NCERT Exemplar Hydrocarbons - Chemistry Class 11 - NEET PDF Download

Table of Contents
1. Multiple Choice Questions - I
2. Multiple Choice Questions - II
3. Part-2
4. Matching Type Questions
5. Assertion and Reason Type
6. Long Answer Type
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Multiple Choice Questions - I

Q.1. Arrange the following in decreasing order of their boiling points.
 (A) n-butane 
(B) 2-methyl butane
 (C) n-pentane 
(D) 2,2-dimethylpropane
 (a) A > B > C > D 
(b) B > C > D > A 
(c) D > C > B > A 
(d) C > B > D > A
 Ans. (d) C > B > D > A
Solution: Boiling point of hydrocarbons increases with molar mass and with surface area available for intermolecular van der Waals interactions. Increased branching reduces surface area and hence lowers boiling point. 
Among the given isomers, n-pentane (linear, C5) has the largest surface area and highest boiling point. 2-methylbutane (a mono-branched C5) has lower boiling point than n-pentane but higher than the more branched 2,2-dimethylpropane (neopentane). n-Butane (C4) has the lowest boiling point. Therefore the order is C > B > D > A.

Q.2. Arrange the halogens F₂, Cl₂, Br₂, I₂, in order of their increasing reactivity with alkanes. 
(a) I₂ < Br₂ < Cl₂ < F₂ 
(b) Br₂ < Cl₂ < F₂ < I₂ 
(c) F₂ < Cl₂ < Br₂ < I₂ 
(d) Br₂ < I₂ < Cl₂ < F₂ 
Ans. (a) I₂ < Br₂ < Cl₂ < F₂

Solution: Reactivity of halogen molecules in free-radical substitution with alkanes correlates with bond dissociation energy and electronegativity. Fluorine has the weakest F-F bond relative to reactivity in this context and is extremely reactive (often violently so). Reactivity decreases down the group; iodine reacts very slowly with alkanes. Thus the increasing reactivity is I₂ < Br₂ < Cl₂ < F₂.

Q.3. The increasing order of reduction of alkyl halides with zinc and dilute HCl is (a) R-Cl < R-I < R-Br (b) R-Cl < R-Br < R-I (c) R-I < R-Br < R-Cl (d) R-Br < R-I < R-Cl Ans. (b) R-Cl < R-Br < R-I

Solution: Reduction (cleavage) of R-X bonds by Zn/HCl is easier when the C-X bond is weaker. Bond strength decreases as the halogen size increases (Cl > Br > I), so reactivity toward reduction increases as R-Cl < R-Br < R-I.

Q.4. The correct IUPAC name of the following alkane is

(a) 3,6 - Diethyl - 2 - methyloctane 
(b) 5 - Isopropyl - 3 - ethyloctane 
(c) 3 - Ethyl - 5 - isopropyl octane 
(d) 3 - Isopropyl - 6 - ethyl octane 
Ans. (a) 3, 6-Diethyl-2-methyl octane
Solution: The longest continuous chain has eight carbons, so the parent is octane. Numbering the chain to give the lowest set of locants to substituents gives substituents at C-2, C-3 and C-6. At C-2 there is a methyl; at C-3 and C-6 there are two ethyl groups. Alphabetical order puts ethyl before methyl; the two ethyl groups are indicated by the prefix di-. Hence the correct name is 3,6-diethyl-2-methyloctane.

Q.5. The addition of HBr to 1-butene gives a mixture of products A, B and C 
(A)

(B)

(C) CH₃ - CH₂ - CH₂ - CH₂ - Br The mixture consists of
(a) A and B as major and C as minor products 
(b) B as major, A and C as minor products 
(c) B as minor, A and C as major products 
(d) A and B as minor and C as major products
 Ans. (a) A and B as major and C as minor products

Solution: 1-Butene is an unsymmetrical alkene. According to Markovnikov's rule, H+ adds to the carbon of the double bond bearing more hydrogen atoms, so the bromide attaches to the more substituted carbon. This produces 2-bromobutane as the major product. 2-Bromobutane has a chiral centre and therefore exists as a racemic mixture of two enantiomers (A and B). 1-bromobutane (C) is the minor product.

