8.8 g of monohydric alcohol added to ethyl magnesium iodide in ether liberates $2240 \mathrm{~cm}^3$ of ethane at STP. This monohydric alcohol when oxidised using pyridinium-chloro-chromate, forms a carbonyl compound that answers silver mirror test (Tollen's test). The monohydric alcohol is

When a tertiary alcohol ' $A^{\prime}\left(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\right)$ reacts with $20 \% \mathrm{H}_3 \mathrm{PO}_4$ at 358 K , it gives a compound ' $B^{\prime}\left(\mathrm{C}_4 \mathrm{H}_8\right)$ as a major product. The IUPAC name of the compound ' $B$ ' is

PCC is

$$\begin{aligned} & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{PCC}}{\longrightarrow} \\\\ & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \end{aligned}$$

Hybridisation change involved at $$\mathrm{C}$$-$$1$$ in the above reaction.