BITSAT Chemistry Test - 2 - JEE MCQ

Explanation:

Br₂ in CCl₄ cannot be used to distinguish between hexanal and 2-hexanone because both compounds will react similarly with bromine to form the corresponding dibromides. Therefore, this reagent is not selective enough to differentiate between the two compounds.
 

Predominantly 2-butene will be formed on elimination of bromine from 2- bromobutane. This is due to Saytzeff's rule. 

Certain haloalkanes can undergo elimination in two different ways giving a mixture of two products. In such reactions, the preferred product is the more highly substituted alkene (i.e. the alkene having lesser number of hydrogens on the doubly bonded carbon atoms). This generalisation is known as Saytzeff’s rule.

1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied.
2. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

Explanation:

Stability of Carbanions:

(IV): This carbanion is stabilized by resonance with the phenyl group, which helps delocalize the negative charge over the aromatic ring, increasing stability.

(II): This carbanion is stabilized by hyperconjugation from the methyl groups, which helps distribute the negative charge and increase stability.

(I): This carbanion is stabilized by three methyl groups, providing even more hyperconjugation and stability compared to (II).

(III): This carbanion has the least stabilization as it has only one alkyl group providing hyperconjugation.

Therefore, the order of decreasing stability for the carbanions is IV > III > II > I. Hence, option B is the correct answer.

Explanation:

Aufbau Rule: The Aufbau principle states that the atomic orbitals are filled in order of increasing energy levels. This means that electrons fill the lowest energy levels first before moving on to higher energy levels.

Although due to +I effect of methyl groups, (CH₃)₂ CHO and (CH₃)₃ CO are stronger bases but due to steric hindrance both are weaker nucleophiles than CH₃O
Further due to resonance, C₆H₅O

is the weakest nucleophile
CH₃O is most reactive (stronger) nucleophile

In a reversible reaction, the catalyst influences both the forward and backward reactions to some extent. This means that the catalyst can increase the rate of both the forward and backward reactions, ultimately speeding up the overall reaction without favoring one direction over the other.

Explanation:

- Effect on Forward Reaction: The catalyst can provide an alternative reaction pathway with lower activation energy for the forward reaction, allowing the reactants to convert into products more quickly.

- Effect on Backward Reaction: Similarly, the catalyst can also lower the activation energy for the backward reaction, facilitating the conversion of products back into reactants at a faster rate.

- Overall Influence: By affecting both the forward and backward reactions simultaneously, the catalyst helps to establish a dynamic equilibrium more quickly. This means that the catalyst does not shift the equilibrium towards the products or reactants, but rather speeds up the attainment of equilibrium by accelerating both reactions.

Plants and living beings as examples of open system
 

• Plants and living beings are examples of open systems because they interact with their environment by exchanging matter and energy.


• They take in nutrients, water, and sunlight from the environment and release waste products and heat back into the environment.



 

Lithium is lightest among all the metals with density 0.534 g/cm³.

Gas evolved on heating ammonium dichromate

- When ammonium dichromate is heated, it undergoes decomposition to form different products.
- One of the gases evolved during this decomposition is nitrogen gas (N2).
 

Actinide Series

- Actinide series is a group of elements in the periodic table that includes elements with atomic numbers from 89 to 103.
- These elements are all radioactive and have similar chemical properties.
- The element U (Uranium) belongs to the actinide series with an atomic number of 92.
- Ta (Tantalum), Li (Lithium), and Y (Yttrium) do not belong to the actinide series.

Explanation:

Reaction Overview:
- When n-propyl iodide is heated with alcoholic KOH, it undergoes an elimination reaction to form propene as one of the products.

Mechanism of Reaction:
- The reaction proceeds through an E2 elimination mechanism.
- Alcoholic KOH acts as a base and abstracts a proton from the beta carbon of n-propyl iodide.
- This results in the formation of a carbanion intermediate.
- The leaving group (iodide ion) departs, leading to the formation of propene.

Product Formation:
- The product formed is propene (C₃H₆), which is an alkene with a double bond between two carbon atoms.
- Propene is the major product due to the stability of the double bond compared to the formation of other possible products.
 

Given the product is a diketone, the likely reagent X is an oxidizing agent that specifically converts the triple bond to a double diketone.

Therefore, the correct option for X is:
Option C: O₃ (ozone)

Explanation:
Dihedral Angle HCH in Staggered Conformation:
- The dihedral angle HCH in the staggered conformation of ethane is the angle between two hydrogen atoms on adjacent carbon atoms.
- In the staggered conformation, the HCH dihedral angle is 60 degrees.
- This angle is formed due to the tetrahedral geometry around each carbon atom, where the hydrogen atoms are positioned at regular intervals around the central carbon atom.

Therefore, the correct answer is B: 60^o.

Covalent Bonds in Diamond

- In diamond, the carbon atoms are held together by covalent bonds.
- Covalent bonds involve the sharing of electrons between atoms.
- Each carbon atom in diamond forms four strong covalent bonds with its neighboring carbon atoms.
- This results in a very rigid and tightly packed crystal lattice structure.
- The strong covalent bonds in diamond make it one of the hardest substances known.