To solve the problem, we need to determine the product formed when a secondary alcohol is oxidized with chromic anhydride (CrO\(_3\)).
1. Understanding Secondary Alcohols:
A secondary alcohol has the general structure R\(_1\)R\(_2\)CHOH, where R\(_1\) and R\(_2\) are alkyl groups, meaning the carbon bearing the -OH group is attached to two other carbons.
2. Identifying the Reagent:
Chromic anhydride (CrO\(_3\)) is a strong oxidizing agent, often used in acidic conditions (e.g., with H\(_2\)SO\(_4\)) to form chromic acid, a common reagent for alcohol oxidation. It can also be used in other forms, such as in the Jones oxidation (CrO\(_3\) in aqueous sulfuric acid and acetone).
3. Oxidation of Secondary Alcohol:
When a secondary alcohol is oxidized with a strong oxidizing agent like CrO\(_3\), the -OH group is converted to a carbonyl group (C=O). Since the carbon is attached to two other carbons, the product is a ketone (R\(_1\)R\(_2\)C=O). For example, if the secondary alcohol is 2-propanol (CH\(_3\)CH(OH)CH\(_3\)), oxidation with CrO\(_3\) yields acetone (CH\(_3\)COCH\(_3\)), a ketone.
4. Considering Reaction Conditions:
CrO\(_3\) in acidic conditions ensures complete oxidation. Secondary alcohols do not oxidize further to carboxylic acids (unlike primary alcohols), as there is no hydrogen on the carbonyl carbon to allow further oxidation to a carboxyl group.
Final Answer:
Oxidation of a secondary alcohol with chromic anhydride (CrO\(_3\)) forms a ketone.
Write the product obtained when D-glucose reacts with \( H_2N - OH \).
Two batteries of emf's \(3V \& 6V\) and internal resistances 0.2 Ω \(\&\) 0.4 Ω are connected in parallel. This combination is connected to a 4 Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination