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WITCHER [35]
2 years ago
3

Complete the mechanism for the following reaction of benzaldehyde dissolved in dry ethanol bubbled with HCl(g) by adding any mis

sing atoms, bonds, charges, non-bonding electrons, and curved arrows. Select the name that best describes the final product.

Chemistry
1 answer:
Bogdan [553]2 years ago
7 0

Answer:Kindly refer the attachment for  structure and mechanism. The name of the product is (diethoxy methyl) benzene.

Explanation:

The reaction between benzaldehyde and dry ethanol  is a example of nucleohilic addition reaction.

Due to the presence of HCl in the reaction the electrophilicity of the carbonyl group in benzaldehyde is increased and ethanol being a weak nucleophile can attack the more electrophilic carbonyl group.

This leads to the formation of ethoxy phenyl methanol.

After the formation of ethoxy phenyl methanol, since there is still HCl present in the medium the alcohol can be protonated and once an alcohol is protonated it becomes an excellent leaving group and leaves as water molecule generating carbocation.

The carbocation formed can be attacked by the another molecule of ethanol present in the system in a substitution reaction which further leads to the formation final product diethoxy methyl benzene.

The name of final product is diethoxy methyl benzene.

Kindly refer attachment for the mechanism of reaction and structure of products.

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Answer:

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6 0
2 years ago
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What is the 2s in 42,256
MissTica

42,256 = 2,000

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Explanation:

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4 0
2 years ago
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na
Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}

valence according to the amount of salt anion  

Input the value :

\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3

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