Balanced equation : 2K + Cl2 > 2KCL.
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
Answer:
62.5%
Explanation:
Number of moles of camphor reacted= 1.6g/152.23 g/mol = 0.011 moles
Since the reaction is 1:1, then 0.011 moles of isoborenol is also produced.
Theoretical yield of isoborenol produced = 154.25 g/mol * 0.011 moles = 1.698 g
% yield = actual yield/theoretical yield * 100
% yield = 1.062 g/1.698 g * 100
% yield = 62.5%