Q.6. Which of the following will not show geometrical isomerism? (a)

(b)

(c)

(d)

Ans. (d)

Solution: Geometrical (cis-trans or E/Z) isomerism around a C=C double bond requires that each double-bonded carbon atom has two different substituents. In structure (d), the two double-bonded carbons have three identical groups and only one different group, so geometrical isomerism is not possible.

Q.7. Arrange the following hydrogen halides in order of their decreasing reactivity with propene. 
(a) HCl > HBr > HI 
(b) HBr > HI > HCl 
(c) HI > HBr > HCl 
(d) HCl > HI > HBr 
Ans. (c) HI > HBr > HCl

Solution: In electrophilic addition to alkenes, the ease of reaction depends on ease of heterolytic bond cleavage to generate H+ and X- (i.e., the H-X bond strength). Bond dissociation energies decrease down the group (HCl > HBr > HI), so the acids react in the order HI > HBr > HCl.

Q.8. Arrange the following carbanions in order of their decreasing stability. 
(A) H₃C - C ≡ C^- 
(B) H - C ≡ C^- 
(C) H₃C - CH₂^- 
(a) A > B > C 
(b) B > A > C 
(c) C > B > A 
(d) C > A > B
 Ans. (b) B > A > C 

Solution: 
Carbanion stability increases with more s-character (sp > sp² > sp³) because the negative charge is held closer to the nucleus.

• (B) H-C≡C⁻ → sp carbanion (most stable)

• (A) CH₃-C≡C⁻ → sp carbanion, but CH₃ is +I (slightly destabilizes vs B)

• (C) CH₃-CH₂⁻ → sp³ carbanion (least stable)

Q.9. Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH. 
(1)

(2) CH₃-CH₂-Br 
(3) CH₃-CH₂-CH₂-Br
 (a) A > B > C 
(b) C > B > A 
(c) B > C > A 
(d) A > C > B 
Ans. (d) A > C > B

Solution: The ease of β-elimination (E2) to form an alkene correlates with the stability of the alkene product. More substituted alkenes (more alkyl groups on the double bond) are more stable (Saytzeff rule) and thus form faster. Structure A leads to the most substituted (and hence most stable) alkene, followed by C, then B. Therefore the order of β-elimination rates is A > C > B.

Q.10. Which of the following reactions of methane is incomplete combustion: 
(a)

(b)

(c) CH₄ + O₂ → C(s) + 2H₂O(l) 
(d) CH₄ + 2O₂ → CO₂(g) + 2H₂O(l) 
Ans. (c)

Solution: Complete combustion of methane in excess oxygen gives carbon dioxide and water (reaction d). Incomplete combustion, caused by insufficient oxygen, gives carbon (soot) and water as shown in (c). Thus (c) represents incomplete combustion.

Multiple Choice Questions - II

In the following questions two or more options may be correct.

Q.11. Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions? 
(a) CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l) 
(b) CH₄ (g) + O₂ (g) → C (s) + 2H₂O (l) 
(c)

(d)

Ans. (c) and (d)

Solution: Combustion of alkanes in excess oxygen gives CO₂ and H₂O (a). Incomplete combustion (b) gives carbon (soot) and water. Controlled oxidation (partial oxidation) of methane under suitable conditions can give formaldehyde (HCHO) or methanol (CH₃OH); these are oxidation products obtained when oxidation is carried out in a controlled manner (not full combustion). Hence options (c) and (d) (which represent partial oxidation products) are controlled oxidations.

