Question

1. Sketch the reaction mechanism (including the final product) corresponding to the following description. Explain the answer thoroughly using concise sentences.

o 1-bromo-1-methylcyclopentane is being reacted with sodium ethoxide in diethyl ether (set up the reaction – next you will get a description of the mechanism to draw).

o Ethoxide will remove a proton from one of the carbons alpha to the carbon bonded to bromine. The electron pair from the carbon-hydrogen bond will move toward the 1 carbon to form a new pi bond. This will push off the bromine, which will leave with its electron pair to form a bromide leaving group. This will form the product 1-methylcyclopentene.

Answer 1

Given dara 1-Bromo - 1 methyl Cyclopentane on eact with Sodium ethoxide ron 19 diethyl ether. The structure of 1 - Bromo -1-Methyl Cyclopentane BY CH3 ReccHon A ug NaOCHS diethyl ether 1 - Bromo 1-Methyl Cyclopent one I - Methyl Cyclopentene The above reaction is elimpnation in presence of base reactant ent molecule 95 Tentary alkyl boromide undergoes elimination reaction an presence of Naoghs to gires alkene By follows El hechen 95m.

The reaction Caraped out inpresence of aprotic silvents then I follow E Mechanism 1 methyl Cyclopentene Socfum ethoxide Acts as base abstract the protan from ring &- carbon to generates Carbonion, an first step. In the second to generates localization of step leaving of bromine unsatonatish by eanbanion to generater alkene In this climination we get a stable alkene de protenation ring carson. F CH2 Minor product toom. Small amout