Step 1: Nucleophilic substitution (\( S_N2 \)) with dimethylamine (\( \text{Me}_2\text{NH} \)): The bromine atom at one of the carbons is replaced by the nucleophile (\( \text{Me}_2\text{N} \)) via an \( S_N2 \) mechanism. This forms an intermediate with a quaternary amine at the adjacent carbon.
Step 2: Deprotonation: The positively charged intermediate loses a proton (\(-\text{H}^+\)) to stabilize the structure, forming an alkene.
Step 3: Second \( S_N2 \) reaction: The second equivalent of dimethylamine attacks the adjacent carbon-bromine bond, substituting the bromine atom with another \( \text{Me}_2\text{N} \) group.
The final product contains two \( \text{Me}_2\text{N} \) groups attached to the cyclopentane ring at adjacent positions.
The above mechanism is valid for both cis and trans isomers. Hence, the products are the same for both.
Nucleophiles attack at that part of the substrate molecule which is electron deficient. The reaction in which a nucleophile replaces an already existing nucleophile in a molecule is called a nucleophilic substitution reaction. Haloalkanes are substrates in these reactions. In this type of reaction, a nucleophile reacts with a haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place, and the halogen atom, called the leaving group, departs as a halide ion. Since the substitution reaction is initiated by a nucleophile, it is called a nucleophilic substitution reaction.
\(Nu^− +C-X →C-Nu+X^−\)
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: