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Propyl lithium reacts with ethene to giv...

Propyl lithium reacts with ethene to give a compound `(A)`, which on reaction with methanal followed by acidic hydrolysis gives compound `(B)`. The compound `(B)` is :

A

`Heptan-1-ol`

B

`Heptan-2-ol`

C

`Hexan-1-ol`

D

`Hexan-2-ol`

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The correct Answer is:
To solve the problem step by step, we will analyze the reactions involving propyl lithium, ethene, and methanal, leading to the final product (B). ### Step 1: Reaction of Propyl Lithium with Ethene Propyl lithium (C3H7Li) is a strong nucleophile. When it reacts with ethene (C2H4), it adds to the double bond of ethene. The reaction can be represented as follows: 1. Propyl lithium reacts with ethene: \[ \text{C}_3\text{H}_7\text{Li} + \text{C}_2\text{H}_4 \rightarrow \text{C}_5\text{H}_{10}\text{Li} \] This reaction results in the formation of a 5-membered ring compound (a cyclic structure) known as a propyl-ethyl compound. ### Step 2: Formation of Compound (A) The product of the reaction between propyl lithium and ethene can be represented as: \[ \text{Compound (A)}: \text{C}_5\text{H}_{10}\text{Li} \] ### Step 3: Reaction of Compound (A) with Methanal Next, compound (A) reacts with methanal (formaldehyde, CH2O). The nucleophilic carbon in compound (A) attacks the carbonyl carbon of methanal, leading to the formation of an alcohol after the reaction: \[ \text{C}_5\text{H}_{10}\text{Li} + \text{CH}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O} + \text{LiOH} \] ### Step 4: Acidic Hydrolysis After the reaction with methanal, the resultant compound undergoes acidic hydrolysis. This step converts the lithium alkoxide into the corresponding alcohol: \[ \text{C}_6\text{H}_{12}\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{14}\text{O} \] ### Step 5: Identification of Compound (B) The product (B) formed after the reaction with methanal and subsequent hydrolysis is hexanol. The structure of hexanol can be represented as: \[ \text{Compound (B)}: \text{C}_6\text{H}_{14}\text{O} \quad (\text{Hexanol}) \] ### Final Answer The compound (B) is hexanol. ---

To solve the problem step by step, we will analyze the reactions involving propyl lithium, ethene, and methanal, leading to the final product (B). ### Step 1: Reaction of Propyl Lithium with Ethene Propyl lithium (C3H7Li) is a strong nucleophile. When it reacts with ethene (C2H4), it adds to the double bond of ethene. The reaction can be represented as follows: 1. Propyl lithium reacts with ethene: \[ \text{C}_3\text{H}_7\text{Li} + \text{C}_2\text{H}_4 \rightarrow \text{C}_5\text{H}_{10}\text{Li} ...
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CENGAGE CHEMISTRY ENGLISH-GRIGNARD REAGENTS AND ORGANOMETALLIC REAGENTS-Exercises (Single Correct)
  1. Propane dithioic acid is prepared by the reaction of the following, fo...

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  2. Propylsulphinic acid is prepared by the reaction of the following, fol...

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  3. .

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  4. Ethanoic propanoic anhydride on reaction with excess of MeMgBr gives t...

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  5. Reactivity of MeMgBr with the following in the decreasing order is : ...

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  6. Product. The major product is :

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  7. Reactivity of with the following G.R in the decreasing order is : (...

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  8. .

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  9. Reactivity of EtMgBr with the following in the decreasing order is : ...

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  10. Reactivity of PhMgBr with the following in the decreasing order is : ...

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  11. Reactivity of PhMgBr with the following in the decreasing order is : ...

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  12. Reactivity of HCHO with the following G.R in the decreasing order is :...

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  13. Which of the following 3^@ alcohols does propyl ester give during reac...

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  14. Ethyl ester reacts with PrMgBr to give 2^@ alcohol. The alcohol is :

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  15. Methyl ester reacts with EtMgBr to give 3^@ alcohol The ester is :

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  16. Propyl ester reacts with isopropyl magnesium bromide to give 2^@ alcoh...

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  17. Propyl lithium reacts with ethene to give a compound (A), which on rea...

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  18. Coupling reaction between RMgX and R'X takes place to give R-R' in the...

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  19. Phenyl isocyanide + Benzyl magnesium bromide overset(1.Ether Delta) un...

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  20. The compound (A) in the previous question is further hydrolysed in dil...

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