromethane and tetra-chloromethane, respectively.

5. Which of the following is a suitable test to identify presence of chlorine atoms or ions?
a) Acidify the given compound and add silver nitrate
b) Add silver nitrate followed by warming the solution
c) Heat the compound strongly
d) Smell the compound
Answer: a
Clarification: The standard, universal qualitative test for detecting presence of chloride is carried out by acidifying the solution and then adding silver nitrate. If chloride ions are present, a thick white ppt. is produced.

6. Why is chlorine bubbled through water during the latter’s treatment?
a) To remove foul odor
b) To kill microorganisms
c) To improve taste
d) To remove hardness
Answer: b
Clarification: Chlorine is a very popular disinfecting agent and is widely used in containers to kill microorganisms. In water treatment plants, chlorine gas may be bubbled through to remove trace quantities of microorganisms.

7. What is the electronic configuration of chloride?
a) 1s²2s²2p⁶3s²3p⁵
b) 1s²2s²2p⁶3s²3p⁶
c) 1s²2s²2p⁵
d) [Ar]4s²4p³
View Answer

Answer: b
Clarification: Chlorine has the atomic number 17. However, chloride has 18 electrons. Consequently, it attains the electronic configuration of argon i.e.1s²2s²2p⁶3s²3p⁶.

8. How is chlorine gas manufactured industrially?
a) Deacon process
b) Ostwald process
c) Chlor-alkali process
d) Haber process
Answer: c
Clarification: Although Deacon process involves release of chlorine gas as a byproduct, the chief purpose of the deacon process is used to manufacture alkalis. Chlorine is best produced by the chloralkali process, which is the electrolysis of brine, producing hydrogen and chlorine gases as by products.

9. Hydrogen chloride as an acid is much stronger than hydrogen fluoride under the same conditions. True or False?
a) True
b) False
Answer: a
Clarification: Fluorine atom is extremely small compared to chlorine. This means that the bond of H – F is much more polar than that of H – Cl, causing the H atom to be more tightly bonded in case of H – F than in H – Cl. Since hydrogen chloride is able to release the H+ much easily, the latter is a stronger acid.

10. Which of the following is a result of bubbling hydrogen chloride gas through gas?
a) H^– + Cl^–
b) H⁺ + Cl^2-
c) H₃O⁺ + Cl^–
d) H⁺ + Cl^–
Answer: c
Clarification: The hydrogen chloride molecule, HCl dissociates in water giving rise to one hydrogen ion and a chloride ion. The extremely small hydrogen ion bearing the positive charge is instantly attracted by the unshared pair of electrons of the oxygen atom in water molecules, consequently giving rise to the hydronium ion, H₃O⁺.

11. What will be the result of subjecting a blue litmus to chlorine gas?
a) Turn to red only
b) Remain blue
c) Turn to white finally
d) Turn white in the first go
Answer: c
Clarification: Chlorine bears bleaching properties. When blue litmus is subjected to chlorine gas, it first turns red (due to the acidic nature) then it turns white (due to chlorine’s ability to bleach materials).

12. To which group does chlorine element belong?
a) Halides
b) Pre inert
c) Chalcogens
d) Halogens
Answer: d
Clarification: Fluorine, chlorine, bromine, iodine and astatine belong to Group 17 of the periodic table. This group is commonly named as Halogens because of their ability to produce salt on reacting with metals.

13. Which of the following is the displays the most basic nature in water?
a) Fluoride
b) Chloride
c) Bromide
d) Iodide
Answer: a
Clarification: The fluoride ion is actually not an oxidizing agent. Due to its high electronegativity and high reactivity, it converts water to form hydroxide ions. The reaction is given by 2F^– + H₂O → H – F – H^– + OH^–.

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Chemistry Online Test for Class 12 on “Nomenclature of Coordination Compounds – 2”.