Q.12. Which of the following alkenes on ozonolysis give a mixture of ketones only? 
(a) CH₃-CH = CH-CH₃ 
(b)

(c)

(d)

Ans. (c) and (d)

Solution: Ozonolysis cleaves a C=C bond to give two carbonyl fragments. If both carbons of the double bond are bonded to alkyl groups (i.e., are internal carbons), the products are ketones. If one carbon is bonded to hydrogen, the product on that side is an aldehyde. Therefore alkenes where both double-bonded carbons bear alkyl groups give ketones only. Options (c) and (d) satisfy this criterion.

Q.13. Which are the correct IUPAC names of the following compounds?

(a) 5-Butyl-4-isopropyldecane 
(b) 5- Ethyl - 4- propyldecane 
(c) 5- sec-Butyl - 4- iso-propyldecane 
(d) 4-(1-methylethyl)- 5 - (1-methylpropyl)-decane 
Ans. (c) and (d)

Solution: The longest chain contains ten carbon atoms (decane). Substituents appear at C-4 and C-5 according to the lowest-set rule. At C4 there is an isopropyl group (IUPAC name 1-methylethyl), and at C5 there is a sec-butyl group (IUPAC name 1-methylpropyl). Thus the correct systematic names are: 5-sec-butyl-4-iso-propyldecane (as a retained trivial form) and the fully systematic name 4-(1-methylethyl)-5-(1-methylpropyl)-decane.

Q.14. Which are the correct IUPAC names of the following compound?

(a) 5 - (2′, 2′-Dimethylpropyl)-decane
(b) 4 - Butyl - 2, 2- dimethylnonane
(c) 2, 2- Dimethyl - 4- pentyloctane
(d) 5 - neo-Pentyldecane
Ans. (a) and (d)

Solution: The parent chain is decane (10 carbons) and the substituent is attached at C-5. The substituent is a neo-pentyl group; its systematic descriptor is 2',2'-dimethylpropyl. Hence the valid names are 5-(2',2'-dimethylpropyl)-decane and the common form 5-neo-pentyldecane.

Q.15. For an electrophilic substitution reaction, the presence of a halogen atom in the benzene ring _______.
 (a) Deactivates the ring by inductive effect.
(b) Deactivates the ring by resonance.
(c) Increases the charge density at ortho and para position relative to meta position by resonance.
(d) Directs the incoming electrophile to meta position by increasing the charge density relative to ortho and para position.
Ans. (a) and (c)

Solution: Halogens are electron-withdrawing by the inductive effect (-I), which makes the ring less reactive overall (deactivating). However, halogens have lone pairs and can donate electron density into the ring by resonance (+R), increasing electron density at the ortho and para positions compared to the meta position. This resonance effect explains why halogen substituents are ortho/para directors despite being deactivating overall.

Q.16. In an electrophilic substitution reaction of nitrobenzene, the presence of nitro group ________. 
(a) Deactivates the ring by inductive effect.
(b) Activates the ring by inductive effect.
(c) Decreases the charge density at ortho and para position of the ring relative to meta position by resonance.
(d) Increases the charge density at meta position relative to the ortho and para positions of the ring by resonance.
Ans. (a) and (c)

Solution: Nitro (-NO₂) is a strongly electron-withdrawing substituent by both -I and -R effects. It decreases electron density on the ring and particularly destabilises intermediates formed by electrophilic attack at ortho and para positions, making meta substitution comparatively more favourable. Thus it deactivates the ring and reduces charge density at ortho/para relative to meta positions.

