$HBr$ reacts with $CH_2=CH-OCH_3$ at room temperature to give

Consider the following four reactions: $(I) C_6H_5-CHCl-CH_3 \xrightarrow{95\% \text{ acetone}, 5\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(II) C_6H_5-CHCl-CH_3 \xrightarrow{90\% \text{ acetone}, 10\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(III) C_6H_5-CHCl-CH_3 \xrightarrow{80\% \text{ acetone}, 20\% \text{ water}} C_6H_5-CH(OH)-CH_3$ $(IV) C_6H_5-CHCl-CH_3 \xrightarrow{100\% \text{ water}} C_6H_5-CH(OH)-CH_3$ Arrange these reactions in decreasing order of the proportion of inverted product and select the correct answer from the codes given below:

Ethyl bromide can be converted into ethyl alcohol by

What is the product of the following reaction?

When $3-$methylbutan$-2-$ol is treated with $HBr$,the following reaction takes place:
$CH_3-CH(CH_3)-CH(OH)-CH_3 \xrightarrow{HBr} CH_3-C(Br)(CH_3)-CH_2-CH_3$
Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step $II$ rearranges to a more stable tertiary carbocation by a hydride ion shift from $3^{rd}$ carbon atom.)

What is the major product obtained when $tert-$butyl bromide is heated with silver fluoride?