1. Identify the correct naming for K₃[Fe(CN)₆].
a) Tripotassium hexacyanidoferrate(III)
b) Potassium hexacyanoferrate(III)
c) Tripotassium hexacyanoferrate(III)
d) Potassium hexacyanidoferrate(III)

Answer: d
Clarification: The number of cations or anions is not denoted in the name of a coordination compound. In this case, the counter ion is named as potassium and not tripotassium. Also, CN is an anionic ligand and should end with -o, with the correct name being cyanido and not cyano.

2. Identify the correct naming for [Zn(OH)₄]^2-.
a) Zinc tetrahydroxide
b) Tetrahydroxozincate(II)
c) Tetrahydroxidozincate(II)
d) Tetrahydroxylzincate(II)
Answer: c
Clarification: The ligand OH is a negative group and should end with -o. The correct name for the OH group is hydroxide and the metal is mentioned after the ligands.

3. Identify the correct naming for [Co(ONO)(NH₃)₅]²⁺.
a) Pentaamminenitritocolbalt(III)
b) Pentaamminenitrito-O-cobalt(III)
c) Pentaamminenitrito-N-cobalt(III)
d) Pentaamminenitrosylcobalt(III)
Answer: b
Clarification: Nitrate is an ambidentate ligand and can bind through either O or N atom. In this case, it is written as (ONO) which implies that it binds through O atom, hence giving it the name nitrito-O.

4. Identify the correct formula for hexaaquamanganese(II) ion.
a) [Mn(H₂O)₆]²⁺
b) [Mn(H₂O)₆]^2-
c) [Mn₂(H₂O)₆]⁺
d) [Mn(H₂O)₆]⁺²
Answer: a
Clarification: The parenthesis after manganese represents its oxidation state (+2) which is indicated outside the square bracket with sign following the number. The H₂O molecule is a neutral ligand named as aqua.

5. Identify the correct formula for potassium tetracyanonickelate(II).
a) K[Ni(CN)₄]
b) K[Ni(CN)₄]₂
c) K₂[Ni(CN)₄]
d) K₂[Ni(CN)₄]₂
Answer: c
Clarification: CN is anionic with charge -1 and there are 4 of them. Ni is given to have oxidation number +2 in the name of the compound. So, the total charge on the compound will be -4 + 2 = -2. This must be balanced by potassium ions which have charge of +1, so two of them will be required. Hence, the formula is K₂[Ni(CN)₄].

6. The term inside the parenthesis in the formula of mercury(I) tetrathiocyanato-S-cobaltate(III) is _________
a) Hg
b) SCN
c) NCS
d) Co
Answer: b
Clarification: When the ligands are polyatomic, its formula is enclosed in parenthesis. The formula for the given compound is Hg[Co(SCN)₄], where the thiocyanate ligand binds through the S atom.

7. Which of the following representation of the complex ion is correct according to IUPAC?
a) [PtBrCl(NH₃)(NO₂)]^–
b) [PtBrCl(NO₂)(NH₃)]^–
c) [Pt(NH₃)BrCl(NO₂)]^–
d) [Pt(NO₂)BrCl(NH₃)]^–
Answer: c
Clarification: The ligands are bromine, chlorine, ammine and nitrite and according to IUPAC nomenclature they should be listed as per alphabetical order which is ammine (NH₃), bromine (Br), chlorine (Cl) and nitrite (NO₂).

8. An imaginary coordination entity named dipalidoqalidoralidometal(IV) according to the IUPAC norms, consists of ligands palide(P), qalide(Q) and ralide(R), all having charge of magnitude 1. The central atom is metal (M). Which of the following options best suits the complex described?
a)
b)
c)
d)
View Answer

Answer: c
Clarification: All the ligands end with -o, which implies they are anionic and have charge -1. The oxidation number of the central atom is given as +4. There are a total of 4 ligands (2 P, 1 Q and 1 R) in the entity. Hence, the overall charge on the complex is +4 – 4 = 0.

