Haloform reaction

27 questions on Haloform reaction from the Aldehydes, Ketones and Carboxylic Acids chapter: 5 NEET level, 14 JEE Main, 8 JEE Advanced. Each one opens with its answer and solution.

• Which of the following compounds gives a yellow precipitate on warming with I2 and aqueous NaOH? NEET level
• Propanone, CH3COCH3, is warmed with excess I2 and NaOH. The two products are NEET level
• Which of the following does NOT give a yellow precipitate in the iodoform test? NEET level
• The pale yellow solid with a characteristic antiseptic smell formed in the iodoform test has the formula NEET level
• Phenyl methyl ketone, C6H5COCH3, is heated with NaOH and I2 and the mixture is then acidified. The products are NEET level
• Which of the following does NOT give iodoform on warming with I2 and NaOH? JEE Main level
• 4-Methylpentan-2-one, (CH3)2CHCH2COCH3, is treated with NaOCl in aqueous NaOH and then acidified. The organic acid obtained is JEE Main level
• 1-(4-Methylphenyl)ethan-1-one is heated with excess NaOCl in aqueous NaOH and then acidified. The organic acid formed is JEE Main level
• 1-Cyclopentylethan-1-one is treated with Br2 in excess aqueous NaOH and then acidified. The organic acid obtained is JEE Main level
• Assertion (A): A methyl ketone treated with excess I2 in aqueous NaOH is converted to the triiodo ketone and cannot be stopped at the monoiodo stage. Reason (R)... JEE Main level
• Why is the haloform reaction with iodine (CHI3) preferred over those giving CHCl3 or CHBr3 as a laboratory test? JEE Main level
• How many moles of I2 are consumed per mole of ethanol in the reaction CH3CH2OH + I2 + NaOH that finally gives CHI3 and HCOONa? JEE Main level
• Which of the following compounds gives a positive iodoform test AND reduces Tollens reagent? JEE Main level
• Ethanal is treated with excess NaOCl in aqueous NaOH. The products are JEE Main level
• Ethanol, 4.6 g, is treated with excess iodine in aqueous NaOH and is completely converted to iodoform and sodium methanoate by CH3CH2OH + 4I2 + 6NaOH -> CHI3 + ... JEE Main level
• Which of the following give a yellow precipitate with I2 and NaOH? JEE Main level
• Butan-2-one is treated with excess NaOCl in aqueous NaOH and the mixture is then acidified. Which statements are correct? JEE Main level
• Which statements about the mechanism of the haloform reaction of a methyl ketone RCOCH3 in aqueous NaOH and I2 are correct? JEE Main level
• Which of the following give iodoform when warmed with I2 and aqueous NaOH? JEE Main level
• Compound A, C5H10O, gives a yellow precipitate with I2/NaOH and does not reduce Tollens reagent. Clemmensen reduction of A gives straight-chain pentane. A is tr... JEE Advanced level
• Benzene is acetylated with CH3COCl/anhydrous AlCl3 to give P. P is treated with NaOCl/NaOH and then acidified to give Q. Q is warmed with SOCl2 to give R. R is JEE Advanced level
• Nine compounds are separately warmed with I2/NaOH: ethanol, propan-2-ol, phenyl methyl ketone, benzaldehyde, pentan-3-one, ethanal, propanal, methanol and butan... JEE Advanced level
• Each of the following is warmed with excess I2/NaOH and then acidified. Which one gives iodoform together with an organic acid of molar mass 128 g/mol? (C 12, H... JEE Advanced level
• Compound X, C8H8O, gives a yellow precipitate with NaOI. On acidification the aqueous layer gives acid Y of molar mass 122 g/mol. X is subjected to Wolff-Kishne... JEE Advanced level
• How many acyclic carbonyl compounds (aldehydes and ketones, ignoring stereoisomers) of molecular formula C5H10O give a positive iodoform test? JEE Advanced level
• Hexane-2,5-dione, CH3COCH2CH2COCH3, is treated with excess NaOCl in aqueous NaOH and then acidified. Which statements are correct? JEE Advanced level
• Butan-2-ol, 14.8 g, is warmed with excess I2 in aqueous NaOH and converted to iodoform in 80% overall yield. Take molar masses: butan-2-ol 74 g/mol, I 127, C 12... JEE Advanced level