Free Alkyl Halides MCQs with Answers
458 Alkyl Halides MCQs from Chemistry, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Alkyl halides are named and related to their carbon-halogen structure, polarity and reactivity. The key reactions are nucleophilic substitution by SN1 and SN2 mechanisms and elimination by E1 and E2 mechanisms, including how substrate structure, nucleophile, base, solvent and temperature influence substitution versus elimination.
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Read the Alkyl Halides notesFree MDCAT chapter notes with key terms458 questions · page 13 of 23
- A. Secondary amine
- B. primary amine
- C. Cyanide
- D. Isocyanide
Explanation: When a primary amine reacts with CHCl3 in the presence of NaOH then a foul smelling gas Isocyanide is formed, which on further reduction…
Correct answer: primary amine- A. alcohols
- B. amines
- C. Sulphates
- D. Salts of ammonium chlorides
Explanation: NH4Cl forms an acidic solution. If this solution is mixed with aq NaHCO3, CO2 will be released and there will be NaCl and freebase NH3…
Correct answer: Salts of ammonium chlorides- A. 2−aminobutane
- B. N−methyl propaneamine
- C. N,N−dimethyl ethaneamine
- D. 1−aminobutanetane
Explanation: In 2-aminobutane (CH3 - CH2 - CHNH2 - CH3), all valency of the 2nd carbon atom is satisfied with four different groups, hence it is a…
Correct answer: 2−aminobutane- A. (B) is less stable than (A) so, it is not acceptable.
- B. (A) is non-aromatic so it is acceptable structure.
- C. Nitrogen of (A) possess 8 valence electrons so, (A) is acceptable.
- D. Nitrogen of (B) possess 10 valence electrons so, Structure (B) is acceptable and (A) is not.
Explanation: Structure (A) is aromatic and has 8 electrons in its outer most shell hence it is the acceptable anilinium ion structure.
Correct answer: Nitrogen of (A) possess 8 valence electrons so, (A) is acceptable.- A. diluteHCl
- B. CuSO4 solution
- C. AgNO3
- D. All of these
Explanation: Amines being basic in nature dissolve in dilute HCl. They can also coordinate with 2Cu2+ +and Ag+ ions to form soluble complexes as they…
Correct answer: All of these- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: Hoffmann's degradation is given by −CONH2 group. Amides on reaction with Br2 in strong alcoholic medium to give reduced product- primary…
Correct answer: Option B- A. 1,3,5−tribromobenzene
- B. p−bromofluorobenzene
- C. p−bromoaniline
- D. 2,4,6−tribromofluorobenzene
Explanation: NH2 is an ortho para directing group, when bromine water reacts with aniline it produces 2,4,6-tribromoaniline.
Correct answer: 2,4,6−tribromofluorobenzene- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: To introduce nitro group in the aniline ring, the reagents used are conc.HNO3 and conc.H2SO4 which are very strong acids.
Correct answer: Option B- A. hydrazobenzene
- B. aniline
- C. azobenzene
- D. N-phenyl hydroxyl amine
Explanation: The reduction of nitrobenzene in the presence of Zn/NH4Cl gives N-phenylhydroxylamine as the product.
Correct answer: N-phenyl hydroxyl amine- A. Sn + HCl
- B. Zn + NH4Cl
- C. Na3AsO3 + NaOH
- D. Zn + NaOH
Explanation: When nitro compounds are reduced, then amines are formed. The bulky nitrobenzene is reduced using zinc, and sodium hydroxide.
Correct answer: Zn + NaOH- A. cannizaro
- B. Claisen
- C. Hoffmannbromamide
- D. Schmidt
Explanation: The Hofmann-Bromamide reaction is a chemical reaction used to convert primary amides to primary amines.
Correct answer: Hoffmannbromamide- A. PCl5
- B. NaOH + Br2
- C. NaOH + Water
- D. HNO3
Explanation: An amide on reaction with Bromine in alcoholic conditions yields primary amines and this reaction proceeds by rearrangement.
Correct answer: NaOH + Br2253. The product X is:
- A. CH2CH3CONH2
- B. CH3CH2 CH2 CH2NH2
- C. C4H10
- D. CH3 CH2NH CH2CH2CH3
Explanation: Alkylcynide is reduced to primary amine in the presence of reducing agent sodium-ethoxide.
Correct answer: CH3CH2 CH2 CH2NH2- A. The reactant was a methyl ether
- B. The reactant was a symmetrical ether
- C. The reactant was a cyclic ether
- D. Both (b) and (c) may be correct
Explanation: An unknown ether reacts with excess of HBr to yield a single product. This implies that the ether must be either symmetrical or cyclic.
Correct answer: Both (b) and (c) may be correct- A. 1° and 2° amine
- B. 1°, 2°, 3° & quaternary
- C. 1°, 2° & 3° amines
- D. 1° & 3° amine
Explanation: Alcoholic solution of ammonia is heated in a scaled tube at 100°C with primary alkyl halide to form primary amine.
Correct answer: 1°, 2°, 3° & quaternary- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: The ease of reaction is: t-alkyl halide > s-alkyl halide > p-alkyl halide So tertiary alkyl halide will undergo the reaction the fastest.
Correct answer: Option D- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: The ease of reaction is: t-alkyl halide > s-alkyl halide > p-alkyl halide So tertiary alkyl halide will undergo the reaction the fastest.
Correct answer: Option A- A. X>Y>Z
- B. X>Z>Y
- C. Y>Z>X
- D. Z>Y>X
Explanation: The correct option is D Z>Y>XIn E2 mechanism, elimination occurs with the hydrogen which is present anti to the leaving group.
Correct answer: Z>Y>X- A. Option A
- B. Option B
- C. Option C
- D. Option D
Explanation: NaI favours substitution in alkyl halide because it is a relatively weak base. So NaI will not favour elimination reaction.
Correct answer: Option D- A. A good leaving group is a strong base
- B. A good leaving group is a weak base
- C. A leaving group must be negatively charged
- D. A leaving group must be a halide
Explanation: A good leaving group is indeed a weak base, as it is easily replaceable by a stronger base and can be easily removed.
Correct answer: A good leaving group is a weak base