Chemistry Test 223

The number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃ is four. This is calculated by the formula 2ⁿ⁺¹. where n is number of chiral centres, which is 1 in this case. So, 2¹⁺¹ = 2²= 4

Br₂ in CCl₄ (a), Br₂ in CH₃ COOH (b) and alk.

KMnO₄ (c) will react with all unsaturated compounds, i.e., 1, 3 and 4 while ammonical AgNO₃ (d) reacts only with terminal alkynes, i.e., 3 and hence compund 3 can be distinguished from 1, 2 and 4 by. ammonical AgNO₃ (d).

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.

All have more than one donor atoms within the same donor group, hence ambident nucleophiles.

A. CaO (used with NaOH as soda lime) is the correct reagent.

Heating a carboxylate salt with soda lime causes decarboxylation, converting the carboxylate into the corresponding alkane and producing a carbonate salt.

The general transformation is: RCOO^-Na⁺ + NaOH + CaO → RH + Na₂CO₃.

This explains formation of the alkane when the carboxylate salt is heated with soda lime.

Brief elimination: DIBAL-H is a selective reducing agent (used to reduce esters/acid derivatives to aldehydes under controlled conditions); B₂H₆ (diborane) is used in hydroboration and related reductions; and Red Phosphorus is used in different reduction/halogenation contexts. None of these effect the decarboxylation of a carboxylate salt the way soda lime (NaOH + CaO) does.

o-Toluidine (ortho-toluidine) is an organic compound with the chemical formula CH₃C₆H₄NH₂. It is the most important of the three isomeric toluidines.