Q.17. Which of the following are correct? 
(a) CH₃-O- CH₂^⊕ is more stable than CH₃- CH₂^⊕
(b) (CH₃)₂ CH^⊕ is less stable than CH₃-CH₂- CH₂^⊕
(c) CH₂ = CH- CH₂^⊕ is more stable than CH₃-CH₂- CH₂^⊕
(d) CH₂ = CH^⊕ is more stable than CH₃- CH₂^⊕
Ans. (a) and (c)
Solution: Stability of carbocations is influenced by electron-donating inductive effects and resonance. In (a) the oxygen bearing group (CH₃-O-) can stabilise the adjacent carbocation by resonance or inductive donation, making CH₃-O-CH₂⁺ more stable than a simple ethyl carbocation CH₃-CH₂⁺. In (c) an allylic carbocation (CH₂=CH-CH₂⁺) is resonance stabilised, so it is more stable than a non-resonance stabilised primary carbocation CH₃-CH₂-CH₂⁺. The statements (b) and (d) are not correct as written.

Q.18. Four structures are given in options (i) to (iv). Examine them and select the aromatic structures. 
(a)

(b)

(c)

(d)

Ans. (a, c)

Solution: Apply Hückel's rule: aromatic systems are planar, fully conjugated rings with (4n + 2) π electrons. Option (a) corresponds to a three-membered aromatic (cyclopropenyl cation) having 2 π electrons (n = 0). Option (c) involves two fused six-membered rings each contributing to a 6 π electron aromatic system. Other options either are non-planar or do not have (4n + 2) π electrons and are therefore non-aromatic.

Q.19. The molecules having dipole moment are __________.
(a) 2,2-Dimethylpropane
(b) trans-Pent-2-ene
(c) cis-Hex-3-ene
(d) 2, 2, 3, 3 - Tetramethylbutane.
Ans. (b, c)
Solution: Saturated symmetrical alkanes such as 2,2-dimethylpropane and 2,2,3,3-tetramethylbutane are nonpolar and have zero dipole moment. In trans-pent-2-ene, although substituents are opposite, the dipole moments due to differing substituents (CH₃ vs CH₂CH₃) do not cancel perfectly and a small net dipole results. In cis-hex-3-ene the two bond dipoles are inclined (~60°) and do not cancel, giving a finite dipole moment.

Part-2

Short Answer Type Questions

Q.20. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain. Ans.

Solution: Both alkenes and arenes are rich in π-electrons and are susceptible to attack by electrophiles. Alkenes are unsaturated (contain a C=C π bond) and addition across the double bond converts the π bond into two σ bonds, which is energetically favourable and does not involve loss of any special stabilisation. 

Arenes (like benzene) possess aromatic stabilization energy (resonance energy, for benzene ≈ 150.4 kJ mol-1). Electrophilic addition to benzene would destroy its aromatic delocalisation and incur a large resonance energy penalty. Therefore benzene prefers electrophilic substitution, where one σ C-H bond is replaced by a σ C-X bond while the aromatic π system is restored, preserving resonance energy.

Q.21. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism? Ans.

Solution: Reduction of 2-butyne by Na/NH₃ (Birch-type radical reduction) gives trans-2-butene. Trans-2-butene is capable of showing geometrical isomerism (cis-trans). 

Q.22. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement. Ans.

Solution: Ethane contains a carbon-carbon sigma (σ) bond. Electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the C - C bond which is not disturbed due to rotation about its axis. This permits free rotation around a C-C single bond. However, rotation around a C - C single bond is not completely free. It is hindered by a small energy barrier due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. 

Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. The energy difference between the two extreme forms is of the order of 12.5 kJ mol^-1, which is very small. It has not been possible to separate and isolate different conformational isomers of ethane.

Q.23. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why? Ans.

Solution: Newman and sawhorse projections show relative positions of the substituents when viewed along the C-C bond. In the staggered conformation the C-H bonds on one carbon are positioned between the C-H bonds on the other carbon, maximising separation and minimising repulsive interactions. In the eclipsed conformation the bonds align and electron cloud repulsions are greater. Therefore the staggered conformation is more stable (lower energy) than the eclipsed one.

Q.24. The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^-1, 363.7 kJ mol^-1 and 296.8 kJ mol^-1 respectively. What will be the order of reactivity of these halogen acids? Ans.

