Draw an arrow-pushing mechanism for the elimination reaction between naoh in ethanol and each of the following haloalkanes: (e1/e2)
(i) 1-bromobutane;
(ii) 2-bromo-2-methylpentane.

The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.

Explanation:

(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.

(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

: )