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the rate law for the following reaction, e.g., CH 3 CH 2 CH 2 CHBrCH 3 + OH   CH 3 CH 2 CH 2 CHOHCH 3 + Br -  ( RBr )

Sagot :

Rambolokbrainidn

Answer:

To determine the rate law for the reaction:

[tex][ \text{CH}_3\text{CH}_2\text{CH}_2\text{CHBrCH}_3 + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CHOHCH}_3 + \text{Br}^- ][/tex]

we need to understand the mechanism of the reaction and the rate-determining step. This reaction is likely an S_N2 (bimolecular nucleophilic substitution) mechanism because it involves a primary or secondary alkyl halide reacting with a strong nucleophile (OH^-).

In an S_N2 mechanism, the rate of reaction depends on the concentration of both the alkyl halide (RBr) and the nucleophile (OH^-). The rate law for an S_N2 reaction can be expressed as:

[tex][ \text{Rate} = k[\text{RBr}][\text{OH}^-] ][/tex]

where:

  • ( k ) is the rate constant.
  • (RBr) is the concentration of the alkyl halide.
  • (OH}^-) is the concentration of the hydroxide ion.

Thus, the rate law for the given reaction would be:

[tex][ \text{Rate} = k[\text{CH}_3\text{CH}_2\text{CH}_2\text{CHBrCH}_3][\text{OH}^-] ][/tex]

This rate law indicates that the reaction rate is directly proportional to the concentrations of both the alkyl halide and the hydroxide ion.

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