NEET Previous Year Questions (2014-2024): Hydrocarbons | Chemistry Class 11 PDF Download

2024

• n-pentane: It is a straight-chain alkane with the formula C₅H₁₂, and it has the highest boiling point among the isomers because of its larger surface area which allows for greater van der Waals forces (dispersion forces).
• isopentane (also known as methylbutane): It has one branch in its carbon chain. The branching reduces the surface area slightly, which slightly weakens the van der Waals forces compared to n-pentane.
• neopentane (also known as dimethylpropane): It has a highly branched structure making it almost spherical. This shape minimizes the surface area significantly, thereby reducing the van der Waals forces drastically compared to the other two isomers.

Thus, the boiling point order correctly reflects the influence of molecular structure and intermolecular forces. Hence, Statement I is correct.
Statement II: This statement elaborates on why branching leads to lower boiling points. The assertion is that increased branching gives the molecule a more spherical shape, which then results in a smaller surface area, and thus weaker intermolecular forces (primarily van der Waals forces). Weaker intermolecular forces correspond with a lower energy requirement for the liquid to gas phase transition, thereby lowering the boiling point. This explanation is coherent with concepts in physical chemistry regarding molecular interactions and phase change. Therefore, Statement II is also correct.
In summary, both Statement I and Statement II are correct, reflecting the accurate relationship between molecular structure, intermolecular forces, and physical properties like boiling points. Therefore, the correct answer is:
Option A: Both Statement I and Statement II are correct.

Q2: Identify the major product C formed in the following reaction sequence :

(a) propylamine
(b) butylamine
(c) butanamide
(d) α-bromobutanoic acid              (NEET 2024)
Ans: (a)

• Step-l is S_N reaction with nucleophile.
• Step-II will give amide.
•  Step-III is Hoffmann bromamide degradation reaction.

2023

Q1: Identify product (A) is the following reaction:         (NEET 2023)

(a)

(b)

(c)

(d)

Ans: (d)

Q2:  Consider the following reaction and identify the product (P).               (NEET 2023)

(a)

(b)

(c)

(d)

Ans: (d)

2022

Q1: The decreasing order of boiling points of the following alkanes is :        (NEET 2022)
(a) heptane
(b) butane
(c) 2-methylbutane
(d) 2-methylpropane
(e) hexane
Choose the correct answer from the options given below:
(a) (a) > (e) > (c) > (b) > (d)
(b) (a) > (c) > (e) > (d) > (b)
(c) (c) > (d) > (a) > (e) > (b)
(d) (a) > (e) > (b) > (c) > (d)
Ans: (a)

• With increase in number of carbons in alkane, boiling point increases
• In case of isomeric alkanes, greater is the number of branches, lesser is the boiling point.

Boiling point order :

Q2: The products A and B in the following reaction sequence are :        (NEET 2022)

(a)

(b)

(c)

(d)

Ans: (c)

Q3: The incorrect method for the synthesis of alkenes is        (NEET 2022)
(a) Treating vicinal dihalides with Zn metal
(b) Treating of alkynes with Na in liquid NH₃
(c) Heating alkyl halides with alcoholic KOH
(d) Treating alkyl halides in aqueous KOH solution
Ans: (d)
Alkenes can be prepared

2021

Q1: The correct structure of 2, 6-dimethyl-dec-4-ene is      (NEET 2021)
(a)

(b)

(c)

(d)

Ans: (a)

2020

Q1: An alkene on ozonolysis gives methanal as one of the product. Its structure is:     (NEET 2020)
A: 
B: 
C: 
D: 
Ans: (a)

2019

Q1: The most suitable reagent for the following conversion, is :    (NEET 2019)

A: Na/liquid NH₃
B: H₂, Pd/C, quinoline
C: Zn/HCl
D: Hg²⁺/H⁺, H₂O
Ans: (b)


Q2: An alkene "A" on reaction with O₃ and Zn–H₂O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene "A" gives "B" as the major product. The structure of product "B" is:    (NEET 2019)
A:
B:
C:
D:
Ans: (c)


