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ChemistryMediumMCQ2014 · 06 Apr

Q57.In the reaction, LiAlH4 PCl5 Alc. KOH −−−CH3COOH → A → B → C, the product C is : (1) Acetaldehyde (2) Acetylene (3) Ethylene (4) Acetyl chloride

What This Question Tests

This question requires tracking a three-step organic synthesis, involving the reduction of a carboxylic acid, replacement of hydroxyl group, and subsequent elimination reaction to form an alkene.

Concepts Tested

Reduction of carboxylic acidSubstitution reactions (PCl₅)Dehydrohalogenation (E2 reaction)

📚 NCERT Sections This Tests

9.9Give The Structures Of A, B And C In The Following Reactions:

Chemistry Class 12 · Chapter 9

80% match

9.9 Give the structures of A, B and C in the following reactions: B NaOH  Br 2 C (i) CH 3 CH 2 I NaCN A Partial OHhydrolysis CuCN H 2 O/H  NH 3 (ii) C 6 H 5 N 2 Cl  A  B  C 4 B HNO 2 C (iii) CH 3 CH 2 Br KCN A LiAlH 0  C  2  2 O/H HCl B H C  (iv) C6 H 5 NO 2 Fe/HCl A NaNO273 K 3 A NaOBr B NaNO 2 /HCl C (v) CH 3 COOH NH 2 B C 6 H 5 OH C (vi) C6 H 5 NO 2 Fe/HCl A HNO273K 279 Amines Reprint 2025-26

6.22What Happens When

Chemistry Class 12 · Chapter 6

78% match

6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN? Answers to Some Intext Questions 6.1 6.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2. 6.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CHCH2CH3 (iv) CH3CCl2CH3 191 Haloalkanes and Haloarenes Reprint 2025-26

6.21Primary Alkyl Halide C4H9Br (A) Reacted With Alcoholic Koh To Give Compound (B).

Chemistry Class 12 · Chapter 6

78% match

6.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.