Chemistry: Topic-wise Test- 5 - NEET MCQ

It is strictly covalent does not shows cationic & anionic form.

Pseudo halides are
CN^–, NC^–, OCN^–, SCN^–, SeCN^–
NCS^–, N₃^–, NCO^–

(NH₄)₂Cr₂O₇ → Cr₂O₃ + H₂O + N₂

NaBH₄ brings about reduction of carbonyls by hydrid e transfer mechanism at planar sp² carbon. Hence, if a chiral carbon is generated, racemic mixture is always produced.

LiAIH₄ reduces both carbonyls and carboxylic groups but does not reduce olefinic bond.

In the above reaction, four stereoisomers, two pairs of enantiomers are formed. Each member of a pair of enantiomer is diastereomer of each member of other pair of enantiomer. Hence, fractional crystallisation would give two fractions, each containing racemic mixture.

Clemmensen reduction is suitable for reductio n of carbonyls containing additional acidic functional group.

Secondary amine forms enamine with aldehydes or ketones.

In acid catalysed aldo! condensation, enol acts as a nucleophile.

In BV oxidation, tertiary alkyl group has greater migrating aptitude than primary and secondary alkyl group.

Free radical coupling reaction of aldehydes and ketones.

This reaction involves the treatment of a methyl ketone (CH₃COCH₃) with nitrous acid (HNO₂) in the presence of ethyl nitrite (C₂H₅ONO) and HCl.

This is a Nef reaction, which converts a methyl ketone into an oxime derivative in the presence of nitrous acid. The mechanism involves the formation of an intermediate diazonium salt, followed by its rearrangement and hydrolysis, leading to the formation of an oxime.

Let's break down the product options:

• Option 1: This structure shows an imine with a hydroxyl group attached directly to nitrogen. However, this is not the expected structure from a Nef reaction.

• Option 2: This shows an oxime (CH₃COCH=NOH), which is the expected major product of the reaction.

• Option 3: This shows an ester-like product (CH₃COCH=NOEt), which is not the expected result of the Nef reaction.

• Option 4: This shows an imine derivative (CH₃COCH=NH), which is not the expected major product in this case.

Conclusion:

The major product of the reaction is the oxime, as shown in Option 2.

The correct answer is:

Option 2.

Glycogen is a branched chain polymer of αD glucose units in which chain is formed by C₁−C₄ glycosidic linkage whereas branchin occurs by the formation of C₁−C₆ glycosidic linkage. Structure of glycogen can be shown below similar to the sturcture amplopectin.

If blood glucose levels become too low, glucagon (pancreatic hormone) is released, which causes the liver to convert the stored glycogen back into glucose, and release it into the bloodstream.

Maltose can be produced from starch by hydrolysis in the presence of the enzyme diastase. It can be broken down into two glucose molecules by hydrolysis. In living organisms, the enzyme maltase can achieve this very rapidly. In the laboratory, heating with a strong acid for several minutes will produce the same result.

Sucrose or cane sugar is the commonest of the disaccharides. It has the molecular formula C₁₂H₂₂O₁₁. It does not form a phenylhydrazone or exhibit carbonyl properties. On hydrolysis with dilute acids, it yields one molecule of D-(+)-glucose and one molecule of D-(—)-fructose. It is not a reducing sugar. 

Lactose: The sugar found naturally in milk, it is a disaccharide composed of one galactose unit and one glucose unit; sometimes called milk sugar.

The correct answer is option D
α helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds (intermolecular and intramolecular hydrogen bond) by twisting into a right-handed screw(helix) with the N−H group of each amino acid residue hydrogen bonded to C=O of an adjacent turn of the helix.
 

Glycine is the simplest amino acid and is the only amino acid that is not optically active (it has no stereoisomers).

The correct answer is Option A.

The correct sentence is option C
Since the half-life period is independent of the changing pressure, the reaction is of the first order.
The integrated rate law is (P_N2​O5​​)_t​=(P​^o_N2​O5​)₀​^e−kt
Thus the partial pressure of N₂​O_5​ decreases exponentially with time and the partial pressure of NO₂​ increases exponentially with time.
Hence, the correct graph is represented by option C.
 

(B) As only A is optically active. So conc. of A at t =20min ∝ 30° 

While concentration of A at t= 50min ∝ mc 15° 

So conc. has deceased to half oils value in 30 min, so t_1/2=30 min.

So volume consumed of H₂0₂ at t =30 min =t_1/2 ,is according to 50% production of B at t= 90 min. production of B=87.5%(Threehalflives)

So volume consumed =