149 questions on Organic Compounds Containing Halogens for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.
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Answer: (B) Opposite side
SN2 = backside attack = Walden inversion. 180° approach.
Quick tip: SN2 = backside = inversion. SN1 = carbocation = racemization. 3° prefers SN1.
NCERT Class 12, Chapter 10, page 296
Answer: (B) 1-Bromopropane
In the presence of peroxides (Kharasch effect), HBr adds to unsymmetrical alkenes via a free-radical mechanism to give anti-Markovnikov addition. Hence, propene reacts with HBr/peroxide to form 1-bromopropane as the major product.
Answer: (C) Is not metabolised rapidly, accumulates in fatty tissues and biomagnifies in food chains
DDT is very persistent; it is stored in fatty tissues and concentrates up food chains. Many insect species have also developed resistance.
NCERT Class 12
Answer: (B) 60% inversion and 40% retention
The observed optical purity of the product is (-3.6° / -18.0°) × 100% = 20% enantiomeric excess of inverted product. Since % excess = % inversion - % retention = 20%, and % inversion + % retention = 100%, solving gives 60% inversion and 40% retention.
Answer: (A) 2-Bromo-2-methylpropane
The product after reaction with D₂O is (CH₃)₃C-D (2-deuterio-2-methylpropane). This indicates that the Grignard reagent 'Y' must be tert-butylmagnesium halide, (CH₃)₃C-MgX. Therefore, the starting haloalkane 'X' must be 2-halo-2-methylpropane, such as 2-bromo-2-methylpropane.
Answer: (C) N,N-Dimethylformamide (DMF)
Polar aprotic solvents like N,N-dimethylformamide (DMF) or acetone do not solvate anions (nucleophiles) effectively via hydrogen bonding. This leaves the nucleophile bare and significantly increases its nucleophilicity and the overall rate of an SN2 reaction. Polar protic solvents like water, ethanol, and acetic acid solvate the nucleophile and decrease its reactivity.
Answer: (A) The secondary carbocation undergoes a 1,2-hydride shift to form a more stable tertiary carbocation
Ionization of the secondary chloride gives a 2° carbocation, (CH₃)₂CH-CH⁺-CH₃. A rapid 1,2-hydride shift converts this into the more stable 3° carbocation, (CH₃)₂C⁺-CH₂CH₃, which is then trapped by ethanol to give the rearranged tertiary ethyl ether as the major product.
Answer: (C) 12
The chemical formula of DDT is C₁₄H₉Cl₅. Its structure consists of two chlorophenyl rings joined to an ethane backbone: (p-Cl-C₆H₄)₂CH-CCl₃. Each benzene ring contains 6 sp²-hybridised carbon atoms, giving 2 × 6 = 12 sp²-hybridised carbons. The remaining 2 carbons in the -CH(CCl₃)- backbone are sp³-hybridised.
Answer: (D) Rises to the upper atmosphere and depletes the ozone layer
CCl₄ reaching the upper atmosphere depletes ozone. Exposure is also linked to liver damage and nervous system effects in humans.
NCERT Class 12
Answer: (A) Both Statement I and Statement II are correct
In many SN1 solvolysis reactions, complete planarization into an entirely free carbocation is not instantaneous; rather, an intimately associated ion-pair is formed first. The departing halide ion physically shields the front side of the carbocation, making solvent attack from the rear face statistically favoured. Hence, the product is not 100% racemic, but shows partial racemization with net inversion.
Answer: (B) 2-Bromobutane
Ozonolysis of alkene (X) produces only ethanal (CH₃CHO), which means alkene (X) is symmetrical and must be but-2-ene (CH₃-CH=CH-CH₃). Dehydrohalogenation of 2-bromobutane with alcoholic KOH yields but-2-ene as the major product according to Saytzeff's rule. Therefore, the starting alkyl bromide is 2-bromobutane.
Answer: (D) Chloroform (CHCl₃)
According to NCERT, the major use of chloroform today is in the production of the Freon refrigerant R-22. Reaction of chloroform with HF in the presence of SbCl₅ yields chlorodifluoromethane (CHClF₂).
Answer: (D) Higher lattice energy and precipitation of sodium chloride in acetone
Sodium chloride is practically insoluble in dry acetone and precipitates out of the solution. According to Le Chatelier's principle, this continuous precipitation drives the forward reaction to completion.
Answer: (A) (I) > (IV) > (III) > (II)
All are primary alkyl halides, so their SN2 reactivity is governed by steric hindrance near the α-carbon. Steric hindrance increases with branching closer to the reacting centre: 1-chlorobutane (least hindrance) > 1-chloro-3-methylbutane (branch at γ) > 1-chloro-2-methylbutane (branch at β) > 1-chloro-2,2-dimethylpropane (neopentyl chloride, severe β-branching). Hence, the order is (I) > (IV) > (III) > (II).
Answer: (C) I > II > III > IV
S_N2 substitution reactions are governed by steric hindrance at the reaction centre. The transition state becomes progressively more crowded with increasing alkyl substitution on the α-carbon. Therefore, the reactivity order decreases as methyl > primary > secondary > tertiary: CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl.