Aldehydes and Ketones Quiz-18
Dear Readers,
The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.
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Q1. The product obtained when is
oxidized with HIO4
Solution
The characteristic property of periodic acid is the oxidative cleavage of bonds with adjacent oxidisable group such as 1, 2-diols, α-hydroxy carbonyl, 1,2-diketones, etc. The reagent does not react with 1, 3- or 1, 4-diols or carbonyl compounds.
Q2.Tartaric acid is not used in :
Solution
Q3. Formaldehyde polymerises from 6 to 100 molecules to form:
Solution
Q4. In a Cannizaro’s reaction, the combination not possible is
- HCHO + HCHO
- C6 H5CHO+HCHO
- CH3CHO+HCHO
-
Solution
As Cannizaro reaction is shown by aldehydes lacking α-hydrogen, hence the combination CH33 CHO+HCHO is not possible.
Q5. Formic acid cannot be halogenated with chlorine in presence of red P, but acetic acid can be halogenated in the same way, because:
Solution
Urea is one of the most important fertilizer as it does not change pH of soil. Urea, after hydrolysis gives ammonia and CO_2. Ammonia is taken up by plants leaving behind CO2.CO2 is a very weak acidic oxide. It doesn’t affect pH of soil
Urea is one of the most important fertilizer as it does not change pH of soil. Urea, after hydrolysis gives ammonia and CO_2. Ammonia is taken up by plants leaving behind CO2.CO2 is a very weak acidic oxide. It doesn’t affect pH of soil
Q6. In question 8 an intermediate involved in step (3) is:
Solution
The intermediate formed during Hofmann’s bromamide reaction is RCH2—N=C=O. Follow mechanism of the reaction.
The intermediate formed during Hofmann’s bromamide reaction is RCH2—N=C=O. Follow mechanism of the reaction.
Q7.0.20 g of a hydrocarbon on combustion gave 0.66 g .The percentage of hydrogen in the hydrocarbon is about :
Solution
Answer-10.
Q8.Carboxylic acids dissolve in aq.NaOH because the acids undergo:
Solution
Q9.Amides may be converted into amines by reaction named after:
Solution
R—CONH2 + Br2+ 4KOH ⟶RNH2 + K2 CO3+2KBr + 2H2 O This is Hofmann’s bromamide reaction.
R—CONH2 + Br2+ 4KOH ⟶RNH2 + K2 CO3+2KBr + 2H2 O This is Hofmann’s bromamide reaction.
Q10. 
In the above reactions ‘A’ and ‘B’ respectively are
Solution:
