JEE Advanced 2018 Paper 1 Chemistry: Paragraph 2

A gives a 2,4-DNP derivative, so it is an aldehyde or ketone. The iodoform positive result requires a $\mathrm{CH_3CO-}$ unit, which excludes pentan-3-one and 2-methylbutanal. This leaves pentan-2-one and 3-methylbutan-2-one. The Clemmensen product of pentan-2-one is n-pentane, which has three monochloro derivatives (1-, 2- and 3-chloropentane). The Clemmensen product of 3-methylbutan-2-one is isopentane, 2-methylbutane, which has four constitutional monochloro derivatives (1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane). The haloform step then gives $\mathrm{(CH_3)_2CHCOOH}$, B. A is therefore 3-methylbutan-2-one, $\mathrm{(CH_3)_2CHCOCH_3}$, and the other three candidates fail one of the tests.

B is isobutyric acid, so the amide is $\mathrm{(CH_3)_2CHCONH_2}$ (molar mass 87) and the amine C is isopropylamine $\mathrm{(CH_3)_2CHNH_2}$ (molar mass 59) with one carbon fewer. Moles of amide $=8.70/87=0.100$ mol. At 80% yield the amine formed is $0.080$ mol, i.e. $0.080\times59=4.72$ g. The stoichiometry uses 4 mol KOH per mole of amide, so the reaction consumes $0.400$ mol KOH $=0.400\times56=22.4$ g (the problem asks for the stoichiometric amount, which is based on the amide taken, not on the yield). This matches option A. Option B quotes the theoretical amine mass, C halves the alkali, and D wrongly applies 60% yield.