Solution: Lower H-X bond dissociation energy means easier heterolytic cleavage and more ready generation of H+. Thus acids react in the order HI > HBr > HCl, the reverse of bond dissociation energies listed.

Q.25. What will be the product obtained as a result of the following reaction and why?

Ans.

Solution: In the presence of anhydrous AlCl₃, propyl chloride can form a carbocation. A less stable primary carbocation rearranges by a hydride shift (or alkyl shift) to give a more stable secondary or tertiary carbocation, which then reacts to give the observed product. The diagrams below show the formation of the primary carbocation and its rearrangement to the more stable carbocation followed by nucleophile capture.

Q.26. How will you convert benzene into 
(1) p-nitrobromobenzene 
(2) m-nitrobromobenzene 
Ans. (1) Benzene into p-nitro bromobenzene:

(2) Benzene into m-nitrobromobenzene:

Solution: Direct bromination of benzene (Br₂/FeBr₃) gives bromobenzene (an ortho/para director). Nitration of bromobenzene (HNO₃/H₂SO₄) places NO₂ at ortho and para positions yielding mainly p-nitrobromobenzene (after separation). To obtain m-nitrobromobenzene, introduce the nitro group first (nitration of benzene gives nitrobenzene; NO₂ is a meta-director) and then brominate nitrobenzene to give m-nitrobromobenzene.

Q.27. Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.

Ans.

Solution: The methoxy group (-OCH₃) is an electron-releasing group by resonance (+R), increasing electron density on the aromatic ring and activating it strongly toward electrophilic substitution; anisole is therefore most reactive.

The chloro group has -I and +R effects; overall it is deactivating but directs ortho/para due to resonance, so chlorobenzene is less reactive than anisole. Nitrobenzene (-NO₂) is strongly deactivating by -I and -R and is least reactive. Thus the order is anisole > chlorobenzene > nitrobenzene.

Q.28. Despite their - I effect, halogens are o- and p-directing in haloarenes. Explain. Ans.

Solution: Halogens exhibit an electron-withdrawing inductive effect (-I) which deactivates the ring, but they also have lone pairs that can participate in resonance (+R) donating electron density to the ring at the ortho and para positions. The resonance donation increases electron density at ortho and para positions relative to meta, so halogens direct electrophiles to ortho/para positions even while being overall deactivating.

Q.29. Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain. Ans.

Solution: Nitro group (-NO₂) withdraws electron density from the ring via both -I and -R effects. This reduces the electron density at the ring, especially at ortho and para positions, making formation of the arenium (σ-complex) intermediate for electrophilic substitution less stable. Therefore nitrobenzene is less reactive than benzene.

Q.30. Suggest a route for the preparation of nitrobenzene starting from acetylene? Ans.

Solution: Acetylene (ethyne) can be converted to benzene by passing through a red-hot iron tube at high temperature (approx. 873 K), which causes cyclic polymerisation to benzene. Benzene can then be nitrated (HNO₃/H₂SO₄) to give nitrobenzene.

Q.31. Predict the major product (s) of the following reactions and explain their formation.

Ans. 

Solution: Addition of HBr to unsymmetrical alkenes follows Markovnikov's rule (in absence of peroxides): the proton adds to the carbon bearing more hydrogens, placing the positive charge (carbocation) on the more substituted carbon, which then combines with Br- to give the Markovnikov product. The mechanism involves electrophilic attack by H+ to form a carbocation intermediate, followed by nucleophilic attack by Br-.

The carbocation (b) is attacked by Br^- ion to form the product as follows:

Solution (peroxide effect): In presence of peroxides, addition of HBr proceeds by a free-radical chain mechanism (anti-Markovnikov) in which bromine radicals add to the less substituted carbon, ultimately giving the anti-Markovnikov product.

