Question: Identify Z in the following series \({C_2}{H_5}OH\xrightarrow{{PB{r_3}}}X\xrightarrow{{alc.KOH}}Y\xr...

Question

Identify Z in the following series ${C_2}{H_5}OH\xrightarrow{{PB{r_3}}}X\xrightarrow{{alc.KOH}}Y\xrightarrow{{dil.{H_2}S{O_4}}}Z$
$(A)C{H_2} = C{H_2}$
$(B)C{H_3}C{H_2}OH$
$(C)C{H_3} - C{H_2} - O - C{H_2} - C{H_3}$
(D)None

Answer 1

Solution

In these types of questions, we need to know step by step the product formed after the reaction in the presence of a catalyst. If we need to find the compound of $Z$ , first we need to know the products formed at $X$ and $Y$ .

Complete answer:
First we see the product formed after the reaction between ethanol and $PB{r_3}$ . This reaction proceeds in two steps. In the first step the alcohol is converted into a good leaving group by forming a bond to $P$ ( $O - P$ bonds are very strong) and displacing $Br$ from $P$ , this is essentially a nucleophilic substitution at phosphorus.
Now that the oxygen has converted to a good leaving group, a substitution reaction at carbon can occur. The bromide ion that is displaced from phosphorus attacks carbon via a backside attack, forming $C - Br$ and left with a new ethyl bromide. So, the $X$ is ${C_2}{H_5}Br$ .

Now, ethyl bromide undergoes reaction in the presence of alcohol $KOH$ , forming ethane. When ethyl bromide is boiled with $KOH$ , the hydrogen atom transfers its electron pair to the adjacent carbon bond, and bromide is removed from the molecule. This forms a double bond between the alpha and beta carbon atoms and gives ethane as a product. So. The $Y$ is $C{H_2} = C{H_2}$ .

When ethane reacts with dilute sulphuric acid, it forms ethanol. All alkenes react with dilute sulphuric acid and give alcohol. This is a hydration reaction. A water molecule is added through double bonds and may give primary, secondary or tertiary alcohol. So the element $Z$ , we get ethanol.

Therefore the correct answer is option B.

Note:
When haloalkane with $\beta$ - hydrogen atoms are boiled with an alcoholic solution of potassium hydroxide, they undergo the elimination of hydrogen halide $(HX)$ resulting in the formation of alkenes.. The dilute sulphuric acid acts as a catalyst in hydrolysis of alkene.