Identify the product $Y$ in the following reaction sequence:
$C_6H_5OH$ $\xrightarrow{Na} X$ $\xrightarrow{CH_3-CH_2-Cl} Y$

An ether $(A)$,$C_5H_{12}O$,when heated with excess of hot concentrated $HI$ produced two alkyl halides which when treated with $NaOH$ yielded compounds $(B)$ and $(C)$. Oxidation of $(B)$ and $(C)$ gave a propanone and an ethanoic acid respectively. The $IUPAC$ name of the ether $(A)$ is

The ether $C_6H_5-O-CH_2-C_6H_5$ when treated with $HI$ produces:

What is the major product obtained when diethyl ether reacts with concentrated $HI$ at $0^o C$?

The $C - O - C$ bond angle in ethers is approximately -

Match the starting materials given in Column-$I$ with the products formed by these (Column-$II$) in the reaction with $HI$.

Column-$I$	Column-$II$
$(A)$ $CH_3-O-CH_3$	$(1)$ $C_6H_5OH + CH_3I$
$(B)$ $(CH_3)_2CH-O-CH_3$	$(2)$ $(CH_3)_3C-I + CH_3OH$
$(C)$ $(CH_3)_3C-O-CH_3$	$(3)$ $C_6H_5I + CH_3OH$
$(D)$ $C_6H_5-O-CH_3$	$(4)$ $CH_3OH + CH_3I$
	$(5)$ $(CH_3)_2CHOH + CH_3I$
	$(6)$ $(CH_3)_2CH-I + CH_3OH$
	$(7)$ $(CH_3)_3COH + CH_3I$