# How to find possible unique monochlorinated isomers?

## write the number of unique mono chlorinated isomers including stereoisomers of 2-methylpentane

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#### Explanation

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#### Explanation:

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anor277 Share
Apr 19, 2018

We start with achiral ${\underbrace{{H}_{3} C - C H \left(C {H}_{3}\right) C {H}_{2} C {H}_{2} C {H}_{3}}}_{\text{2-methylpentane}}$

#### Explanation:

And so we substitute the chain with $X$...and get...

${H}_{2} X C - \stackrel{2}{C} H \left(C {H}_{3}\right) C {H}_{2} C {H}_{2} C {H}_{3}$, $\text{1-halo-2-methylpentane}$, and this will generate a pair of enantiomers...$C 2$ is now stereogenic.

${H}_{3} C - C X \left(C {H}_{3}\right) C {H}_{2} C {H}_{2} C {H}_{3}$, $\text{2-halo-2-methylpentane}$, and this will generate NO enantiomers...$C 2$ is no longer stereogenic because of symmetric dimethyl substitution.

${H}_{3} C - C H \left(C {H}_{3}\right) C H X C {H}_{2} C {H}_{3}$, $\text{3-halo-2-methylpentane}$, and this will generate ONE pair of enantiomers ....

And ${H}_{3} C - C H \left(C {H}_{3}\right) C {H}_{2} C H X C {H}_{3}$, $\text{2-halo-4-methylpentane}$, and this will generate ONE pairs of enantiomers ... carbons 2 is potentially chiral...

And lastly, ${H}_{3} C - C H \left(C {H}_{3}\right) C {H}_{2} C {H}_{2} C {H}_{2} X$, $\text{1-halo-4-methylpentane}$, and this will generate NO enantiomers ....

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