Q3: Among the following, the reaction that proceeds through an electrophilic substitution, is:    (NEET 2019)
A:
B:
C:
D:
Ans: (b)

Generation of electrophile:

2018

Q1: Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is    (NEET 2018)

A: CH $\equiv$ CH 
B: CH₂ = CH₂
C: CH₃ - CH₃
D: CH₄
Ans: (d)

2017

Q1: Which one is the correct order of acidity ?    (NEET 2017)
A:
B:
C:
D:
Ans: (a)
Greater the character of C-atom in hydrocarbons, greater the electronegativity of that carbon and thus greater the acidic nature of the H attached to electronegative carbon.
Thus CH$\equiv$CH > CH₃ $-$C$\equiv$CH > CH₂ $=$ CH₂ > CH₃$-$CH₃ 

Q2: Predict the correct intermediate and product in the following reaction :    (NEET 2017)

A:
B:
C:
D:
Ans: (c)


Q3: With respect to the conformers of ethane, which of the following statements is true?     (NEET 2017)
(a) Bond angle changes but bond length remains same.
(b) Both bond angle and bond length change.
(c) Both bond angle and bond length remain same.
(d) Bond angle remains same but bond length changes.
Ans: (c)
Conformers are the isomers which are formed by rotation about single bonds without any cleavage of any bond. These conformers have same bond angle between them and have same bond length while their dihedral angle changes.

2016

Q1: The pair of electron in the given carbanion, is present in which of the following orbitals?    (NEET 2016 Phase 1)
A: sp
B: 2p
C: sp³
D: sp²
Ans: (a)


Q2: In the reaction    (NEET 2016 Phase 1)

A: X=2-butyne; Y= 3-hexyne
B: X=1-butyne; Y =3-hexyne
C: X=1-butyne ;Y = 2-hexyne
D: X=2-butyne;Y=2-hexyne
Ans: (b)
NaNH₂/liq.NH₃ behaves as a base, so it abstracts a proton from acetylene to form acetylide anion followed by alkylation to give compound (X) i.e 1-butyne (X) further reacts with NaNH₂/Liq NH₃ followed by alkylation with ethyl bromide yields 3-hexyne (Y).


Q3: Consider the nitration of benzene using mixed conc. H₂SO₄ and HNO₃ . If a larger amount of KHSO₄ is added to the mixture the rate of nitration will be :    (NEET 2016 Phase 1)
A: Doubled
B: Faster
C: Slower
D: Unchanged
Ans: (c)
In the nitration of benzene in the presence of conc. H₂SO₄ and HNO₃, benzene is formed.
HNO₃ +H2SO₄ 
If a large amount of KHSO₄ is added to this mixture more HSO₄^- ion furnishes and hence the concentration of electrophile decreases, rate of electrophilic aromatic reaction slows down.

Q4: For the following reactions :       (NEET 2016 Phase 1)

Which of the following statements is correct?
(a) (A) is elimination, (B) and (C) are substitution reactions.
(b) (A) is substitution, (B) and (C) are addition reactions.
(c) (A) and (B) are elimination reactions and (C) is addition reaction.
(d) (A) is elimination, (B) is substitution and (C) is addition reaction.
Ans: (d)

• (A) Saturated compound is converted into unsaturated compound by removal of group of atoms hence, it is an elimination reaction.
• (B) —Br group is replaced by —OH group hence, it is a substitution reaction.
• (C) Addition of Br₂ converts an unsaturated compound into a saturated compound hence, it is an addition reaction.

Q5: Which of the following can be used as the halide component for Friedel-Crafts reaction?
(a) Chlorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride      (NEET 2016 Phase 2)
Ans: (d)
Friedel–Crafts reaction :

Chlorobenzene, bromobenzene and chloroethene are not suitable halide components as lone pair of electrons of halogen are delocalized with π-bonds to attain double bond (C = X) character.