Q.32. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles. 
(i) H₃CO^- 
(ii)
(iii) Cl^.
(iv) Cl₂C:
(v) (H₃C)₃C⁺
(vi) Br^-
(vii) H₃COH
(viii) R-NH-R 

Ans. : Electrophiles are electron-deficient species (neutral or positively charged) that seek electrons:
(iii) Cl^.
(iv) Cl₂C:
(v) (H₃C)₃C⁺ are electrophiles.
Nucleophiles are electron-rich species and may be neutral or negatively charged.
(i) H₃CO^-
(ii) CH₃COO^-
(iv) Br^-
(vi) CH₃OH
(viii) R-NH-R is nucleophiles.

Q.33. The relative reactivity of 1°, 2°, 3° hydrogen's towards chlorination is 1:3.8:5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.

Ans.

(A) Primary (9 positions for Cl group possible)

(B) Secondary (2 positions for Cl group possible at CH)

(c) Tertiary (1 position for Cl group possible) Relative amounts of A, B and C compounds = Number of hydrogen x Relative reactivity.
So, relative amounts = 9 + 7.6 +5 = 21.6Total amount = 9 + 7.6 + 5 = 21.6

Q.34. Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane. Ans.

Solution: Wurtz coupling of alkyl halides with sodium gives symmetrical and mixed alkanes depending on reagents. The possible coupling products are listed in the diagrams below (see placeholders for structures and names):

Q.35. Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons. Ans.

Solution: 2-Methylpropane (isobutane) on radical chlorination gives two radical types:
• tert-butyl radical (3°) formed by removal of a tertiary hydrogen
• primary radicals (1°) formed by removal of a methyl hydrogen. The tertiary radical is more stable due to hyperconjugation and greater alkyl substitution which delocalises electron density and stabilises the radical centre.

Q.36. An alkane C₈H₁₈ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination, this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide. Ans.

The reaction is

The alkane is

The tertiary bromide is

Solution: The given conditions imply a symmetrical alkane that on bromination gives a single tertiary bromide isomer, which identifies the alkane as 2,2,4-trimethylpentane (or an isomeric neopentane derivative) consistent with the provided structures (see images for structures and reasoning).

Q.37. The ring systems having following characteristics are aromatic. (i) Planar ring containing conjugated π bonds. (ii) Complete delocalisation of the π-electrons in ring system i.e. each atom in the ring has unhybridised p-orbital, and (iii) Presence of (4n + 2) π-electrons in the ring where n is an integer (n = 0, 1, 2,...........) [Huckel rule]. Using this information classify the following compounds as aromatic/nonaromatic.

Ans. 

Q.38. Which of the following compounds are aromatic according to Huckel's rule? (A)

(B)

(C)

(D)

(E)

(F)

Ans.

Q.39. Suggest a route to prepare ethyl hydrogen sulphate (CH₃-CH₂-OSO₂-OH) starting from ethanol (C₂H₅OH). Ans.

Solution: Ethyl hydrogen sulphate can be prepared by reaction of ethanol with concentrated sulphuric acid, with removal of water under controlled conditions to form the sulfate ester. The schematic preparation is shown in the image placeholder.

Matching Type Questions

Q.40. Match the reagent from Column I which on reaction with CH3-CH=CH₂ gives some product given in Column II as per the codes given below

Ans. 

Q.41. Match the hydrocarbons in Column I with the boiling points given in Column II.

Ans. (i)→(b); (ii) →(c); (iii) →(a)

Explanation:

Q.42. Match the following reactants in Column I with the corresponding reaction products in Column II.

Ans. (i)→(d); (ii)→(c); (iii)→(b); (iv)→(a)

Solution.

Q.43. Match the reactions given in Column I with the reaction types in Column II.

Ans. (i)→(d); (ii)→(a); (iii)→(b); (iv)→(c)

Solution.

Assertion and Reason Type

In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.

