Question

Hat product will be form when neopentyl alcohol is treated with conc. Hcl in presence of zncl2 brainly

Answer 1

After the protonation of -OH group by HCl and subsequent removal of water molecule, we obtain a secondary carbocation. But a shifting of methyl group(1,2-methyl shift) will prepare a tertiary carbocation and hence the expected results/products are not obtained. The following reactions shows the same:{as shown in the figure}