Q6: In pyrrole the electron density is maximum on      (NEET 2016 Phase 2)

(a) 2 and 3
(b) 3 and 4
(c) 2 and 4
(d) 2 and 5
Ans: (d)
Pyrrole has maximum electron density on 2 and 5. It generally reacts with electrophiles at the C-2 or C-5 due to the highest degree of stability of the protonated intermediate.

Attack at position 3 or 4 yields a carbocation that is a hybrid of structures (I) and (II). Attack at position 2 or 5 yields a carbocation that is a hybrid not only of structures (III) and (IV) (analogous to I and II) but also of structure (V). The extra stabilization conferred by (V) makes this ion the more stable one.
Also, attack at position 2 or 5 is faster because the developing positive charge is accommodated by three atoms of the ring instead of by only two.

Q7: In which of the following molecules, all atoms are coplanar?      (NEET 2016 Phase 2)
(a) 

(b)

(c)

(d)

Ans: (a)
Biphenyl is coplanar as all carbon atoms are sp2 hybridised.

Q8: In the given reaction,

the product P is    (NEET 2016 Phase 2)
(a)

(b)

(c)

(d)

Ans: (c)

Q9: The compound that will react most readily with gaseous bromine has the formula   (NEET 2016 Phase 2)
(a) C₃H₆
(b) C₂H₂
(c) C₄H₁₀
(d) C₂H₄
Ans: (a)
Propene is most reactive towards Br₂ (gaseous) than CH₂ = CH₂, HC $\equiv$ CH and butane due to most electron density. 

2015

Q1: Given

The enthalpy of hydrogenation of these compounds will be in the order as :    (NEET / AIPMT 2015 Cancelled Paper)
A: II > I > III
B: I > II > III
C: III > II > I
D: II > III > I
Ans: (c)
Higher is the stability, lower is the enthalpy of hydrogenation.
(I) is most stable due to aromatic character. Hence it has lowest enthalpy of hydrogenation.
(III) is least stable as no resonance is present. Hence, it has highest enthalpy of hydrogenation.
Thus, the decreasing order of the enthalpy of hydrogenation is III > II > I.

Q2: A single compound of the structure

is obtainable from ozonolysis of which of the following cyclic compounds ?     (NEET / AIPMT 2015 Cancelled Paper)
A:
B:
C:
D:
Ans: (b)

Q3: Which of the following is not the product of dehydration of                  (NEET / AIPMT 2015)

(a)

(b)

(c)

(d)

Ans: (a)

Q4: 2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid?
(a) (CH₃)₃CCH $=$ CH₂ 
(b) (CH₃)₂C $=$ CHCH₂CH₃ 
(c) (CH₃)₂CHCH₂CH $=$ CH₂ 
(d)             (NEET / AIPMT 2015)
Ans: (a)

Q5:  The reaction of C₆H₅CH = CHCH₃ with HBr produces :    (NEET / AIPMT 2015)
A:
B:
C:
D:
Ans: (b)

2014

Q1: Identity Z in the sequence of reactions    (NEET 2014)

A: CH₃ (CH₂)₄ −O−CH₃
B: CH₃CH₂ −CH(CH₃) −O−CH₂CH₃
C: CH₃ − (CH₂)₃ −O−CH₂CH₃
D:(CH₃)₂CH₂ −O−CH₂CH₃
Ans: C


Q2: Which of the following organic compounds has same hybridization as its combustion product (CO₂)?    (NEET 2014)
A: Ethene
B: Ethanol
C: Ethane
D: Ethyne
Ans: D
HC ≡ CH and O = C = O both have sp−hybridised carbon

Q3: What products are formed when the following compound is treated with Br₂ in the presence of FeBr₃?    (NEET 2014)

A:
B:
C:
D:
Ans: (d)
-CH₃ group is o,p-directing. Because of crowding, no substitution occurs at the carbon atom between the two -CH₃ groups in m-Xylene, even though two -CH₃ groups activate that position.