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Question: In the reaction: \(C{H_3} - CH = C{H_2}\, + \,HBr\, \to \,(A)\,\xrightarrow[{{H_2}O}]{{KCN}}\,(B)\) ...

In the reaction: CH3−CH=CH2 + HBr → (A) →H2OKCN (B)C{H_3} - CH = C{H_2}\, + \,HBr\, \to \,(A)\,\xrightarrow[{{H_2}O}]{{KCN}}\,(B) Structure of the compound (B) is:
A) CH3CH2CH2CNC{H_3}C{H_2}C{H_2}CN
B) CH3−CH2−CH2−COOHC{H_3} - C{H_2} - C{H_2} - COOH
C) (CH3)2CH−COOH{(C{H_3})_2}CH - COOH
D) None of these.

Explanation

Solution

In the reaction in which you have given any cyanide and water keep in mind that these conditions are counted for increase of one carbon atom. So, the number of carbons increases in the product. Try to solve firstly by applying the Markovnikov rule as addition of hydrogen bromide and then replace it with cyanide.

Complete step-by-step answer:
We have a double bonded compound as a substrate in which we have to add hydrogen bromide in such a way that Markonikov addition takes place. In Markovnikov addition rule, it says that in a doubly bonded compound the negative part of attacking reagent get attached to the carbon atom having lesser number hydrogen let’s try to solve equation:

Step 1: CH3−CH=CH2 + HBr → CH3−CH2−CH2+ Step\,1:\,C{H_3} - CH = C{H_2}\, + \,HBr\, \to \,C{H_3} - C{H_2} - CH_2^ + \,
Here, addition of hydrogen bromide is in such a way that the negative part which is bromide here gets attached to the carbon center having lesser hydrogen atoms. This step is proved when we see its mechanism where an attack of electrophile takes place. Hydrogen ion is an electrophile by which the carbocation forms is stable.
CH3−CH2−CH2+ →Hydride shift CH3−CH+−CH3 C{H_3} - C{H_2} - CH_2^ + \,\xrightarrow{{Hydride\,shift}}\,C{H_3} - C{H^ + } - C{H_3}\,
As we are seeing that the carbocation formed is primary carbocation so for further stability it will rearrange and forms secondary carbocation. By which we get the structure of (A).

Step2: CH3−CH+−CH3 →Br−CH3−CH(Br)−CH3 →H2OKCN CH3−CH(CN)−CH3Step2:\,C{H_3} - C{H^ + } - C{H_3}\,\xrightarrow{{B{r^ - }}}C{H_3} - CH(Br) - C{H_3}\,\xrightarrow[{{H_2}O}]{{KCN}}\,C{H_3} - CH(CN) - C{H_3}
In this step bromide is replaced by cyanide so that we get an extra carbon atom in the chain as shown in the above reaction. After getting cyanide group in place of bromide, hydrolysis takes place. In hydrolysis, cyanide gets changed to COOHCOOH group.

Step 3:CH3−CH(CN)−CH3→Hydrolysis CH3−CH(COOH)−CH3Step\,3:C{H_3} - CH(CN) - C{H_3}\xrightarrow{{Hydrolysis}}\,C{H_3} - CH(COOH) - C{H_3}

This is the structure of (B) hence option C is the correct answer.

Note: The attack of reagent is always according to Markonikov rule because in that case a stable carbocation forms. But in many cases it was mentioned with attacking reagent a peroxide, which will do anti- Markovnikov attack hence, seeing the regent and finding out which type of attack will it do is very important. In higher classes you will also study the stereochemistry of products.