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Ludmilka [50]
3 years ago
8

Can anyone explain oxidation, and reduction reactions of:

Chemistry
1 answer:
____ [38]3 years ago
4 0

Answer:

Reduction of Aldehydes and Ketones. Hydride reacts with the carbonyl group, C=O, in aldehydes or ketones to give alcohols. ... Reduction of ketones gives secondary alcohols. The acidic work-up converts an intermediate metal alkoxide salt into the desired alcohol via a simple acid base reaction.

The carbon atom of a carboxyl group is in a relatively high oxidation state. Diborane, B2H6, reduces the carboxyl group in a similar fashion. ... Sodium borohydride, NaBH4, does not reduce carboxylic acids; however, hydrogen gas is liberated and salts of the acid are formed.

Primary alcohols can be oxidized to form aldehydes and carboxylic acids; secondary alcohols can be oxidized to give ketones. Tertiary alcohols, in contrast, cannot be oxidized without breaking the molecule's C–C bonds.

A secondary alcohol can be oxidised into a ketone using acidified potassium dichromate and heating under reflux. The orange-red dichromate ion, Cr2O72−, is reduced to the green Cr3+ ion. This reaction was once used in an alcohol breath test.

hope it will help u

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2) Find the volume that 2.0 moles of H2 will occupy at STP?
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Answer:

V = 44.85 L

Explanation:

Given data:

Volume of H₂ = ?

Number of moles of H₂ = 2.0 mol

Given temperature = 273.15 K

Given pressure = 1 atm

Solution:

Formula:

PV = nRT

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V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

By putting values,

1 atm × V = 2.0 mol × 0.0821 atm.L/ mol.K  × 273.15 K

V = 44.85 atm.L / 1 atm

V = 44.85 L

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<u>Explanation:</u>

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