9. Which of the following is not enclosed in parenthesis?
a) Polyatomic ligands in the formula of coordination compounds
b) Ligand abbreviations in the formula of coordination compounds
c) Oxidation number of central metal in the naming of coordination compounds
d) The coordination entity in the formula of coordination compounds
Answer: d
Clarification: The coordination entity consisting of the metal atom/ion followed by ligands is enclosed within square brackets and not in parenthesis.

10. The prefixes bis, tris and tetrakis are used to indicate the number of individual ligands in the coordination entity.
a) True
b) False
Answer: a
Clarification: Mono, di, tri, etc. are used when name of ligands don’t have a numerical prefix. If not, then bis, tris and tetrakis are used as prefixes.

Chemistry Online Test for Class 12,

Chemistry Multiple Choice Questions on “Haloalkanes & Haloarenes – Chemical Reactions – 3”.

1. What is the correct order of reactivity of haloalkanes towards β-elimination reactions?
a) 1°>2°>3°
b) 3°>2°>1°
c) 1°>3°>2°
d) 3°>1°>2°
Answer: b
Clarification: The haloalkanes which will form more stable (highly substituted) alkenes are the ones that will react the fastest. Since tertiary alkyl halides will form the most stable alkenes, they are the most reactive.

2. How many β-carbon atoms does 2-Bromobutane have?
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: A beta carbon is the carbon atom that is adjacent to the carbon atom that is attached to the halogen. In 2-Bromobutane, the halogen is at position C-2, so the alpha carbon has one carbon on either side, making it two beta carbons

3. Which of the following is the suitable medium for preparing Grignard reagents?
a) Dry acetone
b) Dry ether
c) Concentrated HCl
d) Alcoholic KOH
Answer: b
Clarification: When haloalkanes react with magnesium metal in dry ether, special organo-metallic compounds called Grignard reagents are formed.

4. What is formed when ethyl magnesium bromide reacts with water?
a) Grignard reagent
b) Ethane
c) Ethanol
d) Magnesium hydroxide
Answer: b
Clarification: Ethyl magnesium bromide is a Grignard reagent and is highly reactive to any proton giving source because of the polar nature of bonding present in it. It forms hydrocarbons when reacted with water.

5. What will be the product of the following reaction?
2CH₃CH₂Br + 2Na + dry ether = ________
a) Ethane
b) 1-Bromoethane
c) Butane
d) 1-Bromobutane
Answer: c
Clarification: Alkyl halides react with sodium metal in dry ether to form hydrocarbons that have double the number of carbon atoms as that in the alkyl halide. This is called the Wurtz reaction.

6. Compared to haloalkanes, the reactivity of haloarenes towards nucleophilic substitution reactions is _________
a) low
b) high
c) very high
d) equal
Answer: a
Clarification: Haloarenes form resonating structures in which the carbon-halogen bond acquires a partial double bond character and its cleavage is difficult. Hence, they are less reactive towards nucleophilic substitution reactions.

7. The C-Cl bond length in chlorobenzene is less than the C-Cl bond length in chloromethane because of the ________
a) resonance effect in haloarenes
b) difference in hybridisation of carbon in C-Cl bond
c) instability of phenyl cation
d) possible repulsion between nucleophile and chlorobenzene
Answer: b
Clarification: In chlorobenzene, the C of C-Cl bond is sp² hybridised, whereas the C of C-Cl bond in chloromethane is sp³ hybridised. Thus, there is more s character in the carbon of the C-Cl bond in chlorobenzene and as a result holds the electron pair of C-Cl bond more tightly.

8. Which of the following compounds is most easily converted to a phenol when heated with aqueous NaOH solution?
a) Chlorobenzene
b) 4-Chloronitrobenzene
c) 2,4-Dinitrochlorobenzene
d) 2,4,6-Trinitrochlorobenzene
Answer: d
Clarification: The presence of electron withdrawing groups like NO₂ at ortho and para positions with respect to the halogen, increases the reactivity of the haloarene. Chlorobenzene has no such group and converts to phenol at the highest temperature and pressure.

