Characteristics of Halo Compounds - Free MCQ Practice Test with solutions,

• Tert-butyl chloride forms most stable 3° carbocation. So will give white precipitate of AgCl with AgNO₃ solution immediately.
• (CH₃)₃CCl reacts with AgNO₃ because it forms a stable tertiary carbocation upon ionization. This stable carbocation facilitates the reaction, allowing chloride ions (Cl⁻) to combine with Ag⁺ from AgNO₃, resulting in the formation of a white precipitate of AgCl:

(CH₃​)₃​CCl+AgNO_3​→(CH_3​)₃​C⁺+AgCl ↓

Hence, option (B) is the answer.

During racemisation, carbocation intermediate is formed.

Hence option (A) is the answer.

2-bromobutane → CH₃CH = CHCH₃ + CH₃CH₂CH=CH₂

The major product follows Saytzeff rule.

Hence option (A) is the correct answer.

The carbon with which the Br is bonded with another 3 carbon atoms. So haloalkane is Tertiary Haloalkane.

Superimposable (mirror images of each other) substances are achiral because chiral substances cannot have its own mirror image. And achiral substances do not show optical activity nor any optical isomers. Hence, answer is B.

 Because it is Open chain compound /Aliphatic Compound and not aryl halide as it does not contain halogen atom. It's simply  alkane.

In general organic compounds are soluble in organic solvents. Hence alkyl halides are soluble in organic solvents.

It is 1-bromo-3-methylbutane.

C₅H₁₁Br are only very slightly soluble in water. They are more soluble in organic solvents.