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(CH(3))(2)-CH-underset(OH)underset(|)CH-...

`(CH_(3))_(2)-CH-underset(OH)underset(|)CH-CH_(2)-C_(6)H_(5) overset(HBr)rarrP underset("Room Temp")overset(C_(2)H_(5)ONa)rarr Q`.
Then Q is

A

`(CH_(3))_(2)CH-CH=CH-C_(6)H_(5)`

B

`(CH_(3))_(2)CH-underset(OC_(2)H_(5))underset(|)CH-CH_(2)-C_(6)H_(5)`

C

`(CH_(3))_(2)CH-CH_(2)-underset(OC_(2)H_(5))underset(|)CH-C_(6)H_(5)`

D

`CH_(2)=underset(CH_(3))underset(|)C-CH_(2)-CH_(2)-C_(6)H_(5)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given problem step by step, we will analyze the reactions taking place and identify the final product Q. ### Step 1: Identify the Reactant The starting compound is: \[ (CH_3)_2CH-CH(OH)-CH_2-C_6H_5 \] This is a secondary alcohol with a phenyl group attached. ### Step 2: Reaction with HBr When this compound reacts with HBr, the hydroxyl group (-OH) is protonated, making it a better leaving group. The bromide ion (Br-) then attacks the carbon atom bonded to the -OH group, leading to the formation of a bromide compound (let's denote it as P). The reaction can be represented as: \[ (CH_3)_2CH-CH(OH)-CH_2-C_6H_5 + HBr \rightarrow P \] ### Step 3: Formation of Product P After the reaction with HBr, we will have: \[ (CH_3)_2CH-CH(Br)-CH_2-C_6H_5 \] This is the product P, where the -OH group has been replaced by a bromine atom. ### Step 4: Reaction with C2H5ONa Next, the product P reacts with sodium ethoxide (C2H5ONa) at room temperature. Sodium ethoxide acts as a strong base and can facilitate an elimination reaction (dehydrohalogenation). The elimination reaction will remove HBr from P, resulting in the formation of an alkene. The elimination will occur between the carbon bonded to the bromine and the adjacent carbon. ### Step 5: Formation of Product Q The product Q will be: \[ (CH_3)_2C=CH-C_6H_5 \] This is an alkene where the double bond is formed between the second and third carbon atoms of the original structure, resulting in a compound with a phenyl group. ### Final Answer Thus, the final product Q is: \[ (CH_3)_2C=CH-C_6H_5 \] ---
To solve the given problem step by step, we will analyze the reactions taking place and identify the final product Q. ### Step 1: Identify the Reactant The starting compound is: \[ (CH_3)_2CH-CH(OH)-CH_2-C_6H_5 \] This is a secondary alcohol with a phenyl group attached. ...
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