9. 2-Chloronitrobenzene is more reactive than chlorobenzene.
a) True
b) False
Answer: b
Clarification: Although NO₂ is an electron withdrawing group, it is present at meta position with respect to chlorine in 2-Chloronitrobenzene and has no effect on the reactivity of the haloarene. This is because none of the resonating structures of m-nitrobenzene bear the negative charge on the carbon atom bearing the nitro group and as a result the presence of NO₂ at meta position does not stabilize the negative charge.

10. Identify the catalyst in Friedel-Crafts acylation reaction.
a) Anhydrous FeCl₃
b) Anhydrous AlCl₃
c) NaNO₂
d) Alcoholic KOH
Answer: b
Clarification: Friedel-Crafts acylation is carried out by treating a haloarene with and acetyl chloride in the presence of anhydrous aluminium chloride as a catalyst.

11. What is the major product formed when chlorobenzene reacts with nitric acid in concentrated sulphuric acid?
a) Nitrobenzene
b) 1-Chloro-2-nitrobenzene
c) 1-Chloro-3-nitrobenzene
d) 1-Chloro-4-nitrobenzene
Answer: d
Clarification: This is the nitration of chlorobenzene which is an electrophilic substitution reaction. The major product is that where the nitro group is present at a para position to Cl.

12. Identify ‘X’ from the reaction shown.

a) Dilute H₂SO₄
b) Concentrated H₂SO₄
c) Dilute HNO₃
d) Concentrated HNO₃
Answer: b
Clarification: This is the sulphonation of chlorobenzene which results in the formation of ortho and meta chlorobenzenesulfonic acid.

13. Which is the metal involved in Wurtz-Fittig reaction?
a) Iron
b) Magnesium
c) Aluminium
d) Sodium
Answer: d
Clarification: When an aryl halide and alkyl halide and together treated with sodium metal in dry ether, an alkylarene is formed. This is called Wurtz-Fittig reaction.

14. Predict the minor product formed when chlorobenzene reacts with chloromethane in the presence of anhydrous AlCl₃?
a) Toluene
b) m-Chlorotoluene
c) o-Chlorotoluene
d) p-Chlorotoluene
Answer: c
Clarification: This is the Friedel-Crafts alkylation reaction, where o-Chlorotoluene is the minor product and p-Chlorotoluene is the major product. This is because halogen group is ortho and para directing.

15. Fittig reaction results in the formation of a diphenyl.
a) True
b) False
Answer: a
Clarification: Aryl halides give analogous compounds in which two aryl groups are combined when treated with Na in dry ether. This is called Fittig reaction.

Chemistry Multiple Choice Questions on “Aldehydes and Ketones Chemical Reactions – 1”.

1. Aldehydes and ketones undergo __________ reactions.
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution
Answer: c
Clarification: Aldehydes and ketones have a polar CO group which also has a double bond. The incoming nucleophile attacks the sp² hybridised carbon, thus breaking the double bond and converting it into sp³, and forming a tetrahedral alkoxide intermediate. This undergoes another fast step to form an addition product.

2. Which of the following is added to the tetrahedral intermediate when it reacts with a proton from the reaction medium?
a) Nu^–
b) H⁺
c) Both Nu^– and H⁺
d) OH^–
Answer: b
Clarification: A H+ proton is added to the negative O atom of the alkoxide intermediate to form an electrically neutral addition product. The Nu^– is added in the first step itself when it attacks the electrophilic carbon centre in a slow step.

3. The nucleophilic addition reactions of aldehydes are carried out in ________ medium.
a) neutral
b) acidic
c) weakly basic
d) extremely basic

Answer: b
Clarification: When the incoming nucleophile is weak, the acidic medium helps in protonation by increasing the positive charge on the carbonyl carbon so that it is more susceptible to attack by relatively weaker nucleophiles.

