Saytzeff vs Hofmann products

35 questions on Saytzeff vs Hofmann products from the Organic Reaction Mechanisms IV: Elimination chapter: 7 NEET level, 14 JEE Main, 14 JEE Advanced. Each one opens with its answer and solution.

• CH3-CH2-CH(Br)-CH3 (2-bromobutane) is heated with alcoholic KOH. The major product is NEET level
• According to the Saytzeff (Zaitsev) rule, in a dehydrohalogenation the alkene formed in the larger amount is the one that NEET level
• 2-Bromo-2-methylbutane is heated with NaOEt in ethanol. The major product is NEET level
• Ethyltrimethylammonium hydroxide, [CH3CH2N(CH3)3]^+ OH^-, is heated strongly (Hofmann elimination). The hydrocarbon formed is NEET level
• (Butan-2-yl)trimethylammonium hydroxide, CH3CH2CH(CH3)-N^+(CH3)3 OH^-, is heated at about 180 degC. The major alkene is JEE Main level
• 2-Bromo-2-methylbutane is treated with (CH3)3COK in (CH3)3COH. The major alkene is JEE Main level
• 2-Fluorobutane is heated with NaOEt in ethanol. Unlike 2-bromobutane, the major alkene is JEE Main level
• 2-Bromo-2,3-dimethylbutane is heated with alcoholic KOH. The Saytzeff (major) alkene is JEE Main level
• Ethyldimethylpropylammonium hydroxide, CH3CH2CH2-N^+(CH3)2-CH2CH3 OH^-, is heated. Two alkenes can be formed; the one formed in larger amount is JEE Main level
• 2-Bromo-3-methylpentane, CH3-CH(Br)-CH(CH3)-CH2-CH3, undergoes E2 elimination with a base, with no rearrangement. How many distinct alkenes (all constitutional ... JEE Main level
• In the E2 dehydrohalogenation of 2-bromo-2-methylbutane, the proportion of the less substituted alkene 2-methylbut-1-ene (Hofmann product) increases when (selec... JEE Main level
• Trimethyl(1-phenylpropan-2-yl)ammonium hydroxide, C6H5-CH2-CH(CH3)-N^+(CH3)3 OH^-, is heated. The major alkene is JEE Advanced level
• Two stereoisomeric 2-isopropyl-5-methylcyclohexyl chlorides (Cl on C-1, isopropyl on C-2, methyl on C-5) undergo E2 with NaOEt/ethanol. In the more stable chair... JEE Advanced level
• Select all combinations in which the substrate, reagent and major product are correctly matched. JEE Advanced level
• (1S,2S)-1-Bromo-1,2-diphenylpropane is treated with NaOEt/ethanol. The only beta hydrogen is on C-2, and the reaction is a concerted anti E2 elimination. The pr... JEE Advanced level
• A 30.2 g sample (0.200 mol, M = 151 g/mol) of 2-bromo-2-methylbutane is subjected to two separate E2 runs. In run (i), (CH3)3COK/(CH3)3COH converts 95 percent o... JEE Advanced level
• For 2-pentyl-X with a strong base under E2 conditions, which is the correct order of increasing proportion of the Hofmann product, pent-1-ene, for X = Br, F and... JEE Advanced level
• trans-(2-Methylcyclohexyl)trimethylammonium hydroxide (-N^+(CH3)3 on C-1, methyl on C-2, trans) is heated. The major alkene is JEE Advanced level
• 2-Bromobutane is heated with KOH in ethanol. The major alkene is: NEET level
• 2-Bromo-2-methylbutane is treated with (CH3)3COK in (CH3)3COH. The major alkene is: NEET level
• Butan-2-yl trimethylammonium hydroxide, CH3CH2CH(CH3)-N^+(CH3)3 OH^-, is heated. The major alkene is: NEET level
• Which pair correctly gives the major alkene from 2-bromo-2-methylbutane? JEE Main level
• N,N,N,2-Tetramethylbutan-2-aminium hydroxide, CH3CH2C(CH3)2-N^+(CH3)3 OH^-, is heated. The major alkene formed is: JEE Main level
• 2-Fluorobutane is heated with NaOEt in ethanol. Unlike 2-bromobutane, the major alkene is but-1-ene. This is best explained by: JEE Main level
• Potassium tert-butoxide gives a higher proportion of the Hofmann alkene than sodium ethoxide with the same alkyl halide mainly because: JEE Main level
• Which statements are correct? JEE Main level
• 2-Bromo-2-methylbutane gives 2-methylbut-2-ene (70%) and 2-methylbut-1-ene (30%) with NaOEt, but 2-methylbut-2-ene (28%) and 2-methylbut-1-ene (72%) with (CH3)3... JEE Main level
• For the E2 reaction of 2-halobutanes with NaOEt in ethanol the percentages of but-1-ene (rest but-2-ene) in this problem are: F 70, Cl 33, Br 31, I 30. The data... JEE Main level
• 2-Bromo-2,3-dimethylbutane, (CH3)2CBr-CH(CH3)2, is treated separately with NaOEt/ethanol and with (CH3)3COK/(CH3)3COH. Which statements are correct? JEE Advanced level
• Pentan-2-amine is exhaustively methylated with excess CH3I, converted into the hydroxide with Ag2O/H2O, and heated. How many moles of CH3I are consumed per mole... JEE Advanced level
• Ethyl(isopropyl)dimethylammonium hydroxide, (CH3)2CH-N^+(CH3)2(CH2CH3) OH^-, is heated. The major products are: JEE Advanced level
• Which reactions give the less substituted (Hofmann) alkene as the major product? JEE Advanced level
• Heats of hydrogenation are: pent-1-ene 126 kJ/mol, (Z)-pent-2-ene 120 kJ/mol, (E)-pent-2-ene 116 kJ/mol. Pentan-2-ol is dehydrated with conc. H2SO4 under condit... JEE Advanced level
• Piperidine is exhaustively methylated (CH3I in excess, then Ag2O/H2O, heat) and the resulting tertiary amine is subjected to the whole sequence once more. The h... JEE Advanced level
• Which statements about Hofmann elimination (quaternary ammonium hydroxide, heated) are correct? JEE Advanced level