Pinacol (2,3-dimethylbutane-2,3-diol) is warmed with dilute H2SO4 to give a ketone P. P is treated with I2/NaOH to give an acid Q (together with a…

Under acid, one hydroxyl of pinacol is protonated and lost to give a tertiary carbocation. A methyl group shifts from the neighbouring carbinol carbon, and loss of the proton from the remaining $-\mathrm{OH}$ gives pinacolone, $\mathrm{(CH_3)_3C\text{-}CO\text{-}CH_3}$ ($\mathbf{P}$, $\mathrm{C_6H_{12}O}$). It is a methyl ketone, so the haloform reaction gives pivalic acid $\mathrm{(CH_3)_3C\text{-}COOH}$ ($\mathbf{Q}$) and iodoform. $\mathrm{SOCl_2}$ and $\mathrm{NH_3}$ convert it to pivalamide $\mathrm{(CH_3)_3C\text{-}CONH_2}$. In the Hofmann bromamide degradation the $\mathrm{C=O}$ carbon is lost as carbonate and the tert-butyl group migrates to nitrogen with retention, so $\mathbf{R}$ is $\mathrm{(CH_3)_3C\text{-}NH_2}$, $\mathrm{C_4H_{11}N}$. The amide has one carbon more than the amine, so the neopentyl and 2-methylbutan-2-yl amines are excluded.