4. What is the correct order of reactivity of the following towards nucleophilic addition?
a) Methanal > Ethanal > Acetone
b) Acetone > Ethanal > Methanal
c) Methanal > Acetone > Ethanal
d) Ethanal > Methanal > Acetone
Answer: a
Clarification: Methanal is the most reactive among aldehydes and ketones due to electronic and stearic reasons. Methanal (HCHO) does not have any alkyl groups that release electrons toward carbonyl carbon, thus making it more electrophilic. Also, presence of only two H atoms do not provide any hinderance to the attack of incoming nucleophile.

5. Which of the following is least reactive towards a nucleophilic attack?
a) Acetaldehyde
b) Butanone
c) Diisopropyl ketone
d) Ditert-Butyl ketone
Answer: d
Clarification: The stearic effects come into consideration when comparing these compounds. Ditert-Butyl ketone has two very bulky tert-butyl groups, one on either side of the carbonyl carbon. This increases the hinderance to the incoming nucleophile and decreases the reactivity.

6. What is the correct order of reactivity towards nucleophilic addition?
a) Benzaldehyde > Benzophenone > Acetophenone
b) Benzophenone > Benzaldehyde > Acetophenone
c) Acetophenone > Benzaldehyde > Benzophenone
d) Benzaldehyde > Acetophenone > Benzophenone
Answer: d
Clarification: Generally, the aromatic carbonyl compounds are less reactive than corresponding aliphatic compounds. From benzaldehyde to acetophenone to benzophenone, the number of electron releasing groups increases and the magnitude of positive charge on carbonyl carbon reduces, hence decreasing its reactivity towards nucleophile attack.

7. Which of the following is the most reactive towards hydrogen cyanide?
a) Acetophenone
b) Benzaldehyde
c) p-Nitrobenzaldehyde
d) p-Tolualdehyde
Answer: c
Clarification: Addition of hydrogen cyanide is a nucleophilic reaction. CH₃ and COCH₃ are electron releasing groups which decrease the positive charge on carbonyl carbon, whereas NO₂ is an electron withdrawing group which increases the electrophilicity of carbonyl carbon, making it more reactive toward hydrogen cyanide than benzaldehyde.

8. Identify the catalyst in the nucleophilic addition of HCN to acetone.
a) NaOH
b) HCl
c) NaCl
d) NaCN
Answer: a
Clarification: Acetone reacts with HCN to give a cyanohydrin of acetone. But this reaction proceeds very slowly with pure HCN. So, the reaction is carried out in a basic medium which removes the proton from HCN and produces CN^– ion which is a stronger nucleophile and makes the reaction faster.

9. The hydrolysis of the addition product of acetone and methyl magnesium iodide gives _______
a) ethyl alcohol
b) isopropyl alcohol
c) tert-Butyl alcohol
d) phenol
Answer: c
Clarification: The methyl group of Grignard reagent is the nucleophile which initially get attached to the carbonyl carbon of acetone. This results in three methyl groups attached to the carbon atom and one OMgI group. This MgI is replaced by H⁺ on hydrolysis to give tert-Butyl alcohol. Similarly, methanal gives ethyl alcohol (1°) and other aldehydes give 2° alcohols.

10. The following is an addition product of the reaction between _______ and sodium bisulphite.

a) formaldehyde
b) acetaldehyde
c) ethanone
d) acetone
Answer: b
Clarification: Acetaldehyde reacts with NaHSO₃ to form a crystalline bisulphite addition compound. First, the nucleophile HSO₃^– attaches to the carbonyl carbon, after which the hydrogen gets attached to the oxygen atom on proton transfer.

