Question

homework 7
answer all questions
1 multiple choice 2 points
which reagent can be used to dehydrohalogenate an alkyl halide to form an alkene?
nacl
h₂o
hcl
naoh

Explanation:

Step1: Understand the reaction

Dehydrohalogenation of alkyl halide is an elimination reaction. Alkyl halide \(R - X\) (\(X = Cl, Br, I\)) loses \(HX\) to form alkene \(R - CH=CH_2\). This reaction requires a base to abstract a proton adjacent to the carbon - halogen bond.

Step2: Analyze each reagent

• \(NaCl\): Sodium chloride (\(NaCl\)) is a salt. The \(Cl^-\) ion is a very weak base (conjugate base of strong acid \(HCl\)). It cannot abstract a proton from the alkyl halide to initiate dehydrohalogenation.
• \(H_2O\): Water (\(H_2O\)) is a very weak base (\(pK_a\) of \(H_3O^+\) is \(- 1.74\)). The hydroxide ion concentration (\(OH^-\)) from water auto - ionization (\(H_2O

ightleftharpoons H^++OH^-\)) is extremely low (\([OH^-]=1\times10^{-7}M\) at \(25^{\circ}C\)). It is not a strong enough base for dehydrohalogenation of alkyl halides (except for very reactive alkyl halides under special conditions, but in general sense).

• \(HCl\): Hydrochloric acid (\(HCl\)) is an acid. It donates protons (\(H^+\)). In dehydrohalogenation, we need a species that can accept (abstract) protons, so \(HCl\) cannot be used for dehydrohalogenation.
• \(NaOH\): Sodium hydroxide (\(NaOH\)) dissociates in solution (\(NaOH = Na^++OH^-\)). The hydroxide ion (\(OH^-\)) is a strong base (\(pK_a\) of \(H_2O\) is \(15.74\)). It can abstract a proton from the carbon adjacent to the carbon - halogen bond in the alkyl halide. For example, for \(CH_3CH_2Cl\), the reaction is \(CH_3CH_2Cl+NaOH\xrightarrow{\text{alcohol}}CH_2 = CH_2+NaCl + H_2O\) (the reaction is usually carried out in an alcoholic solvent to promote the elimination reaction over substitution in some cases).

Answer:

\(NaOH\)