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1-phenyl-2-chloropropane on treating wit...

1-phenyl-2-chloropropane on treating with alc. `KOH` gives mainly

A

1-phenylpropene

B

2-phenylpropene

C

1-phenylpropane-2-ol

D

1-phenylpropane-1-ol

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The correct Answer is:
To solve the question of what product is formed when 1-phenyl-2-chloropropane is treated with alcoholic KOH, we can follow these steps: ### Step 1: Identify the Reactant 1-Phenyl-2-chloropropane has the structure: - A phenyl group (C6H5) attached to a carbon (C) which is also attached to a chlorine (Cl) and a propyl group (C3H7). This can be represented as: ``` C6H5 | CH-CH-CH3 | Cl ``` ### Step 2: Understand the Reaction Conditions Alcoholic KOH is a strong base that promotes elimination reactions (E1 or E2). In this case, it will lead to the elimination of HCl from the molecule. ### Step 3: Determine Possible Elimination Pathways When 1-phenyl-2-chloropropane reacts with alcoholic KOH, the elimination can occur in two ways: 1. Elimination of H from the carbon adjacent to the chlorine (C2), leading to the formation of a double bond between C1 and C2. 2. Elimination of H from the carbon adjacent to the phenyl group (C1), leading to the formation of a double bond between C2 and C3. ### Step 4: Draw the Possible Products 1. If H is eliminated from C2: - Product: **1-phenylpropene** (C6H5-CH=CH-CH3) 2. If H is eliminated from C1: - Product: **1-phenylpropene** (C6H5-CH2-CH=CH2) ### Step 5: Analyze the Stability of the Products The product formed from the first pathway (elimination from C2) results in a conjugated system: - **C6H5-CH=CH-CH3** has a double bond between C1 and C2, which allows for resonance stabilization due to the phenyl group. The product formed from the second pathway does not have such stabilization: - **C6H5-CH2-CH=CH2** lacks conjugation. ### Step 6: Determine the Major Product Due to the stability provided by conjugation in the first product, **1-phenylpropene** (C6H5-CH=CH-CH3) will be the major product formed when 1-phenyl-2-chloropropane is treated with alcoholic KOH. ### Final Answer The major product formed is **1-phenylpropene**. ---
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ALLEN-ALKYL AND ARYL HALIDE-EXERCISE
  1. The reaction , CH(3)Br +OH^(-)rarrCH(3)OH+Br^(-) obeys the mechanism

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  2. Ethylidene chloride can be prepared by the reaction of HCI and

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  3. 1-phenyl-2-chloropropane on treating with alc. KOH gives mainly

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  4. Grignard reagent is obtained when magnesium is treated with

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  5. Ethylene reacts bromine to form-

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  6. C(2)H(4) overset(Br(2))rarr X overset(KCN)rarr Y, Y is

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  7. Reactivity order of halides of dehydrohalogenation is

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  8. Which of the following is least reactive in a nucleophilic substiution...

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  9. The correct reactivity order of alcohols towards H-X will be (I) CH(...

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  10. Identify 'Z' in the following reaction series, CH(3)CH(2)CH(2)Brover...

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  11. For the reaction, C(2)H(5)Oh +HX overset(ZnX(2))rarr C(2)H(5)X, the ...

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  12. Ethyl alcohol reacts at a faster rate with HI than with HCI in forming...

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  13. What is the order of reactivity of the following compounds towards nuc...

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  14. The order of decreasing nucleophilicity of the following is

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  15. The order of decreasing S(N)1 reactivities of the halides I CH(3)CH...

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  16. Consider the following anions. (I) CF(3)-overset(O)overset(||)unders...

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  17. The basicity of RO^(-),HO^(-),RCOO^(-),ROH and H(2)O are of the order-

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  18. Which of the following are aprotc solvents:

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  19. Which is/are true statements (s):

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  20. Ph-overset(Me)overset(|)underset(H)underset(|)(C)-OH overset(SOCI(2))u...

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