2-Methylpropanal is heated with excess aqueous formaldehyde in the presence of concentrated NaOH. The organic product is

2-Methylpropanal has one $\alpha$-hydrogen, so with $\mathrm{HCHO}$ it undergoes a crossed aldol addition to give $\mathrm{HOCH_2C(CH_3)_2CHO}$. This product has no $\alpha$-hydrogen, so in concentrated alkali with excess formaldehyde it undergoes a crossed Cannizzaro reaction. Formaldehyde, the better hydride donor, is oxidised to formate, while the hindered aldehyde is reduced to $\mathrm{HOCH_2C(CH_3)_2CH_2OH}$, 2,2-dimethylpropane-1,3-diol (neopentyl glycol).