Q.44. Assertion (A): The compound cyclooctane has the following structural formula :

It is cyclic and has conjugated 8π-electron system but it is not an aromatic compound.
Reason (R) : (4n + 2) π electrons rule does not hold good and ring is not planar.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct explanation of A.

Ans. (i) Both A and R are correct and R is the correct explanation of A.
Solution. Aromaticity requires planarity, complete delocalisation of π electrons and (4n + 2) π electrons (Hückel rule). Cyclooctatetraene nominally has 8 π electrons (4n, n=2) and adopts a non-planar tub conformation to avoid antiaromaticity; thus it is not aromatic. Both assertion and reason are correct and reason explains assertion.

Q.45. Assertion (A): Toluene on Friedel Crafts methylation gives o- and p-xylene.
Reason (R) : CH₃-group bonded to benzene ring increases electron density at o- and p- position.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (i) Both A and R are correct and R is the correct explanation of A.

Solution. CH₃ is an electron-donating group by hyperconjugation and weak +I effect; it activates the ring toward electrophilic substitution and increases electron density at ortho and para positions, directing electrophiles there. Hence Friedel-Crafts methylation of toluene preferentially gives ortho and para substituted products.

Q.46. Assertion (A): Nitration of benzene with nitric acid requires the use of concentrated sulphuric acid.
Reason (R): The mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile, NO⁺₂.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (i) Both A and R are correct and R is the correct explanation of A.

Solution. Concentrated sulphuric acid protonates nitric acid and generates the nitronium ion NO₂⁺, the active electrophile for electrophilic aromatic nitration. Therefore the use of conc. H₂SO₄ along with conc. HNO₃ is essential for nitration.

Q.47. Assertion (A): Among isomeric pentanes, 2, 2-dimethylpentane has highest boiling point.
Reason (R): Branching does not affect the boiling point.
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Ans. (iii) Both A and R are not correct.

Solution. Branching decreases surface area and thereby lowers boiling point. Among isomeric pentanes the most branched isomer (e.g., 2,2-dimethylpropane derivatives) will have the lowest boiling point, not the highest. Therefore both the assertion and the reason are incorrect as stated.

Long Answer Type

Q.48. An alkyl halide C₅H₁₁Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br₂ to give a compound 'C', which on dehydro-bromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' gives one mole of the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the reactions involved
Ans.

Q.49. 896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 g at STP. Hydrogenation of 'A' gives 2-methyl pentane. Also 'A' on hydration in the presence of H₂SO₄ and HgSO₄ gives a ketone 'B' having molecular formula C₆H₁₂O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved. Ans.

Q.50. An unsaturated hydrocarbon 'A' adds two molecules of H₂ and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of 'A', write its IUPAC name and explain the reactions involved. Ans.

Solution: Reductive ozonolysis cleaves C=C bonds to give carbonyl fragments. The products butane-1,4-dial (a dialdehyde), ethanal, and propanone indicate cleavage of two double bonds at specific positions. Addition of two molecules of H₂ implies two double bonds are hydrogenated. Combining the fragments yields the parental structure A. The structure and IUPAC name are provided in the image placeholder.

The reaction involved are as follows:

Q.51. In the presence of peroxide addition of HBr to propene takes place according to anti-Markovnikov's rule but peroxide effect is not seen in the case of HCl and HI. Explain. Ans.

Solution: The peroxide (or Kharasch) effect involves a radical chain mechanism and is observed for HBr because the reaction energetics make the radical pathway favourable: the Br· radical adds to the alkene to give the more stable carbon radical, followed by abstraction of H from HBr to complete the chain and give anti-Markovnikov product. For HCl, the H-Cl bond is too strong (high bond dissociation energy) to participate effectively in the radical chain; initiation and propagation steps are unfavourable, so peroxide effect is not observed. For HI, the H-I bond is weak and iodine radicals are readily formed, but iodine radicals rapidly combine to give I₂ and terminate the chain; thus the peroxide effect is not seen for HI either.