11. The following acetal shown is formed by the reaction between ______ and ______ in the presence of dry HCl.

a) methanal; methanol
b) methanal; ethanol
c) ethanal; methanol
d) ethanal; ethanol
Answer: d
Clarification: Ethanal and ethanol react in the presence of dry HCl to initially form an unstable hemiacetal having one ethoxy (OC₂H₅) and one hydroxy (OH) group. This further reacts with ethanol to form the gem-diethoxy compound, acetaldehyde ethyl acetal as shown.

12. Acetone reacts with ethylene glycol in dry HCl gas to form _________
a) hemiacetals
b) acetals
c) cyclic acetals
d) cyclic ketals
Answer: d
Clarification: Dihydric alcohols like ethylene glycol react with aldehydes and ketones to directly form cyclic products known as cyclic acetals and cyclic ketals respectively. The cyclic ketals are also known as ethylene glycol ketals.

13. Which of the following products is formed when an aldehyde reacts with an amine in acidic medium?
a) Imine
b) Schiff’s base
c) Oxime
d) Hydrazone
Answer: b
Clarification: Amine is a nucleophile and a derivative of NH₃ which react with aldehydes and ketones followed by dehydration to form compounds with C-N double bond called Schiff’s base or substituted imines.

14. Identify the N-substituted derivative of carbonyl compounds that are coloured compounds and are useful in the identification of aldehydes and ketones.
a) Hydrazone
b) Phenylhydrazone
c) 2,4-Dinitrophenylhydrazone
d) Semicarbazone
Answer: c
Clarification: 2,4-DNP derivatives are formed from the reaction between aldehydes or ketones and 2,4-Dinitrohydrazine. These products may be yellow, orange or ed coloured and help in characterisation of aldehydes and ketones.

15. Semicarbazones undergo acidic nucleophilic addition followed by dehydration to give N-substituted carbonyl derivatives.
a) True
b) False
Answer: b
Clarification: Semicarbazone is the name of the product formed from the rapid dehydration of the product of the reaction between semicarbazide and aldehyde or ketone in acidic medium. Semicarbazide is the reagent and Semicarbazone is the N-substituted carbonyl derivative.

Chemistry Multiple Choice Questions on “Amines Physical Properties”.

1. Which of the following amines is not a gas?
a) Methylamine
b) Dimethylamine
c) Ethylamine
d) Trimethylamine
Answer: d
Clarification: Lower members of aliphatic amines exist as gases at ordinary temperature. The higher members with three or more carbon atoms are liquid and still higher ones are solid.

2. What is the characteristic odour of relatively lower aliphatic amines?
a) Fruity odour
b) Fishy odour
c) Rotten egg smell
d) Odourless
Answer: b
Clarification: Most of the amines have an unpleasant odour. The smell of lower amines is similar to that of ammonia with a fishy odour.

3. Which of the following best describes aniline in pure form?
a) Colourless liquid
b) White waxy solid
c) Brown gas
d) Yellowish gas
Answer: a
Clarification: Aniline and other aryl amines are usually colourless, but they become coloured on storage due to atmospheric oxygen. Aniline develops a yellow to brown colour dur to this reason.

4. The intermolecular hydrogen bonds, if any, in amines is formed between _______
a) N-H and N-H
b) N and N-H
c) Alkyl carbon and N-H
d) Alkyl H and N-H
Answer: b
Clarification: Primary and secondary amines are involved in intermolecular molecular hydrogen bonding between nitrogen of one and hydrogen of another molecule.

5. What is the correct order of boiling points of the isomeric amines where A=ethylmethylamine, B=propylamine and C=trimethylamine?
a) A > B > C
b) C > B > A
c) B > C > A
d) B > A > C
Answer: d
Clarification: A, B and C are respectively the 2°, 1° and 3° isomers of the compound C₃H₉N. Primary and secondary amines have two and one hydrogen atoms respectively for hydrogen bonding. Whereas, tertiary amines do not have intermolecular association due to absence of H for hydrogen bonding.

