
The product of acid-catalysed hydration of $2-phenylpropene$ is
A. $3-phenyl-2-propanol$
B. $1-phenyl-2-propanol$
C. $2-phenyl-2-propanol$
D. $2-phenyl-1-propanol$
Answer
163.5k+ views
Hint: Acid-catalysed hydration is the addition of water to an alkene that forms alcohol. This is one kind of electrophilic hydration and here the $-OH$ group is attached across the two carbon atoms of a double bond. Here we have a substituted alkene undergoing acid-catalysed hydration to give an alcohol.
Complete Step by Step Answer:
As we know alcohol gets dehydrated to form an alkene, in the same way, alkenes are hydrated to form alcohol. Acid-catalysed hydration of alkenes produces alcohol according to Markownikov’s rule. This process involves the breaking of the pi bond in the alkene and hydroxyl bond ($-OH$) in the water molecule.
$2-phenyl$propene undergoes acid-catalysed hydration to give alcohol by the following step. The first step involves the protonation of alkenes to form the most stable carbocation by the electrophilic attack ${{H}_{3}}{{O}^{+}}$. Here two carbocations are formed but benzylic carbocations are more stable. This is because conjugation with phenyl rings gives them stability.

In the second step, water acts as an incoming nucleophile and hence attacks carbocation.

The final and last step is the deprotonation step. Alcohol named as $2-phenyl-2-propanol$is formed as the major product by deprotonation and $2-phenyl-1-propanol$is the minor product.

Therefore, the product of acid-catalysed hydration $2-phenyl-propanol$.
Thus, option (C) is correct.
Note: The acid-catalysed hydration of alkene is regioselective in nature and also follows Markovnikov’s rule. The regioselectivity of acid-catalysed hydration shows that the hydroxyl group ($-OH$) gets attached to the more substituted carbon in an alkene.
Complete Step by Step Answer:
As we know alcohol gets dehydrated to form an alkene, in the same way, alkenes are hydrated to form alcohol. Acid-catalysed hydration of alkenes produces alcohol according to Markownikov’s rule. This process involves the breaking of the pi bond in the alkene and hydroxyl bond ($-OH$) in the water molecule.
$2-phenyl$propene undergoes acid-catalysed hydration to give alcohol by the following step. The first step involves the protonation of alkenes to form the most stable carbocation by the electrophilic attack ${{H}_{3}}{{O}^{+}}$. Here two carbocations are formed but benzylic carbocations are more stable. This is because conjugation with phenyl rings gives them stability.

In the second step, water acts as an incoming nucleophile and hence attacks carbocation.

The final and last step is the deprotonation step. Alcohol named as $2-phenyl-2-propanol$is formed as the major product by deprotonation and $2-phenyl-1-propanol$is the minor product.

Therefore, the product of acid-catalysed hydration $2-phenyl-propanol$.
Thus, option (C) is correct.
Note: The acid-catalysed hydration of alkene is regioselective in nature and also follows Markovnikov’s rule. The regioselectivity of acid-catalysed hydration shows that the hydroxyl group ($-OH$) gets attached to the more substituted carbon in an alkene.
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