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Chapter 4: Problem 81

The product \(\mathrm{P}\) in the following reaction is \(\mathrm{ClCH}_{2}-\mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{P}\)

Short Answer

Expert verified
Answer: The product P formed in this reaction is 2-hydroxyethanal (H2C=OCH2OH).

Step by step solution

01

Identify the reactive part of the molecule

In this reaction, the hydroxide ion (OH-) acts as a nucleophile, which means it has a high electron density and will look for a somewhat positively charged part of the molecule to react with. In this case, the carbon atom attached to the chlorine atom in ClCH2-CH2OH can acquire a partial positive charge since chlorine is more electronegative than carbon. This makes the carbon atom electrophilic and susceptible to nucleophilic attack by the hydroxide ion.
02

Nucleophilic attack and formation of a possible intermediate

The hydroxide ion (OH-) attacks the somewhat positively charged carbon atom in the molecule, forming a possible intermediate. In this intermediate, the carbon has a single bond with the chlorine atom, a single bond with the CH2OH group, and a single bond with the hydroxide ion.
03

Elimination of chloride ion

In the next step, the chlorine atom leaves the molecule as a chloride ion (Cl-). This is an elimination step, and the C-O bond is now a double bond (C=O). The intermediate transforms into the final product.
04

Identify the product P

After the elimination of the chloride ion, the final product P is formed. The structure of the product is H2C=OCH2OH, which is an aldehyde group (-CHO) attached to the CH2OH group. In IUPAC nomenclature, the compound is named as 2-hydroxyethanal.

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Most popular questions from this chapter

Ethane-1, 2- diol when heated with \(\mathrm{PI}_{3}\) (or excess HI) gives (a) Iodoethane (b) \(1,2-\) di iodoethane (c) 2 - Iodoethanol (d) ethene

\(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{3}+\mathrm{Cl}_{2} \stackrel{800 \mathrm{~K}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{HCl}\). This reaction takes place mainly by (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) free radical mechanism (d) E1 mechanism

What is the product of addition of \(\mathrm{HCl}\) to \(\mathrm{CCl}_{3}-\mathrm{CH}=\mathrm{CH}_{2} ?\)

In the preparation of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\), two methods, namely (i) and (ii) are suggested. Which method is preferable and why? (i) \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{ONa} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{AgF} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}+\mathrm{AgCl}\). This reaction is called (a) Finkelstein reaction (b) Balz-Schiemann reaction (c) Swarts reaction (d) Gattermann reaction
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