6. Which of the following has a lower boiling point than ethanamine?
a) Propane
b) Ethanal
c) Ethanol
d) Methanoic acid
View Answer

Answer: a
Clarification: All given compounds have similar molecular masses and can be compared. Ethanamine has a higher boiling point than propane because it is a polar molecule and forms intermolecular hydrogen bonds. However, it has a lower boiling point than ethanal, ethanol and methanoic acid as the O-H bonds in these compounds is more polar than the N-H bond in ethanamine. This makes the hydrogen bonds stronger.

7. If the boiling point of diethylamine is 329 K, predict the boiling point of ethyldimethylamine.
a) 310 K
b) 329 K
c) 340 K
d) 351 K
Answer: a
Clarification: Diethylamine is a 2° amine whereas, ethyldimethylamine is a 3° amine. In 3° amines, there are no hydrogen atoms available for intermolecular hydrogen bonding. Therefore, the tertiary amines have lowest boiling point among the primary and secondary counterparts.

8. If the boiling point of n-C₄H₉NH₂ is 351 K, what will be the boiling point of n-C₄H₉OH?
a) 329 K
b) 301 K
c) 351 K
d) 390 K
Answer: d
Clarification: Alcohols have higher boiling points than amines of comparable molecular masses. This is because the electronegativity of N is lower than that of O, which results in the OH bond in alcohols being more polar and hence stronger (in hydrogen bonding) than the N-H bond in amines.

9. If A is the boiling point of propanamine and B is the boiling point of butanamine, what is the correct relation between the two?
a) A > B
b) B > A
c) A = B
d) A >> B
Answer: b
Clarification: Butanamine is a larger molecule than ethanamine. As the carbon chain increase, the magnitude of van der Waals forces increases resulting in the increase in boiling point.

10. All aliphatic amines exist as associated molecules due to intermolecular hydrogen bonding.
a) True
b) False
Answer: b
Clarification: Only primary and secondary amines form intermolecular hydrogen bonds as they have individual at least one hydrogen atom to form bonds with. Tertiary amines do not have intermolecular association due to absence of hydrogen atom available for hydrogen bonding.

11. Which of the following amines are insoluble in water?
a) Methanamine
b) Ethanamine
c) Propanamine
d) Benzenamine
Answer: d
Clarification: Lower aliphatic amines are soluble in water as they easily form hydrogen bonds with water molecules. Higher amines containing six or more carbon atoms are insoluble in water due to a large hydrophobic alkyl/aryl part. Therefore, aniline is insoluble in water.

12. If ‘x’ ml of butan-1amine and ‘y’ ml of butan-1-ol is completely soluble in 100ml of water each, what is the relation between x and y?
a) x > y
b) x c) x = y
d) x + y = 100

Answer: b
Clarification: Alcohols are more polar than amines of comparable molecular masses and form stronger intermolecular hydrogen bonds with water. Therefore, the solubility of butan-1-ol is more than that of butan-1-amine.

13. Aniline is insoluble in which of the following solvents?
a) Ethanol
b) Ethoxyethane
c) Benzene
d) Water
View Answer

Answer: d
Clarification: Aromatic amines have a very large hydrocarbon part (benzene ring) which tends to retard the formation of hydrogen bonds with water. However, they are soluble in organic solvents like ethers, alcohols and benzene.

14. Which of the following is least soluble in water?
a) CH₃CH₂NH₂
b) CH₃CH₂OH
c) HCOOH
d) CH₃NH₂
Answer: a
Clarification: Lower amines are more soluble because of the smaller hydrophobic alkyl part. Also, alcohols and carboxylic acids have higher solubility than amines because of the greater polarity of OH bond than NH bond.

15. Primary amines are more soluble than tertiary amines of same formula.
a) True
b) False
Answer: a
Clarification: Solubility in water requires the presence of free hydrogen in the structure of a compound to form hydrogen bonds. Tertiary amines do not have any hydrogen attached to the N atom and are not soluble in water.